IB Analysis and Approaches HL · Differential calculus · End of topic

IB exam questions on differential calculus

287 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

Level

Paper

Solutions

Calculating derivatives

Exercise 1 BasicIB · P1 TZ1 May 2018

Let $y = sin^{2} θ$ , $0 \leq θ \leq π$ .

a) Calculate $\frac{d y}{d θ}$ .

b) Hence, find the values of $θ$ for which $\frac{d y}{d θ} = 2 y$ .

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Answer: a) $\frac{d y}{d θ} = 2 sin θ cos θ$ ; b) $θ = 0, \frac{π}{4}, π$

Show working and answer

a) $y = sin^{2} θ = (sin θ)^{2}$ . By the chain rule: $\frac{d y}{d θ} = 2 sin θ cos θ$ (which is also written $sin 2 θ$ ).

b) 1. We set $2 sin θ cos θ = 2 sin^{2} θ$ . 2. Moving everything to one side and factorising: $2 sin θ (cos θ - sin θ) = 0$ . 3. If $sin θ = 0$ : $θ = 0$ or $θ = π$ . 4. If $cos θ = sin θ$ , then $tan θ = 1$ and, on $[0, π]$ , $θ = \frac{π}{4}$ .

Answer: a) $\frac{d y}{d θ} = 2 sin θ cos θ$ ; b) $θ = 0, \frac{π}{4}, π$

Exercise 2 BasicIB · P1 May 1997

Find $f^{'} (x)$ , given that $f (x) = x ln x + e^{s i n x} + arctan x$ .

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Answer: $f^{'} (x) = ln x + 1 + e^{s i n x} cos x + \frac{1}{1 + x ^{2}}$

Show working and answer

We differentiate term by term.
$x ln x$ (product rule): $ln x + x \cdot \frac{1}{x} = ln x + 1$ .
$e^{s i n x}$ (chain rule): $e^{s i n x} cos x$ .
$arctan x$ : $\frac{1}{1 + x ^{2}}$ .
Adding the results: $f^{'} (x) = ln x + 1 + e^{s i n x} cos x + \frac{1}{1 + x ^{2}}$ .

Answer: $f^{'} (x) = ln x + 1 + e^{s i n x} cos x + \frac{1}{1 + x ^{2}}$

Exercise 3 BasicIB · P1 November 1997

Given that $y = x e^{3 x} + ln x$ , find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ .

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Answer: $\frac{d y}{d x} = e^{3 x} (1 + 3 x) + \frac{1}{x}$ ; $\frac{d ^{2} y}{d x ^{2}} = 3 e^{3 x} (2 + 3 x) - \frac{1}{x ^{2}}$

Show working and answer

We differentiate $y = x e^{3 x} + ln x$ . For $x e^{3 x}$ we use the product and chain rules: $e^{3 x} + 3 x e^{3 x}$ . For $ln x$ : $\frac{1}{x}$ .
$\frac{d y}{d x} = e^{3 x} (1 + 3 x) + \frac{1}{x}$ .
We differentiate again: $\frac{d}{d x} [e^{3 x} (1 + 3 x)] = 3 e^{3 x} (1 + 3 x) + 3 e^{3 x} = 3 e^{3 x} (2 + 3 x)$ (product and chain rules), and $\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}$ .
$\frac{d ^{2} y}{d x ^{2}} = 3 e^{3 x} (2 + 3 x) - \frac{1}{x ^{2}}$ .

Answer: $\frac{d y}{d x} = e^{3 x} (1 + 3 x) + \frac{1}{x}$ ; $\frac{d ^{2} y}{d x ^{2}} = 3 e^{3 x} (2 + 3 x) - \frac{1}{x ^{2}}$

Exercise 4 BasicIB · P1 May 1998

Given that $f (x) = \frac{arcsin x}{ln x}$ , find $f^{'} (x)$ .

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Answer: $f^{'} (x) = \frac{x ln x - 1 - x ^{2} arcsin x}{x 1 - x ^{2} ( ln x ) ^{2}}$

Show working and answer

We apply the quotient rule to $f (x) = \frac{arcsin x}{ln x}$ , with $u = arcsin x$ (so $u^{'} = \frac{1}{1 - x ^{2}}$ ) and $v = ln x$ (so $v^{'} = \frac{1}{x}$ ).
$f^{'} (x) = \frac{\frac{ln x}{1 - x ^{2}} - \frac{arcsin x}{x}}{( ln x ) ^{2}}$ .
We write the numerator over the common denominator $x 1 - x^{2}$ : $\frac{x ln x - 1 - x ^{2} arcsin x}{x 1 - x ^{2}}$ .
$f^{'} (x) = \frac{x ln x - 1 - x ^{2} arcsin x}{x 1 - x ^{2} ( ln x ) ^{2}}$ .

Answer: $f^{'} (x) = \frac{x ln x - 1 - x ^{2} arcsin x}{x 1 - x ^{2} ( ln x ) ^{2}}$

Exercise 5 BasicIB · P1 November 2000

For the function $f : x \to x^{2} ln x$ , $x > 0$ , find the derivative function $f^{'}$ of $f$ with respect to $x$ .

Show answer

Answer: $f^{'} (x) = 2 x ln x + x = x (2 ln x + 1)$

Show working and answer

It is a product: $u = x^{2}$ , $v = ln x$ , with $u^{'} = 2 x$ and $v^{'} = \frac{1}{x}$ .
Product rule: $f^{'} (x) = 2 x ln x + x^{2} \cdot \frac{1}{x} = 2 x ln x + x$ .
Factorising: $f^{'} (x) = x (2 ln x + 1)$ .

Answer: $f^{'} (x) = 2 x ln x + x = x (2 ln x + 1)$

Exercise 6 BasicIB · P1 November 2000

For the function $f : x \mapsto \frac{1}{2} sin 2 x + cos x$ , find the possible values of $sin x$ for which $f^{'} (x) = 0$ .

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Answer: $sin x = \frac{1}{2}$ or $sin x = - 1$

Show working and answer

We differentiate using the chain rule: $f^{'} (x) = \frac{1}{2} \cdot 2 cos 2 x - sin x = cos 2 x - sin x$ .
We use $cos 2 x = 1 - 2 sin^{2} x$ so that there is a single function: $1 - 2 sin^{2} x - sin x = 0$ .
With $s = sin x$ : $2 s^{2} + s - 1 = 0$ , which factorises as $(2 s - 1) (s + 1) = 0$ .
Hence $s = \frac{1}{2}$ or $s = - 1$ (both values lie in $[- 1, 1]$ , so both are possible).

Answer: $sin x = \frac{1}{2}$ or $sin x = - 1$

Exercise 7 BasicIB · P1 November 2001

Consider the function $y = tan x - 8 sin x$ .

a) Find $\frac{d y}{d x}$ .

b) Find the value of $cos x$ when $\frac{d y}{d x} = 0$ .

Show answer

Answer: a) $\frac{d y}{d x} = sec^{2} x - 8 cos x = \frac{1 - 8 cos ^{3} x}{cos ^{2} x}$ ; b) $cos x = \frac{1}{2}$

Show working and answer

a) Differentiate: $\frac{d y}{d x} = sec^{2} x - 8 cos x = \frac{1}{cos ^{2} x} - 8 cos x = \frac{1 - 8 cos ^{3} x}{cos ^{2} x}$ .

b) 1. $\frac{d y}{d x} = 0$ requires $1 - 8 cos^{3} x = 0$ . 2. $cos^{3} x = \frac{1}{8}$ , so $cos x = \frac{1}{2}$ .

Answer: a) $\frac{d y}{d x} = sec^{2} x - 8 cos x = \frac{1 - 8 cos ^{3} x}{cos ^{2} x}$ ; b) $cos x = \frac{1}{2}$

Exercise 8 BasicIB · P1 TZ1 May 2004

The function $f$ is defined by $f : x \mapsto 3^{x^{2}}$ . Find the solution of the equation $f^{'} (x) = 2$ .

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Answer: $x \approx 0.607$

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Write $3^{x^{2}} = e^{x^{2} l n 3}$ . By the chain rule: $f^{'} (x) = 2 x ln 3 \cdot 3^{x^{2}}$ .
The equation $f^{'} (x) = 2$ is $x \cdot 3^{x^{2}} = \frac{1}{ln 3} \approx 0.9102$ .
It cannot be solved exactly. Using the GDC (intersection of $y = 2 x ln 3 \cdot 3^{x^{2}}$ with $y = 2$ ): $x \approx 0.607$ .
It is the only solution, because $f^{'} (x)$ is negative for $x < 0$ and increases for $x > 0$ .

Answer: $x \approx 0.607$

Exercise 9 BasicIB · P1 November 2003

Given the function $f (t) = 3 sec^{2} t + 5 t$

a) Find $f^{'} (t)$ .

b) Find the exact values of

(i) $f (π)$

(ii) $f^{'} (π)$

Show answer

Answer: a) $f^{'} (t) = 6 sec^{2} t tan t + 5 = \frac{6 sin t}{cos ^{3} t} + 5$ ; b) (i) $f (π) = 3 + 5 π$ ; (ii) $f^{'} (π) = 5$

Show working and answer

a) 1. $f (t) = 3 sec^{2} t + 5 t$ . By the chain rule: $\frac{d}{d t} sec^{2} t = 2 sec t \cdot sec t tan t = 2 sec^{2} t tan t$ . 2. $f^{'} (t) = 6 sec^{2} t tan t + 5 = \frac{6 sin t}{cos ^{3} t} + 5$ .

b) (i) $f (π) = 3 sec^{2} π + 5 π = 3 \cdot (- 1)^{2} + 5 π = 3 + 5 π$ .

(ii) $f^{'} (π) = 6 \cdot 1 \cdot tan π + 5 = 6 \cdot 1 \cdot 0 + 5 = 5$ .

Answer: a) $f^{'} (t) = 6 sec^{2} t tan t + 5 = \frac{6 sin t}{cos ^{3} t} + 5$ ; b) (i) $f (π) = 3 + 5 π$ ; (ii) $f^{'} (π) = 5$

Exercise 10 BasicIB · P1 TZ0 November 2004

If $y = ln (2 x - 1)$ , find $\frac{d ^{2} y}{d x ^{2}}$ .

Show answer

Answer: $\frac{d ^{2} y}{d x ^{2}} = - \frac{4}{( 2 x - 1 ) ^{2}}$

Show working and answer

$y = ln (2 x - 1)$ . By the chain rule: $\frac{d y}{d x} = \frac{2}{2 x - 1} = 2 (2 x - 1)^{- 1}$ .
Differentiating again (chain rule): $\frac{d ^{2} y}{d x ^{2}} = - 2 (2 x - 1)^{- 2} \cdot 2 = - \frac{4}{( 2 x - 1 ) ^{2}}$ .

Answer: $\frac{d ^{2} y}{d x ^{2}} = - \frac{4}{( 2 x - 1 ) ^{2}}$

Exercise 11 BasicIB · P1 TZ1 May 2005

Consider the function $f (x) = ln x^{2} + 4$ .

a) Calculate $f^{'} (x)$ .

b) Calculate $f^{''} (x)$ .

Show answer

Answer: a) $f^{'} (x) = \frac{x}{x ^{2} + 4}$ ; b) $f^{''} (x) = \frac{4 - x ^{2}}{( x ^{2} + 4 ) ^{2}}$

Show working and answer

a) 1. By the properties of logarithms: $f (x) = ln (x^{2} + 4)^{1/2} = \frac{1}{2} ln (x^{2} + 4)$ . 2. Chain rule: $f^{'} (x) = \frac{1}{2} \cdot \frac{2 x}{x ^{2} + 4} = \frac{x}{x ^{2} + 4}$ .

b) Quotient rule: $f^{''} (x) = \frac{1 \cdot ( x ^{2} + 4 ) - x \cdot 2 x}{( x ^{2} + 4 ) ^{2}} = \frac{4 - x ^{2}}{( x ^{2} + 4 ) ^{2}} .$

Answer: a) $f^{'} (x) = \frac{x}{x ^{2} + 4}$ ; b) $f^{''} (x) = \frac{4 - x ^{2}}{( x ^{2} + 4 ) ^{2}}$

Exercise 12 BasicIB · P1 TZ0 November 2022

The function $g$ is defined by $g (x) = e^{x^{2} + 1}$ , where $x \in R$ . Find $g^{'} (- 1)$ .

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Answer: $g^{'} (- 1) = - 2 e^{2}$

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Chain rule with $u = x^{2} + 1$ : $g^{'} (x) = 2 x e^{x^{2} + 1}$ .
At $x = - 1$ : $g^{'} (- 1) = 2 (- 1) e^{1 + 1} = - 2 e^{2}$ .

Answer: $g^{'} (- 1) = - 2 e^{2}$

Exercise 13 MediumIB · P1 TZ2 May 2005

Let $f (x) = \frac{x ^{2} + 5 x + 5}{x + 2}$ , $x \neq = - 2$ .

a) Find $f^{'} (x)$ .

b) Solve $f^{'} (x) > 2$ .

Show answer

Answer: a) $f^{'} (x) = \frac{x ^{2} + 4 x + 5}{( x + 2 ) ^{2}}$ ; b) $] - 3, - 1 [- {- 2}$

Show working and answer

a) 1. We differentiate using the quotient rule, $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ , with $u = x^{2} + 5 x + 5$ and $v = x + 2$ . 2. $f^{'} (x) = \frac{( 2 x + 5 ) ( x + 2 ) - ( x ^{2} + 5 x + 5 ) \cdot 1}{( x + 2 ) ^{2}}$ . 3. Expanding the numerator: $2 x^{2} + 9 x + 10 - x^{2} - 5 x - 5 = x^{2} + 4 x + 5$ . 4. Therefore $f^{'} (x) = \frac{x ^{2} + 4 x + 5}{( x + 2 ) ^{2}}$ .

b) 1. We set up $\frac{x ^{2} + 4 x + 5}{( x + 2 ) ^{2}} > 2$ with $x \neq = - 2$ . 2. Since $(x + 2)^{2} > 0$ , we can multiply without changing the direction of the inequality: $x^{2} + 4 x + 5 > 2 (x + 2)^{2} = 2 x^{2} + 8 x + 8$ . 3. Moving everything to one side: $0 > x^{2} + 4 x + 3$ , that is, $(x + 1) (x + 3) < 0$ . 4. An upward-opening parabola is negative between its roots: $- 3 < x < - 1$ . 5. We must exclude $x = - 2$ , which is not in the domain of $f$ .

Answer: a) $f^{'} (x) = \frac{x ^{2} + 4 x + 5}{( x + 2 ) ^{2}}$ ; b) $] - 3, - 1 [- {- 2}$

Exercise 14 MediumIB · P2 TZ0 November 2008Proof

If $y = ln (\frac{1}{3} (1 + e^{- 2 x}))$ verify that $\frac{d y}{d x} = \frac{2}{3} (e^{- y} - 3)$ .

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Answer: proved.

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We differentiate using the chain rule: $y = ln u$ with $u = \frac{1}{3} (1 + e^{- 2 x})$ , so $\frac{d y}{d x} = \frac{u ^{'}}{u} = \frac{- \frac{2}{3} e ^{- 2 x}}{\frac{1}{3} ( 1 + e ^{- 2 x} )} = \frac{- 2 e ^{- 2 x}}{1 + e ^{- 2 x}}$ .
From the statement, $e^{y} = \frac{1}{3} (1 + e^{- 2 x})$ , so $e^{- y} = \frac{3}{1 + e ^{- 2 x}}$ .
We substitute into the right-hand side: $\frac{2}{3} (e^{- y} - 3) = \frac{2}{3} (\frac{3}{1 + e ^{- 2 x}} - 3) = 2 \cdot \frac{1 - ( 1 + e ^{- 2 x} )}{1 + e ^{- 2 x}}$ .
Simplifying: $= \frac{- 2 e ^{- 2 x}}{1 + e ^{- 2 x}}$ , which agrees with $\frac{d y}{d x}$ .

Answer: proved.

Exercise 15 MediumIB · P2 TZ2 May 2009

a) Differentiate $f (x) = arcsin x + 2 1 - x^{2}$ , $x \in [- 1, 1]$

b) Find the coordinates of the point on the graph of $y = f (x)$ in $[- 1, 1]$ at which the gradient of the tangent to the curve is equal to $0$ .

Show answer

Answer: a) $f^{'} (x) = \frac{1 - 2 x}{1 - x ^{2}}$ ; b) $(\frac{1}{2}, \frac{π}{6} + 3)$

Show working and answer

a) 1. $\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$ . 2. For $2 1 - x^{2}$ we use the chain rule: $2 \cdot \frac{- 2 x}{2 1 - x ^{2}} = \frac{- 2 x}{1 - x ^{2}}$ . 3. Adding: $f^{'} (x) = \frac{1 - 2 x}{1 - x ^{2}}$ .

b) 1. Gradient $0$ : $f^{'} (x) = 0$ , and the denominator does not vanish on $] - 1, 1 [$ , so $1 - 2 x = 0$ and $x = \frac{1}{2}$ . 2. At the endpoints $x = \pm 1$ the derivative does not exist, so there are no other candidates. 3. $f (\frac{1}{2}) = arcsin \frac{1}{2} + 2 1 - \frac{1}{4} = \frac{π}{6} + 2 \cdot \frac{3}{2} = \frac{π}{6} + 3$ .

Answer: a) $f^{'} (x) = \frac{1 - 2 x}{1 - x ^{2}}$ ; b) $(\frac{1}{2}, \frac{π}{6} + 3)$

Exercise 16 MediumIB · P1 TZ2 May 2010

The function $f$ is defined by $f (x) = e^{x^{2} - 2 x - 1.5}$ .

a) Find $f^{'} (x)$ .

b) It is given that $y = \frac{f ( x )}{x - 1}$ has a local minimum at $x = a$ , $a > 1$ . Find the value of $a$ .

Show answer

Answer: a) $f^{'} (x) = (2 x - 2) e^{x^{2} - 2 x - 1.5}$ ; b) $a = \frac{2 + 2}{2}$

Show working and answer

a) 1. By the chain rule, $\frac{d}{d x} e^{u} = u^{'} e^{u}$ with $u = x^{2} - 2 x - 1.5$ . 2. $f^{'} (x) = (2 x - 2) e^{x^{2} - 2 x - 1.5}$ .

b) 1. We differentiate $y = \frac{f ( x )}{x - 1}$ using the quotient rule: $y^{'} = \frac{f ^{'} ( x ) ( x - 1 ) - f ( x )}{( x - 1 ) ^{2}} = \frac{2 ( x - 1 ) ^{2} f ( x ) - f ( x )}{( x - 1 ) ^{2}} = \frac{f ( x ) [ 2 ( x - 1 ) ^{2} - 1 ]}{( x - 1 ) ^{2}}$ . 2. Since $f (x) > 0$ , $y^{'} = 0$ when $2 (x - 1)^{2} = 1$ , that is, $x - 1 = \pm \frac{1}{2}$ . 3. As $a > 1$ , $a = 1 + \frac{1}{2} = \frac{2 + 2}{2}$ . 4. It is a minimum: for $1 < x < a$ , $2 (x - 1)^{2} - 1 < 0$ and $y^{'} < 0$ ; for $x > a$ , $y^{'} > 0$ . The derivative changes from negative to positive.

Answer: a) $f^{'} (x) = (2 x - 2) e^{x^{2} - 2 x - 1.5}$ ; b) $a = \frac{2 + 2}{2}$

Exercise 17 MediumIB · P1 TZ0 November 2014

Consider the functions $f$ and $g$ and their derivatives $f^{'}$ and $g^{'}$ .

The following table shows the values of the two functions and of their derivatives at $x = 1$ , $2$ and $3$ .

$x$	$1$	$2$	$3$
$f (x)$	$3$	$1$	$1$
$f^{'} (x)$	$1$	$4$	$2$
$g (x)$	$2$	$1$	$4$
$g^{'} (x)$	$4$	$2$	$3$

Given that $p (x) = f (x) \cdot g (x)$ and $h (x) = (g \circ f) (x)$ , find

a) $p^{'} (3)$

b) $h^{'} (2)$

Show answer

Answer: a) $p^{'} (3) = 11$ ; b) $h^{'} (2) = 16$

Show working and answer

a) 1. Product rule: $p^{'} (x) = f^{'} (x) g (x) + f (x) g^{'} (x)$ . 2. $p^{'} (3) = f^{'} (3) g (3) + f (3) g^{'} (3) = 2 \cdot 4 + 1 \cdot 3 = 11$ .

b) 1. Chain rule: $h^{'} (x) = g^{'} (f (x)) \cdot f^{'} (x)$ . 2. $h^{'} (2) = g^{'} (f (2)) f^{'} (2) = g^{'} (1) \cdot f^{'} (2) = 4 \cdot 4 = 16$ .

Answer: a) $p^{'} (3) = 11$ ; b) $h^{'} (2) = 16$

Exercise 18 MediumIB · P1 TZ1 May 2007

The following table shows the values of two functions $f$ and $g$ and of their first derivatives when $x = 1$ and $x = 0$ .

$x$	$f (x)$	$f^{'} (x)$	$g (x)$	$g^{'} (x)$
$0$	$4$	$1$	$- 4$	$5$
$1$	$- 2$	$3$	$- 1$	$2$

a) Calculate the derivative of $\frac{3 f ( x )}{g ( x ) - 1}$ when $x = 0$ .

b) Calculate the derivative of $f [g (x) + 2 x]$ when $x = 1$ .

Show answer

Answer: a) $- 3$ ; b) $12$

Show working and answer

a) 1. Quotient rule: $\frac{d}{d x} \frac{3 f}{g - 1} = \frac{3 f ^{'} ( g - 1 ) - 3 f g ^{'}}{( g - 1 ) ^{2}}$ . 2. At $x = 0$ : $f = 4$ , $f^{'} = 1$ , $g = - 4$ , $g^{'} = 5$ . Then $\frac{3 \cdot 1 \cdot ( - 5 ) - 3 \cdot 4 \cdot 5}{25} = \frac{- 15 - 60}{25} = - 3$ .

b) 1. Chain rule: $\frac{d}{d x} f [g (x) + 2 x] = f^{'} (g (x) + 2 x) (g^{'} (x) + 2)$ . 2. At $x = 1$ : $g (1) + 2 = - 1 + 2 = 1$ , so we need $f^{'} (1) = 3$ , and $g^{'} (1) + 2 = 4$ . 3. The value is $3 \cdot 4 = 12$ .

Answer: a) $- 3$ ; b) $12$

Exercise 19 MediumIB · P1 May 1996

Given that $f (x) = \frac{x}{x + 2}$ and $g (x) = f (\frac{x - 2}{3})$ , obtain an expression for the derivative of $g (x)$ and find its domain.

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Answer: $g^{'} (x) = \frac{6}{( x + 4 ) ^{2}}$ , with domain $R - {- 4}$

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We simplify $g$ : $g (x) = f (\frac{x - 2}{3}) = \frac{\frac{x - 2}{3}}{\frac{x - 2}{3} + 2} = \frac{x - 2}{x - 2 + 6} = \frac{x - 2}{x + 4}$ .
Quotient rule: $g^{'} (x) = \frac{1 \cdot ( x + 4 ) - ( x - 2 ) \cdot 1}{( x + 4 ) ^{2}} = \frac{6}{( x + 4 ) ^{2}}$ .
The domain of $g^{'}$ : $x \neq = - 4$ (where the denominator vanishes).

Answer: $g^{'} (x) = \frac{6}{( x + 4 ) ^{2}}$ , with domain $R - {- 4}$

Exercise 20 MediumIB · P1 TZ2 May 2018

Consider the curve $y = \frac{1}{1 - x} + \frac{4}{x - 4}$ .

Find the $x$ -coordinates of the points on the curve where the gradient is zero.

Show answer

Answer: $x = 2$ and $x = - 2$ (points $(2, - 3)$ and $(- 2, - \frac{1}{3})$ )

Show working and answer

We differentiate (chain rule on each term): $y^{'} = \frac{1}{( 1 - x ) ^{2}} - \frac{4}{( x - 4 ) ^{2}}$ .
Zero gradient: $\frac{1}{( 1 - x ) ^{2}} = \frac{4}{( x - 4 ) ^{2}} \Rightarrow (x - 4)^{2} = 4 (1 - x)^{2} \Rightarrow x - 4 = \pm 2 (1 - x)$ .
With the $+$ sign: $x - 4 = 2 - 2 x \Rightarrow x = 2$ . With the $-$ sign: $x - 4 = - 2 + 2 x \Rightarrow x = - 2$ .
The $y$ -coordinates are $y (2) = \frac{1}{- 1} + \frac{4}{- 2} = - 3$ and $y (- 2) = \frac{1}{3} + \frac{4}{- 6} = - \frac{1}{3}$ .

Answer: $x = 2$ and $x = - 2$ (points $(2, - 3)$ and $(- 2, - \frac{1}{3})$ )

Exercise 21 MediumIB · P2 TZ0 November 2018

Let $P (x) = 2 x^{4} - 15 x^{3} + a x^{2} + b x + c$ , where $a, b, c \in R$ .

a) Given that $(x - 5)$ is a factor of $P (x)$ , find an expression relating $a$ , $b$ and $c$ .

b) Given that $(x - 5)^{2}$ is a factor of $P (x)$ , write down the value of $P^{'} (5)$ .

c) Given that $(x - 5)^{2}$ is a factor of $P (x)$ and that $a = 2$ , find the values of $b$ and $c$ .

Show answer

Answer: a) $25 a + 5 b + c = 625$ ; b) $P^{'} (5) = 0$ ; c) $b = 105$ , $c = 50$

Show working and answer

a) 1. If $(x - 5)$ is a factor, then $P (5) = 0$ : $2 \cdot 625 - 15 \cdot 125 + 25 a + 5 b + c = 0$ . 2. $1250 - 1875 + 25 a + 5 b + c = 0 \Rightarrow 25 a + 5 b + c = 625$ .

b) 1. If $(x - 5)^{2}$ is a factor, then $P (x) = (x - 5)^{2} Q (x)$ and $P^{'} (x) = 2 (x - 5) Q (x) + (x - 5)^{2} Q^{'} (x)$ . 2. Every term has the factor $(x - 5)$ , so $P^{'} (5) = 0$ .

c) 1. $P^{'} (x) = 8 x^{3} - 45 x^{2} + 2 a x + b$ . $P^{'} (5) = 0$ : $1000 - 1125 + 10 a + b = 0 \Rightarrow b = 125 - 10 a$ . 2. With $a = 2$ : $b = 105$ . 3. From $25 a + 5 b + c = 625$ : $50 + 525 + c = 625 \Rightarrow c = 50$ .

Answer: a) $25 a + 5 b + c = 625$ ; b) $P^{'} (5) = 0$ ; c) $b = 105$ , $c = 50$

Exercise 22 MediumIB · P1 TZ1 May 2016

A curve is given by the equation $y = sin (π cos x)$ .

Find the coordinates of all the points on the curve for which $\frac{d y}{d x} = 0$ , $0 \leq x \leq π$ .

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Answer: $(0, 0)$ , $(π, 0)$ , $(\frac{π}{3}, 1)$ , $(\frac{2 π}{3}, - 1)$

Show working and answer

We differentiate using the chain rule (twice): $\frac{d y}{d x} = cos (π cos x) \cdot (- π sin x)$ .
Setting this equal to zero: $sin x = 0$ or $cos (π cos x) = 0$ .
If $sin x = 0$ , then in $0 \leq x \leq π$ : $x = 0$ and $x = π$ . Values: $y (0) = sin π = 0$ and $y (π) = sin (- π) = 0$ .
If $cos (π cos x) = 0$ , then $π cos x = \pm \frac{π}{2}$ (since $π cos x \in [- π, π]$ ), i.e. $cos x = \pm \frac{1}{2}$ . In $[0, π]$ : $x = \frac{π}{3}$ and $x = \frac{2 π}{3}$ .
Values: $y (\frac{π}{3}) = sin \frac{π}{2} = 1$ and $y (\frac{2 π}{3}) = sin (- \frac{π}{2}) = - 1$ .

Answer: $(0, 0)$ , $(π, 0)$ , $(\frac{π}{3}, 1)$ , $(\frac{2 π}{3}, - 1)$

Exercise 23 MediumIB · P1 TZ2 May 2016Proof

The function $f$ is given by $f (x) = a x^{2} + b x + c$ , where $a, b, c \in R$ .

Hayley makes the following conjecture: $\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}} = \frac{f ^{'} ( x _{2} ) + f ^{'} ( x _{1} )}{2}$ , $x_{1} \neq = x_{2}$ .

Show that Hayley's conjecture is correct.

Show answer

Answer: proved: $\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}} = \frac{f ^{'} ( x _{2} ) + f ^{'} ( x _{1} )}{2} = a (x_{1} + x_{2}) + b$ .

Show working and answer

Let $f (x) = a x^{2} + b x + c$ . Its derivative is $f^{'} (x) = 2 a x + b$ .
Left-hand side: $f (x_{2}) - f (x_{1}) = a (x_{2}^{2} - x_{1}^{2}) + b (x_{2} - x_{1}) = (x_{2} - x_{1}) [a (x_{2} + x_{1}) + b]$ .
Dividing by $x_{2} - x_{1} \neq = 0$ : $\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}} = a (x_{1} + x_{2}) + b$ .
Right-hand side: $\frac{f ^{'} ( x _{2} ) + f ^{'} ( x _{1} )}{2} = \frac{2 a x _{2} + b + 2 a x _{1} + b}{2} = a (x_{1} + x_{2}) + b$ .
Both sides are equal, so Hayley's conjecture is correct.

Answer: proved: $\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}} = \frac{f ^{'} ( x _{2} ) + f ^{'} ( x _{1} )}{2} = a (x_{1} + x_{2}) + b$ .

Exercise 24 MediumIB · P1 May 1999

Differentiate $y = arccos (1 - 2 x^{2})$ with respect to $x$ , and simplify your answer.

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Answer: $\frac{d y}{d x} = \frac{2 x}{x ^{2} - x ^{4}}$

Show working and answer

It is a composite function: $y = arccos u$ with $u = 1 - 2 x^{2}$ . We use the chain rule and $\frac{d}{d u} arccos u = - \frac{1}{1 - u ^{2}}$ .
We differentiate the inner function: $u^{'} = - 4 x$ .
Therefore $\frac{d y}{d x} = - \frac{1}{1 - ( 1 - 2 x ^{2} ) ^{2}} \cdot (- 4 x) = \frac{4 x}{1 - ( 1 - 2 x ^{2} ) ^{2}} .$
We simplify the expression under the root: $1 - (1 - 4 x^{2} + 4 x^{4}) = 4 x^{2} - 4 x^{4} = 4 (x^{2} - x^{4})$ , and $4 (x^{2} - x^{4}) = 2 x^{2} - x^{4}$ .
So $\frac{d y}{d x} = \frac{4 x}{2 x ^{2} - x ^{4}} = \frac{2 x}{x ^{2} - x ^{4}}$ (defined for $0 < ∣ x ∣ < 1$ ).

Answer: $\frac{d y}{d x} = \frac{2 x}{x ^{2} - x ^{4}}$

Exercise 25 MediumIB · P1 November 2003

Let $y = e^{3 x} sin (π x)$ .

a) Find $\frac{d y}{d x}$ .

b) Find the smallest positive value of $x$ for which $\frac{d y}{d x} = 0$ .

Show answer

Answer: a) $\frac{d y}{d x} = e^{3 x} (3 sin π x + π cos π x)$ ; b) $x \approx 0.743$

Show working and answer

a) Product rule (with the chain rule applied to each factor): $\frac{d y}{d x} = 3 e^{3 x} sin (π x) + e^{3 x} π cos (π x) = e^{3 x} [3 sin (π x) + π cos (π x)] .$

b) 1. $\frac{d y}{d x} = 0 ⟺ 3 sin (π x) + π cos (π x) = 0$ (because $e^{3 x} > 0$ ), that is, $tan (π x) = - \frac{π}{3}$ . 2. The principal solution of $tan u = - \frac{π}{3}$ is $u = - arctan \frac{π}{3} \approx - 0.808$ , and the others are $u = - arctan \frac{π}{3} + k π$ . 3. With $u = π x > 0$ , the smallest positive value is $k = 1$ : $π x = π - arctan \frac{π}{3}$ , so $x = 1 - \frac{1}{π} arctan \frac{π}{3} \approx 0.743$ .

Answer: a) $\frac{d y}{d x} = e^{3 x} (3 sin π x + π cos π x)$ ; b) $x \approx 0.743$

Exercise 26 MediumIB · P1 TZ2 May 2004

The function $f$ is given by $f : x \mapsto 3^{x}$ . Solve the equation $f^{''} (x) = 2$ .

Show answer

Answer: $x = \frac{ln ( \frac{2}{( ln 3 ) ^{2}} )}{ln 3} \approx 0.460$

Show working and answer

$f (x) = 3^{x} = e^{x l n 3}$ , so $f^{'} (x) = ln 3 \cdot 3^{x}$ and $f^{''} (x) = (ln 3)^{2} 3^{x}$ .
$f^{''} (x) = 2$ is equivalent to $3^{x} = \frac{2}{( ln 3 ) ^{2}}$ .
Taking natural logarithms: $x ln 3 = ln \frac{2}{( ln 3 ) ^{2}}$ , so $x = \frac{ln ( \frac{2}{( ln 3 ) ^{2}} )}{ln 3} \approx 0.460$ .

Answer: $x = \frac{ln ( \frac{2}{( ln 3 ) ^{2}} )}{ln 3} \approx 0.460$

Exercise 27 MediumIB · P1 TZ0 November 2005

Let $f$ be a cubic polynomial function. Given that $f (0) = 2$ , $f^{'} (0) = - 3$ , $f (1) = f^{'} (1)$ and $f^{''} (- 1) = 6$ , find $f (x)$ .

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Answer: $f (x) = - \frac{1}{5} x^{3} + \frac{12}{5} x^{2} - 3 x + 2$

Show working and answer

Write $f (x) = a x^{3} + b x^{2} + c x + d$ , with $f^{'} (x) = 3 a x^{2} + 2 b x + c$ and $f^{''} (x) = 6 a x + 2 b$ .
$f (0) = 2$ gives $d = 2$ . $f^{'} (0) = - 3$ gives $c = - 3$ .
$f^{''} (- 1) = 6$ : $- 6 a + 2 b = 6$ , so $b = 3 + 3 a$ .
$f (1) = f^{'} (1)$ : $a + b - 3 + 2 = 3 a + 2 b - 3$ , that is, $a + b - 1 = 3 a + 2 b - 3$ , which reduces to $2 a + b = 2$ .
Substituting $b = 3 + 3 a$ : $2 a + 3 + 3 a = 2$ , so $5 a = - 1$ and $a = - \frac{1}{5}$ . Then $b = 3 - \frac{3}{5} = \frac{12}{5}$ .

Answer: $f (x) = - \frac{1}{5} x^{3} + \frac{12}{5} x^{2} - 3 x + 2$

Exercise 28 MediumIB · P2 TZ0 May 2006

Let $f (x) = cos^{3} (4 x + 1)$ , $0 \leq x \leq 1$ .

a) Find $f^{'} (x)$ .

b) Find the exact values of the three roots of $f^{'} (x) = 0$ .

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Answer: a) $f^{'} (x) = - 12 cos^{2} (4 x + 1) sin (4 x + 1)$ ; b) $x = \frac{π - 2}{8}$ , $x = \frac{3 π - 2}{8}$ , $x = \frac{π - 1}{4}$

Show working and answer

a) 1. Write $f (x) = [cos (4 x + 1)]^{3}$ and apply the chain rule twice. 2. $f^{'} (x) = 3 cos^{2} (4 x + 1) \cdot (- sin (4 x + 1)) \cdot 4 = - 12 cos^{2} (4 x + 1) sin (4 x + 1)$ .

b) 1. $f^{'} (x) = 0$ when $cos (4 x + 1) = 0$ or $sin (4 x + 1) = 0$ . 2. If $0 \leq x \leq 1$ , the angle $u = 4 x + 1$ runs over $[1, 5]$ (radians). 3. $cos u = 0$ in $[1, 5]$ : $u = \frac{π}{2} \approx 1.57$ and $u = \frac{3 π}{2} \approx 4.71$ . Then $x = \frac{\frac{π}{2} - 1}{4} = \frac{π - 2}{8}$ and $x = \frac{\frac{3 π}{2} - 1}{4} = \frac{3 π - 2}{8}$ . 4. $sin u = 0$ in $[1, 5]$ : only $u = π \approx 3.14$ (because $2 π \approx 6.28 > 5$ ). Then $x = \frac{π - 1}{4}$ .

Answer: a) $f^{'} (x) = - 12 cos^{2} (4 x + 1) sin (4 x + 1)$ ; b) $x = \frac{π - 2}{8}$ , $x = \frac{3 π - 2}{8}$ , $x = \frac{π - 1}{4}$

Exercise 29 MediumIB · P1 TZ1 May 2007Proof

Let $y = x arcsin x$ , $x \in] - 1, 1 [$ . Show that $\frac{d ^{2} y}{d x ^{2}} = \frac{2 - x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}}$ .

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Answer: proved: $\frac{d ^{2} y}{d x ^{2}} = \frac{2 - x ^{2}}{( 1 - x ^{2} ) ^{3/2}}$ .

Show working and answer

First derivative, using the product rule: $y^{'} = arcsin x + x \cdot \frac{1}{1 - x ^{2}}$ .
Second derivative: $y^{''} = \frac{1}{1 - x ^{2}} + \frac{d}{d x} [x (1 - x^{2})^{- 1/2}]$ .
The last term, again using the product and chain rules: $(1 - x^{2})^{- 1/2} + x \cdot (- \frac{1}{2}) (1 - x^{2})^{- 3/2} (- 2 x) = (1 - x^{2})^{- 1/2} + x^{2} (1 - x^{2})^{- 3/2}$ .
Add the terms: $y^{''} = 2 (1 - x^{2})^{- 1/2} + x^{2} (1 - x^{2})^{- 3/2}$ .
Factor out $(1 - x^{2})^{- 3/2}$ : $y^{''} = \frac{2 ( 1 - x ^{2} ) + x ^{2}}{( 1 - x ^{2} ) ^{3/2}} = \frac{2 - x ^{2}}{( 1 - x ^{2} ) ^{3/2}}$ .

Answer: proved: $\frac{d ^{2} y}{d x ^{2}} = \frac{2 - x ^{2}}{( 1 - x ^{2} ) ^{3/2}}$ .

Exercise 30 MediumIB · P1 TZ0 November 2011Proof

Given that $y = \frac{1}{1 - x}$ , use mathematical induction to prove that $\frac{d ^{n} y}{d x ^{n}} = \frac{n !}{( 1 - x ) ^{n + 1}}$ , $n \in Z^{+}$ .

Show answer

Answer: proven: $\frac{d ^{n} y}{d x ^{n}} = \frac{n !}{( 1 - x ) ^{n + 1}}$ .

Show working and answer

Let $P (n)$ be: $\frac{d ^{n} y}{d x ^{n}} = \frac{n !}{( 1 - x ) ^{n + 1}}$ .
Base case $n = 1$ : $y = (1 - x)^{- 1}$ and, by the chain rule, $y^{'} = (1 - x)^{- 2} = \frac{1 !}{( 1 - x ) ^{2}}$ . It holds.
Inductive hypothesis: assume $P (k)$ is true, $\frac{d ^{k} y}{d x ^{k}} = k! (1 - x)^{- (k + 1)}$ .
Differentiate (chain rule): $\frac{d ^{k + 1} y}{d x ^{k + 1}} = k! \cdot (- (k + 1)) (1 - x)^{- (k + 2)} \cdot (- 1) = (k + 1) k! (1 - x)^{- (k + 2)}$ .
Since $(k + 1) k! = (k + 1)!$ : $\frac{d ^{k + 1} y}{d x ^{k + 1}} = \frac{( k + 1 )!}{( 1 - x ) ^{k + 2}}$ , which is $P (k + 1)$ .
Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by induction $P (n)$ is true for all $n \in Z^{+}$ .

Answer: proven: $\frac{d ^{n} y}{d x ^{n}} = \frac{n !}{( 1 - x ) ^{n + 1}}$ .

Exercise 31 MediumIB · P2 TZ1 May 2015

Find the equation of the normal to the curve $y = \frac{e ^{x} cos x ln ( x + e )}{( x ^{17} + 1 ) ^{5}}$ at the point where $x = 0$ . In your answer, give the value of the gradient of the normal correct to three decimal places.

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Answer: $y = - 0.731 x + 1$

Show working and answer

At $x = 0$ : $y = \frac{1 \cdot 1 \cdot ln e}{( 0 + 1 ) ^{5}} = 1$ . The point is $(0, 1)$ .
To differentiate such a long product and quotient we use logarithmic differentiation: $ln y = x + ln cos x + ln (ln (x + e)) - 5 ln (x^{17} + 1)$ , and differentiating, $\frac{y ^{'}}{y} = 1 - tan x + \frac{1}{( x + e ) ln ( x + e )} - \frac{85 x ^{16}}{x ^{17} + 1} .$
At $x = 0$ : $\frac{y ^{'}}{y} = 1 - 0 + \frac{1}{e \cdot 1} - 0 = 1 + \frac{1}{e}$ , and as $y = 1$ , $y^{'} (0) = 1 + \frac{1}{e} \approx 1.3679$ (this value is also obtained using the GDC).
The gradient of the normal is $- \frac{1}{y ^{'} ( 0 )} \approx - 0.731$ .
The normal through $(0, 1)$ is $y = - 0.731 x + 1$ .

Answer: $y = - 0.731 x + 1$

Exercise 32 MediumIB · P1 TZ0 November 2015Proof

a) Show that $sin (θ + \frac{π}{2}) = cos θ$ .

b) Consider $f (x) = sin (a x)$ , where $a$ is a constant. Prove by mathematical induction that $f^{(n)} (x) = a^{n} sin (a x + \frac{nπ}{2})$ , where $n \in Z^{+}$ and $f^{(n)} (x)$ represents the $n$ th derivative of $f (x)$ .

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Answer: proved.

Show working and answer

a) 1. We use the sine of a sum formula: $sin (θ + \frac{π}{2}) = sin θ cos \frac{π}{2} + cos θ sin \frac{π}{2} = sin θ \cdot 0 + cos θ \cdot 1 = cos θ$ .

b) Proof by induction of $P (n) : f^{(n)} (x) = a^{n} sin (a x + \frac{nπ}{2})$ .

Base case $n = 1$ : $f^{'} (x) = a cos (a x)$ (chain rule). By part a), $a cos (a x) = a sin (a x + \frac{π}{2})$ , which is $P (1)$ .
Hypothesis: suppose $P (k)$ is true for some $k \in Z^{+}$ : $f^{(k)} (x) = a^{k} sin (a x + \frac{k π}{2})$ .
Differentiating (chain rule): $f^{(k + 1)} (x) = a^{k} \cdot a cos (a x + \frac{k π}{2}) = a^{k + 1} cos (a x + \frac{k π}{2})$ .
By part a) with $θ = a x + \frac{k π}{2}$ : $cos (a x + \frac{k π}{2}) = sin (a x + \frac{k π}{2} + \frac{π}{2}) = sin (a x + \frac{( k + 1 ) π}{2})$ .
Therefore $f^{(k + 1)} (x) = a^{k + 1} sin (a x + \frac{( k + 1 ) π}{2})$ , which is $P (k + 1)$ .
As $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , $P (n)$ is true for all $n \in Z^{+}$ .

Answer: proved.

Exercise 33 HardIB · P2 TZ2 May 2005Proof

The function $f$ is defined by $f (x) = e^{p x} (x + 1)$ , with $p \in R$ .

(i) Verify that $f^{'} (x) = e^{p x} (p (x + 1) + 1)$ .

(ii) Let $f^{(n)} (x)$ be the $n$ th derivative of $f (x)$ with respect to $x$ . Prove by mathematical induction that $f^{(n)} (x) = p^{n - 1} e^{p x} (p (x + 1) + n)$ , $n \in Z^{+}$ .

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Answer: proved.

Show working and answer

(i) 1. We apply the product rule to $f (x) = e^{p x} (x + 1)$ . 2. $f^{'} (x) = p e^{p x} (x + 1) + e^{p x} \cdot 1$ . 3. Factoring out $e^{p x}$ : $f^{'} (x) = e^{p x} (p (x + 1) + 1)$ .

(ii) We prove by induction that $f^{(n)} (x) = p^{n - 1} e^{p x} (p (x + 1) + n)$ for $n \in Z^{+}$ .

Base case $n = 1$ : the formula gives $p^{0} e^{p x} (p (x + 1) + 1)$ , which is the result of (i).
Hypothesis: for some $k \in Z^{+}$ , $f^{(k)} (x) = p^{k - 1} e^{p x} (p (x + 1) + k)$ .
Step: we differentiate using the product rule: $f^{(k + 1)} (x) = p^{k - 1} [p e^{p x} (p (x + 1) + k) + e^{p x} \cdot p] .$
Factoring out $p e^{p x}$ : $f^{(k + 1)} (x) = p^{k - 1} \cdot p e^{p x} (p (x + 1) + k + 1) = p^{k} e^{p x} (p (x + 1) + k + 1)$ .
This is the formula for $n = k + 1$ . Since it holds for $n = 1$ and, whenever it holds for $k$ , it also holds for $k + 1$ , it is proved for all $n \in Z^{+}$ .

Answer: proved.

Exercise 34 HardIB · P1 TZ2 May 2014

Consider the functions $h (x) = arctan (x)$ , $x \in R$ and $g (x) = \frac{1}{x}$ , $x \in R$ , $x \neq = 0$ .

a) Sketch approximately the graph of $y = h (x)$ .

b) Find an expression for the composite function $(h \circ g) (x)$ and state its domain.

Given that $f (x) = h (x) + (h \circ g) (x)$ ,

c) i) find $f^{'} (x)$ , giving your answer in simplified form.

(ii) show that $f (x) = \frac{π}{2}$ for $x > 0$ .

Nigel states that $f$ is an odd function, while Tom maintains that $f$ is an even function.

d) i) State who is correct and justify your answer.

(ii) Hence, find the value of $f (x)$ for $x < 0$ .

View solution

a) Graph of y = arctan x: increasing, passing through the origin and approaching two horizontal asymptotes, y = π/2 and y = −π/2

The graph of $y = arctan x$ passes through the origin, is increasing, is symmetric about the origin and has horizontal asymptotes $y = \frac{π}{2}$ (as $x \to + \infty$ ) and $y = - \frac{π}{2}$ (as $x \to - \infty$ ).

b) 1. $(h \circ g) (x) = h (g (x)) = arctan (\frac{1}{x})$ . 2. The domain is that of $g$ (since $arctan$ accepts any real number): $D_{h \circ g} = R - {0}$ .

c) i) 1. $f (x) = arctan x + arctan \frac{1}{x}$ . We differentiate (chain rule on the second term): $f^{'} (x) = \frac{1}{1 + x ^{2}} + \frac{1}{1 + \frac{1}{x ^{2}}} \cdot (- \frac{1}{x ^{2}})$ . 2. Simplifying the second term: $\frac{x ^{2}}{x ^{2} + 1} \cdot (- \frac{1}{x ^{2}}) = - \frac{1}{x ^{2} + 1}$ . 3. Therefore $f^{'} (x) = \frac{1}{1 + x ^{2}} - \frac{1}{1 + x ^{2}} = 0$ .

(ii) 1. As $f^{'} (x) = 0$ on the interval $x > 0$ , $f$ is constant there. 2. We find the constant using $x = 1$ : $f (1) = arctan 1 + arctan 1 = \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$ . 3. Hence $f (x) = \frac{π}{2}$ for $x > 0$ .

d) i) 1. $f (- x) = arctan (- x) + arctan (- \frac{1}{x}) = - arctan x - arctan \frac{1}{x} = - f (x)$ , because $arctan$ is odd. 2. Therefore $f$ is odd: Nigel is right.

(ii) 1. For $x < 0$ we have $- x > 0$ , so $f (- x) = \frac{π}{2}$ . 2. As $f$ is odd, $f (x) = - f (- x) = - \frac{π}{2}$ .

Answer: b) $arctan (\frac{1}{x})$ , domain $R - {0}$ ; c) i) $f^{'} (x) = 0$ ; ii) shown; d) i) Nigel, because $f (- x) = - f (x)$ ; ii) $f (x) = - \frac{π}{2}$ for $x < 0$

Exercise 35 HardIB · P1 TZ2 May 2012Proof

a) Using the definition of the derivative, verify that the derivative of $\frac{1}{2 x + 1}$ is $\frac{- 2}{( 2 x + 1 ) ^{2}}$ .

b) Prove, by mathematical induction, that the $n$ th derivative of $(2 x + 1)^{- 1}$ is $(- 1)^{n} \frac{2 ^{n} n !}{( 2 x + 1 ) ^{n + 1}}$ .

Show answer

Answer: proved.

Show working and answer

a) 1. Definition: $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$ with $f (x) = \frac{1}{2 x + 1}$ . 2. $\frac{f ( x + h ) - f ( x )}{h} = \frac{1}{h} (\frac{1}{2 x + 2 h + 1} - \frac{1}{2 x + 1}) = \frac{1}{h} \cdot \frac{( 2 x + 1 ) - ( 2 x + 2 h + 1 )}{( 2 x + 2 h + 1 ) ( 2 x + 1 )}$ . 3. The numerator is $- 2 h$ : $= \frac{- 2 h}{h ( 2 x + 2 h + 1 ) ( 2 x + 1 )} = \frac{- 2}{( 2 x + 2 h + 1 ) ( 2 x + 1 )}$ . 4. As $h \to 0$ : $f^{'} (x) = \frac{- 2}{( 2 x + 1 ) ^{2}}$ .

b) We prove by induction that $\frac{d ^{n}}{d x ^{n}} (2 x + 1)^{- 1} = (- 1)^{n} \frac{2 ^{n} n !}{( 2 x + 1 ) ^{n + 1}}$ .

Base case $n = 1$ : the formula gives $(- 1) \frac{2 \cdot 1}{( 2 x + 1 ) ^{2}} = \frac{- 2}{( 2 x + 1 ) ^{2}}$ , which is what we obtained in (a).
Hypothesis: $f^{(k)} (x) = (- 1)^{k} 2^{k} k! (2 x + 1)^{- (k + 1)}$ .
Step: differentiating with the chain rule, $f^{(k + 1)} (x) = (- 1)^{k} 2^{k} k! \cdot (- (k + 1)) (2 x + 1)^{- (k + 2)} \cdot 2$ .
Grouping: $= (- 1)^{k + 1} 2^{k + 1} (k + 1) k! (2 x + 1)^{- (k + 2)} = (- 1)^{k + 1} \frac{2 ^{k + 1} ( k + 1 )!}{( 2 x + 1 ) ^{k + 2}}$ .
This is the formula for $n = k + 1$ ; it is proved by induction for all $n \in Z^{+}$ .

Answer: proved.

Exercise 36 HardIB · P2 May 2001Proof

Use mathematical induction to prove that $\frac{d ^{n}}{d x ^{n}} (cos x) = cos (x + \frac{nπ}{2})$ , for all positive integer values of $n$ .

Show answer

Answer: proof complete.

Show working and answer

We want to prove by induction that $P (n) : \frac{d ^{n}}{d x ^{n}} (cos x) = cos (x + \frac{nπ}{2})$ for all integers $n \geq 1$ .

Base case $n = 1$ : $\frac{d}{d x} cos x = - sin x$ , and $cos (x + \frac{π}{2}) = - sin x$ . It holds.
Induction hypothesis: we assume that for some $k \geq 1$ , $\frac{d ^{k}}{d x ^{k}} (cos x) = cos (x + \frac{k π}{2})$ .
Inductive step: we differentiate once more (chain rule): $\frac{d ^{k + 1}}{d x ^{k + 1}} (cos x) = \frac{d}{d x} cos (x + \frac{k π}{2}) = - sin (x + \frac{k π}{2})$ .
Since $- sin α = cos (α + \frac{π}{2})$ , we have $- sin (x + \frac{k π}{2}) = cos (x + \frac{k π}{2} + \frac{π}{2}) = cos (x + \frac{( k + 1 ) π}{2})$ .
That is, $P (k) \Rightarrow P (k + 1)$ . Since $P (1)$ is true, by the principle of mathematical induction $P (n)$ is true for all positive integers $n$ .

Answer: proof complete.

Exercise 37 HardIB · P1 TZ0 November 2017

Consider the function $f_{n} (x) = (cos 2 x) (cos 4 x) \dots (cos 2^{n} x)$ , $n \in Z^{+}$ .

a) Determine whether $f_{n}$ is an even or odd function, justifying your answer.

b) Using mathematical induction, prove that $f_{n} (x) = \frac{sin 2 ^{n + 1} x}{2 ^{n} sin 2 x}$ , $x \neq = \frac{mπ}{2}$ where $m \in Z$ .

c) Using the above or otherwise, calculate an expression for the derivative of $f_{n} (x)$ with respect to $x$ .

d) Show that, for $n > 1$ , the equation of the tangent to the curve $y = f_{n} (x)$ at $x = \frac{π}{4}$ is $4 x - 2 y - π = 0$ .

Show answer

Answer: a) $f_{n}$ is even; b) proved by induction; c) $f_{n}^{'} (x) = \frac{2 ^{n + 1} cos ( 2 ^{n + 1} x ) sin 2 x - 2 cos 2 x sin ( 2 ^{n + 1} x )}{2 ^{n} sin ^{2} 2 x}$ ; d) the tangent is $4 x - 2 y - π = 0$

Show working and answer

a) 1. We calculate $f_{n} (- x) = cos (- 2 x) cos (- 4 x) \dots cos (- 2^{n} x)$ . 2. Since cosine is even, $cos (- 2^{k} x) = cos (2^{k} x)$ for each $k$ , so $f_{n} (- x) = f_{n} (x)$ . 3. $f_{n}$ is an even function.

b) We prove $f_{n} (x) = \frac{sin 2 ^{n + 1} x}{2 ^{n} sin 2 x}$ by induction.

Base case $n = 1$ : $f_{1} (x) = cos 2 x$ and $\frac{sin 4 x}{2 sin 2 x} = \frac{2 sin 2 x cos 2 x}{2 sin 2 x} = cos 2 x$ . It holds.
Hypothesis: we assume $f_{k} (x) = \frac{sin 2 ^{k + 1} x}{2 ^{k} sin 2 x}$ for some $k \in Z^{+}$ .
Step: $f_{k + 1} (x) = f_{k} (x) cos (2^{k + 1} x) = \frac{sin ( 2 ^{k + 1} x ) cos ( 2 ^{k + 1} x )}{2 ^{k} sin 2 x}$ .
We multiply and divide by $2$ and use $2 sin α cos α = sin 2 α$ : $f_{k + 1} (x) = \frac{sin ( 2 \cdot 2 ^{k + 1} x )}{2 ^{k + 1} sin 2 x} = \frac{sin 2 ^{k + 2} x}{2 ^{k + 1} sin 2 x}$ , which is the formula for $n = k + 1$ .
By the principle of induction, the formula is true for all $n \in Z^{+}$ .

c) 1. We differentiate $f_{n} (x) = \frac{sin 2 ^{n + 1} x}{2 ^{n} sin 2 x}$ with the quotient rule (and the chain rule for each sine): numerator $u = sin 2^{n + 1} x$ , with $u^{'} = 2^{n + 1} cos 2^{n + 1} x$ ; denominator $v = 2^{n} sin 2 x$ , with $v^{'} = 2^{n + 1} cos 2 x$ . 2. $f_{n}^{'} (x) = \frac{2 ^{n + 1} cos ( 2 ^{n + 1} x ) \cdot 2 ^{n} sin 2 x - 2 ^{n + 1} cos 2 x \cdot sin ( 2 ^{n + 1} x )}{2 ^{2 n} sin ^{2} 2 x}$ . 3. Simplifying by $2^{n}$ : $f_{n}^{'} (x) = \frac{2 ^{n + 1} cos ( 2 ^{n + 1} x ) sin 2 x - 2 cos 2 x sin ( 2 ^{n + 1} x )}{2 ^{n} sin ^{2} 2 x}$ .

d) 1. At $x = \frac{π}{4}$ : $sin 2 x = sin \frac{π}{2} = 1$ and $cos 2 x = 0$ . 2. Also $2^{n + 1} \cdot \frac{π}{4} = 2^{n - 1} π$ . For $n > 1$ , $2^{n - 1}$ is even, so $sin (2^{n - 1} π) = 0$ and $cos (2^{n - 1} π) = 1$ . 3. Value of the function: $f_{n} (\frac{π}{4}) = \frac{0}{2 ^{n} \cdot 1} = 0$ . 4. Gradient: $f_{n}^{'} (\frac{π}{4}) = \frac{2 ^{n + 1} \cdot 1 \cdot 1 - 0}{2 ^{n} \cdot 1} = 2$ . 5. Tangent: $y - 0 = 2 (x - \frac{π}{4})$ , i.e. $2 x - y - \frac{π}{2} = 0$ , or, multiplying by $2$ : $4 x - 2 y - π = 0$ .

Exercise 38 HardIB · P2 TZ0 November 2007

Given that $y = x e^{- x}$ ,

a) find $\frac{d y}{d x}$ .

b) Use mathematical induction to prove that for all $n \in Z^{+}$ , $\frac{d ^{n} y}{d x ^{n}} = (- 1)^{n + 1} e^{- x} (n - x)$ holds.

Show answer

Answer: a) $\frac{d y}{d x} = e^{- x} (1 - x)$ ; b) proven by induction.

Show working and answer

a) 1. Product rule: $\frac{d y}{d x} = 1 \cdot e^{- x} + x \cdot (- e^{- x}) = e^{- x} (1 - x)$ .

b) 1. Let $P (n)$ be: $\frac{d ^{n} y}{d x ^{n}} = (- 1)^{n + 1} e^{- x} (n - x)$ . 2. Base case $n = 1$ : by part a), $\frac{d y}{d x} = e^{- x} (1 - x)$ , and the formula gives $(- 1)^{2} e^{- x} (1 - x)$ . It holds. 3. Inductive hypothesis: assume $P (k)$ is true, i.e. $\frac{d ^{k} y}{d x ^{k}} = (- 1)^{k + 1} e^{- x} (k - x)$ for some $k \in Z^{+}$ . 4. Differentiate (product rule): $\frac{d ^{k + 1} y}{d x ^{k + 1}} = (- 1)^{k + 1} [- e^{- x} (k - x) + e^{- x} (- 1)] = (- 1)^{k + 1} e^{- x} [- (k - x) - 1]$ . 5. Simplify: $- (k - x) - 1 = - (k + 1 - x)$ , so $\frac{d ^{k + 1} y}{d x ^{k + 1}} = (- 1)^{k + 2} e^{- x} (k + 1 - x)$ , which is $P (k + 1)$ . 6. Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by induction $P (n)$ is true for all $n \in Z^{+}$ .

Answer: a) $\frac{d y}{d x} = e^{- x} (1 - x)$ ; b) proven by induction.

Exercise 39 HardIB · P2 TZ1 May 2009

Let $f (x) = \frac{1 - x}{1 + x}$ and $g (x) = x + 1$ , $x > - 1$ .

Calculate the set of values of $x$ for which $f^{'} (x) \leq f (x) \leq g (x)$ .

Show answer

Answer: $0 \leq x \leq 3$

Show working and answer

Differentiate $f$ using the quotient rule: $f^{'} (x) = \frac{- ( 1 + x ) - ( 1 - x )}{( 1 + x ) ^{2}} = \frac{- 2}{( 1 + x ) ^{2}}$ .
Condition $f^{'} (x) \leq f (x)$ : $\frac{- 2}{( 1 + x ) ^{2}} \leq \frac{1 - x}{1 + x}$ . Since $x > - 1$ , $(1 + x)^{2} > 0$ and we multiply without changing the inequality: $- 2 \leq (1 - x) (1 + x) = 1 - x^{2}$ .
I.e. $x^{2} \leq 3$ , so $- 3 \leq x \leq 3$ . Given $x > - 1$ : $- 1 < x \leq 3$ .
Condition $f (x) \leq g (x)$ : $\frac{1 - x}{1 + x} \leq x + 1$ . Multiplying by $1 + x > 0$ : $1 - x \leq (x + 1)^{3/2}$ .
The function $(x + 1)^{3/2}$ is increasing and $1 - x$ is decreasing, so they intersect at most once. At $x = 0$ both equal $1$ : this is the only intersection. For $x > 0$ , $(x + 1)^{3/2} > 1 - x$ , and for $- 1 < x < 0$ the opposite holds. Therefore the inequality holds for $x \geq 0$ .
Intersection of both conditions: $0 \leq x \leq 3$ .

Answer: $0 \leq x \leq 3$

Exercise 40 HardIB · P2 TZ1 May 2013

The function $f$ has an inverse $f^{- 1} (x)$ and derivative $f^{'} (x)$ for all $x \in R$ .

For all functions with these properties we obtain the result that, for $a \in R$ with $b = f (a)$ and $f^{'} (a) \neq = 0$ , $(f^{- 1})^{'} (b) = \frac{1}{f ^{'} ( a )}$ .

a) Verify that this is true for $f (x) = x^{3} + 1$ and $x = 2$ .

b) Given that $g (x) = x e^{x^{2}}$ , prove that $g^{'} (x) > 0$ for all values of $x$ .

c) Using the result given at the beginning of the question, find the gradient of the function $y = g^{- 1} (x)$ at $x = 2$ .

d) (i) With $f$ and $g$ as defined in parts (a) and (b), solve $(g \circ f) (x) = 2$ .

(ii) Let $h (x) = (g \circ f)^{- 1} (x)$ . Calculate $h^{'} (2)$ .

Show answer

Answer: a) verification; b) $g^{'} (x) = e^{x^{2}} (1 + 2 x^{2}) > 0$ ; c) $0.172$ ; d) (i) $x = - 0.470$ ; (ii) $h^{'} (2) = 0.259$

Show working and answer

a) 1. With $f (x) = x^{3} + 1$ and $a = 2$ : $b = f (2) = 9$ and $f^{'} (x) = 3 x^{2}$ , so $f^{'} (2) = 12$ . 2. The inverse: $y = x^{3} + 1$ implies $x = 3 y - 1$ , that is $f^{- 1} (x) = (x - 1)^{1/3}$ . Its derivative is $(f^{- 1})^{'} (x) = \frac{1}{3} (x - 1)^{- 2/3}$ . 3. At $b = 9$ : $(f^{- 1})^{'} (9) = \frac{1}{3} \cdot 8^{- 2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} = \frac{1}{f ^{'} ( 2 )}$ . It is verified.

b) 1. Product rule: $g^{'} (x) = e^{x^{2}} + x \cdot 2 x e^{x^{2}} = e^{x^{2}} (1 + 2 x^{2})$ . 2. As $e^{x^{2}} > 0$ and $1 + 2 x^{2} \geq 1 > 0$ , $g^{'} (x) > 0$ for all $x \in R$ .

c) 1. We need the $a$ such that $g (a) = 2$ , that is $a e^{a^{2}} = 2$ . Using the GDC: $a \approx 0.8961$ . 2. By the result in the question: $(g^{- 1})^{'} (2) = \frac{1}{g ^{'} ( a )} = \frac{1}{e ^{a^{2}} ( 1 + 2 a ^{2} )} \approx 0.172$ .

d) (i) 1. $(g \circ f) (x) = g (f (x)) = 2$ implies $f (x) = a \approx 0.8961$ , since $g$ is one-to-one (part b: it is strictly increasing). 2. Then $x^{3} + 1 = 0.8961$ , so $x^{3} = - 0.1039$ and $x \approx - 0.470$ .

(ii) 1. Let $x_{0} \approx - 0.470$ be the value above, with $(g \circ f) (x_{0}) = 2$ . By the result in the question applied to $g \circ f$ : $h^{'} (2) = \frac{1}{( g \circ f ) ^{'} ( x _{0} )}$ . 2. Chain rule: $(g \circ f)^{'} (x_{0}) = g^{'} (f (x_{0})) \cdot f^{'} (x_{0}) = g^{'} (a) \cdot 3 x_{0}^{2} \approx 5.816 \cdot 0.6631 \approx 3.857$ . 3. Hence $h^{'} (2) \approx \frac{1}{3.857} \approx 0.259$ .

Answer: a) verification; b) $g^{'} (x) = e^{x^{2}} (1 + 2 x^{2}) > 0$ ; c) $0.172$ ; d) (i) $x = - 0.470$ ; (ii) $h^{'} (2) = 0.259$

Derivative at a point, tangent and normal

Exercise 41 BasicIB · P2 TZ0 November 2018

Differentiate the function $f (x) = 3 x^{3} - x$ using the definition of the derivative.

Show answer

Answer: $f^{'} (x) = 9 x^{2} - 1$

Show working and answer

We use the definition: $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$ .
We calculate $f (x + h) = 3 (x + h)^{3} - (x + h) = 3 x^{3} + 9 x^{2} h + 9 x h^{2} + 3 h^{3} - x - h$ .
We subtract $f (x) = 3 x^{3} - x$ : $f (x + h) - f (x) = 9 x^{2} h + 9 x h^{2} + 3 h^{3} - h$ .
We divide by $h$ (with $h \neq = 0$ ): $\frac{f ( x + h ) - f ( x )}{h} = 9 x^{2} + 9 x h + 3 h^{2} - 1$ .
Letting $h \to 0$ : $f^{'} (x) = 9 x^{2} - 1$ .

Answer: $f^{'} (x) = 9 x^{2} - 1$

Exercise 42 BasicIB · P1 November 1997

A curve is defined by $x = t^{2} + sin 2 t$ and $y = t + sin t$ , where $t$ is a real parameter.

a) Find the gradient of the curve at the point where $t = 0$ .

b) Find the equation of the tangent to the curve at the point where $t = 0$ .

Show answer

Answer: a) gradient $= 1$ ; b) $y = x$

Show working and answer

a) 1. We differentiate each coordinate with respect to $t$ : $\frac{d x}{d t} = 2 t + 2 cos 2 t$ and $\frac{d y}{d t} = 1 + cos t$ . 2. Gradient: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ . 3. At $t = 0$ : $\frac{d x}{d t} = 2$ and $\frac{d y}{d t} = 2$ , so the gradient is $\frac{2}{2} = 1$ .

b) 1. The point at $t = 0$ is $x = 0 + sin 0 = 0$ and $y = 0 + sin 0 = 0$ : the origin. 2. Tangent: $y - 0 = 1 \cdot (x - 0)$ , i.e. $y = x$ .

Answer: a) gradient $= 1$ ; b) $y = x$

Exercise 43 BasicIB · P1 TZ1 May 2015

a) Expand $(x + h)^{3}$ .

b) Hence, find the derivative of $f (x) = x^{3}$ using the definition.

Show answer

Answer: a) $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ ; b) $f^{'} (x) = 3 x^{2}$

Show working and answer

a) 1. We use the binomial theorem with $n = 3$ : $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ .

b) 1. The definition of the derivative: $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$ . 2. We substitute $f (x) = x^{3}$ and the expansion from part a): $\frac{( x + h ) ^{3} - x ^{3}}{h} = \frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} = 3 x^{2} + 3 x h + h^{2} (h \neq = 0) .$ 3. As $h \to 0$ the terms containing $h$ vanish: $f^{'} (x) = 3 x^{2}$ .

Answer: a) $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ ; b) $f^{'} (x) = 3 x^{2}$

Exercise 44 MediumIB · P1 TZ2 May 2008

A normal line to the graph of $y = arctan (x - 1)$ , for $x > 0$ , has equation $y = - 2 x + c$ , where $c \in R$ . Find the value of $c$ .

Show answer

Answer: $c = \frac{π}{4} + 4$

Show working and answer

We differentiate: $y^{'} = \frac{1}{1 + ( x - 1 ) ^{2}}$ (derivative of the arctangent using the chain rule).
The normal has gradient $- 2$ , so the tangent has gradient $\frac{1}{2}$ (perpendicular lines: the product of the gradients is $- 1$ ).
Equating: $\frac{1}{1 + ( x - 1 ) ^{2}} = \frac{1}{2}$ , so $(x - 1)^{2} = 1$ , and $x = 0$ or $x = 2$ .
As $x > 0$ , we take $x = 2$ . Then $y = arctan (1) = \frac{π}{4}$ .
The point $(2, \frac{π}{4})$ lies on the normal $y = - 2 x + c$ : $\frac{π}{4} = - 4 + c$ .
Hence $c = \frac{π}{4} + 4$ .

Answer: $c = \frac{π}{4} + 4$

Exercise 45 MediumIB · P2 TZ2 May 2011

Consider the functions $f (x) = x^{3} + 1$ and $g (x) = \frac{1}{x ^{3} + 1}$ . The graphs of $y = f (x)$ and $y = g (x)$ intersect at the point $(0, 1)$ and at another point called $P$ .

a) Find the coordinates of $P$ .

b) Calculate the size of the angle formed by the tangents to the two graphs at the point $P$ .

Show answer

Answer: a) $P (- 1.26, - 1)$ ; b) $23. 7^{\circ}$

Show working and answer

a) 1. We equate: $x^{3} + 1 = \frac{1}{x ^{3} + 1}$ , that is, $(x^{3} + 1)^{2} = 1$ . 2. Then $x^{3} + 1 = 1$ (which gives $x = 0$ , the point $(0, 1)$ of the statement) or $x^{3} + 1 = - 1$ , that is, $x^{3} = - 2$ . 3. Hence $x = - 32 \approx - 1.26$ and $y = - 2 + 1 = - 1$ . Thus $P (- 1.26, - 1)$ .

b) 1. Gradients: $f^{'} (x) = 3 x^{2}$ and, by the chain rule, $g^{'} (x) = - \frac{3 x ^{2}}{( x ^{3} + 1 ) ^{2}}$ . 2. At $P$ , $x^{2} = 2^{2/3} \approx 1.587$ and $x^{3} + 1 = - 1$ : $f^{'} = 3 \cdot 2^{2/3} \approx 4.762$ and $g^{'} = - 3 \cdot 2^{2/3} \approx - 4.762$ . 3. Each tangent makes with the $x$ -axis an angle of $arctan (4.762) \approx 78.1 4^{\circ}$ (one upwards, one downwards), so between them there is $2 \cdot 78.1 4^{\circ} = 156. 3^{\circ}$ . 4. The acute angle between the lines is the supplementary angle: $18 0^{\circ} - 156. 3^{\circ} \approx 23. 7^{\circ}$ .

Answer: a) $P (- 1.26, - 1)$ ; b) $23. 7^{\circ}$

Exercise 46 MediumIB · P1 TZ0 November 2015

Consider the curve $y = \frac{1}{1 - x}$ , $x \in R$ , $x \neq = 1$ .

a) Find $\frac{d y}{d x}$ .

b) Determine the equation of the normal to the curve at the point where $x = 3$ , in the form $a x + b y + c = 0$ , where $a, b, c \in Z$ .

Show answer

Answer: a) $\frac{d y}{d x} = \frac{1}{( 1 - x ) ^{2}}$ ; b) $8 x + 2 y - 23 = 0$

Show working and answer

a) 1. We rewrite $y = (1 - x)^{- 1}$ and differentiate using the chain rule. 2. $\frac{d y}{d x} = - (1 - x)^{- 2} \cdot (- 1) = \frac{1}{( 1 - x ) ^{2}}$ .

b) 1. At $x = 3$ : $y = \frac{1}{1 - 3} = - \frac{1}{2}$ and $\frac{d y}{d x} = \frac{1}{4}$ . 2. Gradient of the normal: $- 4$ . 3. Normal: $y + \frac{1}{2} = - 4 (x - 3)$ , that is, $y = - 4 x + \frac{23}{2}$ . 4. Multiplying by $2$ and moving everything to one side: $8 x + 2 y - 23 = 0$ .

Answer: a) $\frac{d y}{d x} = \frac{1}{( 1 - x ) ^{2}}$ ; b) $8 x + 2 y - 23 = 0$

Exercise 47 MediumIB · P1 TZ2 May 2013

The curve $C$ is given by $y = \frac{x cos x}{x + cos x}$ , for $x \geq 0$ .

a) Verify that $\frac{d y}{d x} = \frac{cos ^{2} x - x ^{2} sin x}{( x + cos x ) ^{2}}$ , $x \geq 0$ .

b) Find the equation of the tangent to $C$ at the point $(\frac{π}{2}, 0)$ .

Show answer

Answer: b) $y = - x + \frac{π}{2}$

Show working and answer

a) 1. Use the quotient rule with $u = x cos x$ , $v = x + cos x$ . 2. By the product rule, $u^{'} = cos x - x sin x$ , and $v^{'} = 1 - sin x$ . 3. $y^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} = \frac{( cos x - x sin x ) ( x + cos x ) - x cos x ( 1 - sin x )}{( x + cos x ) ^{2}}$ . 4. Expanding the numerator: $x cos x + cos^{2} x - x^{2} sin x - x sin x cos x - x cos x + x sin x cos x = cos^{2} x - x^{2} sin x$ . 5. Hence $\frac{d y}{d x} = \frac{cos ^{2} x - x ^{2} sin x}{( x + cos x ) ^{2}}$ .

b) 1. At $x = \frac{π}{2}$ : $cos \frac{π}{2} = 0$ and $sin \frac{π}{2} = 1$ , so the gradient is $\frac{0 - \frac{π ^{2}}{4}}{( \frac{π}{2} ) ^{2}} = - 1$ . 2. The tangent through $(\frac{π}{2}, 0)$ is $y = - (x - \frac{π}{2})$ .

Answer: b) $y = - x + \frac{π}{2}$

Exercise 48 MediumIB · P1 TZ0 November 2006

Consider the curves $C_{1}$ , $C_{2}$ given by the equations:

$C_{1} : y = x^{2} + k x + k$ , where $k < 0$ is a constant $C_{2} : y = - x^{2} + 2 x - 4$ .

Both curves pass through the point $P$ , and the tangent at $P$ to one of the curves is also tangent at $P$ to the other.

a) Calculate the value of $k$ .

b) Calculate the coordinates of $P$ .

Show answer

Answer: a) $k = - 2$ ; b) $P (1, - 3)$

Show working and answer

a) 1. At $P$ the two curves have the same $y$ -coordinate and the same gradient (same tangent). 2. Gradients: $y_{1}^{'} = 2 x + k$ and $y_{2}^{'} = - 2 x + 2$ . Equating them: $2 x + k = - 2 x + 2 \Rightarrow k = 2 - 4 x$ . 3. Same $y$ -coordinate: $x^{2} + k x + k = - x^{2} + 2 x - 4 \Rightarrow 2 x^{2} + (k - 2) x + k + 4 = 0$ . 4. Substituting $k = 2 - 4 x$ : $2 x^{2} - 4 x^{2} + (6 - 4 x) = 0 \Rightarrow x^{2} + 2 x - 3 = 0 \Rightarrow x = 1$ or $x = - 3$ . 5. $x = 1$ gives $k = - 2$ ; $x = - 3$ gives $k = 14$ , which does not satisfy $k < 0$ . Hence $k = - 2$ .

b) 1. The point is at $x = 1$ . On $C_{2}$ : $y = - 1 + 2 - 4 = - 3$ . 2. Checking on $C_{1}$ : $1 - 2 - 2 = - 3$ . Hence $P (1, - 3)$ .

Answer: a) $k = - 2$ ; b) $P (1, - 3)$

Exercise 49 MediumIB · P2 TZ1 May 2019

Let L be the tangent to the curve $y = x e^{2 x}$ at the point $(1, e^{2})$ .

Calculate the coordinates of the point where L cuts the $x$ -axis.

Show answer

Answer: $(\frac{2}{3}, 0)$

Show working and answer

We differentiate using the product rule: $y^{'} = e^{2 x} + 2 x e^{2 x} = e^{2 x} (1 + 2 x)$ .
At $x = 1$ : $y^{'} (1) = 3 e^{2}$ .
Tangent through $(1, e^{2})$ : $y - e^{2} = 3 e^{2} (x - 1)$ .
Intercept with the $x$ -axis ( $y = 0$ ): $- e^{2} = 3 e^{2} (x - 1) \Rightarrow x - 1 = - \frac{1}{3} \Rightarrow x = \frac{2}{3}$ .

Answer: $(\frac{2}{3}, 0)$

Exercise 50 MediumIB · P1 November 1996

A curve is defined by $x = 2 t - 1$ , $y = \frac{2 t}{t - 1}$ . Find the equation of the tangent to the curve when $x = 5$ .

Show answer

Answer: $y - 3 = - \frac{1}{4} (x - 5)$ , i.e. $y = - \frac{1}{4} x + \frac{17}{4}$

Show working and answer

For $x = 5$ : $2 t - 1 = 5 \Rightarrow t = 3$ , and $y = \frac{6}{2} = 3$ . The point is $(5, 3)$ .
$\frac{d x}{d t} = 2$ . By the quotient rule: $\frac{d y}{d t} = \frac{2 ( t - 1 ) - 2 t}{( t - 1 ) ^{2}} = \frac{- 2}{( t - 1 ) ^{2}}$ , which at $t = 3$ equals $- \frac{1}{2}$ .
The gradient is $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{- 1/2}{2} = - \frac{1}{4}$ .
The tangent is $y - 3 = - \frac{1}{4} (x - 5)$ .

Answer: $y - 3 = - \frac{1}{4} (x - 5)$ , i.e. $y = - \frac{1}{4} x + \frac{17}{4}$

Exercise 51 MediumIB · P1 TZ1 May 2016

Find the $x$ -coordinates of all the points on the curve $y = 2 x^{4} + 6 x^{3} + \frac{7}{2} x^{2} - 5 x + \frac{3}{2}$ at which the tangent to the curve is parallel to the tangent at $(- 1, 6)$ .

Show answer

Answer: $x = \frac{1}{4}$ and $x = - \frac{3}{2}$

Show working and answer

We differentiate: $y^{'} = 8 x^{3} + 18 x^{2} + 7 x - 5$ .
We check that $(- 1, 6)$ lies on the curve: $y (- 1) = 2 - 6 + \frac{7}{2} + 5 + \frac{3}{2} = 6$ . The gradient at $x = - 1$ is $y^{'} (- 1) = - 8 + 18 - 7 - 5 = - 2$ .
We look for the $x$ -values with the same gradient: $8 x^{3} + 18 x^{2} + 7 x - 5 = - 2$ , i.e. $8 x^{3} + 18 x^{2} + 7 x - 3 = 0$ .
$x = - 1$ is a solution ( $- 8 + 18 - 7 - 3 = 0$ ), so we factorise: $(x + 1) (8 x^{2} + 10 x - 3) = 0$ .
Solving $8 x^{2} + 10 x - 3 = 0$ : $x = \frac{- 10 \pm 100 + 96}{16} = \frac{- 10 \pm 14}{16}$ , so $x = \frac{1}{4}$ or $x = - \frac{3}{2}$ .
Besides $x = - 1$ (the given point itself), the tangent is parallel at these two.

Answer: $x = \frac{1}{4}$ and $x = - \frac{3}{2}$

Exercise 52 MediumIB · P1 November 1997

Find the value(s) of $a$ for which the line $9 x - y = 14$ is a tangent to the curve $y = x^{3} - 3 x + a$ at the point on the curve where $x = a$ .

Show answer

Answer: $a = 2$

Show working and answer

The gradient of the line $9 x - y = 14$ is $9$ . The derivative of the curve is $y^{'} = 3 x^{2} - 3$ .
At the point of tangency $x = a$ , the gradient of the curve is $9$ : $3 a^{2} - 3 = 9$ , so $a^{2} = 4$ and $a = \pm 2$ .
Also, the point on the curve with $x = a$ must lie on the line. Its $y$ -coordinate is $y = a^{3} - 3 a + a = a^{3} - 2 a$ , and it must satisfy $y = 9 a - 14$ : $a^{3} - 2 a = 9 a - 14$ , i.e. $a^{3} - 11 a + 14 = 0$ .
Testing: $a = 2$ : $8 - 22 + 14 = 0$ (this works). $a = - 2$ : $- 8 + 22 + 14 = 28 \neq = 0$ (this does not work).
Hence $a = 2$ (the point of tangency is $(2, 4)$ ).

Answer: $a = 2$

Exercise 53 MediumIB · P1 November 1999

Find, in terms of the constant $a$ , the equation of the normal to the curve defined parametrically by ${x = a (θ - sin θ) y = a (1 - cos θ)$ at the point on the curve where $θ = \frac{π}{2}$ .

Show answer

Answer: $y = - x + a \frac{π}{2}$

Show working and answer

Parametric differentiation: $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}$ .
$\frac{d x}{d θ} = a (1 - cos θ)$ and $\frac{d y}{d θ} = a sin θ$ , so $\frac{d y}{d x} = \frac{sin θ}{1 - cos θ}$ .
For $θ = \frac{π}{2}$ : the gradient of the tangent is $\frac{1}{1 - 0} = 1$ , and the gradient of the normal is $- 1$ .
The point is $x = a (\frac{π}{2} - 1)$ , $y = a (1 - 0) = a$ .
Normal: $y - a = - [x - a (\frac{π}{2} - 1)]$ , i.e. $y = - x + a (\frac{π}{2} - 1) + a = - x + \frac{aπ}{2}$ .

Answer: $y = - x + a \frac{π}{2}$

Exercise 54 MediumIB · P2 May 2000

(i) Find, using the definition, the derivative of $f (x) = cos x$ .

(ii) Prove, using mathematical induction, that $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ for all positive integer values of $n$ .

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Answer: i) $f^{'} (x) = - sin x$ ; ii) as required.

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(i) 1. The definition is $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$ . 2. With $f (x) = cos x$ and the formula $cos (x + h) = cos x cos h - sin x sin h$ : $\frac{cos ( x + h ) - cos x}{h} = cos x \cdot \frac{cos h - 1}{h} - sin x \cdot \frac{sin h}{h} .$ 3. We use the known limits $h \to 0 lim \frac{sin h}{h} = 1$ and $h \to 0 lim \frac{cos h - 1}{h} = 0$ . 4. Therefore $f^{'} (x) = cos x \cdot 0 - sin x \cdot 1 = - sin x$ .

(ii) We prove by induction that $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ for $n \in Z^{+}$ .

Base case $n = 1$ : $\frac{d}{d x} (x) = 1 = 1 \cdot x^{0}$ . It holds.
Inductive hypothesis: suppose that for $n = k$ we have $\frac{d}{d x} (x^{k}) = k x^{k - 1}$ .
Inductive step: we write $x^{k + 1} = x \cdot x^{k}$ and apply the product rule and the hypothesis: $\frac{d}{d x} (x^{k + 1}) = 1 \cdot x^{k} + x \cdot k x^{k - 1} = x^{k} + k x^{k} = (k + 1) x^{k} .$
This is the formula for $n = k + 1$ . Since it holds for $n = 1$ and, if it holds for $k$ , it holds for $k + 1$ , the formula is true for all positive integers $n$ (principle of mathematical induction).

Answer: i) $f^{'} (x) = - sin x$ ; ii) as required.

Exercise 55 MediumIB · P1 November 2000

The tangent to the curve $y^{2} = x^{3}$ at the point $P (1, 1)$ meets the $x$ -axis at $Q$ and the $y$ -axis at $R$ .

Find the ratio $PQ : QR$ .

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Answer: $PQ : QR = 2 : 1$

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We differentiate $y^{2} = x^{3}$ implicitly: $2 y \frac{d y}{d x} = 3 x^{2}$ , so $\frac{d y}{d x} = \frac{3 x ^{2}}{2 y}$ .
At $P (1, 1)$ the gradient is $\frac{3}{2}$ , and the tangent is $y - 1 = \frac{3}{2} (x - 1)$ .
Intersection with the $x$ -axis ( $y = 0$ ): $- 1 = \frac{3}{2} (x - 1)$ , so $x = \frac{1}{3}$ , $Q (\frac{1}{3}, 0)$ .
Intersection with the $y$ -axis ( $x = 0$ ): $y = 1 - \frac{3}{2} = - \frac{1}{2}$ , $R (0, - \frac{1}{2})$ .
Distances: $PQ = (\frac{2}{3})^{2} + 1 = \frac{13}{3}$ and $QR = (\frac{1}{3})^{2} + (\frac{1}{2})^{2} = \frac{13}{6}$ .
Ratio: $\frac{PQ}{QR} = \frac{13 /3}{13 /6} = 2$ .

Answer: $PQ : QR = 2 : 1$

Exercise 56 MediumIB · P1 November 2001

The line $y = 16 x - 9$ is a tangent to the curve $y = 2 x^{3} + a x^{2} + b x - 9$ at the point $(1, 7)$ . Find the values of $a$ and $b$ .

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Answer: $a = - 4$ , $b = 18$

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Derivative: $y^{'} = 6 x^{2} + 2 a x + b$ . At $x = 1$ the gradient of the tangent is $16$ : $6 + 2 a + b = 16$ , so $2 a + b = 10$ .
The point $(1, 7)$ lies on the curve: $2 + a + b - 9 = 7$ , so $a + b = 14$ .
Subtract the two equations: $a = 10 - 14 = - 4$ . Then $b = 14 - (- 4) = 18$ .
Check: $y^{'} (1) = 6 - 8 + 18 = 16$ and $y (1) = 2 - 4 + 18 - 9 = 7$ .

Answer: $a = - 4$ , $b = 18$

Exercise 57 MediumIB · P1 November 2001

Consider the tangent to the curve $y = x^{3} + 4 x^{2} + x - 6$ .

a) Find the equation of this tangent at the point where $x = - 1$ .

b) Find the coordinates of the point where the tangent meets the curve again.

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Answer: a) $y = - 4 x - 8$ ; b) $(- 2, 0)$

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a) 1. $y^{'} = 3 x^{2} + 8 x + 1$ ; at $x = - 1$ : $y^{'} = 3 - 8 + 1 = - 4$ . 2. $y (- 1) = - 1 + 4 - 1 - 6 = - 4$ . 3. Tangent: $y + 4 = - 4 (x + 1)$ , i.e. $y = - 4 x - 8$ .

b) 1. Equate the curve and the tangent: $x^{3} + 4 x^{2} + x - 6 = - 4 x - 8$ , i.e. $x^{3} + 4 x^{2} + 5 x + 2 = 0$ . 2. We know $x = - 1$ is a double root (tangency). Factorise: $x^{3} + 4 x^{2} + 5 x + 2 = (x + 1)^{2} (x + 2)$ . 3. The other root is $x = - 2$ , with $y = - 4 (- 2) - 8 = 0$ .

Answer: a) $y = - 4 x - 8$ ; b) $(- 2, 0)$

Exercise 58 MediumIB · P1 November 2002

The tangent to the curve $y = f (x)$ at the point $P (x, y)$ cuts the $x$ -axis at $Q (x - 1, 0)$ . The curve cuts the $y$ -axis at $R (0, 2)$ . Find the equation of the curve.

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Answer: $f (x) = 2 e^{x}$

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The gradient of the tangent at $P (x, y)$ is, since it passes through $Q (x - 1, 0)$ : $\frac{y - 0}{x - ( x - 1 )} = \frac{y}{1} = y$ .
Therefore $\frac{d y}{d x} = y$ : a separable differential equation, $\frac{d y}{y} = d x$ .
Integrating: $ln ∣ y ∣ = x + C$ , that is, $y = A e^{x}$ (with $A$ a constant).
The curve passes through $R (0, 2)$ : $2 = A e^{0}$ , so $A = 2$ .

Answer: $f (x) = 2 e^{x}$

Exercise 59 MediumIB · P1 TZ0 November 2006

Let $f$ be the function defined for $x > - \frac{1}{3}$ by $f (x) = ln (3 x + 1)$ .

a) Find $f^{'} (x)$ .

b) Find the equation of the normal to the curve $y = f (x)$ at the point where $x = 2$ .

Give your answer in the form $y = a x + b$ , where $a, b \in R$ .

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Answer: a) $f^{'} (x) = \frac{3}{3 x + 1}$ ; b) $y = - \frac{7}{3} x + \frac{14}{3} + ln 7$

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a) 1. Chain rule: $f^{'} (x) = \frac{1}{3 x + 1} \cdot 3 = \frac{3}{3 x + 1}$ .

b) 1. At $x = 2$ : $f (2) = ln 7$ and $f^{'} (2) = \frac{3}{7}$ . 2. The gradient of the normal is the negative reciprocal: $- \frac{7}{3}$ . 3. Equation: $y - ln 7 = - \frac{7}{3} (x - 2)$ . 4. Rearranging for $y$ : $y = - \frac{7}{3} x + \frac{14}{3} + ln 7$ .

Answer: a) $f^{'} (x) = \frac{3}{3 x + 1}$ ; b) $y = - \frac{7}{3} x + \frac{14}{3} + ln 7$

Exercise 60 MediumIB · P1 TZ1 May 2009

Consider the functions $f$ and $g$ defined by $f (x) = 2^{\frac{1}{x}}$ and $g (x) = 4 - 2^{\frac{1}{x}}$ , $x \neq = 0$

a) Find the coordinates of $P$ , the point of intersection of the graphs of $f$ and $g$ .

b) Find the equation of the tangent to the graph of $f$ at the point $P$ .

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Answer: a) $P (1, 2)$ ; b) $y - 2 = - 2 ln 2 (x - 1)$

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a) 1. Equate $f (x) = g (x)$ : $2^{1/ x} = 4 - 2^{1/ x}$ . 2. Hence $2 \cdot 2^{1/ x} = 4$ , so $2^{1/ x} = 2$ and $\frac{1}{x} = 1$ , i.e. $x = 1$ . 3. The $y$ -coordinate: $f (1) = 2^{1} = 2$ . Hence $P (1, 2)$ .

b) 1. Differentiate $f (x) = 2^{1/ x}$ using the chain rule: $f^{'} (x) = 2^{1/ x} ln 2 \cdot (- \frac{1}{x ^{2}})$ . 2. At $x = 1$ : $f^{'} (1) = 2 ln 2 \cdot (- 1) = - 2 ln 2$ . 3. Tangent: $y - 2 = - 2 ln 2 (x - 1)$ .

Answer: a) $P (1, 2)$ ; b) $y - 2 = - 2 ln 2 (x - 1)$

Exercise 61 MediumIB · P2 TZ1 May 2012

Given that the graph of $y = x^{3} - 6 x^{2} + k x - 4$ has exactly one point at which the gradient is zero, find the value of $k$ .

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Answer: $k = 12$

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Gradient of the curve: $y^{'} = 3 x^{2} - 12 x + k$ .
The gradient is zero at exactly one point when the equation $3 x^{2} - 12 x + k = 0$ has exactly one solution, i.e. a zero discriminant: $(- 12)^{2} - 4 \cdot 3 \cdot k = 0$ .
$144 - 12 k = 0$ , so $k = 12$ .

Answer: $k = 12$

Exercise 62 MediumIB · P1 TZ2 May 2021

Consider the function $f$ given by $f (x) = ln (x^{2} - 16)$ for $x > 4$ .

The following diagram shows part of the graph of $f$ , which cuts the $x$ -axis at the point $A$ with coordinates $(a, 0)$ . The line $L$ is the tangent to the graph of $f$ at the point $B$ .

Graph of an increasing logarithmic function that cuts the x-axis at point A, with a vertical asymptote and a line L tangent to the curve at point B

a) Find the exact value of $a$ .

b) Given that the gradient of $L$ is equal to $\frac{1}{3}$ , find the $x$ -coordinate of $B$ .

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Answer: a) $a = 17$ ; b) $x = 8$

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a) 1. The point $A$ lies on the $x$ -axis: $f (a) = 0$ , that is $ln (a^{2} - 16) = 0$ . 2. So $a^{2} - 16 = 1$ , $a^{2} = 17$ . As $a > 4$ , $a = 17$ .

b) 1. The gradient of the tangent at $x$ is $f^{'} (x) = \frac{2 x}{x ^{2} - 16}$ (chain rule). 2. Setting this equal to $\frac{1}{3}$ : $\frac{2 x}{x ^{2} - 16} = \frac{1}{3}$ , that is $6 x = x^{2} - 16$ , so $x^{2} - 6 x - 16 = 0$ . 3. Factorising: $(x - 8) (x + 2) = 0$ . As $x > 4$ , $x = 8$ .

Answer: a) $a = 17$ ; b) $x = 8$

Exercise 63 MediumIB · P1 TZ1 May 2021Proof

Consider the functions $f (x) = - (x - h)^{2} + 2 k$ and $g (x) = e^{x - 2} + k$ , where $h, k \in R$ .

a) Calculate $f^{'} (x)$ .

The graphs of $f$ and $g$ have a common tangent at $x = 3$ .

b) Show that $h = \frac{e + 6}{2}$ .

c) Using the above, show that $k = e + \frac{e ^{2}}{4}$ .

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Answer: a) $f^{'} (x) = - 2 (x - h)$ ; b) and c) proofs.

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a) 1. Differentiating (chain rule): $f^{'} (x) = - 2 (x - h)$ .

b) 1. A common tangent at $x = 3$ means that the two graphs have the same gradient at $x = 3$ : $f^{'} (3) = g^{'} (3)$ . 2. $g^{'} (x) = e^{x - 2}$ (chain rule), so $g^{'} (3) = e$ . And $f^{'} (3) = - 2 (3 - h)$ . 3. Equating: $- 2 (3 - h) = e$ , that is $3 - h = - \frac{e}{2}$ , so $h = 3 + \frac{e}{2} = \frac{e + 6}{2}$ .

c) 1. In addition the graphs pass through the same point at $x = 3$ : $f (3) = g (3)$ , that is $- (3 - h)^{2} + 2 k = e + k$ . 2. From b), $3 - h = - \frac{e}{2}$ , so $(3 - h)^{2} = \frac{e ^{2}}{4}$ . 3. Then $- \frac{e ^{2}}{4} + 2 k = e + k$ , from which $k = e + \frac{e ^{2}}{4}$ .

Answer: a) $f^{'} (x) = - 2 (x - h)$ ; b) and c) proofs.

Exercise 64 MediumIB · P1 TZ0 November 2021

The function $f$ is defined for all $x \in R$ .

The line with equation $y = 6 x - 1$ is the tangent to the graph of $f$ at $x = 4$ .

a) Write down the value of $f^{'} (4)$ .

b) Calculate $f (4)$ .

The function $g$ is defined for all $x \in R$ as $g (x) = x^{2} - 3 x$ , and $h (x) = f (g (x))$ .

c) Calculate $h (4)$ .

d) Using the above, calculate the equation of the tangent to the graph of $h$ at $x = 4$ .

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Answer: a) $f^{'} (4) = 6$ ; b) $f (4) = 23$ ; c) $h (4) = 23$ ; d) $y = 30 x - 97$

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a) 1. The gradient of the tangent at $x = 4$ is $f^{'} (4)$ . The line $y = 6 x - 1$ has gradient $6$ : $f^{'} (4) = 6$ .

b) 1. The tangent passes through the point $(4, f (4))$ on the graph, so $f (4) = 6 \cdot 4 - 1 = 23$ .

c) 1. $g (4) = 16 - 12 = 4$ . 2. $h (4) = f (g (4)) = f (4) = 23$ .

d) 1. Chain rule: $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ , with $g^{'} (x) = 2 x - 3$ . 2. At $x = 4$ : $h^{'} (4) = f^{'} (4) \cdot g^{'} (4) = 6 \cdot 5 = 30$ . 3. Tangent through $(4, 23)$ with gradient $30$ : $y - 23 = 30 (x - 4)$ , that is $y = 30 x - 97$ .

Answer: a) $f^{'} (4) = 6$ ; b) $f (4) = 23$ ; c) $h (4) = 23$ ; d) $y = 30 x - 97$

Exercise 65 MediumIB · P1 TZ1 May 2022

Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in R$ and $k \in Q$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Calculate the value of $k$ .

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Answer: $k = 3$

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We differentiate using the product and chain rules: $y^{'} = 2 e^{k x} + (2 x - 1) k e^{k x} = e^{k x} (2 + k (2 x - 1))$ .
At $x = 1$ : $y^{'} (1) = e^{k} (2 + k)$ .
The tangent is parallel to $y = 5 e^{k} x$ , which has gradient $5 e^{k}$ : $e^{k} (2 + k) = 5 e^{k}$ .
Since $e^{k} > 0$ , we divide by it: $2 + k = 5$ , so $k = 3$ .

Answer: $k = 3$

Exercise 66 HardIB · P1 TZ2 May 2006

The normal to the curve $y = k x + ln (x^{2})$ , for $x \neq = 0$ , $k$ real, at the point $x = 2$ , has equation $3 x + 2 y = b$ , where $b$ is real. Find the exact values of $k$ and $b$ .

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Answer: $k = - \frac{1}{3}$ , $b = \frac{14}{3} + 2 ln 4$

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Since $ln (x^{2}) = 2 ln ∣ x ∣$ , the derivative is $y^{'} = k + \frac{2 x}{x ^{2}} = k + \frac{2}{x}$ .
At $x = 2$ the gradient of the tangent is $k + 1$ , and the gradient of the normal is $- \frac{1}{k + 1}$ .
The normal $3 x + 2 y = b$ can be written $y = - \frac{3}{2} x + \frac{b}{2}$ , with gradient $- \frac{3}{2}$ .
Equating the gradients: $- \frac{1}{k + 1} = - \frac{3}{2}$ , so $k + 1 = \frac{2}{3}$ and $k = - \frac{1}{3}$ .
The point on the curve at $x = 2$ is $y = 2 k + ln 4 = - \frac{2}{3} + ln 4$ .
This point lies on the normal: $3 \cdot 2 + 2 (- \frac{2}{3} + ln 4) = b$ .
Therefore $b = 6 - \frac{4}{3} + 2 ln 4 = \frac{14}{3} + 2 ln 4$ .

Answer: $k = - \frac{1}{3}$ , $b = \frac{14}{3} + 2 ln 4$

Exercise 67 HardIB · P1 TZ2 May 2011

The curve $C$ has equation $y = \frac{1}{8} (9 + 8 x^{2} - x^{4})$ .

a) Find the coordinates of the points on $C$ at which $\frac{d y}{d x} = 0$ .

b) The tangent to $C$ at the point $P (1, 2)$ meets the $x$ -axis at the point $T$ . Determine the coordinates of $T$ .

c) The normal to $C$ at the point $P$ meets the $y$ -axis at the point $N$ . Find the area of triangle $PTN$ .

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Answer: a) $(0, \frac{9}{8})$ , $(2, \frac{25}{8})$ , $(- 2, \frac{25}{8})$ ; b) $T (- \frac{1}{3}, 0)$ ; c) $N (0, \frac{8}{3})$ , area $= \frac{13}{9}$

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a) 1. $y = \frac{1}{8} (9 + 8 x^{2} - x^{4})$ , so $\frac{d y}{d x} = \frac{1}{8} (16 x - 4 x^{3}) = \frac{x ( 4 - x ^{2} )}{2}$ . 2. $\frac{d y}{d x} = 0$ when $x = 0$ or $x = \pm 2$ . 3. We calculate $y$ : $y (0) = \frac{9}{8}$ and $y (\pm 2) = \frac{9 + 32 - 16}{8} = \frac{25}{8}$ .

b) 1. We check that $P (1, 2)$ lies on $C$ : $y (1) = \frac{9 + 8 - 1}{8} = 2$ . 2. Gradient of the tangent: $\frac{d y}{d x} (1) = \frac{1 \cdot 3}{2} = \frac{3}{2}$ . 3. Tangent: $y - 2 = \frac{3}{2} (x - 1)$ . 4. It meets the $x$ -axis when $y = 0$ : $- 2 = \frac{3}{2} (x - 1)$ , so $x - 1 = - \frac{4}{3}$ and $x = - \frac{1}{3}$ . Thus $T (- \frac{1}{3}, 0)$ .

c) 1. Gradient of the normal: $- \frac{2}{3}$ . Normal: $y - 2 = - \frac{2}{3} (x - 1)$ . 2. It meets the $y$ -axis at $x = 0$ : $y = 2 + \frac{2}{3} = \frac{8}{3}$ . Thus $N (0, \frac{8}{3})$ . 3. The vectors $PT = (- \frac{4}{3}, - 2)$ and $PN = (- 1, \frac{2}{3})$ are perpendicular (tangent and normal), so the triangle is right-angled at $P$ . 4. Area $= \frac{1}{2} ∣ PT ∣∣ PN ∣ = \frac{1}{2} \frac{16}{9} + 4 \cdot 1 + \frac{4}{9} = \frac{1}{2} \cdot \frac{52}{3} \cdot \frac{13}{3} = \frac{1}{2} \cdot \frac{26}{9} = \frac{13}{9}$ .

Answer: a) $(0, \frac{9}{8})$ , $(2, \frac{25}{8})$ , $(- 2, \frac{25}{8})$ ; b) $T (- \frac{1}{3}, 0)$ ; c) $N (0, \frac{8}{3})$ , area $= \frac{13}{9}$

Exercise 68 HardIB · P2 TZ2 May 2011Proof

The point $P$ , with coordinates $(p, q)$ , lies on the graph of $x^{\frac{1}{2}} + y^{\frac{1}{2}} = a^{\frac{1}{2}}$ , $a > 0$ .

The tangent to the curve at $P$ meets the axes at $(0, m)$ and $(n, 0)$ . Verify that $m + n = a$ .

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Answer: proved that $m + n = a$ .

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We differentiate implicitly $x^{1/2} + y^{1/2} = a^{1/2}$ : $\frac{1}{2 x} + \frac{1}{2 y} \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{y}{x}$ .
At $P (p, q)$ the gradient is $- \frac{q}{p}$ and the tangent is $y - q = - \frac{q}{p} (x - p)$ .
Intersection with the $y$ -axis ( $x = 0$ ): $m = q + \frac{q}{p} p = q + p q$ .
Intersection with the $x$ -axis ( $y = 0$ ): $- q = - \frac{q}{p} (x - p)$ , so $x - p = \frac{q p}{q} = p q$ and $n = p + p q$ .
Adding: $m + n = p + q + 2 p q = (p + q)^{2}$ .
As $P$ lies on the curve, $p + q = a$ , and therefore $m + n = (a)^{2} = a$ .

Answer: proved that $m + n = a$ .

Exercise 69 HardIB · P2 November 2000

a) Sketch the curves $y = x^{2}$ for $- 2 \leq x \leq 2$ , and $y = - \frac{1}{2} ln x$ for $0 < x \leq 2$ .

b) Find the $x$ -coordinate of $P$ , the point of intersection of the two curves.

c) If the tangents to the curves at $P$ cut the $y$ -axis at $Q$ and $R$ , find the area of triangle $PQR$ .

d) Prove that the two tangents, one to each curve, at the points where $x = a$ , $a > 0$ , are perpendicular.

View solution

a) We draw both curves on the same axes: the parabola $y = x^{2}$ for $- 2 \leq x \leq 2$ (vertex at the origin, $y = 4$ at the ends) and $y = - \frac{1}{2} ln x$ for $0 < x \leq 2$ (decreasing, with a vertical asymptote at $x = 0$ , meeting the $x$ -axis at $x = 1$ and reaching $y = - \frac{1}{2} ln 2 \approx - 0.35$ at $x = 2$ ). They intersect at a single point P with $0 < x < 1$ .

Graphs of y = x² for −2 ≤ x ≤ 2 and of y = −(1/2) ln x for 0 < x ≤ 2, with their point of intersection P

b) 1. Equate the two expressions: $x^{2} = - \frac{1}{2} ln x$ , i.e. $2 x^{2} + ln x = 0$ . 2. Using the GDC (intersection of the two graphs): $x \approx 0.548$ , and then $y = x^{2} \approx 0.3005$ . $P (0.548; 0.301)$ .

c) Let $p = 0.5482$ be the $x$ -coordinate of P.

For $y = x^{2}$ : gradient $2 p$ , tangent $y - p^{2} = 2 p (x - p)$ , which cuts the $y$ -axis at $Q (0, - p^{2})$ .
For $y = - \frac{1}{2} ln x$ : gradient $- \frac{1}{2 p}$ , tangent $y - p^{2} = - \frac{1}{2 p} (x - p)$ ; at $x = 0$ : $y = p^{2} + \frac{1}{2}$ , so $R (0, p^{2} + \frac{1}{2})$ (we use the fact that $- \frac{1}{2} ln p = p^{2}$ at P).
$QR = p^{2} + \frac{1}{2} + p^{2} = 2 p^{2} + \frac{1}{2} \approx 1.101$ , and the height of the triangle from P to the line $x = 0$ is $p$ .
Area $= \frac{1}{2} \cdot QR \cdot p = \frac{1}{2} \cdot 1.101 \cdot 0.5482 \approx 0.302$ .

d) 1. For $y = x^{2}$ the gradient of the tangent at $x = a$ is $m_{1} = 2 a$ . 2. For $y = - \frac{1}{2} ln x$ it is $m_{2} = - \frac{1}{2 a}$ . 3. $m_{1} m_{2} = 2 a \cdot (- \frac{1}{2 a}) = - 1$ , so the tangents are perpendicular.

Answer: b) $P (0.548; 0.301)$ ; c) area of triangle $PQR \approx 0.302$ ; d) as required.

Exercise 70 HardIB · P1 TZ2 May 2005

The normal to the curve $y = \frac{k}{x} + ln x^{2}$ , for $x \neq = 0$ , $k \in R$ , at the point where $x = 2$ , has equation $3 x + 2 y = b$ , where $b \in R$ . Find the exact values of $k$ and $b$ .

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Answer: $k = \frac{4}{3}$ , $b = \frac{22}{3} + 2 ln 4$

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Write $y = \frac{k}{x} + 2 ln x$ (for $x > 0$ ; in general $2 ln ∣ x ∣$ , with the same derivative).
Differentiate: $\frac{d y}{d x} = - \frac{k}{x ^{2}} + \frac{2}{x}$ . At $x = 2$ : $y^{'} (2) = - \frac{k}{4} + 1$ .
The normal $3 x + 2 y = b$ has gradient $- \frac{3}{2}$ . The gradient of the normal is $- \frac{1}{y ^{'} ( 2 )}$ , so $y^{'} (2) = \frac{2}{3}$ .
Then $1 - \frac{k}{4} = \frac{2}{3}$ , so $\frac{k}{4} = \frac{1}{3}$ and $k = \frac{4}{3}$ .
Ordinate of the point: $y (2) = \frac{k}{2} + ln 4 = \frac{2}{3} + ln 4$ .
The normal passes through $(2, \frac{2}{3} + ln 4)$ : $b = 3 \cdot 2 + 2 (\frac{2}{3} + ln 4) = 6 + \frac{4}{3} + 2 ln 4 = \frac{22}{3} + 2 ln 4$ .

Answer: $k = \frac{4}{3}$ , $b = \frac{22}{3} + 2 ln 4$

Exercise 71 HardIB · P2 TZ0 November 2005Proof

a) Write down the term $x^{r}$ in the expansion of $(x + h)^{n}$ , where $0 \leq r \leq n$ , $n \in Z^{+}$ .

b) Hence, differentiate $x^{n}$ , $n \in Z^{+}$ , by applying the definition of the derivative.

c) Starting from the result $x^{n} \cdot x^{- n} = 1$ , deduce the derivative of $x^{- n}$ , $n \in Z^{+}$ .

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Answer: proved: a) $(r n) x^{r} h^{n - r}$ ; b) $(x^{n})^{'} = n x^{n - 1}$ ; c) $(x^{- n})^{'} = - n x^{- n - 1}$ .

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a) 1. By the binomial theorem, $(x + h)^{n} = r = 0 \sum n (r n) x^{r} h^{n - r}$ . 2. The term in $x^{r}$ is $(r n) x^{r} h^{n - r}$ (and $(r n) = (n - r n)$ ).

b) 1. By definition, $\frac{d}{d x} x^{n} = h \to 0 lim \frac{( x + h ) ^{n} - x ^{n}}{h}$ . 2. From the expansion, the term $r = n$ is $x^{n}$ , which cancels with $- x^{n}$ . The terms $r = 0, \dots, n - 1$ remain: $\frac{( x + h ) ^{n} - x ^{n}}{h} = r = 0 \sum n - 1 (r n) x^{r} h^{n - r - 1}$ . 3. As $h \to 0$ , all terms with $n - r - 1 \geq 1$ tend to $0$ . Only $r = n - 1$ survives (the power of $h$ equal to $0$ ): $(n - 1 n) x^{n - 1} = n x^{n - 1}$ . 4. Therefore $\frac{d}{d x} x^{n} = n x^{n - 1}$ .

c) 1. Start from $x^{n} \cdot x^{- n} = 1$ (with $x \neq = 0$ ) and differentiate both sides using the product rule: $n x^{n - 1} \cdot x^{- n} + x^{n} \cdot (x^{- n})^{'} = 0$ . 2. Simplify the first term: $n x^{- 1} + x^{n} (x^{- n})^{'} = 0$ . 3. Rearrange: $(x^{- n})^{'} = \frac{- n x ^{- 1}}{x ^{n}} = - n x^{- n - 1}$ .

Answer: proved: a) $(r n) x^{r} h^{n - r}$ ; b) $(x^{n})^{'} = n x^{n - 1}$ ; c) $(x^{- n})^{'} = - n x^{- n - 1}$ .

Exercise 72 HardIB · P2 TZ1 May 2012

The function $f (x) = 3 sin x + 4 cos x$ is defined for $0 < x < 2 π$ .

a) State the coordinates of the minimum point on the graph of $f$ .

b) The points $P (p, 3)$ and $Q (q, 3)$ , $q > p$ , lie on the graph of $y = f (x)$ . Find $p$ and $q$ .

c) Find the coordinates of the point on $y = f (x)$ where the gradient of the graph is 3.

d) Find the coordinates of the point of intersection of the normals to the graph at the points $P$ and $Q$ .

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Answer: a) $(3.79; - 5)$ ; b) $p = 1.57$ , $q = 6.00$ ; c) $(4.43; - 4)$ ; d) $(3.79; 3.55)$

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a) 1. We write $f (x) = 3 sin x + 4 cos x = 5 cos (x - β)$ with $β = arctan \frac{3}{4} \approx 0.6435$ (amplitude $3^{2} + 4^{2} = 5$ ). 2. The minimum is $- 5$ and occurs when $x - β = π$ , i.e. $x = π + β \approx 3.785$ . 3. Minimum: $(3.79; - 5)$ .

b) 1. $f (x) = 3$ is equivalent to $5 cos (x - β) = 3$ , i.e. $cos (x - β) = \frac{3}{5}$ , with $arccos \frac{3}{5} \approx 0.9273$ . 2. Solutions: $x - β = 0.9273$ gives $x \approx 1.571$ (this is $\frac{π}{2}$ ; indeed $f (\frac{π}{2}) = 3$ ). And $x - β = 2 π - 0.9273$ gives $x \approx 5.999 \approx 6.00$ . 3. Hence $p \approx 1.57$ and $q \approx 6.00$ .

c) 1. $f^{'} (x) = 3 cos x - 4 sin x = 3$ . 2. We write $3 cos x - 4 sin x = 5 cos (x + γ)$ with $γ = arctan \frac{4}{3} \approx 0.9273$ : $cos (x + γ) = \frac{3}{5}$ . 3. Solutions: $x + γ = 2 π - 0.9273$ , i.e. $x \approx 4.429$ (the solution $x = 0$ is not in the domain $0 < x < 2 π$ ). 4. The y-coordinate: $f (4.429) \approx - 4$ . Point: $(4.43; - 4)$ .

d) 1. At $P (p, 3)$ with $p = \frac{π}{2}$ : $f^{'} (p) = 3 cos \frac{π}{2} - 4 sin \frac{π}{2} = - 4$ , so the normal has gradient $\frac{1}{4}$ . 2. At $Q (q, 3)$ : $cos q = \frac{24}{25}$ and $sin q = - \frac{7}{25}$ (check: $3 \cdot (- \frac{7}{25}) + 4 \cdot \frac{24}{25} = 3$ ). Then $f^{'} (q) = \frac{72}{25} + \frac{28}{25} = 4$ , and the normal has gradient $- \frac{1}{4}$ . 3. Normal at $P$ : $y - 3 = \frac{1}{4} (x - p)$ . Normal at $Q$ : $y - 3 = - \frac{1}{4} (x - q)$ . 4. Equating: $x - p = - (x - q)$ , so $x = \frac{p + q}{2} \approx \frac{1.5708 + 5.9994}{2} \approx 3.785$ . 5. Substituting: $y = 3 + \frac{1}{4} (3.785 - 1.571) \approx 3.55$ . Point: $(3.79; 3.55)$ .

Answer: a) $(3.79; - 5)$ ; b) $p = 1.57$ , $q = 6.00$ ; c) $(4.43; - 4)$ ; d) $(3.79; 3.55)$

Implicit differentiation

Exercise 73 BasicIB · P1 May 1996

Find the equation of the tangent to the curve $3 x^{2} - 2 x y + y^{2} = 9$ at the point $(2, 3)$ .

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Answer: $y = - 3 x + 9$

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We check that $(2, 3)$ lies on the curve: $3 \cdot 4 - 2 \cdot 2 \cdot 3 + 9 = 12 - 12 + 9 = 9$ . Yes.
We differentiate implicitly (product rule for $x y$ , chain rule for $y^{2}$ ): $6 x - 2 y - 2 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ .
Rearranging: $\frac{d y}{d x} (2 y - 2 x) = 2 y - 6 x \Rightarrow \frac{d y}{d x} = \frac{y - 3 x}{y - x}$ .
At $(2, 3)$ : $\frac{3 - 6}{3 - 2} = - 3$ .
Tangent: $y - 3 = - 3 (x - 2) \Rightarrow y = - 3 x + 9$ .

Answer: $y = - 3 x + 9$

Exercise 74 BasicIB · P1 May 1999

If $2 x^{2} - 3 y^{2} = 2$ , find the two values of $\frac{d y}{d x}$ when $x = 5$ .

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Answer: at $(5, 4)$ : $\frac{d y}{d x} = \frac{5}{6}$ ; at $(5, - 4)$ : $\frac{d y}{d x} = - \frac{5}{6}$

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When $x = 5$ : $2 \cdot 25 - 3 y^{2} = 2$ , so $3 y^{2} = 48$ and $y = \pm 4$ .
We differentiate implicitly (chain rule on $y^{2}$ ): $4 x - 6 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = \frac{2 x}{3 y}$ .
At $(5, 4)$ : $\frac{10}{12} = \frac{5}{6}$ . At $(5, - 4)$ : $\frac{10}{- 12} = - \frac{5}{6}$ .

Answer: at $(5, 4)$ : $\frac{d y}{d x} = \frac{5}{6}$ ; at $(5, - 4)$ : $\frac{d y}{d x} = - \frac{5}{6}$

Exercise 75 BasicIB · P1 May 2001

Find the gradient of the tangent to the curve $3 x^{2} + 4 y^{2} = 7$ at the point where $x = 1$ and $y > 0$ .

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Answer: gradient $= - \frac{3}{4}$

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For $x = 1$ : $3 + 4 y^{2} = 7$ , $y^{2} = 1$ , and since $y > 0$ , $y = 1$ .
We differentiate implicitly: $6 x + 8 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{6 x}{8 y} = - \frac{3 x}{4 y}$ .
At $(1, 1)$ : gradient $= - \frac{3}{4}$ .

Answer: gradient $= - \frac{3}{4}$

Exercise 76 MediumIB · P1 TZ2 May 2008

Consider the curve defined by the equation $x^{2} + x y + y^{2} = 3$ .

a) Find, in terms of $k$ , the gradient of the curve at the point $(- 1, k)$ .

b) Given that at this point the tangent to the curve is parallel to the $x$ -axis, find the value of $k$ .

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Answer: a) $\frac{2 - k}{2 k - 1}$ ; b) $k = 2$

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a) 1. We differentiate implicitly with respect to $x$ (product rule for $x y$ and chain rule for $y^{2}$ ): $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ . 2. Rearranging: $\frac{d y}{d x} (x + 2 y) = - (2 x + y)$ , so $\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y}$ . 3. At the point $(- 1, k)$ : $\frac{d y}{d x} = - \frac{- 2 + k}{- 1 + 2 k} = \frac{2 - k}{2 k - 1}$ .

b) 1. A tangent parallel to the $x$ -axis has gradient $0$ : $\frac{2 - k}{2 k - 1} = 0$ , so $k = 2$ . 2. We check that $(- 1, 2)$ lies on the curve: $1 - 2 + 4 = 3$ . And the denominator $2 k - 1 = 3 \neq = 0$ .

Answer: a) $\frac{2 - k}{2 k - 1}$ ; b) $k = 2$

Exercise 77 MediumIB · P1 TZ0 November 2008

Find the equation of the normal line to the curve $5 x y^{2} - 2 x^{2} = 18$ at the point $(1, 2)$ .

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Answer: $y = \frac{5}{4} x + \frac{3}{4}$

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We check that $(1, 2)$ lies on the curve: $5 \cdot 1 \cdot 4 - 2 = 18$ . Correct.
We differentiate implicitly (product rule for $x y^{2}$ , chain rule for $y^{2}$ ): $5 y^{2} + 10 x y \frac{d y}{d x} - 4 x = 0$ .
At $(1, 2)$ : $20 + 20 \frac{d y}{d x} - 4 = 0$ , so $\frac{d y}{d x} = - \frac{4}{5}$ .
The gradient of the normal is the negative reciprocal: $\frac{5}{4}$ .
Equation: $y - 2 = \frac{5}{4} (x - 1)$ , that is, $y = \frac{5}{4} x + \frac{3}{4}$ .

Answer: $y = \frac{5}{4} x + \frac{3}{4}$

Exercise 78 MediumIB · P2 TZ0 November 2009

Find the gradient of the curve $e^{x y} + ln (y^{2}) + e^{y} = 1 + e$ at the point $(0, 1)$ .

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Answer: $- \frac{1}{2 + e}$

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We check $(0, 1)$ : $e^{0} + ln 1 + e = 1 + e$ . Correct.
We differentiate implicitly, using the chain rule on each term: $e^{x y} (y + x \frac{d y}{d x}) + \frac{2}{y} \frac{d y}{d x} + e^{y} \frac{d y}{d x} = 0$ .
At $(0, 1)$ : $1 \cdot (1 + 0) + 2 \frac{d y}{d x} + e \frac{d y}{d x} = 0$ .
Solving for the gradient: $\frac{d y}{d x} = - \frac{1}{2 + e}$ .

Answer: $- \frac{1}{2 + e}$

Exercise 79 MediumIB · P1 TZ2 May 2019

The curve $C$ is given by the equation $y = x tan (\frac{π x y}{4})$ .

a) Show that, at the point $(1, 1)$ , $\frac{d y}{d x} = \frac{2 + π}{2 - π}$ .

b) Hence, find the equation of the normal to $C$ at the point $(1, 1)$ .

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Answer: (a) shown; (b) $y = \frac{π - 2}{π + 2} x + \frac{4}{π + 2}$

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a) 1. We differentiate implicitly $y = x tan (\frac{π x y}{4})$ (product and chain rules): $\frac{d y}{d x} = tan (\frac{π x y}{4}) + x sec^{2} (\frac{π x y}{4}) \cdot \frac{π}{4} (y + x \frac{d y}{d x})$ . 2. At $(1, 1)$ : $tan \frac{π}{4} = 1$ and $sec^{2} \frac{π}{4} = 2$ . Then $\frac{d y}{d x} = 1 + 2 \cdot \frac{π}{4} (1 + \frac{d y}{d x}) = 1 + \frac{π}{2} + \frac{π}{2} \frac{d y}{d x}$ . 3. Grouping terms: $\frac{d y}{d x} (1 - \frac{π}{2}) = 1 + \frac{π}{2}$ , that is, $\frac{d y}{d x} \cdot \frac{2 - π}{2} = \frac{2 + π}{2}$ . 4. Therefore $\frac{d y}{d x} = \frac{2 + π}{2 - π}$ .

b) 1. The gradient of the normal is $- \frac{2 - π}{2 + π} = \frac{π - 2}{π + 2}$ . 2. Normal: $y - 1 = \frac{π - 2}{π + 2} (x - 1)$ . 3. Therefore $y = \frac{π - 2}{π + 2} x + 1 - \frac{π - 2}{π + 2} = \frac{π - 2}{π + 2} x + \frac{4}{π + 2}$ .

Answer: (a) shown; (b) $y = \frac{π - 2}{π + 2} x + \frac{4}{π + 2}$

Exercise 80 MediumIB · P1 TZ0 November 2015

A curve is given by $x y = y^{2} + 4$ .

a) Show that there is no point at which the tangent to the curve is horizontal.

b) Find the coordinates of those points at which the tangent to the curve is vertical.

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Answer: a) shown; b) $(4, 2)$ and $(- 4, - 2)$

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a) 1. We differentiate implicitly $x y = y^{2} + 4$ (product rule for $x y$ ): $y + x \frac{d y}{d x} = 2 y \frac{d y}{d x}$ . 2. Rearranging: $\frac{d y}{d x} (x - 2 y) = - y$ , so $\frac{d y}{d x} = \frac{y}{2 y - x}$ . 3. A horizontal tangent requires $\frac{d y}{d x} = 0$ , that is, $y = 0$ . 4. But with $y = 0$ the equation of the curve becomes $0 = 0 + 4$ , which is impossible. There is no point on the curve with a horizontal tangent.

b) 1. Vertical tangent: the gradient does not exist, that is, $2 y - x = 0$ (with $y \neq = 0$ ), so $x = 2 y$ . 2. Substituting into the curve: $2 y \cdot y = y^{2} + 4$ , that is, $y^{2} = 4$ and $y = \pm 2$ . 3. Then $x = \pm 4$ . The points are $(4, 2)$ and $(- 4, - 2)$ , and at them $y \neq = 0$ , so the tangent is indeed vertical.

Answer: a) shown; b) $(4, 2)$ and $(- 4, - 2)$

Exercise 81 MediumIB · P1 TZ0 November 2013

A curve has equation $x^{3} y^{2} + x^{3} - y^{3} + 9 y = 0$ . Find the coordinates of the three points on the curve where $\frac{d y}{d x} = 0$ .

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Answer: $(0, 0)$ , $(0, 3)$ , $(0, - 3)$

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Differentiating implicitly with respect to $x$ (product rule for $x^{3} y^{2}$ and chain rule for $y^{2}$ , $y^{3}$ ): $3 x^{2} y^{2} + 2 x^{3} y \frac{d y}{d x} + 3 x^{2} - 3 y^{2} \frac{d y}{d x} + 9 \frac{d y}{d x} = 0$ .
Grouping terms: $\frac{d y}{d x} (2 x^{3} y - 3 y^{2} + 9) = - 3 x^{2} (y^{2} + 1)$ , so $\frac{d y}{d x} = \frac{- 3 x ^{2} ( y ^{2} + 1 )}{2 x ^{3} y - 3 y ^{2} + 9}$ .
$\frac{d y}{d x} = 0$ requires the numerator to be zero: since $y^{2} + 1 > 0$ , this only holds when $x = 0$ .
Substituting $x = 0$ into the curve: $- y^{3} + 9 y = 0 \Rightarrow y (9 - y^{2}) = 0 \Rightarrow y = 0, 3, - 3$ .
We check that the denominator is not zero: at $x = 0$ it equals $9 - 3 y^{2}$ , which is $9$ , $- 18$ and $- 18$ respectively: correct.

Answer: $(0, 0)$ , $(0, 3)$ , $(0, - 3)$

Exercise 82 MediumIB · P2 TZ2 May 2013

The curve $C$ is defined implicitly by the equation $\frac{x ^{2}}{y} - 2 x = ln y$ for $y > 0$ .

a) Express $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the value of $\frac{d y}{d x}$ at the point on $C$ where $y = 1$ and $x > 0$ .

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Answer: a) $\frac{d y}{d x} = \frac{2 x y - 2 y ^{2}}{x ^{2} + y}$ ; b) $\frac{2}{5}$

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a) 1. We differentiate implicitly. The first term is a quotient: $\frac{d}{d x} \frac{x ^{2}}{y} = \frac{2 x y - x ^{2} y ^{'}}{y ^{2}}$ . The rest gives $- 2$ and $\frac{y ^{'}}{y}$ . 2. We get $\frac{2 x y - x ^{2} y ^{'}}{y ^{2}} - 2 = \frac{y ^{'}}{y}$ . Multiplying by $y^{2}$ : $2 x y - x^{2} y^{'} - 2 y^{2} = y y^{'}$ . 3. Grouping the terms: $y^{'} (x^{2} + y) = 2 x y - 2 y^{2}$ . 4. Hence $\frac{d y}{d x} = \frac{2 x y - 2 y ^{2}}{x ^{2} + y}$ .

b) 1. With $y = 1$ : $x^{2} - 2 x = ln 1 = 0 \Rightarrow x (x - 2) = 0$ , and since $x > 0$ , $x = 2$ . 2. $\frac{d y}{d x} = \frac{2 \cdot 2 \cdot 1 - 2 \cdot 1}{4 + 1} = \frac{2}{5}$ .

Answer: a) $\frac{d y}{d x} = \frac{2 x y - 2 y ^{2}}{x ^{2} + y}$ ; b) $\frac{2}{5}$

Exercise 83 MediumIB · P2 TZ2 May 2014

Consider the curve defined by the equation $(x^{2} + y^{2})^{2} = 4 x y^{2}$ .

a) Use implicit differentiation to find an expression for $\frac{d y}{d x}$ .

b) Find the equation of the normal to the curve at the point $(1, 1)$ .

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Answer: a) $\frac{d y}{d x} = \frac{y ^{2} - x ^{3} - x y ^{2}}{x ^{2} y + y ^{3} - 2 x y}$ ; b) $y = 1$

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a) 1. We differentiate implicitly $(x^{2} + y^{2})^{2} = 4 x y^{2}$ . On the left we use the chain rule; on the right, the product rule: $2 (x^{2} + y^{2}) (2 x + 2 y y^{'}) = 4 y^{2} + 8 x y y^{'}$ . 2. We divide by $4$ and expand: $x (x^{2} + y^{2}) + y (x^{2} + y^{2}) y^{'} = y^{2} + 2 x y y^{'}$ . 3. Grouping the terms in $y^{'}$ : $y^{'} (x^{2} y + y^{3} - 2 x y) = y^{2} - x^{3} - x y^{2}$ . 4. Hence $\frac{d y}{d x} = \frac{y ^{2} - x ^{3} - x y ^{2}}{x ^{2} y + y ^{3} - 2 x y}$ .

b) 1. At $(1, 1)$ (which lies on the curve: $(1 + 1)^{2} = 4 = 4 \cdot 1 \cdot 1$ ): the numerator is $1 - 1 - 1 = - 1$ and the denominator is $1 + 1 - 2 = 0$ . 2. Infinite gradient: the tangent is vertical, $x = 1$ , so the normal is horizontal. 3. The normal passes through $(1, 1)$ : $y = 1$ .

Answer: a) $\frac{d y}{d x} = \frac{y ^{2} - x ^{3} - x y ^{2}}{x ^{2} y + y ^{3} - 2 x y}$ ; b) $y = 1$

Exercise 84 MediumIB · P1 TZ1 May 2019

Find the coordinates of the points on the curve $y^{3} + 3 x y^{2} - x^{3} = 27$ at which $\frac{d y}{d x} = 0$ .

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Answer: $(39, 39)$ and $(3, - 3)$

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We differentiate implicitly (product rule for $3 x y^{2}$ , chain rule for $y^{3}$ and $y^{2}$ ): $3 y^{2} \frac{d y}{d x} + 3 y^{2} + 6 x y \frac{d y}{d x} - 3 x^{2} = 0$ .
Solving for the derivative: $\frac{d y}{d x} = \frac{3 x ^{2} - 3 y ^{2}}{3 y ^{2} + 6 x y} = \frac{x ^{2} - y ^{2}}{y ^{2} + 2 x y}$ .
$\frac{d y}{d x} = 0$ requires $x^{2} = y^{2}$ , that is, $y = x$ or $y = - x$ .
If $y = x$ : $x^{3} + 3 x^{3} - x^{3} = 3 x^{3} = 27 \Rightarrow x = 39$ , and the point is $(39, 39)$ . The denominator is $3 x^{2} \neq = 0$ .
If $y = - x$ : $- x^{3} + 3 x^{3} - x^{3} = x^{3} = 27 \Rightarrow x = 3$ , and the point is $(3, - 3)$ . The denominator is $9 - 18 = - 9 \neq = 0$ .

Answer: $(39, 39)$ and $(3, - 3)$

Exercise 85 MediumIB · P1 TZ0 November 2018

Consider the curves $C_{1}$ and $C_{2}$ defined as follows

$C_{1}$ : $x y = 4$ , $x > 0$

$C_{2}$ : $x^{2} - y^{2} = 2$ , $x > 0$

a) Using implicit differentiation or otherwise, find $\frac{d y}{d x}$ in terms of $x$ and $y$ for each of these curves.

Let $P (a, b)$ be the only point at which the curves $C_{1}$ and $C_{2}$ intersect.

b) Show that the tangent to $C_{1}$ at P is perpendicular to the tangent to $C_{2}$ at P.

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Answer: a) For $C_{1}$ : $\frac{d y}{d x} = - \frac{y}{x}$ ; for $C_{2}$ : $\frac{d y}{d x} = \frac{x}{y}$ ; b) the product of the gradients is $- 1$ , so the tangents are perpendicular.

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a) 1. For $C_{1}$ : $x y = 4$ . We differentiate implicitly with respect to $x$ (product rule): $y + x \frac{d y}{d x} = 0$ . 2. We rearrange: $\frac{d y}{d x} = - \frac{y}{x}$ . 3. For $C_{2}$ : $x^{2} - y^{2} = 2$ . We differentiate (chain rule on $y^{2}$ ): $2 x - 2 y \frac{d y}{d x} = 0$ . 4. We rearrange: $\frac{d y}{d x} = \frac{x}{y}$ .

b) 1. At the common point $P (a, b)$ the gradients of the tangents are $m_{1} = - \frac{b}{a}$ (for $C_{1}$ ) and $m_{2} = \frac{a}{b}$ (for $C_{2}$ ). 2. Both exist, because $a > 0$ and $b = \frac{4}{a} \neq = 0$ . 3. Their product is $m_{1} m_{2} = - \frac{b}{a} \cdot \frac{a}{b} = - 1$ . 4. Two lines whose gradients multiply to $- 1$ are perpendicular. This proves the result.

Answer: a) For $C_{1}$ : $\frac{d y}{d x} = - \frac{y}{x}$ ; for $C_{2}$ : $\frac{d y}{d x} = \frac{x}{y}$ ; b) the product of the gradients is $- 1$ , so the tangents are perpendicular.

Exercise 86 MediumIB · P2 TZ1 May 2017

The curve C is defined by the equation $x y - ln y = 1$ , $y > 0$ .

a) Calculate $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the equation of the tangent to C at the point $(\frac{2}{e}, e)$ .

Show answer

Answer: a) $\frac{d y}{d x} = \frac{- y ^{2}}{x y - 1}$ ; b) $y = - e^{2} x + 3 e$

Show working and answer

a) 1. We differentiate $x y - ln y = 1$ with respect to $x$ (product rule on $x y$ and chain rule on $ln y$ ): $y + x \frac{d y}{d x} - \frac{1}{y} \frac{d y}{d x} = 0$ . 2. We group terms: $\frac{d y}{d x} (x - \frac{1}{y}) = - y$ . 3. We solve for the derivative: $\frac{d y}{d x} = \frac{- y}{x - \frac{1}{y}} = \frac{- y ^{2}}{x y - 1}$ .

b) 1. We check that the point lies on C: $\frac{2}{e} \cdot e - ln e = 2 - 1 = 1$ . 2. Gradient at $(\frac{2}{e}, e)$ : $x y = 2$ , so $m = \frac{- e ^{2}}{2 - 1} = - e^{2}$ . 3. Tangent: $y - e = - e^{2} (x - \frac{2}{e})$ , that is, $y = - e^{2} x + 2 e + e$ . 4. Therefore $y = - e^{2} x + 3 e$ .

Answer: a) $\frac{d y}{d x} = \frac{- y ^{2}}{x y - 1}$ ; b) $y = - e^{2} x + 3 e$

Exercise 87 MediumIB · P1 TZ0 November 2017

The folium of Descartes is a curve defined by the equation $x^{3} + y^{3} - 3 x y = 0$ and is shown in the following diagram.

Curve of equation x cubed plus y cubed minus 3xy equals zero (folium of Descartes), with a loop in the first quadrant and two infinite branches, and point P marked on the loop

Determine the exact coordinates of the point P where the tangent to the curve is parallel to the $y$ -axis.

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Answer: $P (34, 32)$

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We differentiate $x^{3} + y^{3} - 3 x y = 0$ implicitly (product in $3 x y$ ): $3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 y - 3 x \frac{d y}{d x} = 0$ .
Rearranging: $\frac{d y}{d x} (y^{2} - x) = y - x^{2}$ , so $\frac{d y}{d x} = \frac{y - x ^{2}}{y ^{2} - x}$ .
The tangent is parallel to the $y$ -axis (vertical) when the denominator is zero and the numerator is not: $y^{2} - x = 0$ , i.e. $x = y^{2}$ .
Substituting into the curve: $y^{6} + y^{3} - 3 y^{3} = 0$ , i.e. $y^{3} (y^{3} - 2) = 0$ .
If $y = 0$ we get the origin, where the numerator is also $0$ (so it is not a vertical tangent). If $y^{3} = 2$ , then $y = 32$ and $x = y^{2} = 34$ . We check that the numerator is not zero: $y - x^{2} = 32 - 316 = - 32 \neq = 0$ .
The point $P$ is the one on the loop to the right: $(34, 32)$ .

Answer: $P (34, 32)$

Exercise 88 MediumIB · P2 TZ2 May 2016

Consider the curve given by the equation $x^{3} + y^{3} = 4 x y$ .

a) Use implicit differentiation to show that $\frac{d y}{d x} = \frac{4 y - 3 x ^{2}}{3 y ^{2} - 4 x}$ .

The tangent to this curve is parallel to the $x$ -axis at the point where $x = k$ , $k > 0$ .

b) Find the value of $k$ .

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Answer: a) proved; b) $k = \frac{4 3 2}{3}$

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a) 1. We differentiate $x^{3} + y^{3} = 4 x y$ implicitly (product in $4 x y$ ): $3 x^{2} + 3 y^{2} \frac{d y}{d x} = 4 y + 4 x \frac{d y}{d x}$ . 2. Grouping: $\frac{d y}{d x} (3 y^{2} - 4 x) = 4 y - 3 x^{2}$ . 3. $\frac{d y}{d x} = \frac{4 y - 3 x ^{2}}{3 y ^{2} - 4 x}$ .

b) 1. Tangent parallel to the $x$ -axis: $\frac{d y}{d x} = 0$ , i.e. $4 y - 3 x^{2} = 0$ , so $y = \frac{3 x ^{2}}{4}$ . 2. Substituting into the curve: $x^{3} + \frac{27 x ^{6}}{64} = 4 x \cdot \frac{3 x ^{2}}{4} = 3 x^{3}$ . 3. Simplifying: $\frac{27 x ^{6}}{64} = 2 x^{3}$ ; with $x > 0$ we divide by $x^{3}$ : $x^{3} = \frac{128}{27}$ . 4. $k = \frac{3 128}{3} = \frac{4 3 2}{3}$ .

Answer: a) proved; b) $k = \frac{4 3 2}{3}$

Exercise 89 MediumIB · P1 TZ0 November 2016

A curve is given by the equation $3 x - 2 y^{2} e^{x - 1} = 2$ .

a) Find an expression for $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the equations of the tangents to this curve at those points where the curve meets the line $x = 1$ .

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Answer: a) $\frac{d y}{d x} = \frac{3 - 2 y ^{2} e ^{x - 1}}{4 y e ^{x - 1}}$ ; b) $y - \frac{1}{2} = \frac{2}{2} (x - 1)$ and $y + \frac{1}{2} = - \frac{2}{2} (x - 1)$

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a) 1. We differentiate $3 x - 2 y^{2} e^{x - 1} = 2$ implicitly (product and chain rules): $3 - 4 y \frac{d y}{d x} e^{x - 1} - 2 y^{2} e^{x - 1} = 0$ . 2. Rearranging: $\frac{d y}{d x} = \frac{3 - 2 y ^{2} e ^{x - 1}}{4 y e ^{x - 1}} = \frac{2 e ^{x - 1} y ^{2} - 3}{- 4 y e ^{x - 1}}$ .

b) 1. With $x = 1$ : $3 - 2 y^{2} = 2$ , so $y^{2} = \frac{1}{2}$ and $y = \pm \frac{1}{2}$ . 2. Gradient at $(1, \frac{1}{2})$ : $\frac{3 - 2 \cdot \frac{1}{2}}{4 \cdot \frac{1}{2}} = \frac{2}{2 2} = \frac{2}{2}$ . Tangent: $y - \frac{1}{2} = \frac{2}{2} (x - 1)$ . 3. Gradient at $(1, - \frac{1}{2})$ : $\frac{2}{- 2 2} = - \frac{2}{2}$ . Tangent: $y + \frac{1}{2} = - \frac{2}{2} (x - 1)$ .

Answer: a) $\frac{d y}{d x} = \frac{3 - 2 y ^{2} e ^{x - 1}}{4 y e ^{x - 1}}$ ; b) $y - \frac{1}{2} = \frac{2}{2} (x - 1)$ and $y + \frac{1}{2} = - \frac{2}{2} (x - 1)$

Exercise 90 MediumIB · P1 May 1997

A curve in the plane has equation $x y^{2} + x^{2} y = 2$ .

a) Calculate the gradient of the curve at the point $(1, 1)$ .

b) Find the equation of the line perpendicular to the curve at the point $(1, 1)$ .

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Answer: a) gradient $= - 1$ ; b) $y = x$

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a) 1. We differentiate $x y^{2} + x^{2} y = 2$ implicitly (product rule on each term): $y^{2} + 2 x y \frac{d y}{d x} + 2 x y + x^{2} \frac{d y}{d x} = 0$ . 2. At $(1, 1)$ : $1 + 2 \frac{d y}{d x} + 2 + \frac{d y}{d x} = 0$ , so $3 \frac{d y}{d x} = - 3$ . 3. The gradient is $- 1$ .

b) 1. The line perpendicular to the curve (the normal) has gradient $\frac{- 1}{- 1} = 1$ . 2. Passing through $(1, 1)$ : $y - 1 = 1 \cdot (x - 1)$ , i.e. $y = x$ .

Answer: a) gradient $= - 1$ ; b) $y = x$

Exercise 91 MediumIB · P1 May 1998

The equation of a curve is $x^{3} + y^{3} = 6 x y$ . Find the equation of the tangent to the curve at the point $(3, 3)$ .

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Answer: $y = - x + 6$

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We check that $(3, 3)$ lies on the curve: $27 + 27 = 54 = 6 \cdot 3 \cdot 3$ .
We differentiate $x^{3} + y^{3} = 6 x y$ implicitly (product rule on $6 x y$ ): $3 x^{2} + 3 y^{2} \frac{d y}{d x} = 6 y + 6 x \frac{d y}{d x}$ .
At $(3, 3)$ : $27 + 27 \frac{d y}{d x} = 18 + 18 \frac{d y}{d x}$ , so $9 \frac{d y}{d x} = - 9$ and the gradient is $- 1$ .
Tangent: $y - 3 = - (x - 3)$ , i.e. $y = - x + 6$ .

Answer: $y = - x + 6$

Exercise 92 MediumIB · P1 May 2002

The equation of a curve is $x y^{3} + 2 x^{2} y = 3$ .

Find the equation of the tangent to this curve at the point $(1, 1)$ .

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Answer: $y = - x + 2$

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We differentiate $x y^{3} + 2 x^{2} y = 3$ implicitly (product rule on each term): $y^{3} + 3 x y^{2} y^{'} + 4 x y + 2 x^{2} y^{'} = 0$ .
At $(1, 1)$ : $1 + 3 y^{'} + 4 + 2 y^{'} = 0$ , so $5 y^{'} = - 5$ and $y^{'} = - 1$ .
Tangent: $y - 1 = - (x - 1)$ , i.e. $y = - x + 2$ .

Answer: $y = - x + 2$

Exercise 93 MediumIB · P1 May 2003

A curve has the equation $x^{3} y^{2} = 8$ . Find the equation of the normal to the curve at the point $(2, 1)$ .

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Answer: $y - 1 = \frac{4}{3} (x - 2)$

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Differentiating $x^{3} y^{2} = 8$ implicitly (product rule): $3 x^{2} y^{2} + 2 x^{3} y \frac{d y}{d x} = 0$ .
At $(2, 1)$ : $12 + 16 \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{3}{4}$ .
The gradient of the normal is the negative reciprocal, $\frac{4}{3}$ .
Normal: $y - 1 = \frac{4}{3} (x - 2)$ , that is, $4 x - 3 y - 5 = 0$ .

Answer: $y - 1 = \frac{4}{3} (x - 2)$

Exercise 94 MediumIB · P1 TZ1 May 2004

The point $P (1, p)$ , where $p > 0$ , lies on the curve $2 y^{2} - x^{3} y = 15$ .

a) Calculate the value of $p$ .

b) Calculate the gradient of the tangent to the curve at $P$ .

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Answer: a) $p = 3$ ; b) gradient $= \frac{9}{11}$

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a) Substituting $x = 1$ into $2 y^{2} - x^{3} y = 15$ : $2 p^{2} - p = 15$ , that is, $2 p^{2} - p - 15 = 0$ . Factorising: $(2 p + 5) (p - 3) = 0$ , so $p = 3$ or $p = - \frac{5}{2}$ . As $p > 0$ , $p = 3$ .

b) 1. Differentiating implicitly: $4 y \frac{d y}{d x} - 3 x^{2} y - x^{3} \frac{d y}{d x} = 0$ . 2. Rearranging: $\frac{d y}{d x} = \frac{3 x ^{2} y}{4 y - x ^{3}}$ . 3. At $P (1, 3)$ : $\frac{9}{12 - 1} = \frac{9}{11}$ .

Answer: a) $p = 3$ ; b) gradient $= \frac{9}{11}$

Exercise 95 MediumIB · P1 November 2003

Consider the equation $2 x y^{2} = x^{2} y + 3$ .

a) Find $y$ when $x = 1$ and $y < 0$ .

b) Find $\frac{d y}{d x}$ when $x = 1$ and $y < 0$ .

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Answer: a) $y = - 1$ ; b) $\frac{d y}{d x} = \frac{4}{5}$

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a) 1. With $x = 1$ : $2 y^{2} = y + 3$ , that is, $2 y^{2} - y - 3 = 0$ . 2. Factorising: $(2 y - 3) (y + 1) = 0$ , so $y = \frac{3}{2}$ or $y = - 1$ . As $y < 0$ , $y = - 1$ .

b) 1. Differentiating implicitly (product rule on all three terms): $2 y^{2} + 4 x y \frac{d y}{d x} = 2 x y + x^{2} \frac{d y}{d x}$ . 2. At $(1, - 1)$ : $2 - 4 \frac{d y}{d x} = - 2 + \frac{d y}{d x}$ . 3. $4 = 5 \frac{d y}{d x}$ , so $\frac{d y}{d x} = \frac{4}{5}$ .

Answer: a) $y = - 1$ ; b) $\frac{d y}{d x} = \frac{4}{5}$

Exercise 96 MediumIB · P1 TZ2 May 2004

The point $P (1, p)$ , with $p > 0$ , lies on the curve $2 x^{2} y + 3 y^{2} = 16$ .

a) Calculate the value of $p$ .

b) Calculate the gradient of the tangent to the curve at $P$ .

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Answer: a) $p = 2$ ; b) gradient $= - \frac{4}{7}$

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a) With $x = 1$ : $2 y + 3 y^{2} = 16$ , that is, $3 p^{2} + 2 p - 16 = 0$ . Then $p = \frac{- 2 \pm 14}{6}$ , so $p = 2$ or $p = - \frac{8}{3}$ . As $p > 0$ , $p = 2$ .

b) 1. Differentiating implicitly: $4 x y + 2 x^{2} \frac{d y}{d x} + 6 y \frac{d y}{d x} = 0$ . 2. At $P (1, 2)$ : $8 + 2 \frac{d y}{d x} + 12 \frac{d y}{d x} = 0$ , so $14 \frac{d y}{d x} = - 8$ . 3. $\frac{d y}{d x} = - \frac{4}{7}$ .

Answer: a) $p = 2$ ; b) gradient $= - \frac{4}{7}$

Exercise 97 MediumIB · P1 TZ0 November 2004

Find the equation of the normal to the curve $x^{3} + y^{3} - 9 x y = 0$ at the point $(2, 4)$ .

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Answer: $y - 4 = - \frac{5}{4} (x - 2)$

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We check that $(2, 4)$ lies on the curve: $8 + 64 - 72 = 0$ .
Differentiating implicitly: $3 x^{2} + 3 y^{2} \frac{d y}{d x} - 9 y - 9 x \frac{d y}{d x} = 0$ .
At $(2, 4)$ : $12 + 48 \frac{d y}{d x} - 36 - 18 \frac{d y}{d x} = 0$ , so $30 \frac{d y}{d x} = 24$ and $\frac{d y}{d x} = \frac{4}{5}$ .
The gradient of the normal is $- \frac{5}{4}$ .
Normal: $y - 4 = - \frac{5}{4} (x - 2)$ , that is, $5 x + 4 y - 26 = 0$ .

Answer: $y - 4 = - \frac{5}{4} (x - 2)$

Exercise 98 MediumIB · P1 TZ1 May 2005

Find the gradient of the curve $2 sin (x y) = 1$ when $y = \frac{1}{2}$ and $π < x < 2 π$ .

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Answer: gradient $= - \frac{3}{10 π}$

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Condition on $x$ : with $y = \frac{1}{2}$ , $2 sin (\frac{x}{2}) = 1$ , so $sin \frac{x}{2} = \frac{1}{2}$ . With $π < x < 2 π$ , $\frac{x}{2} \in] \frac{π}{2}, π [$ , so $\frac{x}{2} = \frac{5 π}{6}$ and $x = \frac{5 π}{3}$ .
Differentiating implicitly $2 sin (x y) = 1$ (chain and product rules): $2 cos (x y) (y + x \frac{d y}{d x}) = 0$ .
At the point, $cos (x y) = cos \frac{5 π}{6} \neq = 0$ , so $y + x \frac{d y}{d x} = 0$ and $\frac{d y}{d x} = - \frac{y}{x}$ .
$\frac{d y}{d x} = - \frac{1/2}{5 π /3} = - \frac{3}{10 π}$ .

Answer: gradient $= - \frac{3}{10 π}$

Exercise 99 MediumIB · P1 TZ0 May 2006

Given that $3^{x + y} = x^{3} + 3 y$ , find $\frac{d y}{d x}$ .

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Answer: $\frac{d y}{d x} = \frac{3 x ^{2} - ( ln 3 ) 3 ^{x + y}}{( ln 3 ) 3 ^{x + y} - 3}$

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Differentiate implicitly with respect to $x$ . On the left-hand side, by the chain rule: $\frac{d}{d x} 3^{x + y} = (ln 3) 3^{x + y} (1 + \frac{d y}{d x})$ .
Right-hand side: $\frac{d}{d x} (x^{3} + 3 y) = 3 x^{2} + 3 \frac{d y}{d x}$ .
Equate: $(ln 3) 3^{x + y} + (ln 3) 3^{x + y} \frac{d y}{d x} = 3 x^{2} + 3 \frac{d y}{d x}$ .
Group the terms in $\frac{d y}{d x}$ : $\frac{d y}{d x} [(ln 3) 3^{x + y} - 3] = 3 x^{2} - (ln 3) 3^{x + y}$ .
Solve for $\frac{d y}{d x}$ .

Answer: $\frac{d y}{d x} = \frac{3 x ^{2} - ( ln 3 ) 3 ^{x + y}}{( ln 3 ) 3 ^{x + y} - 3}$

Exercise 100 MediumIB · P1 TZ1 May 2007

Given that $e^{x y} - y^{2} ln x = e$ for $x \geq 1$ , calculate $\frac{d y}{d x}$ at the point $(1, 1)$ .

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Answer: $\frac{d y}{d x} = \frac{1 - e}{e}$

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Differentiate implicitly with respect to $x$ . The first term by the chain and product rules: $e^{x y} (y + x \frac{d y}{d x})$ . The second, using the product rule: $- (2 y \frac{d y}{d x} ln x + \frac{y ^{2}}{x})$ . The derivative of $e$ is $0$ .
Equation: $e^{x y} (y + x \frac{d y}{d x}) - 2 y ln x \frac{d y}{d x} - \frac{y ^{2}}{x} = 0$ .
Substitute $(1, 1)$ (which satisfies the equation: $e^{1} - 1 \cdot ln 1 = e$ ) and $ln 1 = 0$ : $e (1 + \frac{d y}{d x}) - 0 - 1 = 0$ .
Solve: $\frac{d y}{d x} = \frac{1 - e}{e} = \frac{1}{e} - 1$ .

Answer: $\frac{d y}{d x} = \frac{1 - e}{e}$

Exercise 101 MediumIB · P2 TZ1 May 2008

Find the gradient of the tangent to the curve $x^{3} y^{2} = cos (π y)$ at the point $(- 1, 1)$ .

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Answer: gradient $= \frac{3}{2}$

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Check that $(- 1, 1)$ lies on the curve: $(- 1)^{3} \cdot 1^{2} = - 1$ and $cos π = - 1$ .
Differentiate implicitly. First, the left-hand side using the product rule: $3 x^{2} y^{2} + 2 x^{3} y \frac{d y}{d x}$ . Right-hand side (chain rule): $- π sin (π y) \frac{d y}{d x}$ .
At $(- 1, 1)$ : $sin π = 0$ , so $3 + 2 (- 1) \frac{d y}{d x} = 0$ .
Solving for $\frac{d y}{d x}$ : $\frac{d y}{d x} = \frac{3}{2}$ .

Answer: gradient $= \frac{3}{2}$

Exercise 102 MediumIB · P2 TZ0 November 2010

Find the equation of the normal to the curve $x^{3} y^{3} - x y = 0$ at the point $(1, 1)$ .

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Answer: $y = x$

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Check $(1, 1)$ : $1 - 1 = 0$ .
Differentiate implicitly (product rule for each term): $3 x^{2} y^{3} + 3 x^{3} y^{2} \frac{d y}{d x} - y - x \frac{d y}{d x} = 0$ .
At $(1, 1)$ : $3 + 3 \frac{d y}{d x} - 1 - \frac{d y}{d x} = 0$ , i.e. $2 + 2 \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - 1$ .
The normal has gradient $1$ : $y - 1 = 1 \cdot (x - 1)$ .

Answer: $y = x$

Exercise 103 MediumIB · P1 TZ1 May 2012

The curve $C$ has equation $2 x^{2} + y^{2} = 18$ .

Find the coordinates of the four points of $C$ at which the normal passes through the point $(1, 0)$ .

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Answer: $(3, 0)$ , $(- 3, 0)$ , $(- 1, 4)$ , $(- 1, - 4)$

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Gradient of the tangent at $(a, b)$ on $C$ : we differentiate $2 x^{2} + y^{2} = 18$ implicitly: $4 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{2 x}{y}$ (if $y \neq = 0$ ).
The normal at $(a, b)$ has gradient $\frac{b}{2 a}$ (if $a \neq = 0$ , $b \neq = 0$ ). For it to pass through $(1, 0)$ , the gradient between $(a, b)$ and $(1, 0)$ must be the same: $\frac{b}{a - 1} = \frac{b}{2 a}$ .
Case $b \neq = 0$ : $2 a = a - 1$ , so $a = - 1$ . Then $2 + b^{2} = 18$ and $b = \pm 4$ : points $(- 1, 4)$ and $(- 1, - 4)$ .
Case $b = 0$ : $2 a^{2} = 18$ , so $a = \pm 3$ . At $(\pm 3, 0)$ the tangent is vertical and the normal is the horizontal line $y = 0$ , which passes through $(1, 0)$ : these give $(3, 0)$ and $(- 3, 0)$ .
Case $a = 0$ ( $b = \pm 32$ ): the tangent is horizontal and the normal is $x = 0$ , which does not pass through $(1, 0)$ . Not valid.

Answer: $(3, 0)$ , $(- 3, 0)$ , $(- 1, 4)$ , $(- 1, - 4)$

Exercise 104 MediumIB · P1 TZ1 May 2014

A curve has equation $arctan x^{2} + arctan y^{2} = \frac{π}{4}$ .

a) Calculate $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the gradient of the curve at the point where $x = \frac{1}{2}$ and $y < 0$ .

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Answer: a) $\frac{d y}{d x} = - \frac{x ( 1 + y ^{4} )}{y ( 1 + x ^{4} )}$ ; b) $m = \frac{8 3}{9 2} = \frac{4 6}{9}$

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a) 1. We differentiate implicitly with respect to $x$ , treating $y$ as a function of $x$ . Recall that $\frac{d}{d x} arctan u = \frac{u ^{'}}{1 + u ^{2}}$ . 2. For $arctan x^{2}$ : $u = x^{2}$ , $u^{'} = 2 x$ , which gives $\frac{2 x}{1 + x ^{4}}$ . For $arctan y^{2}$ : $u = y^{2}$ , $u^{'} = 2 y \frac{d y}{d x}$ , which gives $\frac{2 y}{1 + y ^{4}} \frac{d y}{d x}$ . The right-hand side is constant, so its derivative is $0$ : $\frac{2 x}{1 + x ^{4}} + \frac{2 y}{1 + y ^{4}} \frac{d y}{d x} = 0.$ 3. We rearrange: $\frac{d y}{d x} = - \frac{x ( 1 + y ^{4} )}{y ( 1 + x ^{4} )} .$

b) 1. For $x = \frac{1}{2}$ we have $x^{2} = \frac{1}{2}$ . Substituting into the equation: $arctan y^{2} = \frac{π}{4} - arctan \frac{1}{2}$ . 2. We take the tangent of both sides and use $tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ : $y^{2} = tan (\frac{π}{4} - arctan \frac{1}{2}) = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3} .$ 3. Since $y < 0$ , $y = - \frac{1}{3}$ . Also $x^{4} = \frac{1}{4}$ and $y^{4} = \frac{1}{9}$ . 4. We substitute into the expression from part a): $\frac{d y}{d x} = - \frac{\frac{1}{2} \cdot \frac{10}{9}}{- \frac{1}{3} \cdot \frac{5}{4}} = \frac{1}{2} \cdot \frac{10}{9} \cdot \frac{4}{5} \cdot 3 = \frac{8 3}{9 2} = \frac{4 6}{9} .$

Answer: a) $\frac{d y}{d x} = - \frac{x ( 1 + y ^{4} )}{y ( 1 + x ^{4} )}$ ; b) $m = \frac{8 3}{9 2} = \frac{4 6}{9}$

Exercise 105 MediumIB · P1 TZ1 May 2013

A curve is defined by the equation $8 y ln x - 2 x^{2} + 4 y^{2} = 7$ .

Find the equation of the tangent to the curve at the point where $x = 1$ and $y > 0$ .

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Answer: $y = - \frac{2}{3} x + \frac{13}{6}$

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We find $y$ when $x = 1$ : $8 y ln 1 - 2 + 4 y^{2} = 7$ , i.e. $4 y^{2} = 9$ , so $y^{2} = \frac{9}{4}$ . Since $y > 0$ , $y = \frac{3}{2}$ .
We differentiate implicitly with respect to $x$ (product rule on $8 y ln x$ ): $8 y^{'} ln x + \frac{8 y}{x} - 4 x + 8 y y^{'} = 0.$
We substitute $x = 1$ , $y = \frac{3}{2}$ (and $ln 1 = 0$ ): $0 + 12 - 4 + 12 y^{'} = 0$ , so $y^{'} = - \frac{8}{12} = - \frac{2}{3}$ .
The tangent at $(1, \frac{3}{2})$ with gradient $- \frac{2}{3}$ : $y - \frac{3}{2} = - \frac{2}{3} (x - 1)$ , i.e. $y = - \frac{2}{3} x + \frac{13}{6} .$

Answer: $y = - \frac{2}{3} x + \frac{13}{6}$

Exercise 106 MediumIB · P1 TZ0 November 2012

Consider the curve defined by the equation $x^{2} + sin y - x y = 0$ .

a) Find the gradient of the tangent to the curve at the point $(π, π)$ .

b) Hence, show that $tan θ = \frac{1}{1 + 2 π}$ , where $θ$ is the acute angle between the tangent to the curve at $(π, π)$ and the line $y = x$ .

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Answer: a) $\frac{π}{1 + π}$ ; b) proof complete.

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a) 1. We check that $(π, π)$ lies on the curve: $π^{2} + sin π - π^{2} = 0$ . 2. We differentiate implicitly (product rule on $x y$ ): $2 x + cos y \cdot y^{'} - y - x y^{'} = 0.$ 3. At $(π, π)$ : $2 π + (- 1) y^{'} - π - π y^{'} = 0$ , i.e. $π = (1 + π) y^{'}$ . 4. Gradient: $m = \frac{π}{1 + π}$ .

b) 1. The line $y = x$ has gradient $1$ . The acute angle $θ$ between two lines with gradients $m_{1}$ and $m_{2}$ satisfies $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ . 2. With $m_{1} = \frac{π}{1 + π}$ and $m_{2} = 1$ : $m_{1} - m_{2} = \frac{π - 1 - π}{1 + π} = - \frac{1}{1 + π}$ and $1 + m_{1} m_{2} = \frac{1 + π + π}{1 + π} = \frac{1 + 2 π}{1 + π}$ . 3. Therefore $tan θ = \frac{- \frac{1}{1 + π}}{\frac{1 + 2 π}{1 + π}} = \frac{1}{1 + 2 π} .$

Answer: a) $\frac{π}{1 + π}$ ; b) proof complete.

Exercise 107 MediumIB · P2 TZ0 November 2021

Consider the curve $C$ given by $y = x - x y ln (x y)$ , where $x > 0$ , $y > 0$ .

a) Show that $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + ln (x y)) = 1$ .

b) Hence, find the equation of the tangent to $C$ at the point where $x = 1$ .

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Answer: a) proof; b) $y = 1$

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a) 1. We differentiate the equation $y = x - x y ln (x y)$ implicitly with respect to $x$ . For the term $x y ln (x y)$ we use the product rule with $u = x y$ (whose derivative is $y + x \frac{d y}{d x}$ ) and $v = ln (x y)$ (whose derivative is $\frac{1}{x y} (y + x \frac{d y}{d x})$ ): $\frac{d}{d x} [x y ln (x y)] = (y + x \frac{d y}{d x}) ln (x y) + x y \cdot \frac{1}{x y} (y + x \frac{d y}{d x}) = (x \frac{d y}{d x} + y) (ln (x y) + 1) .$ 2. Therefore $\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y) (1 + ln (x y))$ , that is, $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + ln (x y)) = 1.$

b) 1. For $x = 1$ the equation of $C$ is $y = 1 - y ln y$ , that is $y (1 + ln y) = 1$ . Using the GDC (or by checking), $y = 1$ is a solution: $1 \cdot (1 + 0) = 1$ . It is the only one with $y > 0$ : $y (1 + ln y)$ is increasing for $y > e^{- 2}$ and not positive for $y \leq e^{- 1}$ . 2. At $x = 1$ , $y = 1$ (with $ln (x y) = ln 1 = 0$ ), so the relation from part a) gives: $y^{'} + (y^{'} + 1) (1 + 0) = 1$ , that is $2 y^{'} + 1 = 1$ and $y^{'} = 0$ . 3. Tangent through $(1, 1)$ with gradient $0$ : $y = 1$ .

Answer: a) proof; b) $y = 1$

Exercise 108 HardIB · P1 TZ2 May 2010

The normal to the curve $x e^{- y} + e^{y} = 1 + x$ at the point $(c, ln c)$ meets the $y$ -axis at $c^{2} + 1$ . Determine the value of $c$ .

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Answer: $c = e$

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We differentiate implicitly $x e^{- y} + e^{y} = 1 + x$ (product rule for the first term): $e^{- y} - x e^{- y} \frac{d y}{d x} + e^{y} \frac{d y}{d x} = 1$ .
At the point $(c, ln c)$ we have $e^{y} = c$ and $e^{- y} = \frac{1}{c}$ : $\frac{1}{c} - c \cdot \frac{1}{c} \frac{d y}{d x} + c \frac{d y}{d x} = 1$ .
Grouping terms: $\frac{d y}{d x} (c - 1) = 1 - \frac{1}{c} = \frac{c - 1}{c}$ , so, for $c \neq = 1$ , the gradient of the tangent is $\frac{1}{c}$ .
The normal has gradient $- c$ : $y - ln c = - c (x - c)$ .
It meets the $y$ -axis at $x = 0$ : $y = ln c + c^{2}$ .
From the statement this value is $c^{2} + 1$ , so $ln c = 1$ and $c = e$ (which satisfies $c \neq = 1$ ).

Answer: $c = e$

Exercise 109 HardIB · P1 TZ2 May 2012Proof

Let $x^{3} y = a sin n x$ . Using implicit differentiation, verify that:

$x^{3} \frac{d ^{2} y}{d x ^{2}} + 6 x^{2} \frac{d y}{d x} + (n^{2} x^{2} + 6) x y = 0$

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Answer: proved.

Show working and answer

Starting from $x^{3} y = a sin n x$ and differentiating implicitly with respect to $x$ (product rule on the left, chain rule on the right): $3 x^{2} y + x^{3} \frac{d y}{d x} = an cos n x$ .
Differentiating again: $6 x y + 3 x^{2} \frac{d y}{d x} + 3 x^{2} \frac{d y}{d x} + x^{3} \frac{d ^{2} y}{d x ^{2}} = - a n^{2} sin n x$ .
Substituting $a sin n x = x^{3} y$ into the right-hand side: $- a n^{2} sin n x = - n^{2} x^{3} y$ .
Moving all terms to the left and grouping: $x^{3} \frac{d ^{2} y}{d x ^{2}} + 6 x^{2} \frac{d y}{d x} + 6 x y + n^{2} x^{3} y = 0$ .
Factorising $x y$ out of the last two terms: $6 x y + n^{2} x^{3} y = (n^{2} x^{2} + 6) x y$ . This gives $x^{3} \frac{d ^{2} y}{d x ^{2}} + 6 x^{2} \frac{d y}{d x} + (n^{2} x^{2} + 6) x y = 0.$

Answer: proved.

Exercise 110 HardIB · P2 TZ2 May 2018

A curve C is given by the implicit equation $x + y - cos (x y) = 0$ .

a) Show that $\frac{d y}{d x} = - (\frac{1 + y sin ( x y )}{1 + x sin ( x y )})$ .

b) The curve $x y = - \frac{π}{2}$ and C intersect at P and at Q.

(i) Find the coordinates of P and of Q.

(ii) Given that the gradients of the tangents to C at P and at Q are $m_{1}$ and $m_{2}$ respectively, show that $m_{1} \times m_{2} = 1$ .

c) Find the coordinates of the three points on C closest to the origin at which the tangent is parallel to the line $y = - x$ .

Show answer

Answer: b) $P (\frac{π}{2}, - \frac{π}{2})$ , $Q (- \frac{π}{2}, \frac{π}{2})$ ; c) $(0.486; 0.486)$ , $(0, 1)$ and $(1, 0)$

Show working and answer

a) 1. We differentiate implicitly $x + y - cos (x y) = 0$ . For $cos (x y)$ , we use the chain rule and the product rule on $x y$ : $1 + \frac{d y}{d x} + sin (x y) (y + x \frac{d y}{d x}) = 0$ . 2. We group terms: $\frac{d y}{d x} (1 + x sin (x y)) = - (1 + y sin (x y))$ . 3. Therefore $\frac{d y}{d x} = - (\frac{1 + y sin ( x y )}{1 + x sin ( x y )})$ .

b) i) 1. With $x y = - \frac{π}{2}$ : $cos (x y) = cos (- \frac{π}{2}) = 0$ , and the equation of $C$ becomes $x + y = 0$ , that is, $y = - x$ . 2. We substitute: $x (- x) = - \frac{π}{2} \Rightarrow x^{2} = \frac{π}{2} \Rightarrow x = \pm \frac{π}{2}$ . 3. $P (\frac{π}{2}, - \frac{π}{2})$ and $Q (- \frac{π}{2}, \frac{π}{2})$ .

(ii) 1. At both points $sin (x y) = sin (- \frac{π}{2}) = - 1$ , and the gradient is $m = - \frac{1 - y}{1 - x}$ . Let $s = \frac{π}{2}$ . 2. At $P$ ( $x = s$ , $y = - s$ ): $m_{1} = - \frac{1 + s}{1 - s}$ . 3. At $Q$ ( $x = - s$ , $y = s$ ): $m_{2} = - \frac{1 - s}{1 + s}$ . 4. Product: $m_{1} m_{2} = \frac{1 + s}{1 - s} \cdot \frac{1 - s}{1 + s} = 1$ .

c) 1. A tangent parallel to $y = - x$ has $\frac{d y}{d x} = - 1$ , that is, $1 + y sin (x y) = 1 + x sin (x y)$ (the denominator is non-zero). 2. Then $(y - x) sin (x y) = 0$ : $y = x$ or $sin (x y) = 0$ . 3. Case $y = x$ : the curve gives $2 x - cos (x^{2}) = 0$ . Using the GDC, the only solution is $x = 0.486$ . Point $(0.486; 0.486)$ . 4. Case $sin (x y) = 0$ : $x y = nπ$ and $cos (x y) = \pm 1$ , so $x + y = cos (x y) = \pm 1$ . For $n = 0$ : $x y = 0$ and $x + y = 1$ , giving $(0, 1)$ and $(1, 0)$ . For the other possible values of $n$ ( $n < 0$ ) we have $x^{2} + y^{2} = (x + y)^{2} - 2 x y = 1 - 2 nπ \geq 1 + 2 π$ , which is much further from the origin. 5. Distances to the origin: $2 \cdot 0.48 6^{2} = 0.69$ for the first point and $1$ for $(0, 1)$ and $(1, 0)$ , which are smaller than those of any other point.

Answer: b) $P (\frac{π}{2}, - \frac{π}{2})$ , $Q (- \frac{π}{2}, \frac{π}{2})$ ; c) $(0.486; 0.486)$ , $(0, 1)$ and $(1, 0)$

Exercise 111 HardIB · P2 TZ1 May 2016

Consider the curve C defined by the equation $y^{2} - 2 x y = 5 - e^{x}$ .

The point A lies on C and has coordinates $(0, a)$ , $a > 0$ .

a) Calculate the value of $a$ .

b) Show that $\frac{d y}{d x} = \frac{2 y - e ^{x}}{2 ( y - x )}$ .

c) Find the equation of the normal to C at the point A.

d) Find the coordinates of the second point at which the normal found in part (c) intersects C.

e) Given that $v = y^{3}$ , $y > 0$ , calculate $\frac{d v}{d x}$ at $x = 0$ .

Show answer

Answer: a) $a = 2$ ; b) proved; c) $y = - \frac{4}{3} x + 2$ ; d) $(1.56; - 0.0779)$ ; e) $\frac{d v}{d x} = 9$

Show working and answer

a) 1. At $x = 0$ : $y^{2} = 5 - e^{0} = 4$ . 2. Since $a > 0$ , $a = 2$ .

b) 1. We differentiate $y^{2} - 2 x y = 5 - e^{x}$ implicitly (product in $2 x y$ ): $2 y \frac{d y}{d x} - 2 y - 2 x \frac{d y}{d x} = - e^{x}$ . 2. Grouping: $\frac{d y}{d x} (2 y - 2 x) = 2 y - e^{x}$ . 3. $\frac{d y}{d x} = \frac{2 y - e ^{x}}{2 ( y - x )}$ .

c) 1. Gradient of the tangent at $A (0, 2)$ : $\frac{4 - 1}{2 \cdot 2} = \frac{3}{4}$ . 2. The normal has gradient $- \frac{4}{3}$ and passes through $(0, 2)$ : $y = - \frac{4}{3} x + 2$ .

d) 1. We substitute $y = - \frac{4}{3} x + 2$ into the equation of C and solve using the GDC (we obtain $x = 0$ , which is A, and another solution). 2. The other solution is $x \approx 1.56$ , with $y = - \frac{4}{3} (1.56) + 2 \approx - 0.0779$ . 3. The second point is $(1.56; - 0.0779)$ .

e) 1. $v = y^{3}$ , so $\frac{d v}{d x} = 3 y^{2} \frac{d y}{d x}$ (chain rule). 2. At $x = 0$ : $y = 2$ and $\frac{d y}{d x} = \frac{3}{4}$ , so $\frac{d v}{d x} = 3 \cdot 4 \cdot \frac{3}{4} = 9$ .

Answer: a) $a = 2$ ; b) proved; c) $y = - \frac{4}{3} x + 2$ ; d) $(1.56; - 0.0779)$ ; e) $\frac{d v}{d x} = 9$

Exercise 112 HardIB · P2 TZ0 November 2007

a) A curve is defined by the implicit equation $2 x y + 6 x^{2} - 3 y^{2} = 6$ .

Verify that the tangent at the point $A (1, \frac{2}{3})$ has a gradient equal to $\frac{20}{3}$ .

b) The line $x = 1$ intersects the curve at the point $A$ and at the point $B$ . Find, in the form $r = (a b) + s (c d)$

(i) the equation of the tangent at $A$ .

(ii) the equation of the normal at $B$ .

c) Find the angle between the tangent at $A$ and the normal at $B$ .

Show answer

Answer: a) gradient $\frac{20}{3}$ ; b) (i) $(x y) = (1 \frac{2}{3}) + s (320)$ ; (ii) $(x y) = (10) + s (61)$ ; c) $α \approx 7 2^{\circ}$

Show working and answer

a) 1. We check that $A$ lies on the curve: $2 \cdot 1 \cdot \frac{2}{3} + 6 - 3 \cdot \frac{4}{9} = \frac{4}{3} + 6 - \frac{4}{3} = 6$ . 2. Differentiate implicitly (the product $2 x y$ using the product rule): $2 y + 2 x \frac{d y}{d x} + 12 x - 6 y \frac{d y}{d x} = 0$ . 3. Rearrange: $\frac{d y}{d x} = \frac{2 y + 12 x}{6 y - 2 x}$ . 4. At $A (1, \frac{2}{3})$ : $\frac{\frac{4}{3} + 12}{4 - 2} = \frac{40/3}{2} = \frac{20}{3}$ .

b) 1. Substituting $x = 1$ into the equation of the curve: $2 y + 6 - 3 y^{2} = 6$ , i.e. $y (2 - 3 y) = 0$ . So $y = \frac{2}{3}$ (point $A$ ) or $y = 0$ : $B (1, 0)$ .

(i) Tangent at $A$ : direction vector $(3, 20)$ (gradient $\frac{20}{3}$ ). $(x y) = (1 \frac{2}{3}) + s (320)$

(ii) Gradient of the tangent at $B$ : $\frac{0 + 12}{0 - 2} = - 6$ . The normal has gradient $\frac{1}{6}$ , with direction vector $(6, 1)$ . $(x y) = (10) + s (61)$

c) 1. The angle between the lines is the angle between their direction vectors: $cos α = \frac{∣ ( 3 , 20 ) \cdot ( 6 , 1 ) ∣}{409 37} = \frac{38}{409 37} \approx 0.309$ . 2. So $α \approx 7 2^{\circ}$ .

Answer: a) gradient $\frac{20}{3}$ ; b) (i) $(x y) = (1 \frac{2}{3}) + s (320)$ ; (ii) $(x y) = (10) + s (61)$ ; c) $α \approx 7 2^{\circ}$

Exercise 113 HardIB · P1 TZ1 May 2011

Prove that the points $(0, 0)$ and $(2 π, - 2 π)$ on the curve $e^{(x + y)} = cos (x y)$ have a common tangent.

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Answer: proven: the common tangent is $y = - x$ .

Show working and answer

We check that both points lie on the curve $e^{x + y} = cos (x y)$ . At $(0, 0)$ : $e^{0} = 1 = cos 0$ . At $(2 π, - 2 π)$ : $x + y = 0$ , so $e^{0} = 1$ , and $x y = - 2 π$ , with $cos (- 2 π) = 1$ .
Differentiate implicitly: $e^{x + y} (1 + y^{'}) = - sin (x y) (y + x y^{'})$ .
At $(0, 0)$ : $sin 0 = 0$ , so $1 \cdot (1 + y^{'}) = 0$ and $y^{'} = - 1$ .
At the second point: $sin (- 2 π) = 0$ , and $e^{0} (1 + y^{'}) = 0$ , so again $y^{'} = - 1$ .
Tangent at $(0, 0)$ : $y = - x$ . Tangent at $(2 π, - 2 π)$ : $y + 2 π = - (x - 2 π)$ , i.e. $y = - x$ .
Both tangents are the same line.

Answer: proven: the common tangent is $y = - x$ .

Exercise 114 HardIB · P2 TZ2 May 2015

A curve is defined by $x^{2} - 5 x y + y^{2} = 7$ .

a) Show that $\frac{d y}{d x} = \frac{5 y - 2 x}{2 y - 5 x}$ .

b) Find the equation of the normal to the curve at the point $(6, 1)$ .

c) Find the distance between the two points on the curve at which the tangent is parallel to the line $y = x$ .

Show answer

Answer: a) proof; b) $y = - 4 x + 25$ ; c) $22$

Show working and answer

a) 1. We differentiate implicitly with respect to $x$ (product rule on $5 x y$ ): $2 x - 5 y - 5 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0.$ 2. We collect the terms in $\frac{d y}{d x}$ : $(2 y - 5 x) \frac{d y}{d x} = 5 y - 2 x$ . 3. Rearranging: $\frac{d y}{d x} = \frac{5 y - 2 x}{2 y - 5 x}$ .

b) 1. We verify that $(6, 1)$ lies on the curve: $36 - 30 + 1 = 7$ . 2. Gradient of the tangent: $\frac{5 - 12}{2 - 30} = \frac{- 7}{- 28} = \frac{1}{4}$ . 3. The normal has gradient $- 4$ (negative reciprocal). Equation: $y - 1 = - 4 (x - 6)$ , i.e. $y = - 4 x + 25$ .

c) 1. Tangent parallel to $y = x$ : $\frac{d y}{d x} = 1$ , i.e. $5 y - 2 x = 2 y - 5 x$ , so $3 y = - 3 x$ and $y = - x$ . 2. We substitute into the curve: $x^{2} - 5 x (- x) + x^{2} = 7$ , i.e. $7 x^{2} = 7$ , so $x = \pm 1$ . 3. Points: $(1, - 1)$ and $(- 1, 1)$ (at both points $2 y - 5 x \neq = 0$ ). 4. Distance: $(1 - (- 1))^{2} + (- 1 - 1)^{2} = 8 = 22$ .

Answer: a) proof; b) $y = - 4 x + 25$ ; c) $22$

Exercise 115 HardIB · P1 TZ0 November 2022

Consider the curve defined by the equation $(x^{2} + y^{2}) y^{2} = 4 x^{2}$ , where $x \geq 0$ and $- 2 < y < 2$ .

Show that the curve has no local maximum or local minimum point for $x > 0$ .

Show answer

Answer: $\frac{d y}{d x} = \frac{8 x - 2 x y ^{2}}{4 y ^{3} + 2 y x ^{2}}$ is zero only if $x = 0$ or $y = \pm 2$ , values which are excluded; there are no local maxima or minima.

Show working and answer

We expand: $x^{2} y^{2} + y^{4} = 4 x^{2}$ .
We differentiate implicitly with respect to $x$ (product rule on $x^{2} y^{2}$ ): $2 x y^{2} + 2 x^{2} y \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x} = 8 x .$
We solve for the derivative: $\frac{d y}{d x} (2 x^{2} y + 4 y^{3}) = 8 x - 2 x y^{2}$ , so $\frac{d y}{d x} = \frac{8 x - 2 x y ^{2}}{4 y ^{3} + 2 y x ^{2}} = \frac{2 x ( 4 - y ^{2} )}{2 y ( x ^{2} + 2 y ^{2} )} .$
A local maximum or minimum requires $\frac{d y}{d x} = 0$ , that is $2 x (4 - y^{2}) = 0$ : $x = 0$ or $y = \pm 2$ .
But the question requires $x > 0$ and $- 2 < y < 2$ , values which exclude both possibilities. (For $x > 0$ the denominator does not vanish: $y = 0$ would imply $0 = 4 x^{2}$ , that is $x = 0$ .)
Therefore $\frac{d y}{d x} \neq = 0$ for $x > 0$ , and the curve has no local maxima or minima.

Answer: $\frac{d y}{d x} = \frac{8 x - 2 x y ^{2}}{4 y ^{3} + 2 y x ^{2}}$ is zero only if $x = 0$ or $y = \pm 2$ , values which are excluded; there are no local maxima or minima.

Studying functions: extrema, concavity and inflection

Exercise 116 BasicIB · P1 TZ2 May 2011

The quadratic function $f (x) = p + q x - x^{2}$ reaches a maximum value, equal to $5$ , at $x = 3$ .

a) Find the value of $p$ and the value of $q$ .

b) The graph of $f (x)$ is translated $3$ units in the direction parallel to the $x$ -axis and in the positive direction. Determine the equation of the new graph.

Show answer

Answer: a) $p = - 4$ , $q = 6$ ; b) $y = - x^{2} + 12 x - 31$

Show working and answer

a) 1. A maximum at $x = 3$ requires $f^{'} (3) = 0$ . With $f^{'} (x) = q - 2 x$ : $q - 6 = 0$ , so $q = 6$ . 2. The maximum value is $5$ : $f (3) = p + 6 \cdot 3 - 9 = p + 9 = 5$ , so $p = - 4$ .

b) 1. A translation of $3$ units in the positive direction of the $x$ -axis replaces $x$ by $x - 3$ : $y = f (x - 3)$ . 2. $y = - 4 + 6 (x - 3) - (x - 3)^{2} = - 4 + 6 x - 18 - x^{2} + 6 x - 9$ . 3. $y = - x^{2} + 12 x - 31$ .

Answer: a) $p = - 4$ , $q = 6$ ; b) $y = - x^{2} + 12 x - 31$

Exercise 117 BasicIB · P1 May 1996

Find the interval between $0$ and $2 π$ over which the function $y (x) = x - 2 sin x$ is increasing.

Show answer

Answer: increasing on $] \frac{π}{3}, \frac{5 π}{3} [$

Show working and answer

$y^{'} = 1 - 2 cos x$ .
$y$ is increasing where $y^{'} > 0$ : $1 - 2 cos x > 0 \Rightarrow cos x < \frac{1}{2}$ .
On $[0, 2 π]$ , $cos x = \frac{1}{2}$ at $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ , and $cos x < \frac{1}{2}$ between these two values.
The function is increasing on $] \frac{π}{3}, \frac{5 π}{3} [$ .

Answer: increasing on $] \frac{π}{3}, \frac{5 π}{3} [$

Exercise 118 BasicIB · P2 TZ1 May 2015

The function $f$ is defined by $f (x) = x^{3} + e^{x} + 1$ , $x \in R$ . Considering $f^{'} (x)$ , determine whether $f$ is a one-to-one function or a many-to-one function.

Show answer

Answer: $f$ is strictly increasing on $R$ , so it is one-to-one (injective).

Show working and answer

Differentiating: $f^{'} (x) = 3 x^{2} + e^{x}$ .
For all $x \in R$ : $3 x^{2} \geq 0$ and $e^{x} > 0$ , so $f^{'} (x) > 0$ .
A function with a positive derivative everywhere is strictly increasing on $R$ ; therefore each value of $y$ is reached only once.
Consequently $f$ is a one-to-one (injective) function.

Answer: $f$ is strictly increasing on $R$ , so it is one-to-one (injective).

Exercise 119 MediumIB · P1 TZ2 May 2015

Consider the function defined by $f (x) = x^{3} - 3 x^{2} + 4$ .

a) Determine the values of $x$ for which $f (x)$ is a decreasing function.

On the curve $y = f (x)$ there is a point of inflexion, $P$ .

b) Find the coordinates of $P$ .

Show answer

Answer: a) $0 < x < 2$ ; b) $P (1, 2)$

Show working and answer

a) 1. $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ . 2. $f$ is decreasing where $f^{'} (x) < 0$ . As $3 x (x - 2)$ is an upward-opening parabola, it is negative between its roots $0$ and $2$ . 3. Therefore $f$ is decreasing for $0 < x < 2$ .

b) 1. A point of inflexion requires $f^{''} (x) = 0$ with a change of sign: $f^{''} (x) = 6 x - 6 = 0$ , so $x = 1$ . 2. $f^{''} < 0$ for $x < 1$ and $f^{''} > 0$ for $x > 1$ : there is a change of sign, so it is a point of inflexion. 3. $f (1) = 1 - 3 + 4 = 2$ .

Answer: a) $0 < x < 2$ ; b) $P (1, 2)$

Exercise 120 MediumIB · P1 TZ2 May 2005

The function $f$ is given by $f (x) = \frac{x ^{5} + 2}{x}$ , $x \neq = 0$ . The graph of $f$ has a point of inflexion at $P$ . Find the coordinates of $P$ .

Show answer

Answer: $P (- \frac{1}{5 3}, - \frac{5 5 3}{3})$

Show working and answer

We rewrite $f (x) = \frac{x ^{5} + 2}{x} = x^{4} + 2 x^{- 1}$ .
$f^{'} (x) = 4 x^{3} - 2 x^{- 2}$ and $f^{''} (x) = 12 x^{2} + 4 x^{- 3} = \frac{12 x ^{5} + 4}{x ^{3}}$ .
A point of inflexion requires $f^{''} (x) = 0$ : $12 x^{5} + 4 = 0$ , that is, $x^{5} = - \frac{1}{3}$ , so $x = - \frac{1}{5 3}$ .
There is a change of sign of $f^{''}$ at that value (the numerator $12 x^{5} + 4$ changes sign and the denominator $x^{3}$ is negative on both sides), so it is a point of inflexion.
The $y$ -coordinate: $f (x) = x^{4} + \frac{2}{x} = 3^{- 4/5} - 2 \cdot 3^{1/5} = 3^{1/5} (\frac{1}{3} - 2) = - \frac{5 5 3}{3}$ .

Answer: $P (- \frac{1}{5 3}, - \frac{5 5 3}{3})$

Exercise 121 MediumIB · P2 TZ2 May 2008

Consider the curve defined by the equation $f (x) = e^{- 2 x^{2}}$ for $x < 0$ .

Find the coordinates of the point of inflexion and justify why it is a point of inflexion.

Show answer

Answer: $(- \frac{1}{2}, e^{- \frac{1}{2}})$

Show working and answer

$f (x) = e^{- 2 x^{2}}$ . By the chain rule, $f^{'} (x) = - 4 x e^{- 2 x^{2}}$ .
Applying the product rule: $f^{''} (x) = - 4 e^{- 2 x^{2}} + (- 4 x) (- 4 x) e^{- 2 x^{2}} = (16 x^{2} - 4) e^{- 2 x^{2}} = 4 (4 x^{2} - 1) e^{- 2 x^{2}}$ .
$f^{''} (x) = 0$ when $4 x^{2} = 1$ , that is, $x = \pm \frac{1}{2}$ . As $x < 0$ , $x = - \frac{1}{2}$ .
$f (- \frac{1}{2}) = e^{- 2 \cdot \frac{1}{4}} = e^{- 1/2}$ .
Justification: as $e^{- 2 x^{2}} > 0$ , the sign of $f^{''}$ is that of $4 x^{2} - 1$ . For $x < - \frac{1}{2}$ , $f^{''} > 0$ (convex), and for $- \frac{1}{2} < x < 0$ , $f^{''} < 0$ (concave). There is a change of sign of $f^{''}$ , so it is a point of inflexion.

Answer: $(- \frac{1}{2}, e^{- \frac{1}{2}})$

Exercise 122 MediumIB · P2 TZ0 November 2009

Consider the function $f$ , defined by $f (x) = x - a x$ , where $x \geq 0$ , $a \in R^{+}$ .

a) Find, in terms of $a$ ,

(i) the zeros of $f$

(ii) the values of $x$ for which $f$ is decreasing

(iii) the values of $x$ for which $f$ is increasing

(iv) the range of $f$

b) State the concavity of the graph of $f$ .

Show answer

Answer: a) (i) $x = 0$ , $x = a^{2}$ ; (ii) $] 0, \frac{a ^{2}}{4} [$ ; (iii) $] \frac{a ^{2}}{4}, + \infty [$ ; (iv) $[- \frac{a ^{2}}{4}, + \infty [$ ; b) concave up on $] 0, + \infty [$

Show working and answer

a) (i) 1. $f (x) = 0$ : $x - a x = x (x - a) = 0$ . 2. Then $x = 0$ or $x = a$ , that is, $x = 0$ or $x = a^{2}$ .

(ii) and (iii) 1. $f^{'} (x) = 1 - \frac{a}{2 x}$ for $x > 0$ . 2. $f^{'} (x) < 0 ⟺ \frac{a}{2 x} > 1 ⟺ x < \frac{a}{2} ⟺ x < \frac{a ^{2}}{4}$ . Hence $f$ is decreasing for $0 < x < \frac{a ^{2}}{4}$ . 3. $f^{'} (x) > 0 ⟺ x > \frac{a ^{2}}{4}$ . Hence $f$ is increasing for $x > \frac{a ^{2}}{4}$ .

(iv) 1. From the above, $f$ has its minimum at $x = \frac{a ^{2}}{4}$ : $f (\frac{a ^{2}}{4}) = \frac{a ^{2}}{4} - a \cdot \frac{a}{2} = - \frac{a ^{2}}{4}$ . 2. As $f$ increases without bound for $x \to + \infty$ , the range is $[- \frac{a ^{2}}{4}, + \infty [$ .

b) 1. $f^{''} (x) = \frac{a}{4} x^{- 3/2} = \frac{a}{4 x x}$ . 2. With $a > 0$ and $x > 0$ , $f^{''} (x) > 0$ : the graph is concave up on $] 0, + \infty [$ .

Answer: a) (i) $x = 0$ , $x = a^{2}$ ; (ii) $] 0, \frac{a ^{2}}{4} [$ ; (iii) $] \frac{a ^{2}}{4}, + \infty [$ ; (iv) $[- \frac{a ^{2}}{4}, + \infty [$ ; b) concave up on $] 0, + \infty [$

Exercise 123 MediumIB · P2 TZ2 May 2004

Let $f (x) = x \cdot cos (x)$ , for $x$ in the interval $[0, π]$ . The graph of $f (x)$ has a local maximum at $x = a$ and a point of inflexion at $x = b$ .

a) Find the value of

(i) $a$

(ii) $b$

b) Sketch the graph of $f (x)$ .

View solution

a) Using the GDC, solving the equations of the derivatives on $[0, π]$ .

(i) 1. $f^{'} (x) = cos x - x sin x$ (product rule). 2. Local maximum: $f^{'} (x) = 0$ , that is, $cos x = x sin x$ ; using the GDC $x \approx 0.8603$ . 3. It is a maximum: $f^{'} > 0$ before (for example $f^{'} (0) = 1$ ) and $f^{'} < 0$ after (for example $f^{'} (\frac{π}{2}) = - \frac{π}{2}$ ). So $a \approx 0.86$ .

(ii) 1. $f^{''} (x) = - sin x - sin x - x cos x = - 2 sin x - x cos x$ . 2. Point of inflexion: $f^{''} (x) = 0$ , using the GDC $x \approx 2.2889$ , and $f^{''}$ changes sign there. So $b \approx 2.29$ .

b) Features of the graph: it starts at $(0, 0)$ with gradient $f^{'} (0) = 1$ , rises to the maximum at $x \approx 0.86$ (with $f \approx 0.56$ ), cuts the $x$ -axis at $x = \frac{π}{2}$ and falls; it changes from concave down to concave up at $x \approx 2.29$ and ends at $(π, - π)$ .

Graph of y = x cos x on the interval 0 to π: rises from the origin to a maximum, crosses the x-axis and falls to the right-hand endpoint

Answer: (a) (i) $a \approx 0.86$ ; (ii) $b \approx 2.29$ ; (b) graph sketched with a maximum at $x \approx 0.86$ and a point of inflexion at $x \approx 2.29$

Exercise 124 MediumIB · P1 TZ2 May 2013

Consider the complex numbers $z_{1} = 1 + 3 i$ and $z_{2} = - 1 - i$ .

a) Write down the exact value of $∣ z_{1} ∣$ and of $ar g (z_{2})$ .

b) Find the minimum value of $∣ z_{1} + α z_{2} ∣$ , where $α \in R$ .

Show answer

Answer: a) $∣ z_{1} ∣ = 10$ , $ar g (z_{2}) = \frac{5 π}{4}$ ; b) minimum $2$ when $α = 2$

Show working and answer

a) 1. The modulus is $∣ z_{1} ∣ = 1^{2} + 3^{2} = 10$ . 2. $z_{2} = - 1 - i$ lies in the third quadrant (negative real and imaginary parts). Its reference angle is $arctan \frac{1}{1} = \frac{π}{4}$ . 3. Therefore $ar g (z_{2}) = π + \frac{π}{4} = \frac{5 π}{4}$ (or, equivalently, $- \frac{3 π}{4}$ ).

b) 1. We calculate the sum: $z_{1} + α z_{2} = (1 - α) + (3 - α) i$ . 2. Its modulus squared is $∣ z_{1} + α z_{2} ∣^{2} = (1 - α)^{2} + (3 - α)^{2} = 2 α^{2} - 8 α + 10$ . 3. Completing the square: $2 α^{2} - 8 α + 10 = 2 (α - 2)^{2} + 2$ , which is a minimum when $α = 2$ and equals $2$ . 4. As the modulus is the square root of this value, the minimum is $2$ .

Answer: a) $∣ z_{1} ∣ = 10$ , $ar g (z_{2}) = \frac{5 π}{4}$ ; b) minimum $2$ when $α = 2$

Exercise 125 MediumIB · P1 TZ2 May 2007

The function $f$ is defined as $f (x) = \frac{2 x}{x ^{2} + 6}$ for $x \geq b$ , $b \in R$ .

a) Verify that $f^{'} (x) = \frac{12 - 2 x ^{2}}{( x ^{2} + 6 ) ^{2}}$ .

b) Hence, find the smallest exact value of $b$ for which the inverse function $f^{- 1}$ exists. Justify your answer.

Show answer

Answer: b) $b = 6$

Show working and answer

a) 1. Quotient rule with $u = 2 x$ , $v = x^{2} + 6$ : $f^{'} (x) = \frac{2 ( x ^{2} + 6 ) - 2 x \cdot 2 x}{( x ^{2} + 6 ) ^{2}}$ . 2. Simplifying the numerator: $2 x^{2} + 12 - 4 x^{2} = 12 - 2 x^{2}$ . 3. Hence $f^{'} (x) = \frac{12 - 2 x ^{2}}{( x ^{2} + 6 ) ^{2}}$ .

b) 1. $f^{'} (x) = 0 \Rightarrow x^{2} = 6 \Rightarrow x = \pm 6$ . The sign of $f^{'}$ is given by $12 - 2 x^{2}$ : negative if $x < - 6$ , positive if $- 6 < x < 6$ , and negative if $x > 6$ . 2. A function has an inverse if it is injective, that is, strictly monotonic on its domain $[b, \infty [$ . 3. If $b < 6$ , the domain contains a neighbourhood of $x = 6$ , where $f$ changes from increasing to decreasing. There is a local maximum, and values close to the maximum are attained twice, so $f$ is not injective. 4. If $b = 6$ , $f$ is strictly decreasing on $[6, \infty [$ and therefore injective. 5. Hence the smallest value is $b = 6$ .

Answer: b) $b = 6$

Exercise 126 MediumIB · P1 May 1997

Graph of a periodic sawtooth-shaped function with three equal sections, of period three, between x = -3 and x = 6

The diagram above shows the graph of a periodic function on the interval $- 3 \leq x \leq 6$ .

a) What is the period?

b) Assuming that the graph continues to have the same shape over the interval $- 30 \leq x \leq 60$ , find $f (52)$ .

c) Find $f^{'} (\frac{53}{2})$ .

Show answer

Answer: a) $P = 3$ ; b) $f (52) = 1$ ; c) $f^{'} (\frac{53}{2}) = - 2$

Show working and answer

a) 1. The pattern of the graph repeats every $3$ units: for example $f (0) = f (3) = f (6) = 0$ , and the maxima $f (- 1) = 2$ , $f (2) = 2$ , $f (5) = 2$ are $3$ units apart. 2. The period is $3$ .

b) 1. Since the period is $3$ , $f (52) = f (52 - 3 \cdot 17) = f (1)$ . 2. Between $(0, 0)$ and $(2, 2)$ the graph is the line $y = x$ , so $f (1) = 1$ . 3. $f (52) = 1$ .

c) 1. $\frac{53}{2} = 26.5$ ; we subtract $24$ (a multiple of $3$ ): $26.5 - 24 = 2.5$ , so $f^{'} (\frac{53}{2}) = f^{'} (2.5)$ . 2. On $] 2, 3 [$ the graph is the segment from $(2, 2)$ to $(3, 0)$ , with gradient $\frac{0 - 2}{3 - 2} = - 2$ . 3. $f^{'} (\frac{53}{2}) = - 2$ .

Answer: a) $P = 3$ ; b) $f (52) = 1$ ; c) $f^{'} (\frac{53}{2}) = - 2$

Exercise 127 MediumIB · P1 November 1997

The graph of $f^{'} (x)$ , the derivative of $f (x)$ , is shown below.

Graph of the derivative of f: a parabola that cuts the X-axis at x = -1 and x = 3, with its minimum at x = 1

Given that $f (- 1) = 5$ , $f (0) = 0$ and $f (3) = - 27$ , sketch the graph of $y = f (x)$ on the same set of axes.

View solution

From the graph of $f^{'}$ : it cuts the $X$ -axis at $x = - 1$ and $x = 3$ ; it is positive if $x < - 1$ or $x > 3$ and negative if $- 1 < x < 3$ ; it has its minimum at $x = 1$ . (Its equation is $f^{'} (x) = 3 x^{2} - 6 x - 9$ .)
Sign of $f^{'}$ $\Rightarrow$ behaviour of $f$ : $f$ increases for $x < - 1$ , decreases on $] - 1, 3 [$ and increases for $x > 3$ . Hence $f$ has a local maximum at $x = - 1$ and a local minimum at $x = 3$ .
Since $f^{'}$ has a minimum at $x = 1$ , $f^{'}$ is decreasing before it and increasing after it, and $f$ has a point of inflexion at $x = 1$ (concave down for $x < 1$ and concave up for $x > 1$ ).
Points we know: the local maximum $(- 1, 5)$ , the origin $(0, 0)$ and the local minimum $(3, - 27)$ .
Integrating $f^{'}$ and using $f (0) = 0$ gives $f (x) = x^{3} - 3 x^{2} - 9 x$ (it agrees with the three given values). The sketch passes through these three points, with the shape of a cubic with a maximum and a minimum.

Graphs of f and of its derivative on the same axes: f has a local maximum, passes through the origin and has a local minimum at the zeros of the derivative

Answer: $f$ has a local maximum at $(- 1, 5)$ and a local minimum at $(3, - 27)$ , and passes through $(0, 0)$ ; $f (x) = x^{3} - 3 x^{2} - 9 x$ .

Exercise 128 MediumIB · P1 May 2000

Let $f : x \mapsto e^{s i n x}$ .

a) Find $f^{'} (x)$ .

There is a point of inflexion on the graph of $f$ for $0 < x < 1$ .

b) Write down, without solving, an equation in terms of $x$ which can be used to find the value of $x$ at this point of inflexion.

Show answer

Answer: a) $f^{'} (x) = cos x e^{s i n x}$ ; b) $cos^{2} x - sin x = 0$

Show working and answer

a) It is $f (x) = e^{u}$ with $u = sin x$ . By the chain rule: $f^{'} (x) = cos x \cdot e^{s i n x}$ .

b) 1. A point of inflexion requires $f^{''} (x) = 0$ . 2. We differentiate $f^{'}$ using the product and chain rules: $f^{''} (x) = - sin x e^{s i n x} + cos x \cdot cos x e^{s i n x} = e^{s i n x} (cos^{2} x - sin x)$ . 3. Since $e^{s i n x} > 0$ , the equation is $cos^{2} x - sin x = 0$ .

Answer: a) $f^{'} (x) = cos x e^{s i n x}$ ; b) $cos^{2} x - sin x = 0$

Exercise 129 MediumIB · P1 May 2000

The diagram shows the graph of $y = f^{'} (x)$ .

Graph of y = f'(x): a cubic-shaped curve that cuts the x-axis at three points, with a local maximum to the left of the y-axis and a local minimum to the right

Indicate and clearly label on the graph:

a) the points where $y = f (x)$ has minimum points.

b) the points where $y = f (x)$ has maximum points.

c) the points where $y = f (x)$ has points of inflexion.

View solution

The turning points of $f$ are where $f^{'} (x) = 0$ , and their nature is decided by the sign of $f^{'}$ ; the points of inflexion of $f$ are where $f^{'}$ has a local extremum (because $f^{''} = (f^{'})^{'}$ changes sign).

Let $a < 0 < d$ be the three points where the graph of $f^{'}$ cuts the $x$ -axis: $x = a$ , $x = 0$ and $x = d$ . For $x < a$ , $f^{'} < 0$ (the cubic comes from below); between $a$ and $0$ , $f^{'} > 0$ ; between $0$ and $d$ , $f^{'} < 0$ ; for $x > d$ , $f^{'} > 0$ .

a) Minimum points of $f$ : $f^{'}$ changes from negative to positive at $x = a$ and at $x = d$ .

b) Maximum point of $f$ : $f^{'}$ changes from positive to negative at $x = 0$ .

c) Points of inflexion of $f$ : they are the $x$ -coordinates of the local extrema of $f^{'}$ , the maximum at $x = b$ (between $a$ and $0$ ) and the minimum at $x = c$ (between $0$ and $d$ ), because $f^{''}$ changes sign there.

![Graph of f'(x) with the points a, b, O, c and d marked on the x-axis: a and d are the zeros on either side, b and c are the extrema of f']](/recursos/figuras/aa-t10-ib-151-s.svg)

Answer: a) minimum points at $x = a$ and $x = d$ ; b) maximum point at $x = 0$ ; c) points of inflexion at $x = b$ and $x = c$ .

Exercise 130 MediumIB · P2 May 2000

The diagram shows the graphs of $y = - x^{3} + 3 x^{2}$ and $y = g (x)$ , where $g (x)$ is a cubic polynomial.

Graphs of y = −x³ + 3x² (continuous line) and of y = g(x), a cubic polynomial (dashed line), with the points A and A′

a) If $g (- 2) = 0$ , $g (0) = - 4$ , $g^{'} (- 2) = 0$ and $g^{'} (0) = 0$ , show that $g (x) = x^{3} + 3 x^{2} - 4$ .

The graph of $y = - x^{3} + 3 x^{2}$ is reflected in the $y$ -axis and translated by the vector $(- 1 - 1)$ to obtain the graph of $y = h (x)$ .

b) Write $h (x)$ in the form $h (x) = a x^{3} + b x^{2} + c x + d$ .

The graph of $y = x^{3} + 3 x^{2} - 4$ is obtained by applying two transformations to the graph of $y = - x^{3} + 3 x^{2}$ .

c) State the two transformations whose composition transforms the graph of $y = - x^{3} + 3 x^{2}$ into the graph of $y = x^{3} + 3 x^{2} - 4$ , and such that the point A is transformed to the point A′.

Show answer

Answer: b) $h (x) = x^{3} + 6 x^{2} + 9 x + 3$ ; c) reflection in the $x$ -axis and translation by the vector $(- 2 0)$ .

Show working and answer

a) 1. Let $g (x) = a x^{3} + b x^{2} + c x + d$ , so $g^{'} (x) = 3 a x^{2} + 2 b x + c$ . 2. $g (0) = - 4$ gives $d = - 4$ . $g^{'} (0) = 0$ gives $c = 0$ . 3. $g^{'} (- 2) = 0$ : $12 a - 4 b = 0$ , so $b = 3 a$ . 4. $g (- 2) = 0$ : $- 8 a + 4 b - 4 = 0$ . Substituting $b = 3 a$ : $- 8 a + 12 a - 4 = 0$ , i.e. $a = 1$ and $b = 3$ . 5. Therefore $g (x) = x^{3} + 3 x^{2} - 4$ , as required.

b) 1. Reflecting in the $y$ -axis is equivalent to replacing $x$ with $- x$ : $y = - (- x)^{3} + 3 (- x)^{2} = x^{3} + 3 x^{2}$ . 2. The translation by the vector $(- 1 - 1)$ replaces $x$ with $x + 1$ and subtracts $1$ : $h (x) = (x + 1)^{3} + 3 (x + 1)^{2} - 1$ . 3. Expanding: $x^{3} + 3 x^{2} + 3 x + 1 + 3 x^{2} + 6 x + 3 - 1 = x^{3} + 6 x^{2} + 9 x + 3$ .

c) 1. In $y = - x^{3} + 3 x^{2}$ , the local maximum A is found from $- 3 x^{2} + 6 x = 0$ , i.e. $x = 2$ : $A (2, 4)$ . 2. $A^{'}$ is the local minimum of $g$ on the $y$ -axis: $A^{'} (0, - 4)$ (from $g^{'} (x) = 3 x^{2} + 6 x = 0$ ). 3. Reflection in the $x$ -axis: $y = x^{3} - 3 x^{2}$ , and A goes to $(2, - 4)$ . 4. Translation by the vector $(- 2 0)$ : $y = (x + 2)^{3} - 3 (x + 2)^{2} = x^{3} + 6 x^{2} + 12 x + 8 - 3 x^{2} - 12 x - 12 = x^{3} + 3 x^{2} - 4$ , and $(2, - 4)$ goes to $(0, - 4) = A^{'}$ . This is what was required.

Answer: b) $h (x) = x^{3} + 6 x^{2} + 9 x + 3$ ; c) reflection in the $x$ -axis and translation by the vector $(- 2 0)$ .

Exercise 131 MediumIB · P1 November 2002

Find the $x$ -coordinate of the point of inflexion of the graph of $y = x e^{x}$ , $- 3 \leq x \leq 1$ .

Show answer

Answer: $x = - 2$

Show working and answer

$y = x e^{x}$ . Product rule: $y^{'} = e^{x} + x e^{x} = (1 + x) e^{x}$ .
Differentiating again: $y^{''} = e^{x} + (1 + x) e^{x} = (x + 2) e^{x}$ .
$y^{''} = 0 ⟺ x = - 2$ , which lies in $[- 3, 1]$ . As $e^{x} > 0$ , $y^{''}$ changes from negative to positive at $x = - 2$ : there is a change of concavity.

Answer: $x = - 2$

Exercise 132 MediumIB · P1 May 2003

The diagram below shows the curve $y_{1} = f (x)$ .

Graph of y₁ = f(x): two separate branches divided by a dashed vertical asymptote; the left branch forms an arc that descends at both ends, and the right branch rises from below, forms an arc and approaches the x-axis

On the axes below, sketch the graph of $y_{2} = ∣ f^{'} (x) ∣$ .

Empty axes of y₂ against x, with the same dashed vertical line, for sketching the graph of y₂ = |f′(x)|

View solution

The graph of $y_{2} = ∣ f^{'} (x) ∣$ is obtained from that of $f^{'}$ by reflecting the negative parts upwards. One must read $f^{'}$ from the graph of $f$ : the gradient of $f$ on each section.
At each maximum of $f$ the tangent is horizontal: $f^{'} = 0$ , and $∣ f^{'} ∣$ touches the $x$ -axis (with value $0$ ). As $f^{'}$ changes sign there, the graph of $∣ f^{'} ∣$ reaches the $x$ -axis forming a peak ("bounces").
Near the vertical asymptote of $f$ the gradient becomes very large in absolute value: $∣ f^{'} ∣ \to \infty$ on both sides of the asymptote.
On the right-hand branch, $f$ decreases towards the $x$ -axis and flattens out: $f^{'} \to 0$ and $∣ f^{'} ∣ \to 0$ (the graph approaches the $x$ -axis from above). Between the asymptote and the maximum of $f$ , $∣ f^{'} ∣$ falls from very large values to $0$ ; then $f$ decreases, $∣ f^{'} ∣$ rises to a small maximum (where $f$ falls most steeply) and then decreases towards $0$ .
On the left-hand branch, similarly: $∣ f^{'} ∣$ starts large, falls to $0$ at the maximum of $f$ and rises again without limit towards the asymptote.

Graph of y₂ = |f′(x)|: on each branch it falls to touch the x-axis at the maximum of f, rises without limit towards the vertical asymptote, and on the right it has a small arc before approaching the x-axis

Answer: see the graph of $y_{2} = ∣ f^{'} (x) ∣$ : it touches the $x$ -axis at the maxima of $f$ , tends to $\infty$ near the vertical asymptote and to $0$ as $x \to + \infty$ .

Exercise 133 MediumIB · P1 TZ1 May 2004

Let $f (x) = x^{3} cos x$ , $0 \leq x \leq \frac{π}{2}$ .

a) Find $f^{'} (x)$ .

b) Calculate the value of $x$ for which $f (x)$ is a maximum.

c) Find the $x$ -coordinate of the point of inflexion on the graph of $f (x)$ .

Show answer

Answer: a) $f^{'} (x) = x^{2} (3 cos x - x sin x)$ ; b) $x \approx 1.19$ ; c) $x \approx 0.823$

Show working and answer

a) Product rule: $f^{'} (x) = 3 x^{2} cos x - x^{3} sin x = x^{2} (3 cos x - x sin x)$ .

b) 1. $f^{'} (x) = 0$ with $0 < x < \frac{π}{2}$ requires $3 cos x = x sin x$ , that is, $x tan x = 3$ . 2. Using the GDC: $x \approx 1.19$ . 3. It is a maximum: for smaller $x$ , $f^{'} > 0$ (for example, at $x = 1$ , $3 cos 1 - sin 1 \approx 0.78 > 0$ ) and for larger $x$ , $f^{'} < 0$ (at $x = 1.5$ , $3 cos 1.5 - 1.5 sin 1.5 \approx - 1.28 < 0$ ). $f^{'}$ changes from positive to negative.

c) 1. Differentiating $f^{'}$ : $f^{''} (x) = 6 x cos x - 3 x^{2} sin x - 3 x^{2} sin x - x^{3} cos x = x (6 cos x - 6 x sin x - x^{2} cos x)$ . 2. For $x > 0$ we solve $6 cos x - 6 x sin x - x^{2} cos x = 0$ using the GDC: $x \approx 0.823$ . 3. $f^{''}$ changes sign at this value (positive before, negative after), so it is a point of inflexion.

Answer: a) $f^{'} (x) = x^{2} (3 cos x - x sin x)$ ; b) $x \approx 1.19$ ; c) $x \approx 0.823$

Exercise 134 MediumIB · P1 TZ0 November 2005

The curve $y = \frac{x ^{3}}{3} - x^{2} - 3 x + 4$ has a local maximum at the point $P$ and a local minimum at the point $Q$ .

Determine the equation of the line passing through $P$ and $Q$ , and express it in the form $a x + b y + c = 0$ , where $a, b, c \in R$ .

Show answer

Answer: $8 x + 3 y - 9 = 0$

Show working and answer

Differentiating: $y^{'} = x^{2} - 2 x - 3 = (x - 3) (x + 1)$ . The critical points are $x = - 1$ and $x = 3$ .
Classify: $y^{''} = 2 x - 2$ . At $x = - 1$ , $y^{''} = - 4 < 0$ : local maximum (this is $P$ ). At $x = 3$ , $y^{''} = 4 > 0$ : local minimum (this is $Q$ ).
Ordinates: $y (- 1) = - \frac{1}{3} - 1 + 3 + 4 = \frac{17}{3}$ , so $P (- 1, \frac{17}{3})$ . $y (3) = 9 - 9 - 9 + 4 = - 5$ , so $Q (3, - 5)$ .
Gradient of $PQ$ : $\frac{- 5 - \frac{17}{3}}{3 - ( - 1 )} = \frac{- \frac{32}{3}}{4} = - \frac{8}{3}$ .
Line through $Q$ : $y + 5 = - \frac{8}{3} (x - 3)$ . Multiplying by 3: $3 y + 15 = - 8 x + 24$ .
Moving everything to one side: $8 x + 3 y - 9 = 0$ .

Answer: $8 x + 3 y - 9 = 0$

Exercise 135 MediumIB · P1 TZ0 November 2007

Given that $y = e^{- x^{2}}$ , find

a) $\frac{d ^{2} y}{d x ^{2}}$ .

b) The exact values of the $x$ -coordinates of the points of inflexion of the graph of $y = e^{- x^{2}}$ , justifying why they are points of inflexion.

Show answer

Answer: a) $\frac{d ^{2} y}{d x ^{2}} = - 2 e^{- x^{2}} (1 - 2 x^{2})$ ; b) $x = \pm \frac{1}{2}$

Show working and answer

a) 1. First derivative (chain rule): $\frac{d y}{d x} = - 2 x e^{- x^{2}}$ . 2. Second derivative (product rule): $\frac{d ^{2} y}{d x ^{2}} = - 2 e^{- x^{2}} - 2 x \cdot (- 2 x) e^{- x^{2}} = - 2 e^{- x^{2}} (1 - 2 x^{2})$ .

b) 1. The candidates satisfy $\frac{d ^{2} y}{d x ^{2}} = 0$ . Since $e^{- x^{2}} > 0$ , we need $1 - 2 x^{2} = 0$ , so $x = \pm \frac{1}{2}$ . 2. Justification: the sign of $y^{''} = 2 e^{- x^{2}} (2 x^{2} - 1)$ is that of $2 x^{2} - 1$ . It is positive if $∣ x ∣ > \frac{1}{2}$ and negative if $∣ x ∣ < \frac{1}{2}$ . The second derivative changes sign at both values, so they are points of inflexion.

$x$	$x < - \frac{1}{2}$	$- \frac{1}{2} < x < \frac{1}{2}$	$x > \frac{1}{2}$
Concavity	concave up	concave down	concave up

Answer: a) $\frac{d ^{2} y}{d x ^{2}} = - 2 e^{- x^{2}} (1 - 2 x^{2})$ ; b) $x = \pm \frac{1}{2}$

Exercise 136 MediumIB · P1 TZ1 May 2008

If $f (x) = x - 3 x^{\frac{2}{3}}$ , $x > 0$

a) find the $x$ -coordinate of the point $P$ where $f^{'} (x) = 0$ ;

b) determine whether $P$ is a maximum or minimum point.

Show answer

Answer: a) $x = 8$ ; b) $P$ is a minimum.

Show working and answer

a) 1. Write $f (x) = x - 3 x^{2/3}$ and differentiate: $f^{'} (x) = 1 - 3 \cdot \frac{2}{3} x^{- 1/3} = 1 - \frac{2}{x ^{1/3}}$ . 2. Set equal to $0$ : $x^{1/3} = 2$ , so $x = 8$ .

b) 1. Second derivative: $f^{''} (x) = \frac{2}{3} x^{- 4/3}$ . 2. At $x = 8$ : $f^{''} (8) = \frac{2}{3} \cdot \frac{1}{16} = \frac{1}{24} > 0$ , so $P$ is a minimum point.

Answer: a) $x = 8$ ; b) $P$ is a minimum.

Exercise 137 MediumIB · P2 TZ1 May 2011

Consider the function $f (x) = x^{3} - 3 x^{2} - 9 x + 10$ , $x \in R$

a) Find the equation of the line passing through the maximum and minimum points of the graph of $y = f (x)$ .

b) Show that the point of inflexion of the graph of $y = f (x)$ lies on this line.

Show answer

Answer: a) $y = - 8 x + 7$ ; b) the point of inflexion $(1, - 1)$ lies on the line.

Show working and answer

a) 1. Differentiate: $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ . Stationary points: $x = - 1$ and $x = 3$ . 2. $f^{''} (x) = 6 x - 6$ : $f^{''} (- 1) = - 12 < 0$ (maximum) and $f^{''} (3) = 12 > 0$ (minimum). 3. The $y$ -coordinates: $f (- 1) = - 1 - 3 + 9 + 10 = 15$ and $f (3) = 27 - 27 - 27 + 10 = - 17$ . Maximum $(- 1, 15)$ , minimum $(3, - 17)$ . 4. Gradient of the line joining them: $\frac{- 17 - 15}{3 - ( - 1 )} = - 8$ . 5. Equation: $y - 15 = - 8 (x + 1)$ , i.e. $y = - 8 x + 7$ .

b) 1. $f^{''} (x) = 6 x - 6 = 0$ gives $x = 1$ , and $f^{''}$ changes sign there (it is linear with positive gradient): point of inflexion. 2. $f (1) = 1 - 3 - 9 + 10 = - 1$ , so the point is $(1, - 1)$ . 3. On the line: $y = - 8 \cdot 1 + 7 = - 1$ . So the point of inflexion lies on the line.

Answer: a) $y = - 8 x + 7$ ; b) the point of inflexion $(1, - 1)$ lies on the line.

Exercise 138 MediumIB · P1 TZ0 November 2012

The diagram shows the graph of the function defined by $y = x (ln x)^{2}$ for $x > 0$ .

Graph of a curve that rises from the y-axis to a local maximum A, falls to a local minimum B on the x-axis and rises again

The function has a local maximum at the point $A$ and a local minimum at the point $B$ .

a) Find the coordinates of the points $A$ and $B$ .

b) Given that the graph of the function has exactly one point of inflexion, find its coordinates.

Show answer

Answer: a) $A (e^{- 2}, 4 e^{- 2})$ , $B (1, 0)$ ; b) $(e^{- 1}, e^{- 1})$

Show working and answer

a) 1. Local extrema occur where $y^{'} = 0$ . We differentiate with the product rule, using $\frac{d}{d x} (ln x)^{2} = 2 ln x \cdot \frac{1}{x}$ (chain rule): $y^{'} = (ln x)^{2} + x \cdot \frac{2 ln x}{x} = (ln x)^{2} + 2 ln x = ln x (ln x + 2) .$ 2. $y^{'} = 0$ if $ln x = 0$ , i.e. $x = 1$ , or if $ln x = - 2$ , i.e. $x = e^{- 2}$ . 3. The y-coordinates: $y (1) = 1 \cdot 0 = 0$ and $y (e^{- 2}) = e^{- 2} (- 2)^{2} = 4 e^{- 2}$ . 4. In the graph, the local maximum ( $A$ ) is on the left and the local minimum ( $B$ ), on the $x$ -axis, is on the right. Thus $A (e^{- 2}, 4 e^{- 2}), B (1, 0) .$

b) 1. The point of inflexion satisfies $y^{''} = 0$ . We differentiate $y^{'} = (ln x)^{2} + 2 ln x$ : $y^{''} = \frac{2 ln x}{x} + \frac{2}{x} = \frac{2 ( ln x + 1 )}{x} .$ 2. $y^{''} = 0$ if $ln x = - 1$ , i.e. $x = e^{- 1}$ . As $y^{''}$ changes sign ( $< 0$ if $x < e^{- 1}$ and $> 0$ if $x > e^{- 1}$ ), it is a point of inflexion. 3. Its y-coordinate: $y (e^{- 1}) = e^{- 1} (- 1)^{2} = e^{- 1}$ .

Answer: a) $A (e^{- 2}, 4 e^{- 2})$ , $B (1, 0)$ ; b) $(e^{- 1}, e^{- 1})$

Exercise 139 MediumIB · P1 TZ2 May 2022

The function $f$ is defined by $f (x) = x 1 - x^{2}$ , where $- 1 \leq x \leq 1$ .

The graph of $y = f (x)$ is shown below.

Graph of an odd function defined between -1 and 1: it falls to a local minimum, rises through the origin to a local maximum, and returns to the x-axis at the right-hand end

a) Show that $f$ is an odd function.

The range of $f$ is $a \leq y \leq b$ , where $a, b \in R$ .

b) Find the value of $a$ and the value of $b$ .

Show answer

Answer: a) $f (- x) = - f (x)$ ; b) $a = - \frac{1}{2}$ , $b = \frac{1}{2}$

Show working and answer

a) 1. For $- 1 \leq x \leq 1$ we also have $- x \in [- 1, 1]$ , and $f (- x) = (- x) 1 - (- x)^{2} = - x 1 - x^{2} = - f (x)$ . 2. Therefore $f$ is odd.

b) 1. We look for the extreme values of $f$ . Differentiating with the product rule: $f^{'} (x) = 1 - x^{2} + x \cdot \frac{- x}{1 - x ^{2}} = \frac{1 - x ^{2} - x ^{2}}{1 - x ^{2}} = \frac{1 - 2 x ^{2}}{1 - x ^{2}} .$ 2. $f^{'} (x) = 0$ implies $x = \pm \frac{1}{2}$ . 3. $f (\frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$ (local maximum) and, since $f$ is odd, $f (- \frac{1}{2}) = - \frac{1}{2}$ (local minimum). 4. At the endpoints of the domain $f (\pm 1) = 0$ , so the extreme values are $- \frac{1}{2}$ and $\frac{1}{2}$ : $a = - \frac{1}{2}$ and $b = \frac{1}{2}$ .

Answer: a) $f (- x) = - f (x)$ ; b) $a = - \frac{1}{2}$ , $b = \frac{1}{2}$

Exercise 140 HardIB · P2 TZ2 May 2009

The cubic curve defined by $y = 8 x^{3} + b x^{2} + c x + d$ contains two distinct points $P$ and $Q$ at which the gradient is equal to zero.

a) Verify that $b^{2} > 24 c$ .

b) Given that the coordinates of $P$ and $Q$ are $(\frac{1}{2}, - 12)$ and $(- \frac{3}{2}, 20)$ respectively, find the values of $b$ , $c$ and $d$ .

Show answer

Answer: a) it is proved that $b^{2} > 24 c$ ; b) $b = 12$ , $c = - 18$ , $d = - 7$

Show working and answer

a) 1. We differentiate: $y^{'} = 24 x^{2} + 2 b x + c$ . 2. There are two distinct points with zero gradient if and only if the equation $24 x^{2} + 2 b x + c = 0$ has two distinct real solutions, that is, a positive discriminant. 3. $Δ = (2 b)^{2} - 4 \cdot 24 \cdot c = 4 b^{2} - 96 c > 0$ . 4. Dividing by $4$ : $b^{2} - 24 c > 0$ , that is, $b^{2} > 24 c$ .

b) 1. The $x$ -coordinates of $P$ and $Q$ are the roots of $y^{'}$ : $x = \frac{1}{2}$ and $x = - \frac{3}{2}$ . Hence $y^{'} = 24 (x - \frac{1}{2}) (x + \frac{3}{2}) = 24 x^{2} + 24 x - 18$ . 2. Comparing with $24 x^{2} + 2 b x + c$ : $2 b = 24$ and $c = - 18$ , so $b = 12$ and $c = - 18$ . 3. $P$ lies on the curve: $8 \cdot \frac{1}{8} + 12 \cdot \frac{1}{4} - 18 \cdot \frac{1}{2} + d = - 12$ , that is, $1 + 3 - 9 + d = - 12$ and $d = - 7$ . 4. Check with $Q$ : $8 (- \frac{27}{8}) + 12 \cdot \frac{9}{4} + 27 - 7 = - 27 + 27 + 27 - 7 = 20$ . Correct.

Answer: a) it is proved that $b^{2} > 24 c$ ; b) $b = 12$ , $c = - 18$ , $d = - 7$

Exercise 141 HardIB · P1 TZ0 November 2009

A tangent line to the graph of $y = ln x$ passes through the origin.

a) Sketch approximately, on the same set of axes, the graphs of $y = ln x$ and of the tangent line. Hence, find the equation of the tangent line.

b) Use your approximate graph to explain why $ln x \leq \frac{x}{e}$ for $x > 0$ .

c) Verify that $x^{e} \leq e^{x}$ for $x > 0$ . Determine which is greater: $π^{e}$ or $e^{π}$ .

View solution

a) 1. The tangent line to $y = ln x$ at the point with abscissa $a$ is $y - ln a = \frac{1}{a} (x - a)$ . 2. It passes through the origin: $- ln a = - 1$ , so $ln a = 1$ and $a = e$ . 3. Substituting: $y - 1 = \frac{1}{e} (x - e)$ , that is, $y = \frac{x}{e}$ .

Graphs of y = ln x and of a line through the origin that touches the curve at one point

In the sketch: the curve $ln x$ crosses the $x$ -axis at $(1, 0)$ , has the $y$ -axis as a vertical asymptote and increases ever more slowly. The line $y = \frac{x}{e}$ starts at the origin and touches the curve only at $(e, 1)$ .

b) 1. In the sketch, the tangent line lies above the curve for all $x > 0$ (the curve is concave: $(ln x)^{''} = - \frac{1}{x ^{2}} < 0$ ), and they touch only at $x = e$ . 2. Therefore $ln x \leq \frac{x}{e}$ for $x > 0$ .

c) 1. Multiplying the previous inequality by $e > 0$ : $e ln x \leq x$ , that is, $ln (x^{e}) \leq x = ln (e^{x})$ . 2. Since $ln$ is increasing, $x^{e} \leq e^{x}$ for $x > 0$ . 3. With $x = π$ : $π^{e} \leq e^{π}$ , and equality holds only at $x = e$ ; as $π \neq = e$ , the inequality is strict. 4. $e^{π}$ is greater.

Answer: a) $y = \frac{x}{e}$ ; b) the line lies above the curve; c) $π^{e} < e^{π}$

Exercise 142 HardIB · P2 TZ2 May 2011

Sketch approximately the graph of $f (x) = x + \frac{8 x}{x ^{2} - 9}$ . Clearly indicate the coordinates of the two maxima and of the two minima. Clearly indicate and give the equations of the vertical asymptotes and of the oblique asymptote.

View solution

We study $f (x) = x + \frac{8 x}{x ^{2} - 9}$ .

Domain and symmetry: $x^{2} - 9 \neq = 0$ , so $D_{f} = R - {- 3, 3}$ . As $f (- x) = - f (x)$ , the graph is symmetric about the origin.
Intercepts with the axes: $f (0) = 0$ . And $f (x) = x (1 + \frac{8}{x ^{2} - 9}) = x \cdot \frac{x ^{2} - 1}{x ^{2} - 9} = 0$ gives $x = 0, \pm 1$ . Intercepts: $(0, 0)$ , $(1, 0)$ , $(- 1, 0)$ .
Vertical asymptotes: at $x = \pm 3$ the denominator is $0$ and the numerator $8 x \neq = 0$ : $x = 3$ and $x = - 3$ .
Oblique asymptote: $f (x) - x = \frac{8 x}{x ^{2} - 9} \to 0$ as $x \to \pm \infty$ , so $y = x$ .
Extrema: $f^{'} (x) = 1 + \frac{8 ( x ^{2} - 9 ) - 8 x \cdot 2 x}{( x ^{2} - 9 ) ^{2}} = 1 - \frac{8 ( x ^{2} + 9 )}{( x ^{2} - 9 ) ^{2}}$ (quotient rule). $f^{'} (x) = 0$ gives $(x^{2} - 9)^{2} = 8 (x^{2} + 9)$ , that is, $x^{4} - 26 x^{2} + 9 = 0$ , and $x^{2} = 13 \pm 410$ . Using the GDC, $x \approx \pm 0.592$ and $x \approx \pm 5.06$ .
Values and type: $f (0.592) \approx 0.04$ and $f (5.06) \approx 7.5$ . On $] - 3, 3 [$ , $f$ increases from $0$ to the maximum at $x \approx 0.592$ and then decreases to $- \infty$ as $x \to 3^{-}$ ; on $] 3, + \infty [$ the function comes from $+ \infty$ , decreases to the minimum at $x \approx 5.06$ and then increases, approaching the line $y = x$ . By symmetry, $(- 0.592; - 0.04)$ is a minimum and $(- 5.06; - 7.5)$ a maximum.

Graph of a rational function with two vertical asymptotes, an oblique asymptote, two maxima and two minima, symmetric about the origin

Answer: asymptotes $x = 3$ , $x = - 3$ , $y = x$ ; maxima $(0.592; 0.04)$ and $(- 5.06; - 7.5)$ ; minima $(5.06; 7.5)$ and $(- 0.592; - 0.04)$

Exercise 143 HardIB · P1 TZ0 November 2009

The diagram below shows an approximate sketch of $f^{'} (x)$ , the gradient function of the curve $f (x)$ .

Approximate sketch of the gradient function y = f′(x): starts at a positive value, touches the x-axis at x = 1, rises to a maximum at x = 2, cuts the x-axis at x = 3 and continues below the axis

Sketch approximately, on the same axes, the curve $y = f (x)$ given that $f (0) = 0$ . Clearly indicate on the graph all maxima, minima and points of inflexion.

View solution

What the graph of $f^{'}$ tells us about $f$ :

$f^{'} (x) > 0$ on $[0, 3 [$ (except at $x = 1$ , where $f^{'} (1) = 0$ but without changing sign): $f$ is increasing from $x = 0$ to $x = 3$ , starting from $f (0) = 0$ with a gradient slightly greater than $3$ .
At $x = 3$ , $f^{'}$ cuts the $x$ -axis changing from positive to negative: $f$ has a maximum at $x = 3$ and decreases for $x > 3$ .
The concavity of $f$ is given by how $f^{'}$ increases or decreases: on $] 0, 1 [$ , $f^{'}$ decreases, so $f^{''} < 0$ and $f$ is concave down. On $] 1, 2 [$ , $f^{'}$ increases, so $f^{''} > 0$ and $f$ is concave up. On $] 2, 4 [$ , $f^{'}$ decreases, so $f^{''} < 0$ and $f$ is concave down.
The points where $f^{'}$ has a local minimum ( $x = 1$ ) and a local maximum ( $x = 2$ ) are the points of inflexion of $f$ : at $x = 1$ there is a horizontal tangent with an inflexion, and at $x = 2$ there is an inflexion with maximum gradient ( $\approx 2.3$ ).

The graph of $f$ is: it starts at the origin and rises concave down to $x = 1$ (horizontal tangent, inflexion), keeps rising concave up to $x = 2$ (inflexion), and continues rising concave down to the maximum at $x = 3$ ; after that it falls.

Graph of y = f(x) with f(0) = 0: increasing up to x = 3, with points of inflexion at x = 1 and x = 2, a maximum at x = 3 and decreasing afterwards

Answer: $f$ passes through the origin, is increasing up to $x = 3$ with points of inflexion at $x = 1$ and $x = 2$ and a maximum at $x = 3$ , and decreases after that.

Exercise 144 HardIB · P2 TZ2 May 2009

Consider the graphs of $y = e^{- x}$ and $y = e^{- x} sin 4 x$ , for $0 \leq x \leq \frac{5 π}{4}$ .

a) Using the same axes, draw accurately on graph paper these two graphs for $0 \leq x \leq \frac{5 π}{4}$ . Use the following scale: on the $x$ -axis, $1$ cm represents $\frac{π}{8}$ , and on the $y$ -axis, $5$ cm represent $1$ unit.

b) Verify that the $x$ -intercepts of the graph of $y = e^{- x} sin 4 x$ are $\frac{nπ}{4}$ , $n = 0, 1, 2, 3, 4, 5$ .

c) Find the $x$ -coordinates of the points at which the graph of $y = e^{- x} sin 4 x$ and the graph of $y = e^{- x}$ meet. Give your answers in terms of $π$ .

d) i) Verify that when the graph of $y = e^{- x} sin 4 x$ meets the graph of $y = e^{- x}$ , both graphs have the same gradient.

(ii) Hence, explain why these three points of intersection are not local maxima of the graph of $y = e^{- x} sin 4 x$ .

View solution

a) 1. With the required scale, the $x$ -axis is marked every $\frac{π}{8}$ (1 cm) up to $\frac{5 π}{4}$ (10 cm), and on the $y$ -axis each unit is 5 cm. 2. $y = e^{- x}$ is a decreasing exponential passing through $(0, 1)$ and falling to $e^{- 5 π /4} \approx 0.02$ . 3. $y = e^{- x} sin 4 x$ oscillates with decreasing amplitude, enclosed between $y = e^{- x}$ and $y = - e^{- x}$ , with period $\frac{π}{2}$ , and touches the exponential at $x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}$ .

Graphs of y = e⁻ˣ and y = e⁻ˣ sin 4x for x from 0 to 5π/4: the second oscillates with decreasing amplitude and touches the first at three points

b) 1. The $x$ -intercepts satisfy $e^{- x} sin 4 x = 0$ . 2. As $e^{- x} > 0$ always, we need $sin 4 x = 0$ , that is, $4 x = nπ$ . 3. Hence $x = \frac{nπ}{4}$ , and for $0 \leq x \leq \frac{5 π}{4}$ we get $n = 0, 1, 2, 3, 4, 5$ .

c) 1. Equating: $e^{- x} sin 4 x = e^{- x}$ . Dividing by $e^{- x} \neq = 0$ : $sin 4 x = 1$ . 2. Hence $4 x = \frac{π}{2} + 2 nπ$ , that is, $x = \frac{π}{8} + \frac{nπ}{2}$ . 3. In the given interval: $x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}$ .

d) i) 1. Differentiating $y = e^{- x} sin 4 x$ with the product rule: $y^{'} = - e^{- x} sin 4 x + 4 e^{- x} cos 4 x$ . 2. At the points of intersection $sin 4 x = 1$ and therefore $cos 4 x = 0$ . Then $y^{'} = - e^{- x}$ . 3. The gradient of $y = e^{- x}$ is also $- e^{- x}$ . At $x = \frac{π}{8}$ , $\frac{5 π}{8}$ and $\frac{9 π}{8}$ both equal $- e^{- π /8}$ , $- e^{- 5 π /8}$ and $- e^{- 9 π /8}$ respectively: they coincide.

(ii) 1. At a local maximum the derivative is zero. 2. Here the derivative equals $- e^{- x} < 0$ (strictly negative), so the graph is decreasing at those points and they cannot be local maxima.

Answer: b) $x = \frac{nπ}{4}$ ; c) $x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}$ ; d) at the three points both gradients equal $- e^{- x} \neq = 0$ , so they are not local maxima.

Exercise 145 HardIB · P2 TZ2 May 2012

a) Sketch approximately the curve $y = \frac{cos x}{x ^{2} + 1}$ , $- 4 \leq x \leq 4$ , clearly showing the coordinates of the $x$ -intercepts, and all maxima and minima.

b) Write down the gradient of the curve at $x = 1$ .

c) Find the equation of the normal to the curve at $x = 1$ .

View solution

a) 1. Using the GDC, we plot $y = \frac{cos x}{x ^{2} + 1}$ for $- 4 \leq x \leq 4$ . The function is even: its graph is symmetric about the $y$ -axis. 2. $x$ -intercepts: $cos x = 0$ , that is, $x = - \frac{π}{2} = - 1.57$ and $x = \frac{π}{2} = 1.57$ . 3. Maximum at $(0, 1)$ and minima at $(- 2.84; - 0.317)$ and $(2.84; - 0.317)$ (calculated using the GDC).

Graph of y = cos x / √(x² + 1) for x between −4 and 4: curve symmetric about the y-axis, with a maximum at x = 0, x-intercepts and two minima

b) Using the GDC, the gradient at $x = 1$ is $y^{'} (1) = - 0.786$ .

c) 1. The normal has gradient $- \frac{1}{- 0.786} = 1.272$ . 2. The point on the curve is $(1, \frac{cos 1}{2}) = (1; 0.382)$ . 3. The normal is $y - 0.382 = 1.272 (x - 1)$ , that is, $y = 1.27 x - 0.89$ .

Answer: a) $x$ -intercepts $x = \pm 1.57$ ; maximum $(0, 1)$ ; minima $(\pm 2.84; - 0.317)$ ; b) $- 0.786$ ; c) $y = 1.27 x - 0.89$

Exercise 146 HardIB · P1 TZ2 May 2012

The diagram shows the graph of $y = f (x)$ , where $A$ is a local maximum and $D$ is a local minimum.

Graph of y = f(x) with a local maximum A(−1, 4), the point B(0, 3), the x-intercepts C(1, 0) and E(3, 0) and a local minimum D(2, −3); the curve increases without bound to the right of E

a) On a set of axes, sketch the graph of $y = \frac{1}{f ( x )}$ , showing clearly the coordinates of the images of the points $A$ , $B$ and $D$ , labelling them $A^{'}$ , $B^{'}$ and $D^{'}$ respectively, and the equations of all vertical asymptotes.

b) On a set of axes, sketch the graph of the derivative function $y = f^{'} (x)$ , showing clearly the coordinates of the images of the points $A$ and $D$ , and labelling them $A^{''}$ and $D^{''}$ respectively.

View solution

a) 1. The points are transformed using $y \mapsto \frac{1}{y}$ : $A (- 1, 4) \to A^{'} (- 1, \frac{1}{4})$ , $B (0, 3) \to B^{'} (0, \frac{1}{3})$ and $D (2, - 3) \to D^{'} (2, - \frac{1}{3})$ . 2. The zeros of $f$ , $x = 1$ ( $C$ ) and $x = 3$ ( $E$ ), give vertical asymptotes of $\frac{1}{f}$ : $x = 1$ and $x = 3$ . 3. The local maximum of $f$ at $A$ (with a positive value) becomes a local minimum $A^{'}$ ; the local minimum at $D$ (with a negative value) becomes a local maximum $D^{'}$ . 4. As $f$ grows large (to the right of $E$ ), $\frac{1}{f} \to 0^{+}$ : the $x$ -axis is a horizontal asymptote.

Graph of y = 1/f(x) with vertical asymptotes x = 1 and x = 3, and the points A′(−1, 1/4), B′(0, 1/3) and D′(2, −1/3)

b) 1. At local extrema the derivative is zero: $A^{''} (- 1, 0)$ and $D^{''} (2, 0)$ . 2. Where $f$ is increasing ( $x < - 1$ and $x > 2$ ), $f^{'} > 0$ ; where it is decreasing ( $- 1 < x < 2$ ), $f^{'} < 0$ . In addition, $f^{'}$ increases rapidly for large $x$ , because $f$ increases very rapidly.

Graph of the derivative function y = f ′(x), which cuts the x-axis at the points A″(−1, 0) and D″(2, 0); it is negative between them

Answer: a) $A^{'} (- 1, \frac{1}{4})$ , $B^{'} (0, \frac{1}{3})$ , $D^{'} (2, - \frac{1}{3})$ ; asymptotes $x = 1$ , $x = 3$ ; b) $A^{''} (- 1, 0)$ , $D^{''} (2, 0)$

Exercise 147 HardIB · P1 TZ2 May 2013

The function $f$ has the form $f (x) = \frac{2 x - 1}{x + 2}$ and its domain is $D = {x : - 1 \leq x \leq 8}$ .

a) Express $f (x)$ in the form $A + \frac{B}{x + 2}$ , where $A$ and $B \in Z$ .

b) Hence, verify that $f^{'} (x) > 0$ on $D$ .

c) State the range of $f$ .

d) i) Find an expression for $f^{- 1} (x)$ .

(ii) Sketch the graph of $y = f (x)$ , showing the intercepts with both axes.

(iii) On the same diagram, sketch the graph of $y = f^{- 1} (x)$ .

e) i) On another diagram, sketch the graph of $y = f (∣ x ∣)$ , where $x \in D$ .

(ii) Find all the solutions of the equation $f (∣ x ∣) = - \frac{1}{4}$ .

View solution

a) 1. We write the numerator in terms of $x + 2$ : $2 x - 1 = 2 (x + 2) - 5$ . 2. Hence $f (x) = \frac{2 ( x + 2 ) - 5}{x + 2} = 2 - \frac{5}{x + 2}$ , with $A = 2$ and $B = - 5$ .

b) 1. We differentiate $f (x) = 2 - 5 (x + 2)^{- 1}$ (power and chain rules): $f^{'} (x) = 5 (x + 2)^{- 2} = \frac{5}{( x + 2 ) ^{2}}$ . 2. On $D = [- 1, 8]$ , $x + 2 > 0$ , so $(x + 2)^{2} > 0$ and $f^{'} (x) > 0$ .

c) 1. Since $f^{'} > 0$ , $f$ is strictly increasing on $D$ : the minimum is at $x = - 1$ and the maximum at $x = 8$ . 2. $f (- 1) = \frac{- 3}{1} = - 3$ and $f (8) = \frac{15}{10} = \frac{3}{2}$ . 3. The range is $[- 3, \frac{3}{2}]$ .

d) i) 1. We write $y = \frac{2 x - 1}{x + 2}$ and solve for $x$ : $y (x + 2) = 2 x - 1$ . 2. $x y + 2 y = 2 x - 1 \Rightarrow x (y - 2) = - 1 - 2 y \Rightarrow x = \frac{1 + 2 y}{2 - y}$ . 3. Hence $f^{- 1} (x) = \frac{2 x + 1}{2 - x}$ .

(ii) 1. $f$ cuts the $y$ -axis at $f (0) = - \frac{1}{2}$ and the $x$ -axis when $2 x - 1 = 0$ , i.e. at $x = \frac{1}{2}$ . 2. The graph is increasing and runs from $(- 1, - 3)$ to $(8, \frac{3}{2})$ .

(iii) The graph of $f^{- 1}$ is the reflection of that of $f$ in the line $y = x$ : it runs from $(- 3, - 1)$ to $(\frac{3}{2}, 8)$ and cuts the axes at $(0, \frac{1}{2})$ and $(- \frac{1}{2}, 0)$ .

Graphs of y = f(x) on the domain (−1, 8) and of its inverse y = f⁻¹(x), symmetric about the line y = x (dashed); the intercepts with the axes are marked

e) i) 1. $f (∣ x ∣)$ is even: for $x \geq 0$ it coincides with $f (x)$ , and for $x < 0$ it is the reflection of that in the $y$ -axis. 2. With $x \in D = [- 1, 8]$ , the graph has a minimum at $(0, - \frac{1}{2})$ and cuts the $x$ -axis at $x = \pm \frac{1}{2}$ .

Graph of y = f(|x|) for x between −1 and 8: the part for negative x is a reflection of the part for positive x, with a minimum at the point (0, −1/2) and x-axis intercepts at x = ±1/2

(ii) 1. For $x \geq 0$ : $f (x) = - \frac{1}{4} \Rightarrow 2 - \frac{5}{x + 2} = - \frac{1}{4} \Rightarrow \frac{5}{x + 2} = \frac{9}{4} \Rightarrow x + 2 = \frac{20}{9} \Rightarrow x = \frac{2}{9}$ . 2. By symmetry, $x = - \frac{2}{9}$ is also a solution, and it lies in $D$ .

Answer: a) $f (x) = 2 - \frac{5}{x + 2}$ ; c) $[- 3, \frac{3}{2}]$ ; d) i) $f^{- 1} (x) = \frac{2 x + 1}{2 - x}$ ; e) ii) $x = \pm \frac{2}{9}$

Exercise 148 HardIB · P2 TZ0 May 2019

Consider the polynomial $P (z) = z^{4} - 6 z^{3} - 2 z^{2} + 58 z - 51$ , $z \in C$ .

a) Express $P (z)$ in the form $(z^{2} + a z + b) (z^{2} + cz + d)$ , where $a, b, c, d \in R$ .

b) Sketch the graph of $y = x^{4} - 6 x^{3} - 2 x^{2} + 58 x - 51$ , clearly indicating the coordinates of all maxima, minima and intercepts with the axes.

c) Hence, or otherwise, state the condition that $k \in R$ must satisfy for all the roots of the equation $P (z) = k$ to be real.

View solution

a) 1. We look for real quadratic factors. Trying integer roots: $P (1) = 1 - 6 - 2 + 58 - 51 = 0$ and $P (- 3) = 81 + 162 - 18 - 174 - 51 = 0$ . 2. The factors $(z - 1) (z + 3) = z^{2} + 2 z - 3$ divide $P$ . Dividing: $P (z) = (z^{2} + 2 z - 3) (z^{2} - 8 z + 17)$ . 3. Checking by expanding: $z^{4} - 8 z^{3} + 17 z^{2} + 2 z^{3} - 16 z^{2} + 34 z - 3 z^{2} + 24 z - 51 = z^{4} - 6 z^{3} - 2 z^{2} + 58 z - 51$ .

b) 1. $x$ -intercepts: $x = 1$ and $x = - 3$ (the factor $x^{2} - 8 x + 17$ has no real roots: discriminant $64 - 68 < 0$ ). $y$ -intercept: $P (0) = - 51$ . 2. Turning points: using the derivative $4 x^{3} - 18 x^{2} - 4 x + 58 = 0$ and the GDC (or from the graph), there are three local extrema. Minimum at $(- 1.62; - 117.8)$ , maximum at $(2.4; 26.9)$ and minimum at $(3.72; 19.7)$ . 3. The graph falls from $+ \infty$ on the left, passes through $(- 3, 0)$ , reaches the minimum, rises passing through $(0, - 51)$ and $(1, 0)$ to the maximum, falls to the second minimum and rises again towards $+ \infty$ .

Graph of a quartic polynomial with two local minima and a local maximum between them, with the intercepts with the axes and the turning points marked with their coordinates

c) 1. The roots of $P (z) = k$ are the $x$ -coordinates of the points where the graph meets the horizontal line $y = k$ (when they are real). 2. For all four roots to be real, the line must cut the graph four times (counting tangencies as double roots). 3. This happens when $k$ lies between the higher local minimum and the local maximum: the minimum at $(3.72; 19.7)$ and the maximum $26.9$ . 4. Hence $19.7 \leq k \leq 26.9$ .

Answer: a) $P (z) = (z^{2} + 2 z - 3) (z^{2} - 8 z + 17)$ ; b) intercepts $(- 3, 0)$ , $(1, 0)$ , $(0, - 51)$ ; max. $(2.4; 26.9)$ ; min. $(- 1.62; - 117.8)$ and $(3.72; 19.7)$ ; c) $19.7 \leq k \leq 26.9$

Exercise 149 HardIB · P1 TZ2 May 2019

Consider the functions $f$ and $g$ defined by

$f (x) = ln ∣ x ∣$ , $x \in R - {0}$ and $g (x) = ln ∣ x + k ∣$ , $x \in R - {- k}$ , where $k \in R$ , $k > 2$ .

a) Describe the transformation that maps $f (x)$ onto $g (x)$ .

b) State the range of $g$ .

c) On the same set of axes, sketch the graphs of $y = f (x)$ and $y = g (x)$ , clearly indicating all points of intersection with the axes.

The graphs of $f$ and $g$ intersect at the point P.

d) Find the coordinates of P.

The tangent to $y = f (x)$ at P passes through the origin $(0, 0)$ .

e) Determine the value of $k$ .

View solution

a) $g (x) = ln ∣ x + k ∣ = f (x + k)$ : this is a horizontal translation of $k$ units to the left (vector $(- k 0)$ ).

b) $∣ x + k ∣$ takes all positive values, so $ln ∣ x + k ∣$ takes all real values: $R_{g} = R$ .

c) 1. For $f$ : vertical asymptote $x = 0$ ; $ln ∣ x ∣ = 0 \Rightarrow ∣ x ∣ = 1$ , so the intercepts are $(1, 0)$ and $(- 1, 0)$ ; it does not cut the $y$ -axis. It is symmetric about the $y$ -axis. 2. For $g$ : vertical asymptote $x = - k$ ; intercepts with the $x$ -axis when $x + k = \pm 1$ , that is, $(- k + 1, 0)$ and $(- k - 1, 0)$ ; intercept with the $y$ -axis at $(0, ln k)$ .

Graphs of f(x) = ln|x| and g(x) = ln|x + k| on the same axes, with their vertical asymptotes, the intercepts with the axes and the point of intersection P of the two curves

d) 1. Equate the two functions: $ln ∣ x ∣ = ln ∣ x + k ∣ \Rightarrow ∣ x ∣ = ∣ x + k ∣$ . 2. $x = x + k$ is impossible ( $k > 2$ ), so $x = - (x + k) \Rightarrow x = - \frac{k}{2}$ . 3. The $y$ -coordinate is $ln - \frac{k}{2} = ln \frac{k}{2}$ . Therefore $P (- \frac{k}{2}, ln \frac{k}{2})$ .

e) 1. For $x < 0$ , $f (x) = ln (- x)$ and $f^{'} (x) = \frac{1}{x}$ . At P the gradient is $m = \frac{1}{- k /2} = - \frac{2}{k}$ . 2. The tangent at P passes through the origin, so $\frac{y _{P} - 0}{x _{P} - 0} = m$ , that is, $ln \frac{k}{2} = m \cdot (- \frac{k}{2}) = - \frac{2}{k} \cdot (- \frac{k}{2}) = 1$ . 3. $ln \frac{k}{2} = 1 \Rightarrow \frac{k}{2} = e \Rightarrow k = 2 e$ (which satisfies $k > 2$ ).

Answer: a) horizontal translation of $k$ units to the left; b) $R_{g} = R$ ; d) $P (- \frac{k}{2}, ln \frac{k}{2})$ ; e) $k = 2 e$

Exercise 150 HardIB · P1 TZ1 May 2018

Let $f (x) = \frac{2 - 3 x ^{5}}{2 x ^{3}}$ , $x \in R$ , $x \neq = 0$ .

a) The graph of $y = f (x)$ has a local maximum at A. Find the coordinates of A.

b) (i) Show that there is exactly one point of inflexion, B, on the graph of $y = f (x)$ .

(ii) The coordinates of B can be written in the form $B (2^{a}, b \cdot 2^{- 3 a})$ , where $a, b \in Q$ . Find the value of $a$ and the value of $b$ .

c) Sketch the graph of $y = f (x)$ , clearly showing the positions of the points A and B.

View solution

a) 1. We rewrite $f (x) = x^{- 3} - \frac{3}{2} x^{2}$ . 2. $f^{'} (x) = - 3 x^{- 4} - 3 x = - \frac{3}{x ^{4}} - 3 x$ . 3. $f^{'} (x) = 0 \Rightarrow - 3 x^{- 4} = 3 x \Rightarrow x^{5} = - 1 \Rightarrow x = - 1$ . 4. $f^{''} (x) = 12 x^{- 5} - 3$ and $f^{''} (- 1) = - 15 < 0$ : it is a local maximum. 5. $f (- 1) = - 1 - \frac{3}{2} = - \frac{5}{2}$ . Therefore $A (- 1, - \frac{5}{2})$ .

b) i) 1. $f^{''} (x) = 0 \Rightarrow \frac{12}{x ^{5}} = 3 \Rightarrow x^{5} = 4 \Rightarrow x = 4^{1/5} = 2^{2/5}$ , the only real solution. 2. Sign change: if $x < 0$ , $f^{''} (x) = \frac{12}{x ^{5}} - 3 < 0$ ; if $0 < x < 2^{2/5}$ , $\frac{12}{x ^{5}} > 3$ and $f^{''} > 0$ ; if $x > 2^{2/5}$ , $f^{''} < 0$ . At $x = 2^{2/5}$ the sign changes from $+$ to $-$ (and $x = 0$ is not in the domain). 3. Hence there is exactly one point of inflexion $B$ .

(ii) 1. $a = \frac{2}{5}$ (abscissa $2^{2/5}$ ). 2. Ordinate: $f (2^{2/5}) = 2^{- 6/5} - \frac{3}{2} \cdot 2^{4/5} = 2^{- 6/5} - 3 \cdot 2^{- 1/5} = 2^{- 6/5} - 6 \cdot 2^{- 6/5} = - 5 \cdot 2^{- 6/5}$ . 3. With $- 3 a = - \frac{6}{5}$ we get $b = - 5$ . Therefore $a = \frac{2}{5}$ , $b = - 5$ .

c) 1. Vertical asymptote $x = 0$ : as $x \to 0^{+}$ , $f \to + \infty$ , and as $x \to 0^{-}$ , $f \to - \infty$ . For large $∣ x ∣$ , $f \approx - \frac{3}{2} x^{2} \to - \infty$ . 2. Intercept with the $x$ -axis: $2 - 3 x^{5} = 0 \Rightarrow x = (\frac{2}{3})^{1/5} \approx 0.92$ . It does not cut the $y$ -axis. 3. Local maximum $A (- 1; - 2.5)$ and point of inflexion $B (2^{2/5}; - 5 \cdot 2^{- 6/5}) \approx (1.32; - 2.18)$ .

Graph of the function with a vertical asymptote along the y-axis, the local maximum A on the negative side of the x-axis and the point of inflexion B on the positive side

Answer: a) $A (- 1, - \frac{5}{2})$ ; b) $B (2^{2/5}, - 5 \cdot 2^{- 6/5})$ , with $a = \frac{2}{5}$ and $b = - 5$

Exercise 151 HardIB · P2 TZ0 November 2018

The function $f$ is defined by $f (x) = \frac{2 ln x + 1}{x - 3}$ , $0 < x < 3$ .

a) Find $f^{'} (x)$ .

b) Hence, or otherwise, find the coordinates of the point of inflexion on the graph of $y = f (x)$ .

c) Draw a set of Cartesian axes on which $x$ and $y$ range from $- 3$ to $3$ . On the same axes,

(i) sketch the graph of $y = f (x)$ , clearly showing any intercepts with the axes and giving the equations of any asymptotes.

(ii) sketch the graph of $y = f^{- 1} (x)$ , clearly showing any intercepts with the axes and giving the equations of any asymptotes.

d) Hence, or otherwise, solve the inequality $f (x) > f^{- 1} (x)$ .

View solution

a) 1. We apply the quotient rule with $u = 2 ln x + 1$ (so $u^{'} = \frac{2}{x}$ ) and $v = x - 3$ (so $v^{'} = 1$ ): $f^{'} (x) = \frac{\frac{2}{x} ( x - 3 ) - ( 2 ln x + 1 )}{( x - 3 ) ^{2}}$ . 2. We simplify the numerator: $2 - \frac{6}{x} - 2 ln x - 1 = \frac{x - 6 - 2 x ln x}{x}$ . 3. Therefore $f^{'} (x) = \frac{x - 6 - 2 x ln x}{x ( x - 3 ) ^{2}}$ .

b) 1. A point of inflexion occurs where $f^{''} (x) = 0$ and $f^{''}$ changes sign, that is, where $f^{'}$ has an extremum. 2. Using the GDC, we draw $y = f^{'} (x)$ and find its minimum (or solve $f^{''} (x) = 0$ for $0 < x < 3$ ): $x \approx 0.899$ . 3. We calculate $y = f (0.899) \approx - 0.375$ . 4. The point of inflexion is $(0.899; - 0.375)$ .

c) (i) 1. Vertical asymptotes: as $x \to 0^{+}$ , $ln x \to - \infty$ and the numerator tends to $- \infty$ while the denominator tends to $- 3$ , so $f \to + \infty$ . As $x \to 3^{-}$ , the numerator is $2 ln 3 + 1 > 0$ and the denominator tends to $0^{-}$ , so $f \to - \infty$ . Asymptotes: $x = 0$ and $x = 3$ . 2. Intercept with the $x$ -axis: $2 ln x + 1 = 0 \Rightarrow x = e^{- 1/2} \approx 0.607$ . Point $(0.607; 0)$ . There is no intercept with the $y$ -axis because $0$ is not in the domain. 3. The function is decreasing over its whole domain (seen from $f^{'}$ ).

(ii) 1. The graph of $f^{- 1}$ is the reflection of the graph of $f$ in the line $y = x$ . 2. The vertical asymptotes of $f$ become horizontal asymptotes: $y = 0$ and $y = 3$ . 3. The intercept of $f$ with the $x$ -axis becomes an intercept with the $y$ -axis: $(0; 0.607)$ .

Graphs of f and of its inverse on a square of side six centred at the origin, with the asymptotes, the line y = x and the intercepts with the axes

d) 1. The two curves intersect on $y = x$ : we solve $f (x) = x$ using the GDC and obtain $x \approx 0.372$ . 2. For $0 < x < 0.372$ the graph of $f$ lies above that of $f^{- 1}$ (for example, $f (0.1) \approx 1.24$ and $f^{- 1} (0.1) \approx 0.54$ ); for $x > 0.372$ it lies below. Also, $f$ is only defined for $x > 0$ . 3. Therefore $f (x) > f^{- 1} (x)$ if $0 < x < 0.372$ .

Answer: a) $f^{'} (x) = \frac{x - 6 - 2 x ln x}{x ( x - 3 ) ^{2}}$ ; b) point of inflexion $(0.899; - 0.375)$ ; c) $f$ : vertical asymptotes $x = 0$ and $x = 3$ , intercept $(0.607; 0)$ ; $f^{- 1}$ : horizontal asymptotes $y = 0$ and $y = 3$ , intercept $(0; 0.607)$ ; d) $0 < x < 0.372$

Exercise 152 HardIB · P1 TZ1 May 2017

Consider the polynomial $P (z) = z^{5} - 10 z^{2} + 15 z - 6$ , $z \in C$ .

a) Write down the sum and the product of the roots of $P (z) = 0$ .

b) Show that $(z - 1)$ is a factor of $P (z)$ .

The polynomial can be written in the form $P (z) = (z - 1)^{3} (z^{2} + b z + c)$ .

c) Calculate the value of $b$ and the value of $c$ .

d) Using the above, find the complex roots of $P (z) = 0$ .

Consider the function $q (x) = x^{5} - 10 x^{2} + 15 x - 6$ , $x \in R$ .

e) (i) Show that the graph of $y = q (x)$ is concave up for $x > 1$ .

(ii) Sketch the graph of $y = q (x)$ , clearly showing any intercepts with the axes.

View solution

a) 1. For a monic polynomial of degree 5, $z^{5} + a_{4} z^{4} + \dots + a_{0}$ , the sum of the roots is $- a_{4}$ and the product is $(- 1)^{5} a_{0} = - a_{0}$ . 2. Here $a_{4} = 0$ and $a_{0} = - 6$ : sum $= 0$ and product $= 6$ .

b) 1. We calculate $P (1) = 1 - 10 + 15 - 6 = 0$ . 2. By the factor theorem, $(z - 1)$ is a factor of $P (z)$ .

c) 1. We expand $(z - 1)^{3} = z^{3} - 3 z^{2} + 3 z - 1$ and multiply by $(z^{2} + b z + c)$ . The coefficient of $z^{4}$ is $b - 3$ , which must equal $0$ , so $b = 3$ . 2. The coefficient of $z^{3}$ is $c - 3 b + 3 = 0$ ; with $b = 3$ , $c - 9 + 3 = 0$ , so $c = 6$ . 3. We check the $z^{2}$ coefficient: $- 3 c + 3 b - 1 = - 18 + 9 - 1 = - 10$ , and the $z$ coefficient: $3 c - b = 18 - 3 = 15$ . These agree with $P$ .

d) 1. The roots of $P$ are $z = 1$ (triple) and the roots of $z^{2} + 3 z + 6 = 0$ . 2. By the general formula: $z = \frac{- 3 \pm 9 - 24}{2} = \frac{- 3 \pm 15 i}{2}$ . 3. The complex roots are $- \frac{3}{2} + \frac{15}{2} i$ and $- \frac{3}{2} - \frac{15}{2} i$ .

e) (i) 1. We differentiate twice: $q^{'} (x) = 5 x^{4} - 20 x + 15$ and $q^{''} (x) = 20 x^{3} - 20 = 20 (x^{3} - 1)$ . 2. If $x > 1$ , then $x^{3} > 1$ , so $q^{''} (x) > 0$ and the graph is concave up.

(ii) 1. Intercept with the $y$ -axis: $q (0) = - 6$ , point $(0, - 6)$ . 2. Intercepts with the $x$ -axis: $q (x) = (x - 1)^{3} (x^{2} + 3 x + 6)$ and $x^{2} + 3 x + 6 > 0$ always (negative discriminant), so the only intercept is $(1, 0)$ . 3. Moreover $q^{'} (x) = 5 (x - 1)^{2} (x^{2} + 2 x + 3) \geq 0$ : the function is increasing, and at $x = 1$ the tangent is horizontal. Since $q^{''}$ changes sign at $x = 1$ , there is a point of inflexion.

Graph of a fifth-degree polynomial that is increasing, with a flattening at the intercept with the x-axis and the intercept with the y-axis marked

Answer: a) sum $= 0$ , product $= 6$ ; b) $P (1) = 0$ ; c) $b = 3$ , $c = 6$ ; d) $- \frac{3}{2} \pm \frac{15}{2} i$ ; e) (i) $q^{''} (x) = 20 (x^{3} - 1) > 0$ for $x > 1$ ; (ii) intercepts with the axes $(1, 0)$ and $(0, - 6)$

Exercise 153 HardIB · P2 TZ1 May 2017

Consider $f (x) = - 1 + ln (x^{2} - 1)$ .

a) Find the largest possible domain $D$ for which $f$ is a function.

The function $f$ is defined by $f (x) = - 1 + ln (x^{2} - 1)$ , $x \in D$ .

b) Sketch the graph of $y = f (x)$ , clearly showing the equations of the asymptotes and the coordinates of any intercepts with the axes.

c) Explain why $f$ is an even function.

d) Explain why the inverse function $f^{- 1}$ does not exist.

The function $g$ is defined by $g (x) = - 1 + ln (x^{2} - 1)$ , $x \in] 1, \infty [$ .

e) Find the inverse function $g^{- 1}$ and write down its domain.

f) Calculate $g^{'} (x)$ .

g) Using the above, show that there are no solutions for

(i) $g^{'} (x) = 0$ ;

(ii) $(g^{- 1})^{'} (x) = 0$ .

View solution

a) 1. We need $x^{2} - 1 > 0$ (the square root must exist and the argument of the logarithm must be positive). 2. This holds if $x < - 1$ or $x > 1$ . 3. $D =] - \infty, - 1 [\cup] 1, + \infty [$ .

b) 1. As $x \to 1^{+}$ (or $x \to - 1^{-}$ ), $x^{2} - 1 \to 0^{+}$ and $ln x^{2} - 1 \to - \infty$ : vertical asymptotes $x = 1$ and $x = - 1$ . 2. Intercept with the $x$ -axis: $ln x^{2} - 1 = 1 \Rightarrow x^{2} - 1 = e \Rightarrow x^{2} = 1 + e^{2}$ , that is, $x = \pm 1 + e^{2} \approx \pm 2.9$ . 3. There is no intercept with the $y$ -axis, because $0 \in / D$ . 4. The graph is symmetric about the $y$ -axis, and in each branch it grows slowly (the right-hand branch is increasing).

Graph symmetric about the y-axis with two separate branches divided by two vertical asymptotes, and the two intercepts with the x-axis marked

c) 1. The domain is symmetric about $0$ . 2. $f (- x) = - 1 + ln (- x)^{2} - 1 = - 1 + ln x^{2} - 1 = f (x)$ . 3. Hence $f$ is even (its graph is symmetric about the $y$ -axis).

d) 1. Since $f$ is even, $f (- x) = f (x)$ with $- x \neq = x$ for any $x \neq = 0$ : for example $f (2) = f (- 2)$ . 2. $f$ is not one-to-one (a horizontal line cuts the graph at two points), so it has no inverse.

e) 1. We write $y = - 1 + ln x^{2} - 1$ and solve for $x$ : $y + 1 = ln x^{2} - 1 = \frac{1}{2} ln (x^{2} - 1)$ . 2. Hence $e^{2 y + 2} = x^{2} - 1$ , so $x^{2} = e^{2 y + 2} + 1$ and, since $x > 1$ , $x = e^{2 y + 2} + 1$ . 3. $g^{- 1} (x) = e^{2 x + 2} + 1$ . Its domain is the range of $g$ : since $g$ is continuous and increases from $- \infty$ to $+ \infty$ on $] 1, + \infty [$ , the domain is $R$ .

f) 1. $g (x) = - 1 + \frac{1}{2} ln (x^{2} - 1)$ . We differentiate (chain rule): $g^{'} (x) = \frac{1}{2} \cdot \frac{2 x}{x ^{2} - 1} = \frac{x}{x ^{2} - 1}$ .

g) (i) $g^{'} (x) = 0 \Rightarrow x = 0$ , but $0 \in /] 1, + \infty [$ . There is no solution.

(ii) We differentiate $g^{- 1} (x) = (e^{2 x + 2} + 1)^{1/2}$ with the chain rule: $(g^{- 1})^{'} (x) = \frac{e ^{2 x + 2}}{e ^{2 x + 2} + 1}$ . The numerator is $e^{2 x + 2} > 0$ for all $x \in R$ , so it never vanishes. There is no solution.

Answer: a) $D =] - \infty, - 1 [\cup] 1, + \infty [$ ; b) vertical asymptotes $x = 1$ and $x = - 1$ ; intercepts $(\pm 2.9; 0)$ ; c) $f (- x) = f (x)$ ; d) $f$ is not one-to-one; e) $g^{- 1} (x) = e^{2 x + 2} + 1$ , with domain $R$ ; f) $g^{'} (x) = \frac{x}{x ^{2} - 1}$ ; g) (i) $x = 0 \in /] 1, + \infty [$ ; (ii) $e^{2 x + 2} > 0$ for all $x$

Exercise 154 HardIB · P2 TZ0 November 2017

Consider the function $f (x) = 2 sin^{2} x + 7 sin 2 x + tan x - 9$ , $0 \leq x < \frac{π}{2}$ .

a) (i) Find an expression for $f^{'} (x)$ in terms of $x$ .

(ii) Sketch the graph of $y = f^{'} (x)$ for $0 \leq x < \frac{π}{2}$ .

(iii) Calculate the $x$ -coordinate(s) of the point(s) of inflexion of the graph of $y = f (x)$ , labelling them clearly on the graph of $y = f^{'} (x)$ .

b) Let $u = tan x$

(i) Express $sin x$ in terms of $u$ .

(ii) Express $sin 2 x$ in terms of $u$ .

(iii) Using the above, show that $f (x) = 0$ can be expressed as $u^{3} - 7 u^{2} + 15 u - 9 = 0$ .

c) Solve the equation $f (x) = 0$ , giving your answers in the form $arctan k$ where $k \in Z$ .

View solution

a) (i) 1. We differentiate each term: $\frac{d}{d x} (2 sin^{2} x) = 4 sin x cos x$ (chain rule), $\frac{d}{d x} (7 sin 2 x) = 14 cos 2 x$ and $\frac{d}{d x} tan x = \frac{1}{cos ^{2} x}$ . 2. $f^{'} (x) = 4 sin x cos x + 14 cos 2 x + \frac{1}{cos ^{2} x}$ .

(ii) Using the GDC we sketch $y = f^{'} (x)$ on $0 \leq x < \frac{π}{2}$ : it starts at $f^{'} (0) = 15$ , has a local maximum, decreases to a local minimum and then increases without bound towards the vertical asymptote $x = \frac{π}{2}$ .

Graph of the derivative in the interval zero to pi over two: local maximum near the origin, local minimum further right and unbounded increase towards the vertical asymptote x = pi/2

(iii) 1. The points of inflexion of $f$ are where $f^{''} (x) = 0$ and $f^{''}$ changes sign, i.e. where $f^{'}$ has a local maximum or minimum. 2. Using the GDC, on the graph of $f^{'}$ : the local maximum is at $x \approx 0.0736$ and the local minimum at $x \approx 1.13$ . 3. The points of inflexion of $f$ have $x \approx 0.0736$ and $x \approx 1.13$ .

b) (i) 1. From $u = tan x$ and $1 + tan^{2} x = \frac{1}{cos ^{2} x}$ : $cos^{2} x = \frac{1}{1 + u ^{2}}$ . 2. Then $sin^{2} x = 1 - cos^{2} x = \frac{u ^{2}}{1 + u ^{2}}$ and, as $sin x > 0$ on the interval, $sin x = \frac{u}{u ^{2} + 1}$ .

(ii) 1. $sin 2 x = 2 sin x cos x = 2 tan x cos^{2} x$ . 2. Therefore $sin 2 x = \frac{2 u}{u ^{2} + 1}$ .

(iii) 1. Substituting into $f (x) = 0$ : $2 \cdot \frac{u ^{2}}{u ^{2} + 1} + 7 \cdot \frac{2 u}{u ^{2} + 1} + u - 9 = 0$ . 2. Multiplying by $u^{2} + 1$ : $2 u^{2} + 14 u + (u - 9) (u^{2} + 1) = 0$ . 3. Expanding $(u - 9) (u^{2} + 1) = u^{3} - 9 u^{2} + u - 9$ and adding: $u^{3} - 7 u^{2} + 15 u - 9 = 0$ .

c) 1. Factorise: $u = 1$ is a root ( $1 - 7 + 15 - 9 = 0$ ), and $u^{3} - 7 u^{2} + 15 u - 9 = (u - 1) (u - 3)^{2}$ . 2. So $u = 1$ or $u = 3$ , i.e. $tan x = 1$ or $tan x = 3$ . 3. On $0 \leq x < \frac{π}{2}$ each equation has one solution: $x = arctan 1$ and $x = arctan 3$ .

Answer: a) (i) $f^{'} (x) = 4 sin x cos x + 14 cos 2 x + \frac{1}{cos ^{2} x}$ ; (iii) $x \approx 0.0736$ and $x \approx 1.13$ ; b) (i) $sin x = \frac{u}{u ^{2} + 1}$ ; (ii) $sin 2 x = \frac{2 u}{u ^{2} + 1}$ ; c) $x = arctan 1$ and $x = arctan 3$

Exercise 155 HardIB · P2 TZ1 May 2016

Let $f (x) = x^{4} + 0.2 x^{3} - 5.8 x^{2} - x + 4$ , $x \in R$ .

a) Find the solutions of $f (x) > 0$ .

b) For the curve $y = f (x)$ :

(i) Find the coordinates of the local minimum points.

(ii) Find the $x$ -coordinates of the points of inflexion.

The domain of $f$ is now restricted to $[0, a]$ .

c) (i) Write down the greatest value of $a$ for which $f$ has an inverse. Give your answer correct to three significant figures.

(ii) For that value of $a$ , sketch the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ on the same axes, clearly showing the coordinates of the endpoints of each curve.

(iii) Solve $f^{- 1} (x) = 1$ .

Let $g (x) = 2 sin (x - 1) - 3$ , $- \frac{π}{2} + 1 \leq x \leq \frac{π}{2} + 1$ .

d) (i) Find an expression for $g^{- 1} (x)$ , stating the domain.

(ii) Solve $(f^{- 1} \circ g) (x) < 1$ .

View solution

a) 1. Using the GDC we solve $f (x) = 0$ : $x = - 2.24$ , $x = - 1$ , $x = 0.8$ and $x = 2.24$ (that is, $\pm 5$ , $- 1$ and $\frac{4}{5}$ ). 2. Since the coefficient of $x^{4}$ is positive, $f > 0$ at the extremes and the sign alternates at each root. 3. $f (x) > 0$ if $x \in] - \infty, - 2.24 [\cup] - 1, 0.8 [\cup] 2.24, + \infty [$ .

b) (i) 1. Using the GDC we solve $f^{'} (x) = 0$ : $x \approx - 1.74$ , $x \approx - 0.0860$ and $x \approx 1.67$ . 2. The local minima are those where $f^{''} > 0$ (or as seen on the graph): $x \approx - 1.74$ and $x \approx 1.67$ ; the middle one is a local maximum. 3. We calculate the $y$ -values: $f (- 1.74) \approx - 3.71$ and $f (1.67) \approx - 5.14$ . 4. Local minima: $(- 1.74; - 3.71)$ and $(1.67; - 5.14)$ .

(ii) 1. Using the GDC we solve $f^{''} (x) = 0$ : $x \approx 0.934$ and $x \approx - 1.03$ (with $f^{''}$ changing sign at both).

c) (i) 1. For $f$ to have an inverse on $[0, a]$ it must be injective, i.e. monotonic. 2. On $[0, 1.67]$ it is decreasing (the local maximum is at $x < 0$ and the local minimum at $x \approx 1.67$ ); if $a$ were larger, it would increase again. 3. The greatest value is $a \approx 1.67$ .

(ii) 1. $y = f (x)$ on $[0, 1.67]$ : decreasing from $(0, 4)$ to $(1.67; - 5.14)$ , passing through $(0.8; 0)$ . 2. $y = f^{- 1} (x)$ is its reflection in $y = x$ : it goes from $(- 5.14; 1.67)$ to $(4, 0)$ .

Graphs of f restricted to an interval and of its inverse, symmetric about the line y = x, with the endpoints of each curve marked

(iii) 1. $f^{- 1} (x) = 1 ⟺ x = f (1)$ . 2. $f (1) = 1 + 0.2 - 5.8 - 1 + 4 = - 1.6$ , so $x = - 1.6$ .

d) (i) 1. We write $y = 2 sin (x - 1) - 3$ and rearrange: $sin (x - 1) = \frac{y + 3}{2}$ . 2. On the given interval $x - 1 \in [- \frac{π}{2}, \frac{π}{2}]$ , where the arcsine reverses the sine: $x = arcsin (\frac{y + 3}{2}) + 1$ . 3. $g^{- 1} (x) = arcsin (\frac{x + 3}{2}) + 1$ , with domain equal to the range of $g$ : $[- 5, - 1]$ .

(ii) 1. $f^{- 1}$ is decreasing, so $f^{- 1} (w) < 1 ⟺ w > f (1) = - 1.6$ (the inequality is reversed). 2. With $w = g (x)$ : $2 sin (x - 1) - 3 > - 1.6$ , i.e. $sin (x - 1) > 0.7$ . 3. In the domain of $g$ : $x - 1 > arcsin 0.7 \approx 0.7754$ , so $x > 1.775$ . 4. Combining with the domain of $g$ : $1.78 < x \leq \frac{π}{2} + 1$ (the endpoint $x = \frac{π}{2} + 1$ is included in the domain of $g$ and satisfies the inequality).

Answer: a) $x \in] - \infty, - 2.24 [\cup] - 1, 0.8 [\cup] 2.24, + \infty [$ ; b) (i) $(- 1.74; - 3.71)$ and $(1.67; - 5.14)$ ; (ii) $x \approx 0.934$ and $x \approx - 1.03$ ; c) (i) $a \approx 1.67$ ; (iii) $x = - 1.6$ ; d) (i) $g^{- 1} (x) = arcsin (\frac{x + 3}{2}) + 1$ , domain $[- 5, - 1]$ ; (ii) $1.78 < x \leq \frac{π}{2} + 1$

Exercise 156 HardIB · P2 May 1997

The diagram shows the graph of $y = f (x)$ , where $f (x) = \frac{a x + b}{x ^{2} + d x + e}$ , with $f (- \frac{5}{2}) = 0$ and $f (0) = \frac{5}{4}$ . The lines $x = - 4$ , $x = - 1$ and the $x$ -axis are all asymptotes of the graph.

Graph of a rational function with two vertical asymptotes, x = -4 and x = -1, and the x-axis as horizontal asymptote; it passes through the points (-5/2, 0) and (0, 5/4)

a) Using the information given, find the values of $a$ , $b$ , $d$ and $e$ .

b) Find the interval where $f^{'} (x) < 0$ .

c) Using the values of $a$ , $b$ , $d$ and $e$ found in part (a), express $f (x)$ in partial fractions.

d) Prove that $(- \frac{5}{2}, 0)$ is a point of inflexion.

e) Find the interval where $f^{''} (x) > 0$ .

Show answer

Answer: a) $a = 2$ , $b = 5$ , $d = 5$ , $e = 4$ ; b) $] - \infty, - 4 [\cup] - 4, - 1 [\cup] - 1, + \infty [$ ; c) $\frac{2 x + 5}{x ^{2} + 5 x + 4} = \frac{1}{x + 1} + \frac{1}{x + 4}$ ; d) as required; e) $] - 4, - \frac{5}{2} [\cup] - 1, + \infty [$

Show working and answer

a) 1. The vertical asymptotes $x = - 4$ and $x = - 1$ are the roots of the denominator: $x^{2} + d x + e = (x + 4) (x + 1) = x^{2} + 5 x + 4$ . Hence $d = 5$ and $e = 4$ . 2. $f (0) = \frac{b}{4} = \frac{5}{4}$ , so $b = 5$ . 3. $f (- \frac{5}{2}) = 0$ implies $- \frac{5}{2} a + 5 = 0$ , so $a = 2$ . 4. $f (x) = \frac{2 x + 5}{x ^{2} + 5 x + 4}$ . (The $x$ -axis is an asymptote because the degree of the numerator is less than that of the denominator.)

b) 1. From part (c), $f (x) = \frac{1}{x + 1} + \frac{1}{x + 4}$ , so $f^{'} (x) = - \frac{1}{( x + 1 ) ^{2}} - \frac{1}{( x + 4 ) ^{2}}$ . 2. This expression is negative for all $x$ in the domain, since it is the sum of two negative terms. 3. $f^{'} (x) < 0$ on $] - \infty, - 4 [\cup] - 4, - 1 [\cup] - 1, + \infty [$ .

c) 1. We decompose: $\frac{2 x + 5}{( x + 1 ) ( x + 4 )} = \frac{A}{x + 1} + \frac{B}{x + 4}$ , with $2 x + 5 = A (x + 4) + B (x + 1)$ . 2. With $x = - 1$ : $3 = 3 A$ , so $A = 1$ . With $x = - 4$ : $- 3 = - 3 B$ , so $B = 1$ . 3. $f (x) = \frac{1}{x + 1} + \frac{1}{x + 4}$ .

d) 1. We differentiate again: $f^{''} (x) = \frac{2}{( x + 1 ) ^{3}} + \frac{2}{( x + 4 ) ^{3}} = \frac{2 [ ( x + 4 ) ^{3} + ( x + 1 ) ^{3} ]}{( x + 1 ) ^{3} ( x + 4 ) ^{3}}$ . 2. We use $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ : $(x + 4)^{3} + (x + 1)^{3} = (2 x + 5) (x^{2} + 5 x + 13)$ , with $x^{2} + 5 x + 13 > 0$ for all $x$ (discriminant $25 - 52 < 0$ ). 3. So $f^{''} (x) = \frac{2 ( 2 x + 5 ) ( x ^{2} + 5 x + 13 )}{( x + 1 ) ^{3} ( x + 4 ) ^{3}}$ , which is zero at $x = - \frac{5}{2}$ . 4. Near $x = - \frac{5}{2}$ (in $] - 4, - 1 [$ ): $(x + 1)^{3} < 0$ and $(x + 4)^{3} > 0$ , so the denominator is negative. If $x < - \frac{5}{2}$ , then $2 x + 5 < 0$ and $f^{''} > 0$ ; if $x > - \frac{5}{2}$ , then $2 x + 5 > 0$ and $f^{''} < 0$ . 5. $f^{''}$ changes sign and $f (- \frac{5}{2}) = 0$ , so the point $(- \frac{5}{2}, 0)$ is a point of inflexion.

e) 1. We study the sign of $f^{''} (x) = \frac{2 ( 2 x + 5 ) ( x ^{2} + 5 x + 13 )}{( x + 1 ) ^{3} ( x + 4 ) ^{3}}$ , where the factor $x^{2} + 5 x + 13$ is always positive:

$x < - 4$ : $2 x + 5 < 0$ , $(x + 1)^{3} < 0$ and $(x + 4)^{3} < 0$ , so the denominator is positive and $f^{''} < 0$ .
$- 4 < x < - \frac{5}{2}$ : numerator negative and denominator negative, so $f^{''} > 0$ .
$- \frac{5}{2} < x < - 1$ : numerator positive and denominator negative, so $f^{''} < 0$ .
$x > - 1$ : numerator and denominator positive, so $f^{''} > 0$ .

$f^{''} (x) > 0$ on $] - 4, - \frac{5}{2} [\cup] - 1, + \infty [$ .

Exercise 157 HardIB · P2 May 1999

Give exact answers to this question.

The temperature $g (t)$ at time $t$ of a given point on a heated iron bar is given by $g (t) = \frac{ln t}{t}$ , where $t > 0$ .

a) Find the interval where $g^{'} (t) > 0$ .

b) Find the interval where $g^{''} (t) > 0$ and the interval where $g^{''} (t) < 0$ .

c) Find the value of $t$ at which the graph of $g (t)$ has a point of inflexion.

d) Let $t^{*}$ be the value of $t$ for which $g^{'} (t^{*}) = 0$ and $g^{''} (t^{*}) < 0$ . Find $t^{*}$ .

e) Find the point where the normal to the graph of $g (t)$ at the point $(t^{*}, g (t^{*}))$ meets the $x$ -axis.

Show answer

Answer: a) $0 < t < e^{2}$ ; b) $g^{''} > 0$ if $t > e^{8/3}$ and $g^{''} < 0$ if $0 < t < e^{8/3}$ ; c) $t = e^{8/3}$ ; d) $t^{*} = e^{2}$ ; e) $(e^{2}, 0)$

Show working and answer

We first calculate the derivatives of $g (t) = \frac{ln t}{t} = t^{- 1/2} ln t$ .

Product rule: $g^{'} (t) = - \frac{1}{2} t^{- 3/2} ln t + t^{- 1/2} \cdot \frac{1}{t} = \frac{2 - ln t}{2 t ^{3/2}}$ .
We differentiate again (product rule): $g^{''} (t) = \frac{1}{2} [- \frac{1}{t} t^{- 3/2} - \frac{3}{2} (2 - ln t) t^{- 5/2}] = \frac{3 ln t - 8}{4 t ^{5/2}}$ .

a) $g^{'} (t) > 0$ is equivalent to $2 - ln t > 0$ (the denominator is positive), i.e. $ln t < 2$ . Since $t > 0$ : $0 < t < e^{2}$ .

b) The denominator of $g^{''}$ is positive, so the sign is given by $3 ln t - 8$ . $g^{''} (t) > 0 ⟺ ln t > \frac{8}{3} ⟺ t > e^{8/3}$ , and $g^{''} (t) < 0 ⟺ 0 < t < e^{8/3}$ .

c) There is a point of inflexion where $g^{''}$ is zero and changes sign: $t = e^{8/3}$ (the change of sign can be seen in part b).

d) $g^{'} (t^{*}) = 0 ⟺ ln t^{*} = 2$ , so $t^{*} = e^{2}$ . As $e^{2} < e^{8/3}$ , $g^{''} (t^{*}) < 0$ : it is a maximum.

e) 1. $g (t^{*}) = \frac{2}{e}$ , and the tangent at $(e^{2}, \frac{2}{e})$ has gradient $g^{'} (t^{*}) = 0$ : it is horizontal. 2. The normal is perpendicular to it, i.e. the vertical line $t = e^{2}$ . 3. It meets the $x$ -axis at $(e^{2}, 0)$ .

Answer: a) $0 < t < e^{2}$ ; b) $g^{''} > 0$ if $t > e^{8/3}$ and $g^{''} < 0$ if $0 < t < e^{8/3}$ ; c) $t = e^{8/3}$ ; d) $t^{*} = e^{2}$ ; e) $(e^{2}, 0)$

Exercise 158 HardIB · P1 May 2001

The function $f$ is given by $f : x \mapsto e^{(1 + s i n π x)}$ , $x \geq 0$ .

a) Find $f^{'} (x)$ .

Let $x_{n}$ be the value of $x$ at which the $n + 1$ th maximum or minimum occurs, $n \in N$ (that is, $x_{0}$ is the value of $x$ at which the first maximum or minimum occurs, $x_{1}$ is the value of $x$ at which the second maximum or minimum occurs, etc.).

b) Find $x_{n}$ in terms of $n$ .

Show answer

Answer: a) $f^{'} (x) = π cos (π x) e^{1 + s i n π x}$ ; b) $x_{n} = \frac{1}{2} + n$

Show working and answer

a) Chain rule: the derivative of the exponent $1 + sin π x$ is $π cos π x$ . Hence $f^{'} (x) = π cos (π x) e^{1 + s i n π x}$ .

b) 1. Maxima and minima occur where $f^{'} (x) = 0$ . Since $π e^{1 + s i n π x} > 0$ , we need $cos π x = 0$ . 2. $π x = \frac{π}{2} + k π$ , with $k \in Z$ , so $x = \frac{1}{2} + k$ . 3. For $x \geq 0$ the values are $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots$ ; at each of them $f^{'}$ changes sign (the cosine changes sign), so they are all extrema. 4. The first ( $n = 0$ ) is $\frac{1}{2}$ , the second ( $n = 1$ ) is $\frac{3}{2}$ , and so on.

Answer: a) $f^{'} (x) = π cos (π x) e^{1 + s i n π x}$ ; b) $x_{n} = \frac{1}{2} + n$

Exercise 159 HardIB · P2 May 2001

Let $f (x) = x 3 (x^{2} - 1)^{2}$ , $- 1.4 \leq x \leq 1.4$

a) Sketch the graph of $f (x)$ . (An accurate scale diagram is not required.)

On your graph, indicate the approximate positions of

(i) the zeros

(ii) the maximum points

(iii) the minimum points

b) (i) Find $f^{'} (x)$ , clearly stating its domain.

(ii) Find the $x$ -coordinates of the maximum and minimum points of $f (x)$ , for $- 1 < x < 1$ .

c) Find the $x$ -coordinate of the point of inflexion of $f (x)$ , where $x > 0$ , giving your answer correct to four decimal places.

View solution

a) We have $f (x) = x (x^{2} - 1)^{2/3}$ .

Zeros: $x = 0$ and $x^{2} = 1$ , i.e. $x = - 1, 0, 1$ .
From the analysis in part (b): $f$ is increasing on $] - 0.655; 0.655 [$ and decreasing on $] - 1; - 0.655 [$ and on $] 0.655; 1 [$ . At $x = \pm 1$ there are sharp points (the derivative tends to infinity).
For $x < - 1$ , $f < 0$ and $f$ rises towards $0$ ( $f (- 1.4) \approx - 1.36$ ); for $x > 1$ , $f > 0$ and increases to $f (1.4) \approx 1.36$ .

Graph of f(x) = x times the cube root of (x²−1)² on the interval −1.4 to 1.4: it cuts the x-axis at −1, 0 and 1, with a local maximum and a local minimum between −1 and 1

(i) Zeros: $x = - 1$ , $x = 0$ , $x = 1$ .

(ii) Maxima: $x = - 1$ (a sharp point) and $x \approx 0.655$ .

(iii) Minima: $x = 1$ (a sharp point) and $x \approx - 0.655$ .

b) (i) 1. Product and chain rules: $f^{'} (x) = (x^{2} - 1)^{2/3} + x \cdot \frac{2}{3} (x^{2} - 1)^{- 1/3} \cdot 2 x$ . 2. Factor out $(x^{2} - 1)^{- 1/3}$ : $f^{'} (x) = (x^{2} - 1)^{- 1/3} [x^{2} - 1 + \frac{4}{3} x^{2}] = \frac{7 x ^{2} - 3}{3 ( x ^{2} - 1 ) ^{1/3}}$ . 3. The denominator is zero at $x = \pm 1$ , where $f^{'}$ does not exist: domain $[- 1.4; 1.4] ∖ {- 1, 1}$ .

(ii) 1. $f^{'} (x) = 0 ⟺ 7 x^{2} = 3 ⟺ x = \pm \frac{3}{7} \approx \pm 0.655$ . 2. On $] - 1, 1 [$ the denominator is negative, so $f^{'}$ has the opposite sign to $7 x^{2} - 3$ : $f^{'} < 0$ for $∣ x ∣ > 0.655$ and $f^{'} > 0$ for $∣ x ∣ < 0.655$ . 3. Hence $x = - 0.655$ is a minimum and $x = 0.655$ is a maximum.

c) 1. Differentiate $f^{'}$ (quotient/product rule). We get $f^{''} (x) = \frac{4 x ( 7 x ^{2} - 9 )}{9 ( x ^{2} - 1 ) ^{4/3}}$ . 2. For $x > 0$ , $f^{''} = 0 ⟺ 7 x^{2} = 9 ⟺ x = \frac{3}{7}$ , and the sign of $f^{''}$ changes from $-$ to $+$ (we can also find this using the GDC, by solving $f^{''} (x) = 0$ ). 3. $x \approx 1.1339$ .

Answer: a) (i) $x = - 1, 0, 1$ ; (ii) maxima at $x = - 1$ and $x \approx 0.655$ ; (iii) minima at $x = 1$ and $x \approx - 0.655$ ; b) (i) $f^{'} (x) = \frac{7 x ^{2} - 3}{3 ( x ^{2} - 1 ) ^{1/3}}$ , with $x \neq = \pm 1$ ; (ii) $x = \pm \frac{3}{7} \approx \pm 0.655$ ; c) $x \approx 1.1339$

Exercise 160 HardIB · P2 May 2002

The function $f$ is defined by $f (x) = \frac{x ^{2} - x + 1}{x ^{2} + x + 1}$ .

a) (i) Find an expression for $f^{'} (x)$ , simplifying your answer.

(ii) The tangents to the graph of $f (x)$ at the points $A$ and $B$ are parallel to the $x$ -axis. Find the coordinates of $A$ and $B$ .

b) (i) Sketch the graph of $y = f^{'} (x)$ .

(ii) Find the $x$ -coordinates of the three points of inflexion on the graph of $f$ .

c) Find the range of

(i) $f$

(ii) the composite function $f \circ f$ .

View solution

a) (i) Quotient rule: $f^{'} (x) = \frac{( 2 x - 1 ) ( x ^{2} + x + 1 ) - ( x ^{2} - x + 1 ) ( 2 x + 1 )}{( x ^{2} + x + 1 ) ^{2}} .$ The numerator is $(2 x^{3} + x^{2} + x - 1) - (2 x^{3} - x^{2} + x + 1) = 2 x^{2} - 2$ . So $f^{'} (x) = \frac{2 ( x ^{2} - 1 )}{( x ^{2} + x + 1 ) ^{2}} .$

(ii) 1. A tangent parallel to the $x$ -axis means $f^{'} (x) = 0$ , so $x^{2} = 1$ , $x = \pm 1$ (the denominator never vanishes, since $x^{2} + x + 1 > 0$ ). 2. $f (1) = \frac{1}{3}$ and $f (- 1) = \frac{3}{1} = 3$ . The points are $A (1, \frac{1}{3})$ and $B (- 1, 3)$ .

b) (i) Features of $y = f^{'} (x)$ : it cuts the $x$ -axis at $x = - 1$ and $x = 1$ ; $f^{'} (0) = - 2$ ; it is positive for $∣ x ∣ > 1$ and negative on $] - 1, 1 [$ ; it tends to $0$ from above as $x \to \pm \infty$ (the $x$ -axis is a horizontal asymptote). It has two local maxima (at $x \approx - 1.53$ and $x \approx 1.88$ ) and one local minimum at $x \approx - 0.347$ .

Approximate graph of y = f′(x) = 2(x²−1)/(x²+x+1)² for −6.5 ≤ x ≤ 6.5, with its zeros and its three local extrema marked

(ii) 1. The points of inflexion of $f$ are the extrema of $f^{'}$ , that is, the roots of $f^{''}$ where the sign changes. Differentiating $f^{'}$ with the quotient rule: $f^{''} (x) = \frac{4 ( - x ^{3} + 3 x + 1 )}{( x ^{2} + x + 1 ) ^{3}} .$ 2. $f^{''} = 0 ⟺ x^{3} - 3 x - 1 = 0$ . Using the GDC, the three real roots are $x \approx - 1.53$ , $x \approx - 0.347$ and $x \approx 1.88$ (exactly $2 cos \frac{7 π}{9}$ , $2 cos \frac{5 π}{9}$ and $2 cos \frac{π}{9}$ ). At all three, $f^{''}$ changes sign.

c) (i) 1. $f (x) \to 1$ as $x \to \pm \infty$ . From the study of $f^{'}$ : $f$ increases on $] - \infty, - 1 [$ , decreases on $] - 1, 1 [$ and increases on $] 1, \infty [$ . 2. Maximum $f (- 1) = 3$ and minimum $f (1) = \frac{1}{3}$ , and the values at infinity (the limit $1$ ) lie between these two. As $f$ is continuous, $R_{f} = [\frac{1}{3}, 3]$ .

(ii) 1. $f \circ f$ applies $f$ to values of $f$ , that is, to $[\frac{1}{3}, 3]$ . The range of $f \circ f$ is the set of values of $f$ on that interval. 2. On $[\frac{1}{3}, 3]$ , $f$ decreases until $x = 1$ and then increases, so the minimum is $f (1) = \frac{1}{3}$ . 3. The endpoints of the interval give $f (\frac{1}{3}) = \frac{7/9}{13/9} = \frac{7}{13}$ and $f (3) = \frac{7}{13}$ . The maximum is $\frac{7}{13}$ . 4. $R_{f \circ f} = [\frac{1}{3}, \frac{7}{13}]$ .

Answer: a) (i) $f^{'} (x) = \frac{2 ( x ^{2} - 1 )}{( x ^{2} + x + 1 ) ^{2}}$ ; (ii) $A (1, \frac{1}{3})$ , $B (- 1, 3)$ ; b) (i) graph as described; (ii) $x \approx - 1.53$ , $x \approx - 0.347$ , $x \approx 1.88$ ; c) (i) $R_{f} = [\frac{1}{3}, 3]$ ; (ii) $R_{f \circ f} = [\frac{1}{3}, \frac{7}{13}]$

Exercise 161 HardIB · P2 May 2003

The function $f$ is defined by $f (x) = \frac{x ^{2}}{2 ^{x}}$ , for $x > 0$ .

a) (i) Show that $f^{'} (x) = \frac{2 x - x ^{2} ln 2}{2 ^{x}}$ .

(ii) Find an expression for $f^{''} (x)$ . Simplify your answer as far as possible.

b) (i) Find the exact value of $x$ that satisfies the equation $f^{'} (x) = 0$ .

(ii) Show that for this value a maximum of $f (x)$ is obtained.

c) Find the $x$ -coordinates of the two points of inflexion on the graph of $f$ .

Show answer

Answer: a) (i) proved; (ii) $f^{''} (x) = \frac{2 - 4 x ln 2 + x ^{2} ( ln 2 ) ^{2}}{2 ^{x}}$ ; b) (i) $x = \frac{2}{ln 2}$ ; (ii) $f^{''} (\frac{2}{ln 2}) < 0$ , maximum; c) $x = \frac{2 - 2}{ln 2} \approx 0.845$ and $x = \frac{2 + 2}{ln 2} \approx 4.93$

Show working and answer

a) (i) 1. Quotient rule with $u = x^{2}$ and $v = 2^{x}$ , where $v^{'} = 2^{x} ln 2$ . 2. $f^{'} (x) = \frac{2 x \cdot 2 ^{x} - x ^{2} \cdot 2 ^{x} ln 2}{( 2 ^{x} ) ^{2}} = \frac{2 ^{x} ( 2 x - x ^{2} ln 2 )}{2 ^{2 x}} = \frac{2 x - x ^{2} ln 2}{2 ^{x}}$ .

(ii) 1. Using the quotient rule again, with $N = 2 x - x^{2} ln 2$ ( $N^{'} = 2 - 2 x ln 2$ ) and $v = 2^{x}$ : $f^{''} (x) = \frac{( 2 - 2 x ln 2 ) 2 ^{x} - ( 2 x - x ^{2} ln 2 ) 2 ^{x} ln 2}{2 ^{2 x}} = \frac{2 - 2 x ln 2 - 2 x ln 2 + x ^{2} ( ln 2 ) ^{2}}{2 ^{x}} .$ 2. $f^{''} (x) = \frac{2 - 4 x ln 2 + x ^{2} ( ln 2 ) ^{2}}{2 ^{x}}$ .

b) (i) $f^{'} (x) = 0 ⟺ x (2 - x ln 2) = 0$ . As $x > 0$ , $x = \frac{2}{ln 2}$ .

(ii) 1. Substituting into $f^{''}$ : $x ln 2 = 2$ , so the numerator equals $2 - 8 + 4 = - 2$ . 2. $f^{''} (\frac{2}{ln 2}) = \frac{- 2}{2 ^{2/ l n 2}} < 0$ , so it is a maximum.

c) 1. A point of inflexion requires $f^{''} = 0$ : $x^{2} (ln 2)^{2} - 4 x ln 2 + 2 = 0$ . 2. With $u = x ln 2$ : $u^{2} - 4 u + 2 = 0$ , so $u = 2 \pm 2$ . 3. $x = \frac{2 \pm 2}{ln 2}$ , that is, $x \approx 0.845$ and $x \approx 4.93$ . Both are positive, and $f^{''}$ changes sign at each one (simple roots).

Exercise 162 HardIB · P1 TZ1 May 2005

A function $f$ is defined by $f (x) = a x^{3} + b x^{2} + 30 x + c$ , where $a$ , $b$ and $c$ are constants. The graph of $f$ has a maximum at $(1, 7)$ and a point of inflexion when $x = 3$ . Find the values of $a$ , $b$ and $c$ .

Show answer

Answer: $a = 2$ , $b = - 18$ , $c = - 7$

Show working and answer

Differentiating: $f^{'} (x) = 3 a x^{2} + 2 b x + 30$ and $f^{''} (x) = 6 a x + 2 b$ .
There is a maximum at $(1, 7)$ , so $f^{'} (1) = 0$ : $3 a + 2 b + 30 = 0$ .
There is a point of inflexion at $x = 3$ , so $f^{''} (3) = 0$ : $18 a + 2 b = 0$ , that is, $b = - 9 a$ .
Substituting into the first equation: $3 a - 18 a + 30 = 0$ , so $- 15 a = - 30$ and $a = 2$ . Then $b = - 18$ .
The curve passes through $(1, 7)$ : $f (1) = a + b + 30 + c = 2 - 18 + 30 + c = 7$ , so $c = - 7$ .
Check that it is a maximum: $f^{''} (1) = 6 \cdot 2 \cdot 1 + 2 (- 18) = - 24 < 0$ .

Answer: $a = 2$ , $b = - 18$ , $c = - 7$

Exercise 163 HardIB · P2 TZ0 November 2007

The function $f$ is defined by $f (x) = csc x + tan 2 x$ .

a) Sketch an approximate graph of $f$ for $- \frac{π}{2} \leq x \leq \frac{π}{2}$ .

Hence, state

(i) the $x$ -intercepts.

(ii) the equations of the asymptotes.

(iii) the coordinates of the maximum and minimum points.

b) Verify that the roots of $f (x) = 0$ satisfy the equation $2 cos^{3} x - 2 cos^{2} x - 2 cos x + 1 = 0$ .

c) Verify that the $x$ -coordinates of the maximum and minimum points of the curve satisfy the equation $4 cos^{5} x - 4 cos^{3} x + 2 cos^{2} x + cos x - 2 = 0$ .

d) Verify that $f (π - x) + f (π + x) = 0$ .

View solution

a) 1. Using the GDC, we plot $f (x) = \frac{1}{sin x} + tan 2 x$ on $[- \frac{π}{2}, \frac{π}{2}]$ . Vertical asymptotes appear where $sin x = 0$ (i.e. $x = 0$ ) and where $cos 2 x = 0$ (i.e. $x = \pm \frac{π}{4}$ ). At the endpoints, $f (\pm \frac{π}{2}) = \pm 1 + tan (\pm π) = \pm 1$ .

Graph of f(x) = cosec x + tan 2x for −π/2 ≤ x ≤ π/2, with dashed vertical asymptotes, the x-intercepts and a local maximum and a local minimum marked

The function is odd: the graph is symmetric about the origin. On $(0, π /4)$ it comes down from $+ \infty$ as $x \to 0^{+}$ , falls to a minimum and rises again to $+ \infty$ as $x \to \frac{π}{4}^{-}$ . On $(\frac{π}{4}, \frac{π}{2})$ it starts from $- \infty$ , crosses the $x$ -axis and reaches $1$ at $\frac{π}{2}$ .

(i) $x$ -intercepts (using the GDC): $(- 1.16; 0)$ and $(1.16; 0)$ .

(ii) Vertical asymptotes: $x = - \frac{π}{4}$ , $x = 0$ and $x = \frac{π}{4}$ .

(iii) Turning points (using the GDC): local maximum at $(- 0.441; - 3.56)$ and local minimum at $(0.441; 3.56)$ .

b) 1. We write $f (x) = 0$ : $\frac{1}{sin x} + \frac{sin 2 x}{cos 2 x} = 0$ . 2. Multiply by $sin x cos 2 x$ : $cos 2 x + sin x sin 2 x = 0$ . 3. We use $sin 2 x = 2 sin x cos x$ and $cos 2 x = 2 cos^{2} x - 1$ : $2 cos^{2} x - 1 + 2 sin^{2} x cos x = 0$ . 4. Using $sin^{2} x = 1 - cos^{2} x$ : $2 cos^{2} x - 1 + 2 cos x - 2 cos^{3} x = 0$ . 5. Multiply by $- 1$ and rearrange: $2 cos^{3} x - 2 cos^{2} x - 2 cos x + 1 = 0$ .

c) 1. Differentiate (chain rule, and $\frac{d}{d x} csc x = - csc x cot x$ ): $f^{'} (x) = - \frac{cos x}{sin ^{2} x} + \frac{2}{cos ^{2} 2 x}$ . 2. At an extremum $f^{'} (x) = 0$ : $2 sin^{2} x = cos x cos^{2} 2 x$ . 3. Let $c = cos x$ . Then $sin^{2} x = 1 - c^{2}$ and $cos 2 x = 2 c^{2} - 1$ , so $2 (1 - c^{2}) = c (2 c^{2} - 1)^{2} = c (4 c^{4} - 4 c^{2} + 1)$ . 4. Expanding: $2 - 2 c^{2} = 4 c^{5} - 4 c^{3} + c$ , and moving all terms to one side: $4 c^{5} - 4 c^{3} + 2 c^{2} + c - 2 = 0$ , i.e. $4 cos^{5} x - 4 cos^{3} x + 2 cos^{2} x + cos x - 2 = 0$ .

d) 1. $f (π - x) = csc (π - x) + tan (2 π - 2 x) = csc x - tan 2 x$ (since $sin (π - x) = sin x$ , and $tan$ is odd with period $π$ ). 2. $f (π + x) = csc (π + x) + tan (2 π + 2 x) = - csc x + tan 2 x$ (since $sin (π + x) = - sin x$ ). 3. Adding: $f (π - x) + f (π + x) = 0$ .

Answer: a) graph; (i) $(- 1.16; 0)$ and $(1.16; 0)$ ; (ii) $x = 0$ , $x = - \frac{π}{4}$ , $x = \frac{π}{4}$ ; (iii) maximum $(- 0.441; - 3.56)$ , minimum $(0.441; 3.56)$ ; b), c), d) verified.

Exercise 164 HardIB · P1 TZ1 May 2008

The function $f$ is defined by $f (x) = x e^{2 x}$ .

It can be shown that $f^{(n)} (x) = (2^{n} x + n 2^{n - 1}) e^{2 x}$ for all $n \in Z^{+}$ , where $f^{(n)} (x)$ represents the $n$ th derivative of $f (x)$ .

a) Using $f^{(n)} (x)$ for $n = 1$ and $n = 2$ , show that there is a minimum point $P$ on the graph of $f$ , and find the coordinates of $P$ .

b) Show that $f$ has a point of inflexion $Q$ at $x = - 1$ .

c) Determine the intervals in the domain of $f$ where $f$ is

(i) concave up;

(ii) concave down.

d) Sketch $f$ , clearly showing any intercepts, asymptotes and the points $P$ and $Q$ .

e) Use mathematical induction to prove that $f^{(n)} (x) = (2^{n} x + n 2^{n - 1}) e^{2 x}$ for all $n \in Z^{+}$ , where $f^{(n)} (x)$ represents the $n$ th derivative of $f (x)$ .

View solution

a) 1. For $n = 1$ : $f^{'} (x) = (2 x + 1) e^{2 x}$ . For $n = 2$ : $f^{''} (x) = (4 x + 4) e^{2 x}$ . 2. $f^{'} (x) = 0$ implies $2 x + 1 = 0$ (since $e^{2 x} > 0$ ): $x = - \frac{1}{2}$ . 3. $f^{''} (- \frac{1}{2}) = (- 2 + 4) e^{- 1} = \frac{2}{e} > 0$ : it is a minimum. 4. The $y$ -coordinate: $f (- \frac{1}{2}) = - \frac{1}{2} e^{- 1}$ . Hence $P (- \frac{1}{2}, - \frac{1}{2 e})$ .

b) 1. $f^{''} (x) = 4 (x + 1) e^{2 x} = 0$ when $x = - 1$ . 2. $f^{''}$ is negative for $x < - 1$ and positive for $x > - 1$ (it changes sign), so there is a point of inflexion $Q$ at $x = - 1$ , with $f (- 1) = - e^{- 2}$ .

c) (i) $f^{''} (x) > 0$ when $x > - 1$ : concave up on $] - 1, + \infty [$ .

(ii) $f^{''} (x) < 0$ when $x < - 1$ : concave down on $] - \infty, - 1 [$ .

d) 1. Intercepts: $f (x) = 0$ only when $x = 0$ , so $(0, 0)$ (the only intercept with both axes). 2. As $x \to - \infty$ , $x e^{2 x} \to 0^{-}$ (the exponential dominates): horizontal asymptote $y = 0$ on the left. As $x \to + \infty$ , $f \to + \infty$ rapidly (no asymptote). 3. Key points: minimum $P (- \frac{1}{2}, - \frac{1}{2 e})$ and point of inflexion $Q (- 1, - e^{- 2})$ .

Graph of f(x) = x·e^(2x): passes through the origin, has a horizontal asymptote y = 0 on the left, and the points P (minimum) and Q (inflexion) marked

e) 1. Let $P (n)$ be: $f^{(n)} (x) = (2^{n} x + n 2^{n - 1}) e^{2 x}$ . 2. Base case $n = 1$ : $f^{'} (x) = e^{2 x} + 2 x e^{2 x} = (2 x + 1) e^{2 x}$ , and the formula gives $(2 x + 1 \cdot 2^{0}) e^{2 x}$ . It holds. 3. Inductive hypothesis: assume $P (k)$ is true, i.e. $f^{(k)} (x) = (2^{k} x + k 2^{k - 1}) e^{2 x}$ . 4. Differentiate (product rule): $f^{(k + 1)} (x) = 2^{k} e^{2 x} + 2 (2^{k} x + k 2^{k - 1}) e^{2 x} = (2^{k} + 2^{k + 1} x + k 2^{k}) e^{2 x}$ . 5. Grouping terms: $2^{k} + k 2^{k} = (k + 1) 2^{k}$ . Hence $f^{(k + 1)} (x) = (2^{k + 1} x + (k + 1) 2^{k}) e^{2 x}$ , which is $P (k + 1)$ . 6. Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , the result is proven by induction for all $n \in Z^{+}$ .

Answer: a) $P (- \frac{1}{2}, - \frac{1}{2 e})$ is a minimum; b) inflexion at $x = - 1$ ; c) (i) concave up on $] - 1, + \infty [$ ; (ii) concave down on $] - \infty, - 1 [$ ; d) graph; e) proven.

Exercise 165 HardIB · P2 TZ1 May 2008

A family of cubic functions is defined by $f_{k} (x) = k^{2} x^{3} - k x^{2} + x$ , $k \in Z^{+}$ .

a) Express in terms of $k$

(i) $f_{k}^{'} (x)$ and $f_{k}^{''} (x)$

(ii) the coordinates of the points of inflexion $P_{k}$ of the graphs of $f_{k}$ .

b) Show that all the $P_{k}$ lie on a straight line and state its equation.

c) Prove that, for all values of $k$ , the tangents to the graphs of $f_{k}$ at $P_{k}$ are parallel, and calculate the equations of the tangent lines.

Show answer

Answer: a) (i) $f_{k}^{'} (x) = 3 k^{2} x^{2} - 2 k x + 1$ , $f_{k}^{''} (x) = 6 k^{2} x - 2 k$ ; (ii) $P_{k} (\frac{1}{3 k}, \frac{7}{27 k})$ ; b) $y = \frac{7}{9} x$ ; c) $y = \frac{2}{3} x + \frac{1}{27 k}$

Show working and answer

a) (i) 1. Differentiate: $f_{k}^{'} (x) = 3 k^{2} x^{2} - 2 k x + 1$ . 2. Differentiate again: $f_{k}^{''} (x) = 6 k^{2} x - 2 k$ .

(ii) 1. A point of inflexion requires $f_{k}^{''} (x) = 0$ : $6 k^{2} x = 2 k$ , so $x = \frac{1}{3 k}$ . $f_{k}^{''}$ is a linear function with gradient $6 k^{2} > 0$ , so it changes sign: this is an inflexion point. 2. The $y$ -coordinate: $f_{k} (\frac{1}{3 k}) = k^{2} \cdot \frac{1}{27 k ^{3}} - k \cdot \frac{1}{9 k ^{2}} + \frac{1}{3 k} = \frac{1 - 3 + 9}{27 k} = \frac{7}{27 k}$ . 3. Hence $P_{k} (\frac{1}{3 k}, \frac{7}{27 k})$ .

b) 1. For any $k$ : $\frac{y}{x} = \frac{7/ ( 27 k )}{1/ ( 3 k )} = \frac{7}{27 k} \cdot 3 k = \frac{7}{9}$ . 2. Hence all the $P_{k}$ satisfy $y = \frac{7}{9} x$ : they lie on a straight line through the origin.

c) 1. Gradient of the tangent at $P_{k}$ : $f_{k}^{'} (\frac{1}{3 k}) = 3 k^{2} \cdot \frac{1}{9 k ^{2}} - 2 k \cdot \frac{1}{3 k} + 1 = \frac{1}{3} - \frac{2}{3} + 1 = \frac{2}{3}$ . 2. It does not depend on $k$ : all the tangents at $P_{k}$ have gradient $\frac{2}{3}$ , so they are parallel. 3. Equation: $y - \frac{7}{27 k} = \frac{2}{3} (x - \frac{1}{3 k})$ , so $y = \frac{2}{3} x - \frac{2}{9 k} + \frac{7}{27 k} = \frac{2}{3} x + \frac{1}{27 k}$ .

Exercise 166 HardIB · P1 TZ1 May 2009

Let $f$ be a function defined by $f (x) = x - arctan x$ , $x \in R$ .

a) Find $f (1)$ and $f (- 3)$ .

b) Show that $f (- x) = - f (x)$ for $x \in R$ .

c) Show that $x - \frac{π}{2} < f (x) < x + \frac{π}{2}$ for $x \in R$ .

d) Calculate expressions for $f^{'} (x)$ and $f^{''} (x)$ . Hence describe the behaviour of the graph of $f$ at the origin and justify your answer.

e) Sketch the graph of $f$ , clearly showing the asymptotes.

f) Justify that the inverse of $f$ is defined for all $x \in R$ , and sketch its graph.

View solution

a) 1. $f (1) = 1 - arctan 1 = 1 - \frac{π}{4}$ . 2. $f (- 3) = - 3 - arctan (- 3) = - 3 + \frac{π}{3}$ .

b) 1. $f (- x) = - x - arctan (- x)$ . 2. Since $arctan$ is odd, $arctan (- x) = - arctan x$ , so $f (- x) = - x + arctan x = - (x - arctan x) = - f (x)$ .

c) 1. We know that $- \frac{π}{2} < arctan x < \frac{π}{2}$ for all $x$ , i.e. $- \frac{π}{2} < - arctan x < \frac{π}{2}$ . 2. Add $x$ to all three parts: $x - \frac{π}{2} < x - arctan x < x + \frac{π}{2}$ , i.e. $x - \frac{π}{2} < f (x) < x + \frac{π}{2}$ .

d) 1. $f^{'} (x) = 1 - \frac{1}{1 + x ^{2}} = \frac{x ^{2}}{1 + x ^{2}}$ . 2. Using the quotient rule: $f^{''} (x) = \frac{2 x ( 1 + x ^{2} ) - x ^{2} \cdot 2 x}{( 1 + x ^{2} ) ^{2}} = \frac{2 x}{( 1 + x ^{2} ) ^{2}}$ . 3. At the origin: $f^{'} (0) = 0$ and $f^{''} (0) = 0$ . As $f^{'} (x) \geq 0$ for all $x$ (and equals $0$ only at $x = 0$ ), $f$ is strictly increasing, including at $x = 0$ (where the tangent is horizontal). 4. $f^{''} (x)$ is negative for $x < 0$ and positive for $x > 0$ : it changes sign at $0$ . Hence $(0, 0)$ is a point of inflexion (with a horizontal tangent).

e) 1. By part c), the graph lies between the lines $y = x - \frac{π}{2}$ and $y = x + \frac{π}{2}$ . As $x \to + \infty$ , $arctan x \to \frac{π}{2}$ and $f (x) - (x - \frac{π}{2}) \to 0$ : oblique asymptote $y = x - \frac{π}{2}$ . As $x \to - \infty$ , $arctan x \to - \frac{π}{2}$ : oblique asymptote $y = x + \frac{π}{2}$ . 2. The graph is symmetric about the origin (odd function), increasing, with a point of inflexion at the origin.

Graph of f(x) = x − arctan x, an increasing curve passing through the origin that approaches the two oblique lines y = x − π/2 and y = x + π/2 drawn dashed

f) 1. Since $f$ is strictly increasing on all of $R$ , it is one-to-one and therefore has an inverse. 2. It is continuous and, by the oblique asymptotes, $f (x) \to \pm \infty$ as $x \to \pm \infty$ , so its range is $R$ . Hence $D_{f^{- 1}} = R_{f} = R$ : the inverse is defined for all $x \in R$ . 3. The graph of $f^{- 1}$ is the reflection of that of $f$ in the line $y = x$ : it passes through the origin with a vertical tangent, and its asymptotes are the same oblique lines $y = x \pm \frac{π}{2}$ .

Graphs of y = f(x) = x − arctan x (blue) and of its inverse y = f⁻¹(x) (orange), symmetric about the line y = x, with the lines y = x ± π/2 dashed

Answer: a) $f (1) = 1 - \frac{π}{4}$ , $f (- 3) = - 3 + \frac{π}{3}$ ; b) and c) proven; d) $f^{'} (x) = \frac{x ^{2}}{1 + x ^{2}}$ , $f^{''} (x) = \frac{2 x}{( 1 + x ^{2} ) ^{2}}$ , strictly increasing at $x = 0$ and point of inflexion at $(0, 0)$ ; e) graph with oblique asymptotes $y = x \mp \frac{π}{2}$ ; f) $f$ is one-to-one and $D_{f^{- 1}} = R_{f} = R$ .

Exercise 167 HardIB · P1 TZ1 May 2010

Consider $f (x) = \frac{x ^{2} - 5 x + 4}{x ^{2} + 5 x + 4}$ .

a) Find the equations of all the asymptotes of the graph of $f$ .

b) Find the coordinates of the points where the graph of $f$ cuts the $x$ -axis and the $y$ -axis.

c) Find the coordinates of

(i) the maximum point, justifying your answer;

(ii) the minimum point, justifying your answer.

d) Sketch the graph of $f$ , clearly showing all the features above.

e) Hence, write down the number of points of inflexion of the graph of $f$ .

View solution

a) 1. We factorise: $f (x) = \frac{( x - 1 ) ( x - 4 )}{( x + 1 ) ( x + 4 )}$ . 2. Vertical asymptotes: where the denominator is zero without the numerator being zero, $x = - 1$ and $x = - 4$ . 3. Horizontal asymptote: $x \to \pm \infty lim f (x) = 1$ (same degree, and the ratio of the leading coefficients is $1/1$ ): $y = 1$ .

b) 1. $x$ -axis: $f (x) = 0$ implies $x = 1$ or $x = 4$ : $(1, 0)$ and $(4, 0)$ . 2. $y$ -axis: $f (0) = \frac{4}{4} = 1$ : $(0, 1)$ .

c) 1. Quotient rule: $f^{'} (x) = \frac{( 2 x - 5 ) ( x ^{2} + 5 x + 4 ) - ( x ^{2} - 5 x + 4 ) ( 2 x + 5 )}{( x ^{2} + 5 x + 4 ) ^{2}}$ . 2. Expanding the numerator: $(2 x^{3} + 5 x^{2} - 17 x - 20) - (2 x^{3} - 5 x^{2} - 17 x + 20) = 10 x^{2} - 40$ . 3. Hence $f^{'} (x) = \frac{10 ( x ^{2} - 4 )}{( x ^{2} + 5 x + 4 ) ^{2}}$ , which is zero at $x = \pm 2$ . 4. Sign of $f^{'}$ : the denominator is positive, so the sign is that of $x^{2} - 4$ : positive for $∣ x ∣ > 2$ and negative for $∣ x ∣ < 2$ .

(i) At $x = - 2$ , $f^{'}$ changes from positive to negative: maximum. $f (- 2) = \frac{4 + 10 + 4}{4 - 10 + 4} = - 9$ . Maximum $(- 2, - 9)$ .

(ii) At $x = 2$ , $f^{'}$ changes from negative to positive: minimum. $f (2) = \frac{4 - 10 + 4}{4 + 10 + 4} = - \frac{1}{9}$ . Minimum $(2, - \frac{1}{9})$ .

d) 1. Behaviour by regions. For $x < - 4$ : the graph rises from the asymptote $y = 1$ (above it) to $+ \infty$ at $x = - 4$ . Between $- 4$ and $- 1$ : it goes from $- \infty$ to $- \infty$ , passing through the maximum $(- 2, - 9)$ . For $x > - 1$ : it falls from $+ \infty$ , passes through $(0, 1)$ and $(1, 0)$ , reaches the minimum $(2, - \frac{1}{9})$ , cuts the axis at $(4, 0)$ and rises, approaching $y = 1$ from below.

Graph of a rational function with two vertical asymptotes, a horizontal asymptote and two marked local extrema

e) 1. From the graph: for $x < - 4$ the curve is concave up (it rises to $+ \infty$ ); between $- 4$ and $- 1$ it is concave down (a hump between two $- \infty$ ends); for $x > - 1$ it is concave up while it falls from $+ \infty$ until it passes the minimum, and concave down as it approaches $y = 1$ from below. 2. The concavity only changes within $x > - 1$ , to the right of the minimum (the vertical asymptotes are not points of the graph). There is exactly one point of inflexion.

Answer: a) $x = - 1$ , $x = - 4$ , $y = 1$ ; b) $(1, 0)$ , $(4, 0)$ , $(0, 1)$ ; c) (i) maximum $(- 2, - 9)$ ; (ii) minimum $(2, - \frac{1}{9})$ ; d) graph; e) one point of inflexion.

Exercise 168 HardIB · P1 TZ1 May 2011

Consider the function $f (x) = \frac{ln x}{x}$ , $0 < x < e^{2}$ .

a) (i) Solve the equation $f^{'} (x) = 0$ .

(ii) Hence, show that the graph of $f$ has a local maximum.

(iii) State the range of the function $f$ .

b) Show that there is a point of inflexion on the graph and determine its coordinates.

c) Sketch the graph of $y = f (x)$ , clearly indicating the asymptote, the intersection with the $x$ -axis and the local maximum.

d) Now consider the functions $g (x) = \frac{ln ∣ x ∣}{x}$ and $h (x) = \frac{ln ∣ x ∣}{∣ x ∣}$ , where $0 < ∣ x ∣ < e^{2}$ .

(i) Sketch the graph of $y = g (x)$ .

(ii) State the range of $g$ .

(iii) Find the values of $x$ such that $h (x) > g (x)$ .

View solution

a) (i) 1. We differentiate with the quotient rule: $f^{'} (x) = \frac{\frac{1}{x} \cdot x - ln x \cdot 1}{x ^{2}} = \frac{1 - ln x}{x ^{2}}$ . 2. $f^{'} (x) = 0$ if $1 - ln x = 0$ , i.e. $x = e$ (which lies in $] 0, e^{2} [$ ).

(ii) 1. We differentiate again with the quotient rule: $f^{''} (x) = \frac{- \frac{1}{x} \cdot x ^{2} - ( 1 - ln x ) \cdot 2 x}{x ^{4}} = \frac{- 3 + 2 ln x}{x ^{3}} .$ 2. At $x = e$ : $f^{''} (e) = \frac{- 3 + 2}{e ^{3}} = - \frac{1}{e ^{3}} < 0$ . As $f^{'} (e) = 0$ and $f^{''} (e) < 0$ , there is a local maximum.

(iii) 1. The maximum value is $f (e) = \frac{1}{e}$ . For $x < e$ the function increases, and $f (x) \to - \infty$ as $x \to 0^{+}$ ; for $e < x < e^{2}$ it decreases to $f (e^{2}) = \frac{2}{e ^{2}}$ (which is not the maximum). 2. Range: $R_{f} =] - \infty, \frac{1}{e}]$ .

b) 1. Points of inflexion satisfy $f^{''} (x) = 0$ : $- 3 + 2 ln x = 0$ , so $ln x = \frac{3}{2}$ and $x = e^{3/2}$ (inside the domain, since $\frac{3}{2} < 2$ ). 2. The sign of $f^{''}$ changes: negative if $x < e^{3/2}$ and positive if $x > e^{3/2}$ (the denominator $x^{3} > 0$ ). So there is a point of inflexion. 3. Its y-coordinate: $f (e^{3/2}) = \frac{3/2}{e ^{3/2}} = \frac{3}{2 e ^{3/2}}$ .

c) The graph has the vertical asymptote $x = 0$ (as $x \to 0^{+}$ , $f \to - \infty$ ), cuts the $x$ -axis at $(1, 0)$ (since $ln x = 0$ ), rises to the local maximum $(e, \frac{1}{e})$ and then falls gently (concave up after $x = e^{3/2}$ ).

Graph of a logarithmic function divided by x on an open interval, with a vertical asymptote on the y-axis, an intersection with the x-axis and a local maximum

d) (i) For $x > 0$ , $g (x) = \frac{ln x}{x} = f (x)$ . For $x < 0$ , $g (- x) = \frac{ln ∣ x ∣}{- x} = - g (x)$ : $g$ is odd, so its graph for $x < 0$ is the reflection of the graph for $x > 0$ in the origin. It has vertical asymptote $x = 0$ , intercepts at $(- 1, 0)$ and $(1, 0)$ , a local maximum at $(e, \frac{1}{e})$ and a local minimum at $(- e, - \frac{1}{e})$ .

Graph symmetric about the origin of a logarithmic function of the absolute value divided by x, with a vertical asymptote on the y-axis, two intercepts with the x-axis, a local maximum and a local minimum

(ii) For $x > 0$ the range is $] - \infty, \frac{1}{e}]$ and for $x < 0$ it is $[- \frac{1}{e}, + \infty [$ . The union is $R_{g} = R$ .

(iii) 1. For $x > 0$ , $∣ x ∣ = x$ and $h (x) = g (x)$ : the condition $h > g$ is never satisfied. 2. For $x < 0$ let $u = - x > 0$ . Then $h (x) = \frac{ln u}{u}$ and $g (x) = - \frac{ln u}{u}$ , and $h > g$ is equivalent to $\frac{2 ln u}{u} > 0$ , i.e. $u > 1$ . 3. That is, $x < - 1$ ; together with $0 < ∣ x ∣ < e^{2}$ this gives $- e^{2} < x < - 1$ .

Answer: a) (i) $x = e$ ; (ii) $f^{''} (e) < 0$ ; (iii) $] - \infty, \frac{1}{e}]$ ; b) $(e^{3/2}, \frac{3}{2 e ^{3/2}})$ ; c) graph; d) (i) graph; (ii) $R$ ; (iii) $] - e^{2}, - 1 [$

Exercise 169 HardIB · P2 TZ1 May 2014

Let $f (x) = \frac{e ^{2 x} + 1}{e ^{x} - 2}$ .

a) Find the equations of the horizontal and vertical asymptotes of the curve $y = f (x)$ .

b) (i) Find $f^{'} (x)$ .

(ii) Show that the curve has exactly one point where its tangent is horizontal.

(iii) Calculate the coordinates of this point.

c) Find the equation of $L_{1}$ , the normal to the curve at the point where it cuts the $y$ -axis.

The line $L_{2}$ is parallel to $L_{1}$ and is a tangent to the curve $y = f (x)$ .

d) Find the equation of the line $L_{2}$ .

Show answer

Answer: a) $x = ln 2$ , $y = - \frac{1}{2}$ (on the left); b) (i) $f^{'} (x) = \frac{e ^{x} ( e ^{2 x} - 4 e ^{x} - 1 )}{( e ^{x} - 2 ) ^{2}}$ ; (ii) $x = ln (2 + 5)$ ; (iii) $(ln (2 + 5), 25 + 4)$ ; c) $y = \frac{1}{4} x - 2$ ; d) $y = \frac{1}{4} x + 8.11$

Show working and answer

a) 1. Vertical asymptote: the denominator is zero when $e^{x} = 2$ , i.e. $x = ln 2$ (the numerator equals $5 \neq = 0$ there). Vertical asymptote: $x = ln 2$ . 2. Horizontal asymptote: as $x \to - \infty$ , $e^{x} \to 0$ and $f (x) \to \frac{0 + 1}{0 - 2} = - \frac{1}{2}$ . As $x \to + \infty$ , $f (x) \approx e^{x} \to + \infty$ , so there is no horizontal asymptote on the right. 3. Horizontal asymptote: $y = - \frac{1}{2}$ (on the left).

b) (i) 1. Quotient rule with $u = e^{2 x} + 1$ and $v = e^{x} - 2$ : $f^{'} (x) = \frac{2 e ^{2 x} ( e ^{x} - 2 ) - e ^{x} ( e ^{2 x} + 1 )}{( e ^{x} - 2 ) ^{2}} = \frac{e ^{3 x} - 4 e ^{2 x} - e ^{x}}{( e ^{x} - 2 ) ^{2}} = \frac{e ^{x} ( e ^{2 x} - 4 e ^{x} - 1 )}{( e ^{x} - 2 ) ^{2}} .$

(ii) 1. Horizontal tangent: $f^{'} (x) = 0$ . As $e^{x} > 0$ , we need $e^{2 x} - 4 e^{x} - 1 = 0$ . 2. Let $u = e^{x} > 0$ : $u^{2} - 4 u - 1 = 0$ gives $u = 2 \pm 5$ . The root $2 - 5 < 0$ is not valid (since $e^{x} > 0$ ), so only $e^{x} = 2 + 5$ remains. 3. Hence there is exactly one such point: $x = ln (2 + 5)$ .

(iii) 1. With $u = 2 + 5$ : $u^{2} = 9 + 45$ , so $f = \frac{u ^{2} + 1}{u - 2} = \frac{10 + 4 5}{5} = 25 + 4 \approx 8.47$ . 2. Point: $(ln (2 + 5), 25 + 4)$ .

c) 1. Intercept with the $y$ -axis: $f (0) = \frac{1 + 1}{1 - 2} = - 2$ , point $(0, - 2)$ . 2. Gradient of the tangent: $f^{'} (0) = \frac{1 \cdot ( 1 - 4 - 1 )}{( 1 - 2 ) ^{2}} = - 4$ . 3. The normal has gradient $\frac{1}{4}$ : $y + 2 = \frac{1}{4} x$ , i.e. $L_{1} : y = \frac{1}{4} x - 2$ .

d) 1. $L_{2}$ has gradient $\frac{1}{4}$ , so we look for $f^{'} (x) = \frac{1}{4}$ . Using the GDC (intersection of $y = f^{'} (x)$ with $y = \frac{1}{4}$ ) we obtain a unique solution: $x \approx 1.46$ . (For $x < ln 2$ , $f^{'} < 0$ ; for $x > ln 2$ , $f^{'}$ increases to $+ \infty$ , passing through $0.25$ only once.) 2. The y-coordinate: $f (1.4595) \approx 8.474$ . 3. $L_{2}$ : $y - 8.474 = \frac{1}{4} (x - 1.4595)$ , i.e. $y = \frac{1}{4} x + 8.11$ .

Exercise 170 HardIB · P1 TZ1 May 2015

Let $y (x) = x e^{3 x}$ , $x \in R$ .

a) Calculate $\frac{d y}{d x}$ .

b) Prove by induction that $\frac{d ^{n} y}{d x ^{n}} = n 3^{n - 1} e^{3 x} + x 3^{n} e^{3 x}$ , for $n \in Z^{+}$ .

c) Calculate the coordinates of a local maximum or minimum point on the graph of $y (x)$ . Justify whether this point is a maximum or a minimum.

d) Find the coordinates of a point of inflexion on the graph of $y (x)$ . Justify whether this point is a point of inflexion.

e) Using the above, sketch the graph of $y (x)$ , clearly indicating the points found in parts (c) and (d) and any intercepts with the axes.

View solution

a) 1. Product rule with $u = x$ , $v = e^{3 x}$ (and $v^{'} = 3 e^{3 x}$ , by the chain rule): $\frac{d y}{d x} = e^{3 x} + 3 x e^{3 x} .$

b) Proof by induction of $P (n) : \frac{d ^{n} y}{d x ^{n}} = n 3^{n - 1} e^{3 x} + x 3^{n} e^{3 x}$ .

Base case $n = 1$ : $1 \cdot 3^{0} e^{3 x} + x \cdot 3 e^{3 x} = e^{3 x} + 3 x e^{3 x}$ , which is the result from part a). So $P (1)$ holds.
Hypothesis: suppose $P (k)$ is true for some $k \in Z^{+}$ : $\frac{d ^{k} y}{d x ^{k}} = k 3^{k - 1} e^{3 x} + x 3^{k} e^{3 x}$ .
We differentiate once more (product rule on the second term): $\frac{d ^{k + 1} y}{d x ^{k + 1}} = k 3^{k - 1} \cdot 3 e^{3 x} + 3^{k} (e^{3 x} + 3 x e^{3 x}) = k 3^{k} e^{3 x} + 3^{k} e^{3 x} + x 3^{k + 1} e^{3 x} .$
Grouping terms: $= (k + 1) 3^{k} e^{3 x} + x 3^{k + 1} e^{3 x}$ , which is $P (k + 1)$ .
Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by the principle of mathematical induction $P (n)$ is true for all $n \in Z^{+}$ .

c) 1. $\frac{d y}{d x} = e^{3 x} (1 + 3 x) = 0$ implies $x = - \frac{1}{3}$ (since $e^{3 x} > 0$ ). Then $y = - \frac{1}{3} e^{- 1} = - \frac{1}{3 e}$ . 2. Using the formula from b) for $n = 2$ : $y^{''} = 6 e^{3 x} + 9 x e^{3 x} = e^{3 x} (6 + 9 x)$ . 3. $y^{''} (- \frac{1}{3}) = e^{- 1} \cdot 3 > 0$ . As $y^{'} = 0$ and $y^{''} > 0$ , it is a local minimum: $(- \frac{1}{3}, - \frac{1}{3 e})$ .

d) 1. $y^{''} = 0$ implies $6 + 9 x = 0$ , so $x = - \frac{2}{3}$ and $y = - \frac{2}{3} e^{- 2} = - \frac{2}{3 e ^{2}}$ . 2. We check that it is a point of inflexion: with $n = 3$ , $y^{'''} = 27 e^{3 x} + 27 x e^{3 x} = 27 e^{3 x} (1 + x)$ , and $y^{'''} (- \frac{2}{3}) = 9 e^{- 2} \neq = 0$ . As $y^{''} = 0$ and $y^{'''} \neq = 0$ , $y^{''}$ changes sign: it is a point of inflexion $(- \frac{2}{3}, - \frac{2}{3 e ^{2}})$ .

e) The graph meets the axes only at the origin ( $y = 0$ implies $x = 0$ ). As $x \to - \infty$ , $y \to 0^{-}$ , so $y = 0$ is a horizontal asymptote on the left. The graph goes down to the local minimum, rises passing through the origin and increases very rapidly for $x > 0$ . It is concave down for $x < - \frac{2}{3}$ and concave up after that.

Graph of the product of x and an exponential function, with a local minimum and a point of inflexion marked to the left of the origin, and very rapid growth to the right

Answer: a) $\frac{d y}{d x} = e^{3 x} + 3 x e^{3 x}$ ; b) proof; c) minimum $(- \frac{1}{3}, - \frac{1}{3 e})$ ; d) inflexion $(- \frac{2}{3}, - \frac{2}{3 e ^{2}})$ ; e) graph.

Exercise 171 HardIB · P2 TZ2 May 2021

The function $f$ is given by $f (x) = arcsin (\frac{x ^{2} - 1}{x ^{2} + 1})$ , $x \in R$ .

a) Show that $f$ is an even function.

b) Using limits, show that the graph of $y = f (x)$ has a horizontal asymptote. State the equation of this asymptote.

c) (i) Show that $f^{'} (x) = \frac{2 x}{x ^{2} ( x ^{2} + 1 )}$ for $x \in R$ , $x \neq = 0$ .

(ii) Using the expression for $f^{'} (x)$ and the result $x^{2} = ∣ x ∣$ , show that $f$ is decreasing for $x < 0$ .

The function $g$ is given by $g (x) = arcsin (\frac{x ^{2} - 1}{x ^{2} + 1})$ , $x \in R$ , $x \geq 0$ .

d) Find an expression for $g^{- 1} (x)$ . Justify your answer.

e) State the domain of $g^{- 1}$ .

f) Sketch the approximate graph of $g^{- 1} (x)$ , clearly indicating all asymptotes, with their equations, and the coordinates of all intercepts with the axes.

View solution

a) 1. $f (- x) = arcsin (\frac{( - x ) ^{2} - 1}{( - x ) ^{2} + 1}) = arcsin (\frac{x ^{2} - 1}{x ^{2} + 1}) = f (x)$ for all $x \in R$ , so $f$ is even.

b) 1. $\frac{x ^{2} - 1}{x ^{2} + 1} = \frac{1 - 1/ x ^{2}}{1 + 1/ x ^{2}} \to 1$ as $x \to \pm \infty$ . 2. By the continuity of $arcsin$ : $x \to \pm \infty lim f (x) = arcsin 1 = \frac{π}{2}$ . 3. Therefore $y = \frac{π}{2}$ is a horizontal asymptote (as $x$ tends to both $+ \infty$ and $- \infty$ ).

c) (i) 1. With $u = \frac{x ^{2} - 1}{x ^{2} + 1}$ , $f^{'} (x) = \frac{u ^{'}}{1 - u ^{2}}$ . By the quotient rule: $u^{'} = \frac{2 x ( x ^{2} + 1 ) - 2 x ( x ^{2} - 1 )}{( x ^{2} + 1 ) ^{2}} = \frac{4 x}{( x ^{2} + 1 ) ^{2}}$ . 2. Also $1 - u^{2} = \frac{( x ^{2} + 1 ) ^{2} - ( x ^{2} - 1 ) ^{2}}{( x ^{2} + 1 ) ^{2}} = \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$ , so $1 - u^{2} = \frac{2 x ^{2}}{x ^{2} + 1}$ (for $x \neq = 0$ ). 3. Then $f^{'} (x) = \frac{4 x}{( x ^{2} + 1 ) ^{2}} \cdot \frac{x ^{2} + 1}{2 x ^{2}} = \frac{2 x}{x ^{2} ( x ^{2} + 1 )} .$

(ii) 1. If $x < 0$ , then $x^{2} = ∣ x ∣ = - x$ , and so $f^{'} (x) = \frac{2 x}{- x ( x ^{2} + 1 )} = - \frac{2}{x ^{2} + 1} < 0$ . 2. As $f^{'} (x) < 0$ for $x < 0$ , $f$ is decreasing on that interval.

d) 1. $g$ is the restriction of $f$ to $x \geq 0$ , where it is strictly increasing ( $f^{'} (x) = \frac{2}{x ^{2} + 1} > 0$ ), so it is one-to-one and has an inverse. 2. We write $y = arcsin \frac{x ^{2} - 1}{x ^{2} + 1}$ , that is $sin y = \frac{x ^{2} - 1}{x ^{2} + 1}$ , and solve for $x$ : $sin y (x^{2} + 1) = x^{2} - 1$ implies $x^{2} (sin y - 1) = - 1 - sin y$ , so $x^{2} = \frac{1 + sin y}{1 - sin y}$ . 3. As the domain of $g$ is $x \geq 0$ , we take the positive root (the values of $g^{- 1}$ are the $x \geq 0$ ): $g^{- 1} (x) = \frac{1 + sin x}{1 - sin x} .$

e) 1. The domain of $g^{- 1}$ is the range of $g$ . $g (0) = arcsin (- 1) = - \frac{π}{2}$ , $g$ is increasing and $g (x) \to \frac{π}{2}$ without reaching it. 2. $D_{g^{- 1}} = [- \frac{π}{2}, \frac{π}{2} [$ .

f) 1. As $x \to \frac{π}{2}^{-}$ , $sin x \to 1$ and the denominator tends to $0^{+}$ , so $g^{- 1} (x) \to + \infty$ : vertical asymptote $x = \frac{π}{2}$ . 2. Intercepts: $g^{- 1} (- \frac{π}{2}) = \frac{1 - 1}{1 + 1} = 0$ , giving the point $(- \frac{π}{2}, 0)$ ; $g^{- 1} (0) = 1$ , giving the point $(0, 1)$ . 3. The graph starts at $(- \frac{π}{2}, 0)$ , passes through $(0, 1)$ and increases without limit as it approaches the asymptote.

Graph of a function involving the square root of a quotient of sines, defined on a half-open interval, with a vertical asymptote at its right-hand end and two intercepts with the axes

Answer: a) proof; b) $y = \frac{π}{2}$ ; c) proofs; d) $g^{- 1} (x) = \frac{1 + sin x}{1 - sin x}$ ; e) $[- \frac{π}{2}, \frac{π}{2} [$ ; f) asymptote $x = \frac{π}{2}$ , intercepts $(- \frac{π}{2}, 0)$ and $(0, 1)$ , graph.

Exercise 172 HardIB · P3 TZ2 May 2021

In this question you will be asked to analyse the behaviour and some key features of the function $f_{n} (x) = x^{n} (a - x)^{n}$ , where $a \in R^{+}$ and $n \in Z^{+}$ .

In parts (a) and (b), consider only the case $a = 2$ .

Consider $f_{1} (x) = x (2 - x)$ .

a) Sketch the approximate graph of $y = f_{1} (x)$ , and state the coordinates of all intercepts with the axes and of all local maxima and minima.

Consider $f_{n} (x) = x^{n} (2 - x)^{n}$ , where $n \in Z^{+}$ , $n > 1$ .

b) Use the graphic display calculator (GDC) to analyse the graph of $y = f_{n} (x)$ corresponding to:

The odd values $n = 3$ and $n = 5$
The even values $n = 2$ and $n = 4$

Hence, copy and complete the following table.

	Number of local maxima	Number of local minima	Number of points of inflexion with zero gradient
$n = 3$ and $n = 5$
$n = 2$ and $n = 4$

Now consider $f_{n} (x) = x^{n} (a - x)^{n}$ , where $a \in R^{+}$ and $n \in Z^{+}$ , $n > 1$ .

c) Show that $f_{n}^{'} (x) = n x^{n - 1} (a - 2 x) (a - x)^{n - 1}$ .

d) State the three solutions of the equation $f_{n}^{'} (x) = 0$ .

e) Show that the point $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ on the graph of $y = f_{n} (x)$ always lies above the horizontal axis.

f) Hence, or otherwise, show that $f_{n}^{'} (\frac{a}{4}) > 0$ , for $n \in Z^{+}$ .

g) Using the result in part (f) and considering the sign of $f_{n}^{'} (- 1)$ , show that the point $(0, 0)$ on the graph of $y = f_{n} (x)$ is:

(i) A local minimum for even values of $n$ , where $n > 1$ and $a \in R^{+}$ .

(ii) A point of inflexion with zero gradient for odd values of $n$ , where $n > 1$ and $a \in R^{+}$ .

Consider the graph of $y = x^{n} (a - x)^{n} - k$ , where $n \in Z^{+}$ , $a \in R^{+}$ and $k \in R$ .

h) State the conditions that $n$ and $k$ must satisfy for the equation $x^{n} (a - x)^{n} = k$ to have four solutions for $x$ .

View solution

a) 1. $f_{1} (x) = x (2 - x) = 2 x - x^{2}$ is a parabola opening downwards. Intercepts with the $x$ -axis: $x = 0$ and $x = 2$ , that is $(0, 0)$ and $(2, 0)$ ; with the $y$ -axis: $(0, 0)$ . 2. $f_{1}^{'} (x) = 2 - 2 x = 0$ gives $x = 1$ and $f_{1} (1) = 1$ : local maximum at $(1, 1)$ .

Graph of a parabola opening downwards, with two intercepts with the x-axis and a local maximum

b) Using the GDC, the graphs of $y = f_{n} (x)$ are drawn for the requested values. For $n = 3$ and $n = 5$ the graph has a single local maximum (at $x = 1$ ), and at $x = 0$ and $x = 2$ the tangent is horizontal but the curve does not change direction (points of inflexion with zero gradient). For $n = 2$ and $n = 4$ there is a local maximum at $x = 1$ and local minima at $x = 0$ and $x = 2$ .

	Number of local maxima	Number of local minima	Number of points of inflexion with zero gradient
$n = 3$ and $n = 5$	1	0	2
$n = 2$ and $n = 4$	1	2	0

c) 1. Product rule with $u = x^{n}$ and $v = (a - x)^{n}$ (chain rule on $v$ ): $f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n} - n x^{n} (a - x)^{n - 1} .$ 2. Factoring out the common factor $n x^{n - 1} (a - x)^{n - 1}$ : $f_{n}^{'} (x) = n x^{n - 1} (a - x)^{n - 1} [(a - x) - x] = n x^{n - 1} (a - 2 x) (a - x)^{n - 1} .$

d) 1. $f_{n}^{'} (x) = 0$ if $x^{n - 1} = 0$ , $a - 2 x = 0$ or $(a - x)^{n - 1} = 0$ . 2. The three solutions are $x = 0$ , $x = \frac{a}{2}$ and $x = a$ .

e) 1. $f_{n} (\frac{a}{2}) = (\frac{a}{2})^{n} (a - \frac{a}{2})^{n} = (\frac{a}{2})^{2 n}$ . 2. As $a > 0$ , $(\frac{a}{2})^{2 n} > 0$ : the point always lies above the horizontal axis.

f) 1. We evaluate at $x = \frac{a}{4}$ : $f_{n}^{'} (\frac{a}{4}) = n (\frac{a}{4})^{n - 1} (a - \frac{a}{2}) (\frac{3 a}{4})^{n - 1}$ . 2. All the factors are positive ( $n > 0$ , $a > 0$ ), so $f_{n}^{'} (\frac{a}{4}) > 0$ .

g) 1. The only zeros of $f_{n}^{'}$ are $0$ , $\frac{a}{2}$ and $a$ , so $f_{n}^{'}$ does not change sign on $] - \infty, 0 [$ or on $] 0, \frac{a}{2} [$ . The sign of $f_{n}^{'}$ to the right of $0$ is given by $f_{n}^{'} (\frac{a}{4}) > 0$ (part f), because $0 < \frac{a}{4} < \frac{a}{2}$ . 2. To the left of $0$ we take $x = - 1$ : $f_{n}^{'} (- 1) = n (- 1)^{n - 1} (a + 2) (a + 1)^{n - 1}$ . The factors $n$ , $a + 2$ and $(a + 1)^{n - 1}$ are positive, so the sign is that of $(- 1)^{n - 1}$ .

(i) If $n$ is even, $(- 1)^{n - 1} = - 1$ and $f_{n}^{'} (- 1) < 0$ : $f_{n}^{'}$ changes from negative to positive at $x = 0$ (with $f_{n}^{'} (0) = 0$ ). It is a local minimum at $(0, 0)$ .

(ii) If $n$ is odd, $(- 1)^{n - 1} = + 1$ and $f_{n}^{'} (- 1) > 0$ : $f_{n}^{'}$ is positive on both sides of $x = 0$ and $f_{n}^{'} (0) = 0$ (since $n > 1$ ). The curve keeps increasing, so $(0, 0)$ is a point of inflexion with zero gradient.

h) 1. There are four solutions of $f_{n} (x) = k$ when the horizontal line $y = k$ cuts the graph four times. This only happens with the "W" shape of even $n$ : two minima with value $0$ and a local maximum with value $(\frac{a}{2})^{2 n}$ . 2. For odd $n$ the line cuts the graph at most twice (the graph tends to $- \infty$ on both sides). 3. Conditions: $n$ even and $0 < k < (\frac{a}{2})^{2 n}$ .

Answer: a) intercepts $(0, 0)$ and $(2, 0)$ , local maximum $(1, 1)$ ; b) table; c) proof; d) $x = 0$ , $x = \frac{a}{2}$ , $x = a$ ; e), f) and g) proofs; h) $n$ even and $0 < k < (\frac{a}{2})^{2 n}$

Exercise 173 HardIB · P2 TZ1 May 2021

The function $f$ is defined by $f (x) = \frac{3 x + 2}{4 x ^{2} - 1}$ , for $x \in R$ , $x \neq = p$ , $x \neq = q$ .

a) Calculate the value of $p$ and the value of $q$ .

b) Calculate an expression for $f^{'} (x)$ .

The graph of $y = f (x)$ has exactly one point of inflexion.

c) Calculate the $x$ -coordinate of the point of inflexion.

d) Sketch the graph of $y = f (x)$ for $- 3 \leq x \leq 3$ , show the values of any intercepts with the axes, the coordinates of any local maximum and local minimum, and write down the equation of any asymptote.

The function $g (x)$ is defined by $g (x) = \frac{4 x ^{2} - 1}{3 x + 2}$ , for $x \in R$ , $x \neq = - \frac{2}{3}$ .

e) Calculate the equations of all asymptotes of the graph of $y = g (x)$ .

f) Considering the graph of $y = g (x) - f (x)$ , or otherwise, solve $f (x) < g (x)$ for $x \in R$ .

View solution

a) 1. $f$ is not defined if $4 x^{2} - 1 = 0$ , that is $x = \pm \frac{1}{2}$ . 2. With $p < q$ : $p = - \frac{1}{2}$ and $q = \frac{1}{2}$ .

b) 1. Quotient rule with $u = 3 x + 2$ and $v = 4 x^{2} - 1$ : $f^{'} (x) = \frac{3 ( 4 x ^{2} - 1 ) - ( 3 x + 2 ) \cdot 8 x}{( 4 x ^{2} - 1 ) ^{2}} = \frac{12 x ^{2} - 3 - 24 x ^{2} - 16 x}{( 4 x ^{2} - 1 ) ^{2}} = \frac{- 12 x ^{2} - 16 x - 3}{( 4 x ^{2} - 1 ) ^{2}} .$

c) 1. The point of inflexion occurs where $f^{''} (x) = 0$ and $f^{''}$ changes sign. Using the GDC we plot $f^{''} (x)$ (or the derivative of $f^{'}$ ) and find its zero. 2. The only solution is $x \approx - 1.60$ .

d) 1. Intercepts with the axes: $x = 0$ gives $f (0) = - 2$ , point $(0, - 2)$ ; $y = 0$ gives $3 x + 2 = 0$ , so $x = - \frac{2}{3}$ , point $(- 0.667, 0)$ . 2. Vertical asymptotes: $x = - \frac{1}{2}$ and $x = \frac{1}{2}$ . As $f (x) \to 0$ when $x \to \pm \infty$ , the $x$ -axis ( $y = 0$ ) is a horizontal asymptote. 3. Turning points: $f^{'} (x) = 0$ implies $12 x^{2} + 16 x + 3 = 0$ , so $x = \frac{- 4 \pm 7}{6}$ , that is $x \approx - 1.108$ and $x \approx - 0.226$ . As the sign of $f^{'}$ is that of $- (12 x^{2} + 16 x + 3)$ (the denominator is positive), $f$ decreases, then increases, then decreases again: local minimum at $(- 1.11, - 0.339)$ and local maximum at $(- 0.226, - 1.66)$ . 4. Shape: to the left of $x = - 0.5$ the graph starts just below the $x$ -axis, falls to the minimum, rises crossing the axis at $- 0.667$ and goes to $+ \infty$ as $x \to - 0. 5^{-}$ ; between the two vertical asymptotes it comes up from $- \infty$ , rises to the local maximum, passes through $(0, - 2)$ and falls to $- \infty$ as $x \to 0. 5^{-}$ ; to the right of $x = 0.5$ it falls from $+ \infty$ , approaching $y = 0$ .

Graph of a rational function with two vertical asymptotes, the x-axis as horizontal asymptote, two intercepts with the axes, one local maximum and one local minimum

e) 1. Vertical asymptote of $g (x) = \frac{4 x ^{2} - 1}{3 x + 2}$ : the denominator vanishes at $x = - \frac{2}{3}$ (the numerator is $\frac{7}{9} \neq = 0$ ), so $x = - \frac{2}{3}$ . 2. As the degree of the numerator (2) exceeds that of the denominator by one, there is no horizontal asymptote but an oblique one. Dividing: $4 x^{2} - 1 = (\frac{4}{3} x - \frac{8}{9}) (3 x + 2) + \frac{7}{9}$ , so $g (x) = \frac{4}{3} x - \frac{8}{9} + \frac{7/9}{3 x + 2}$ . 3. The oblique asymptote is $y = \frac{4}{3} x - \frac{8}{9}$ .

f) 1. We study the sign of $g - f$ . We reduce to a common denominator and factorise using the difference of two squares: $g - f = \frac{( 4 x ^{2} - 1 ) ^{2} - ( 3 x + 2 ) ^{2}}{( 3 x + 2 ) ( 4 x ^{2} - 1 )} = \frac{( 4 x ^{2} - 3 x - 3 ) ( 4 x ^{2} + 3 x + 1 )}{( 3 x + 2 ) ( 4 x ^{2} - 1 )} .$ 2. $4 x^{2} + 3 x + 1 > 0$ always (discriminant $9 - 16 < 0$ ), so the sign is that of $\frac{4 x ^{2} - 3 x - 3}{( 3 x + 2 ) ( 2 x - 1 ) ( 2 x + 1 )}$ . The roots of the numerator are $x = \frac{3 \pm 57}{8} \approx 1.32$ and $- 0.569$ . 3. We solve $f (x) < g (x)$ , that is $g - f > 0$ , with a sign table using the points $- \frac{2}{3}$ , $- 0.569$ , $- \frac{1}{2}$ , $\frac{1}{2}$ and $1.32$ (also seen on the GDC with the graph of $y = g (x) - f (x)$ ):

Interval	numerator	denominator	quotient
$x < - \frac{2}{3}$	$+$	$-$	$-$
$- \frac{2}{3} < x < - 0.569$	$+$	$+$	$+$
$- 0.569 < x < - \frac{1}{2}$	$-$	$+$	$-$
$- \frac{1}{2} < x < \frac{1}{2}$	$-$	$-$	$+$
$\frac{1}{2} < x < 1.32$	$-$	$+$	$-$
$x > 1.32$	$+$	$+$	$+$

The solution is $] - \frac{2}{3}, - 0.569 [\cup] - \frac{1}{2}, \frac{1}{2} [\cup] 1.32, + \infty [$ .

Answer: a) $p = - \frac{1}{2}$ , $q = \frac{1}{2}$ ; b) $f^{'} (x) = \frac{- 12 x ^{2} - 16 x - 3}{( 4 x ^{2} - 1 ) ^{2}}$ ; c) $x = - 1.60$ ; d) intercepts $(0, - 2)$ and $(- 0.667, 0)$ , graph; e) $x = - \frac{2}{3}$ , $y = \frac{4}{3} x - \frac{8}{9}$ ; f) $] - \frac{2}{3}, - 0.569 [\cup] - \frac{1}{2}, \frac{1}{2} [\cup] 1.32, + \infty [$

Exercise 174 HardIB · P1 TZ2 May 2022

The function is defined by $f (x) = \frac{1}{x ^{2} - 2 x - 3}$ , where $x \in R$ , $x \neq = - 1$ , $x \neq = 3$ .

a) Sketch the curve $y = f (x)$ , clearly indicating all asymptotes with their equations. Also state the coordinates of all local maximum and minimum points and of all points where the curve intersects the axes.

The function $g$ is defined by $g (x) = \frac{1}{x ^{2} - 2 x - 3}$ , where $x \in R$ , $x > 3$ .

b) The inverse of $g$ is $g^{- 1}$ .

(i) Show that $g^{- 1} (x) = 1 + \frac{4 x ^{2} + x}{x}$ .

(ii) State the domain of $g^{- 1}$ .

The function $h$ is defined by $h (x) = arctan \frac{x}{2}$ , where $x \in R$ .

c) Given that $(h \circ g) (a) = \frac{π}{4}$ , find the value of $a$ .

Give your answer in the form $p + \frac{q}{2} r$ , where $p, q, r \in Z^{+}$ .

View solution

a) 1. We write $f (x) = \frac{1}{( x - 3 ) ( x + 1 )}$ . Vertical asymptotes: $x = - 1$ and $x = 3$ . Since $f (x) \to 0$ as $x \to \pm \infty$ , the horizontal asymptote is $y = 0$ . 2. The curve does not cut the $x$ -axis (the numerator is $1$ ). Intersection with the $y$ -axis: $f (0) = - \frac{1}{3}$ , giving the point $(0, - \frac{1}{3})$ . 3. Differentiating: $f^{'} (x) = - \frac{2 x - 2}{( x ^{2} - 2 x - 3 ) ^{2}}$ . $f^{'} (x) = 0$ implies $x = 1$ , with $f (1) = \frac{1}{1 - 2 - 3} = - \frac{1}{4}$ . The sign of $f^{'}$ is that of $- (2 x - 2)$ : positive for $x < 1$ and negative for $x > 1$ , so $(1, - \frac{1}{4})$ is a local maximum. 4. Shape: for $x < - 1$ and $x > 3$ the graph is above the $x$ -axis, approaching $y = 0$ at the ends and $+ \infty$ near the asymptotes; between $x = - 1$ and $x = 3$ it is below the axis, with $- \infty$ near the asymptotes and the local maximum at $(1, - \frac{1}{4})$ .

Graph of a rational function with two vertical asymptotes and the x-axis as horizontal asymptote, with a local maximum between the two vertical asymptotes

b) (i) 1. Let $y = \frac{1}{x ^{2} - 2 x - 3}$ with $x > 3$ . We solve for $x$ : $x^{2} - 2 x - 3 = \frac{1}{y}$ . 2. Completing the square: $(x - 1)^{2} = 4 + \frac{1}{y}$ . 3. Since $x > 3$ , $x - 1 > 0$ , so we take the positive root: $x = 1 + 4 + \frac{1}{y}$ . For $y > 0$ : $4 + \frac{1}{y} = \frac{4 y + 1}{y} = \frac{y ( 4 y + 1 )}{y} = \frac{4 y ^{2} + y}{y}$ . 4. Swapping $y$ for $x$ : $g^{- 1} (x) = 1 + \frac{4 x ^{2} + x}{x}$ .

(ii) 1. The domain of $g^{- 1}$ is the range of $g$ . For $x > 3$ , $x^{2} - 2 x - 3 = (x - 3) (x + 1)$ takes all positive values (tending to $0^{+}$ as $x \to 3^{+}$ and to $+ \infty$ as $x \to + \infty$ ), so $g$ takes all values in $] 0, + \infty [$ . 2. $D_{g^{- 1}} = R_{g} =] 0, + \infty [$ .

c) 1. $(h \circ g) (a) = arctan \frac{g ( a )}{2} = \frac{π}{4}$ , so $\frac{g ( a )}{2} = tan \frac{π}{4} = 1$ and $g (a) = 2$ . 2. Therefore $a = g^{- 1} (2) = 1 + \frac{16 + 2}{2} = 1 + \frac{3 2}{2}$ (with $a > 3$ ). 3. In the required form: $a = 1 + \frac{3}{2} 2$ , with $p = 1$ , $q = 3$ , $r = 2$ .

Answer: a) asymptotes $x = - 1$ , $x = 3$ , $y = 0$ ; maximum $(1, - \frac{1}{4})$ ; intercept $(0, - \frac{1}{3})$ ; graph; b) (i) proof; (ii) $] 0, + \infty [$ ; c) $a = 1 + \frac{3}{2} 2$

Exercise 175 HardIB · P2 TZ2 May 2022

A scientist carried out a nine-week experiment with two plants ( $A$ and $B$ ) of the same species. She wanted to find out what effect a new fertiliser has. Plant $A$ received the fertiliser regularly, whereas plant $B$ did not receive it.

The scientist found that the height of plant $A$ ( $h_{A}$ cm) after $t$ weeks can be modelled by the function $h_{A} (t) = sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ . The scientist found that the height of plant $B$ ( $h_{B}$ cm) after $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

a) Use the scientist's models to find the initial height of:

(i) Plant $B$

(ii) Plant $A$ , rounded to three significant figures.

b) Find the values of $t$ for which $h_{A} (t) = h_{B} (t)$ .

c) For $t > 6$ , prove that plant $A$ was always taller than plant $B$ .

d) For $0 \leq t \leq 9$ , find the total time during which the growth rate of plant $B$ was greater than the growth rate of plant $A$ .

Show answer

Answer: a) (i) $32$ cm; (ii) $26.7$ cm; b) $t = 4.01$ , $t = 4.70$ and $t = 5.88$ weeks; c) proof; d) $1.19 < t < 2.24$ , $4.33 < t < 5.38$ , $7.47 < t < 8.52$ ; total $3.14$ weeks

Show working and answer

a) (i) $h_{B} (0) = 8 \cdot 0 + 32 = 32$ cm.

(ii) $h_{A} (0) = sin 6 + 27 \approx - 0.2794 + 27 = 26.72 \dots$ , that is $26.7$ cm (3 significant figures; the sine is taken with the argument in radians).

b) 1. We equate: $sin (2 t + 6) + 9 t + 27 = 8 t + 32$ , that is $sin (2 t + 6) + t - 5 = 0$ . 2. Using the GDC (intersection of $y = sin (2 t + 6) + t - 5$ with the $x$ -axis, for $0 \leq t \leq 9$ ), we obtain three solutions: $t \approx 4.01$ , $t \approx 4.70$ and $t \approx 5.88$ weeks.

c) 1. The difference in heights is $h_{A} (t) - h_{B} (t) = sin (2 t + 6) + t - 5$ . 2. Since $sin (2 t + 6) \geq - 1$ , for $t > 6$ : $h_{A} (t) - h_{B} (t) \geq - 1 + t - 5 = t - 6 > 0$ . 3. Therefore $h_{A} (t) > h_{B} (t)$ for all $t > 6$ : plant $A$ is always taller.

d) 1. The growth rates are the derivatives: $h_{A}^{'} (t) = 2 cos (2 t + 6) + 9$ and $h_{B}^{'} (t) = 8$ . 2. $h_{B}^{'} > h_{A}^{'}$ is equivalent to $2 cos (2 t + 6) + 9 < 8$ , that is $cos (2 t + 6) < - \frac{1}{2}$ . 3. Using the GDC (intersections of $y = 2 cos (2 t + 6) + 9$ with $y = 8$ ) for $0 \leq t \leq 9$ , the inequality holds on the intervals $1.19 < t < 2.24$ , $4.33 < t < 5.38$ and $7.47 < t < 8.52$ . (Algebraically: $\frac{2 π}{3} < 2 t + 6 - 2 π n < \frac{4 π}{3}$ for $n = 1, 2, 3$ ; each interval lasts $\frac{π}{3}$ .) 4. Total time: $3 \cdot \frac{π}{3} = π \approx 3.14$ weeks.

Answer: a) (i) $32$ cm; (ii) $26.7$ cm; b) $t = 4.01$ , $t = 4.70$ and $t = 5.88$ weeks; c) proof; d) $1.19 < t < 2.24$ , $4.33 < t < 5.38$ , $7.47 < t < 8.52$ ; total $3.14$ weeks

Exercise 176 HardIB · P1 TZ0 November 2022

The function $f$ is defined by $f (x) = cos^{2} x - 3 sin^{2} x$ , $0 \leq x \leq π$ .

a) Find the roots of the equation $f (x) = 0$ .

b) (i) Find $f^{'} (x)$ .

(ii) Hence, find the coordinates of the points on the graph of $y = f (x)$ at which $f^{'} (x) = 0$ .

c) Sketch the graph of $y = ∣ f (x) ∣$ , clearly showing the coordinates of all points at which $f^{'} (x) = 0$ and of all points where the graph touches the axes.

d) Using the above, or otherwise, solve the inequality $∣ f (x) ∣ > 1$ .

View solution

a) 1. $f (x) = 0$ implies $cos^{2} x = 3 sin^{2} x$ , that is $tan^{2} x = \frac{1}{3}$ and $tan x = \pm \frac{1}{3}$ . 2. On $[0, π]$ : $x = \frac{π}{6}$ (where $tan x = \frac{1}{3}$ ) and $x = \frac{5 π}{6}$ (where $tan x = - \frac{1}{3}$ ).

b) (i) 1. Using $cos^{2} x = \frac{1 + cos 2 x}{2}$ and $sin^{2} x = \frac{1 - cos 2 x}{2}$ : $f (x) = \frac{1 + cos 2 x}{2} - \frac{3 ( 1 - cos 2 x )}{2} = 2 cos 2 x - 1$ . 2. Differentiating (chain rule): $f^{'} (x) = - 4 sin 2 x$ .

(ii) 1. $f^{'} (x) = 0$ implies $sin 2 x = 0$ , with $2 x \in [0, 2 π]$ : $2 x = 0, π, 2 π$ , that is $x = 0, \frac{π}{2}, π$ . 2. The $y$ -coordinates: $f (0) = 2 - 1 = 1$ , $f (\frac{π}{2}) = - 2 - 1 = - 3$ and $f (π) = 1$ . 3. The points are $(0, 1)$ , $(\frac{π}{2}, - 3)$ and $(π, 1)$ .

c) 1. $f (x) = 2 cos 2 x - 1$ starts at $1$ when $x = 0$ , falls to $- 3$ at $\frac{π}{2}$ and returns to $1$ at $π$ , crossing the $x$ -axis at $\frac{π}{6}$ and $\frac{5 π}{6}$ . 2. For $∣ f ∣$ we reflect the negative part (between $\frac{π}{6}$ and $\frac{5 π}{6}$ ) above the $x$ -axis: the minimum $(\frac{π}{2}, - 3)$ becomes a maximum $(\frac{π}{2}, 3)$ , and at $x = \frac{π}{6}$ and $x = \frac{5 π}{6}$ the graph touches the $x$ -axis forming peaks. The endpoints are $(0, 1)$ and $(π, 1)$ .

Graph of the absolute value of a trigonometric function between 0 and pi, with two peaks touching the x-axis, a maximum in the middle and a dashed horizontal line

d) 1. $∣ f (x) ∣ > 1$ is equivalent to $f (x) > 1$ or $f (x) < - 1$ . 2. $f (x) > 1$ : $2 cos 2 x - 1 > 1$ implies $cos 2 x > 1$ , which is impossible. 3. $f (x) < - 1$ : $2 cos 2 x - 1 < - 1$ implies $cos 2 x < 0$ , with $2 x \in [0, 2 π]$ : $\frac{π}{2} < 2 x < \frac{3 π}{2}$ . 4. That is, $\frac{π}{4} < x < \frac{3 π}{4}$ (seen in the graph from c: it is where $∣ f ∣$ is above the line $y = 1$ ).

Answer: a) $x = \frac{π}{6}$ , $x = \frac{5 π}{6}$ ; b) (i) $f^{'} (x) = - 4 sin 2 x$ ; (ii) $(0, 1)$ , $(\frac{π}{2}, - 3)$ , $(π, 1)$ ; c) graph; d) $\frac{π}{4} < x < \frac{3 π}{4}$

Optimisation

Exercise 177 BasicIB · P1 May 2001

An astronaut on the Moon throws a ball vertically upwards. The height, $s$ metres, of the ball after $t$ seconds is given by the equation $s = 40 t + 0.5 a t^{2}$ , where $a$ is a constant. If the ball reaches its maximum height when $t = 25$ , find the value of $a$ .

Show answer

Answer: $a = - \frac{8}{5} = - 1.6$

Show working and answer

The height is maximum when the vertical velocity is zero: $\frac{d s}{d t} = 40 + a t = 0$ .
At $t = 25$ : $40 + 25 a = 0$ , so $a = - \frac{40}{25} = - \frac{8}{5}$ .
Check that it is a maximum: $\frac{d ^{2} s}{d t ^{2}} = a = - \frac{8}{5} < 0$ .

Answer: $a = - \frac{8}{5} = - 1.6$

Exercise 178 MediumIB · P1 TZ0 November 2014

A point $P$ , relative to the origin $O$ , has position vector $OP = 1 + s 3 + 2 s 1 - s$ , $s \in R$ .

Find the minimum length of $OP$ .

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Answer: $5$ (for $s = - 1$ )

Show working and answer

The square of the length is $∣ OP ∣^{2} = (1 + s)^{2} + (3 + 2 s)^{2} + (1 - s)^{2}$ .
Expanding: $1 + 2 s + s^{2} + 9 + 12 s + 4 s^{2} + 1 - 2 s + s^{2} = 6 s^{2} + 12 s + 11$ .
Completing the square: $6 s^{2} + 12 s + 11 = 6 (s + 1)^{2} + 5$ , which is a minimum when $s = - 1$ and equals $5$ . (Using derivatives: the derivative $12 s + 12$ vanishes at $s = - 1$ and the second derivative $12 > 0$ .)
The minimum length is $5$ .

Answer: $5$ (for $s = - 1$ )

Exercise 179 MediumIB · P1 TZ0 November 2014

A tranquiliser is injected into a muscle, from where it passes into the bloodstream. The concentration $C$ in $mg l^{- 1}$ of tranquiliser present in the bloodstream can be modelled by the function $C (t) = \frac{2 t}{3 + t ^{2}}$ , $t \geq 0$ , where $t$ is the number of minutes elapsed since the injection. Find the maximum concentration of tranquiliser in the bloodstream.

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Answer: $C (3) = \frac{3}{3} mg l^{- 1}$

Show working and answer

We differentiate using the quotient rule: $C^{'} (t) = \frac{2 ( 3 + t ^{2} ) - 2 t \cdot 2 t}{( 3 + t ^{2} ) ^{2}} = \frac{6 - 2 t ^{2}}{( 3 + t ^{2} ) ^{2}}$ .
$C^{'} (t) = 0 \Rightarrow t^{2} = 3 \Rightarrow t = 3$ (with $t \geq 0$ ).
It is a maximum: $C^{'} > 0$ for $0 \leq t < 3$ and $C^{'} < 0$ for $t > 3$ (the sign is given by $6 - 2 t^{2}$ ).
$C (3) = \frac{2 3}{3 + 3} = \frac{3}{3}$ .

Answer: $C (3) = \frac{3}{3} mg l^{- 1}$

Exercise 180 MediumIB · P1 May 1996

The sum of the perimeter of a square of side $a$ cm and the circumference of a circle of radius $r$ cm is $240$ cm. What is the value of $r$ if the sum of the areas is a minimum?

Show answer

Answer: $r = \frac{120}{π + 4} \approx 16.8$ cm

Show working and answer

Perimeter of the square $4 a$ plus circumference $2 π r$ : $4 a + 2 π r = 240$ , so $a = 60 - \frac{π r}{2}$ .
Sum of areas: $A = a^{2} + π r^{2} = (60 - \frac{π r}{2})^{2} + π r^{2}$ .
Differentiating (chain rule): $\frac{d A}{d r} = 2 (60 - \frac{π r}{2}) (- \frac{π}{2}) + 2 π r = - π (60 - \frac{π r}{2}) + 2 π r$ .
Setting this equal to zero and dividing by $π$ : $- 60 + \frac{π r}{2} + 2 r = 0 \Rightarrow r (\frac{π}{2} + 2) = 60 \Rightarrow r = \frac{120}{π + 4}$ .
It is a minimum: $\frac{d ^{2} A}{d r ^{2}} = \frac{π ^{2}}{2} + 2 π > 0$ .

Answer: $r = \frac{120}{π + 4} \approx 16.8$ cm

Exercise 181 MediumIB · P1 November 1996

Find the points on the curve $y^{2} - x^{2} = 4$ that are closest to the point $(2, 0)$ .

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Answer: $(1, 5)$ and $(1, - 5)$

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Let $(x, y)$ be a point on the curve $y^{2} = x^{2} + 4$ . The square of its distance to $(2, 0)$ is $d^{2} = (x - 2)^{2} + y^{2}$ .
Substituting for $y^{2}$ : $d^{2} = (x - 2)^{2} + x^{2} + 4 = 2 x^{2} - 4 x + 8$ .
We minimise $d^{2}$ (it has the same minimum as $d$ ): $\frac{d ( d ^{2} )}{d x} = 4 x - 4 = 0 \Rightarrow x = 1$ ; it is a minimum because the second derivative is $4 > 0$ .
For $x = 1$ : $y^{2} = 1 + 4 = 5$ , so $y = \pm 5$ .

Answer: $(1, 5)$ and $(1, - 5)$

Exercise 182 MediumIB · P1 November 2001

The point $P (x, x^{2})$ lies on the curve $y = x^{2}$ .

Find the minimum distance from the point $A (2, - \frac{1}{2})$ to the point $P$ .

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Answer: minimum distance $\approx 1.63$

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The square of the distance between $P (x, x^{2})$ and $A (2, - \frac{1}{2})$ is $D (x) = (x - 2)^{2} + (x^{2} + \frac{1}{2})^{2} = x^{4} + 2 x^{2} - 4 x + \frac{17}{4}$ . Minimising the distance is equivalent to minimising $D$ .
$D^{'} (x) = 4 x^{3} + 4 x - 4 = 0$ , i.e. $x^{3} + x - 1 = 0$ .
This equation has no simple root: using the GDC, $x \approx 0.6823$ (it is the only real root, because $x^{3} + x - 1$ is increasing).
It is a minimum: $D^{''} (x) = 12 x^{2} + 4 > 0$ .
$D (0.6823) \approx 2.669$ , so the minimum distance is $2.669 \approx 1.63$ .

Answer: minimum distance $\approx 1.63$

Exercise 183 MediumIB · P1 May 2002

The function $f$ is defined on the domain $[0, π]$ by $f (θ) = 4 cos θ + 3 sin θ$ .

a) Express $f (θ)$ in the form $R cos (θ - α)$ where $0 < α < \frac{π}{2}$ .

b) Hence, or otherwise, write down the value of $θ$ for which $f (θ)$ is a maximum.

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Answer: a) $f (θ) = 5 cos (θ - arctan \frac{3}{4})$ ; b) $θ = arctan \frac{3}{4} \approx 0.644$

Show working and answer

a) 1. We expand $R cos (θ - α) = R cos θ cos α + R sin θ sin α$ and equate coefficients with $4 cos θ + 3 sin θ$ : $R cos α = 4$ and $R sin α = 3$ . 2. $R = 4^{2} + 3^{2} = 5$ and $tan α = \frac{3}{4}$ , so $α = arctan \frac{3}{4} \approx 0.644$ , which lies in $] 0, \frac{π}{2} [$ . 3. $f (θ) = 5 cos (θ - arctan \frac{3}{4})$ .

b) $f$ is a maximum when $cos (θ - α) = 1$ , i.e. $θ = α = arctan \frac{3}{4} \approx 0.644$ (which belongs to $[0, π]$ ). The maximum value is $5$ .

Answer: a) $f (θ) = 5 cos (θ - arctan \frac{3}{4})$ ; b) $θ = arctan \frac{3}{4} \approx 0.644$

Exercise 184 MediumIB · P1 November 2002

The point $B (a, b)$ lies on the curve $f (x) = x^{2}$ and is such that $B$ is the point closest to $A (6, 0)$ .

Find the value of $a$ .

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Answer: $a \approx 1.33$

Show working and answer

The square of the distance from $B (a, a^{2})$ to $A (6, 0)$ is $D = (a - 6)^{2} + (a^{2})^{2} = (a - 6)^{2} + a^{4}$ . Minimising the distance is equivalent to minimising $D$ .
$D^{'} = 2 (a - 6) + 4 a^{3} = 0$ , so $2 a^{3} + a - 6 = 0$ .
This equation has no simple rational root: using the GDC, $a \approx 1.33$ . It is the only real root because $2 a^{3} + a - 6$ is strictly increasing (its derivative $6 a^{2} + 1 > 0$ ).
It is a minimum: $D^{''} = 2 + 12 a^{2} > 0$ .

Answer: $a \approx 1.33$

Exercise 185 MediumIB · P1 May 2002

A rectangle is drawn such that its vertices lie on the $x$ -axis and its upper vertices lie on the curve $y = e^{- x^{2}}$ . The area of this rectangle is denoted by $A$ .

a) Write down an expression for $A$ in terms of $x$ .

b) Find the maximum value of $A$ .

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Answer: a) $A = 2 x e^{- x^{2}}$ ; b) $A_{m a x} = \frac{2}{e} \approx 0.858$

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a) 1. The upper vertices are $(x, e^{- x^{2}})$ and $(- x, e^{- x^{2}})$ , with $x > 0$ . The base has length $2 x$ and the height is $e^{- x^{2}}$ . 2. $A = 2 x e^{- x^{2}}$ .

b) 1. Product rule: $A^{'} = 2 e^{- x^{2}} + 2 x (- 2 x) e^{- x^{2}} = 2 e^{- x^{2}} (1 - 2 x^{2})$ . 2. $A^{'} = 0 ⟺ x = \frac{1}{2}$ (with $x > 0$ ). $A^{'} > 0$ before and $A^{'} < 0$ after, so it is a maximum. 3. $A_{m a x} = 2 \cdot \frac{1}{2} e^{- 1/2} = 2 e^{- 1/2} = \frac{2}{e} \approx 0.858$ .

Answer: a) $A = 2 x e^{- x^{2}}$ ; b) $A_{m a x} = \frac{2}{e} \approx 0.858$

Exercise 186 MediumIB · P1 TZ0 November 2004

A closed cylindrical can has a volume of $500 c m^{3}$ . The height of the can is $h$ cm and the radius of the base is $r$ cm.

a) Find an expression for the total surface area $A$ of the can in terms of $r$ .

b) Given that $A$ has a minimum value for $r > 0$ , find this value of $r$ .

Show answer

Answer: a) $A = 2 π r^{2} + \frac{1000}{r}$ ; b) $r = 5 3 \frac{2}{π} \approx 4.30$ cm

Show working and answer

a) 1. Total surface area: two lids of area $π r^{2}$ each plus the curved surface $2 π r h$ : $A = 2 π r^{2} + 2 π r h$ . 2. Eliminate $h$ using the volume $π r^{2} h = 500$ , that is, $h = \frac{500}{π r ^{2}}$ . 3. $A = 2 π r^{2} + 2 π r \cdot \frac{500}{π r ^{2}} = 2 π r^{2} + \frac{1000}{r}$ .

b) 1. $A^{'} (r) = 4 π r - \frac{1000}{r ^{2}} = 0$ , so $r^{3} = \frac{250}{π}$ . 2. $r = 3 \frac{250}{π} = 5 3 \frac{2}{π} \approx 4.30$ cm. 3. It is a minimum: $A^{''} (r) = 4 π + \frac{2000}{r ^{3}} > 0$ for $r > 0$ .

Answer: a) $A = 2 π r^{2} + \frac{1000}{r}$ ; b) $r = 5 3 \frac{2}{π} \approx 4.30$ cm

Exercise 187 MediumIB · P1 TZ0 November 2005

Given that the maximum value of $\frac{1}{4 sin θ + 3 cos θ + k}$ is 2, for $0^{\circ} \leq θ \leq 36 0^{\circ}$ , find the value of $k$ .

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Answer: $k = \frac{11}{2}$

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Let $D (θ) = 4 sin θ + 3 cos θ + k$ . The expression is $\frac{1}{D}$ .
As $4 sin θ + 3 cos θ = 5 sin (θ + α)$ with $α = arctan \frac{3}{4}$ (amplitude $4^{2} + 3^{2} = 5$ ), $D$ takes all values between $k - 5$ and $k + 5$ .
If $D$ were to reach $0$ , $\frac{1}{D}$ would not be bounded; since there is a maximum, $D > 0$ always and the maximum of $\frac{1}{D}$ occurs where $D$ is smallest: $D_{m i n} = k - 5$ .
Condition: $\frac{1}{k - 5} = 2$ , so $k - 5 = \frac{1}{2}$ and $k = \frac{11}{2}$ .
Check: with $k = \frac{11}{2}$ , $D$ runs from $\frac{1}{2}$ to $\frac{21}{2}$ , always positive.

Answer: $k = \frac{11}{2}$

Exercise 188 MediumIB · P1 TZ2 May 2015

In triangle $ABC$ , $BC = 3$ cm, $ABC = θ$ and $BCA = \frac{π}{3}$ .

a) Show that the length $AB = \frac{3}{3 cos θ + sin θ}$ .

b) Given that $AB$ reaches a minimum value, determine the value of $θ$ for which this occurs.

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Answer: a) shown; b) $θ = \frac{π}{6}$

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a) 1. In triangle $ABC$ , $BAC = π - θ - \frac{π}{3} = \frac{2 π}{3} - θ$ . 2. Sine rule: $\frac{AB}{sin \frac{π}{3}} = \frac{BC}{sin ( \frac{2 π}{3} - θ )}$ , so $AB = \frac{3 \cdot \frac{3}{2}}{sin ( \frac{2 π}{3} - θ )} = \frac{3/2}{sin ( \frac{2 π}{3} - θ )}$ . 3. We expand the sine of a difference: $sin (\frac{2 π}{3} - θ) = sin \frac{2 π}{3} cos θ - cos \frac{2 π}{3} sin θ = \frac{3}{2} cos θ + \frac{1}{2} sin θ$ . 4. We substitute: $AB = \frac{3/2}{\frac{1}{2} ( 3 cos θ + sin θ )} = \frac{3}{3 cos θ + sin θ}$ .

b) 1. $AB$ is a minimum when the denominator $D (θ) = 3 cos θ + sin θ$ is a maximum (for $0 < θ < \frac{2 π}{3}$ the denominator is positive). 2. $D^{'} (θ) = - 3 sin θ + cos θ = 0$ implies $tan θ = \frac{1}{3}$ , so $θ = \frac{π}{6}$ . 3. $D^{''} (θ) = - 3 cos θ - sin θ = - D (θ) < 0$ at $θ = \frac{π}{6}$ , since $D (\frac{π}{6}) = \frac{3}{2} + \frac{1}{2} = 2 > 0$ : this is a maximum of $D$ , and therefore a minimum of $AB$ (which equals $\frac{3}{2}$ cm).

Answer: a) shown; b) $θ = \frac{π}{6}$

Exercise 189 HardIB · P1 TZ0 November 2008

A packaging company manufactures boxes for chocolates. An example of one of these boxes is shown below. The box has a lid, and the top and bottom faces of the box are identical regular hexagons of side $x$ cm.

Box with a lid in the shape of a regular hexagonal prism; the edge of the hexagonal base measures x cm

a) verify that the area of each hexagon is equal to $\frac{3 3 x ^{2}}{2} cm^{2}$ .

b) Given that the volume of the box is equal to $90 cm^{3}$ , verify that when $x = 320$ the total surface area of the box reaches a minimum, justifying why this value gives a minimum.

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Answer: verified: $S = 33 x^{2} + \frac{360}{3 x}$ , $S^{'} = 0$ at $x = 320$ and $S^{''} > 0$ , so it is a minimum.

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a) 1. A regular hexagon of side $x$ is divided into $6$ equilateral triangles of side $x$ . 2. Each triangle has area $\frac{1}{2} x \cdot x sin 6 0^{\circ} = \frac{3}{4} x^{2}$ . 3. The hexagon: $6 \cdot \frac{3}{4} x^{2} = \frac{3 3 x ^{2}}{2} cm^{2}$ .

b) 1. Let $h$ be the height of the box. Volume: $\frac{3 3 x ^{2}}{2} h = 90$ , so $h = \frac{180}{3 3 x ^{2}} = \frac{60}{3 x ^{2}}$ . 2. Total surface area: two hexagons and six rectangular faces of $x \cdot h$ : $S = 2 \cdot \frac{3 3 x ^{2}}{2} + 6 x h = 33 x^{2} + \frac{360}{3 x}$ . 3. Differentiating: $\frac{d S}{d x} = 63 x - \frac{360}{3 x ^{2}}$ . 4. Setting it equal to $0$ : $63 x = \frac{360}{3 x ^{2}}$ , so $x^{3} = \frac{360}{6 \cdot 3} = 20$ , and $x = 320$ . 5. Second derivative: $\frac{d ^{2} S}{d x ^{2}} = 63 + \frac{720}{3 x ^{3}} > 0$ for all $x > 0$ . 6. As $\frac{d ^{2} S}{d x ^{2}} > 0$ at $x = 320$ , the surface area has a minimum there (in fact, it is the only critical point and the curve is convex for $x > 0$ ).

Answer: verified: $S = 33 x^{2} + \frac{360}{3 x}$ , $S^{'} = 0$ at $x = 320$ and $S^{''} > 0$ , so it is a minimum.

Exercise 190 HardIB · P2 TZ0 November 2009

The diagram below shows two concentric circles with centre $O$ and radii $2$ cm and $4$ cm.

Two concentric circles with centre O and radii 2 cm and 4 cm; P and Q lie on the larger circle, and the region bounded by the chord PQ, the radii OP and OQ and the arc of the smaller circle is shaded

Points $P$ and $Q$ lie on the circle of larger radius and $POQ = x$ , where $0 < x < \frac{π}{2}$ .

a) Verify that the area of the shaded region is equal to $8 sin x - 2 x$ .

b) Find the maximum value of the area of the shaded region.

Show answer

Answer: a) verified: $A = 8 sin x - 2 x$ ; b) $x \approx 1.32$ rad; maximum area $\approx 5.11 cm^{2}$

Show working and answer

a) 1. The shaded region is the triangle $OPQ$ minus the circular sector of the smaller circle (radius $2$ , angle $x$ ). 2. Area of the triangle with two sides of $4$ cm and angle $x$ between them: $\frac{1}{2} \cdot 4 \cdot 4 sin x = 8 sin x$ . 3. Area of the sector of radius $2$ : $\frac{1}{2} \cdot 2^{2} \cdot x = 2 x$ . 4. Shaded area $A = 8 sin x - 2 x$ .

b) 1. $\frac{d A}{d x} = 8 cos x - 2$ . Setting $\frac{d A}{d x} = 0$ : $cos x = \frac{1}{4}$ . 2. In $] 0, \frac{π}{2} [$ : $x = arccos \frac{1}{4} \approx 1.32$ rad. 3. $\frac{d ^{2} A}{d x ^{2}} = - 8 sin x < 0$ on that interval, so it is a maximum (also because $\frac{d A}{d x}$ changes from positive to negative). 4. $A_{m a x} = 8 \cdot \frac{15}{4} - 2 \cdot 1.318 \approx 7.746 - 2.636 = 5.11 cm^{2}$ (using the GDC).

Answer: a) verified: $A = 8 sin x - 2 x$ ; b) $x \approx 1.32$ rad; maximum area $\approx 5.11 cm^{2}$

Exercise 191 HardIB · P1 TZ2 May 2008

André wants to go from a point $A$ situated in the sea to the point $Y$ , situated on a straight stretch of beach. $P$ is the point on the beach closest to $A$ , such that $AP = 2$ km and $PY = 2$ km. André makes this journey by first swimming in a straight line to a point $Q$ on the beach and then running to $Y$ .

Diagram of a straight beach: point A in the sea 2 km from point P on the beach, point Q between P and Y at x km from P, and P and Y 2 km apart

When André swims, he covers a distance of $1$ km in $55$ minutes. When he runs, he covers a distance of $1$ km in $5$ minutes.

a) If $PQ = x$ km, $0 \leq x \leq 2$ , find an expression for the time $T$ in minutes that André takes to reach the point $Y$ .

b) Verify that $\frac{d T}{d x} = \frac{5 5 x}{x ^{2} + 4} - 5$ .

c) (i) Solve $\frac{d T}{d x} = 0$

(ii) Use the value of $x$ found in part (c)(i) to determine the time, $T$ , in minutes, that André takes to reach the point $Y$ .

(iii) Verify that $\frac{d ^{2} T}{d x ^{2}} = \frac{20 5}{( x ^{2} + 4 ) ^{\frac{3}{2}}}$ and, hence, verify that the time you obtained in part (c)(ii) is a minimum.

Show answer

Answer: a) $T = 55 4 + x^{2} + 5 (2 - x)$ ; b) verified; c) (i) $x = 1$ ; (ii) $T = 30$ minutes; (iii) verified, $\frac{d ^{2} T}{d x ^{2}} (1) = 4 > 0$ , so it is a minimum

Show working and answer

a) 1. With $PQ = x$ , the swimming route is $AQ = 4 + x^{2}$ km (Pythagoras in triangle $APQ$ ) and the running route is $QY = 2 - x$ km. 2. Swimming time: $55 4 + x^{2}$ min. Running time: $5 (2 - x)$ min. 3. $T = 55 x^{2} + 4 + 5 (2 - x)$ .

b) 1. Differentiating with the chain rule: $\frac{d T}{d x} = 55 \cdot \frac{2 x}{2 x ^{2} + 4} - 5$ . 2. Simplifying: $\frac{d T}{d x} = \frac{5 5 x}{x ^{2} + 4} - 5$ .

c) (i) 1. $\frac{d T}{d x} = 0$ : $\frac{5 5 x}{x ^{2} + 4} = 5$ , that is, $5 x = x^{2} + 4$ . 2. Squaring: $5 x^{2} = x^{2} + 4$ , so $x^{2} = 1$ and, with $0 \leq x \leq 2$ , $x = 1$ .

(ii) 1. $T (1) = 55 \cdot 5 + 5 \cdot 1 = 25 + 5 = 30$ minutes.

(iii) 1. Differentiating $\frac{d T}{d x}$ with the quotient rule: $\frac{d ^{2} T}{d x ^{2}} = 55 \cdot \frac{x ^{2} + 4 - x \cdot \frac{x}{x ^{2} + 4}}{x ^{2} + 4} = 55 \cdot \frac{( x ^{2} + 4 ) - x ^{2}}{( x ^{2} + 4 ) ^{3/2}}$ . 2. This gives $\frac{d ^{2} T}{d x ^{2}} = \frac{20 5}{( x ^{2} + 4 ) ^{3/2}}$ . 3. At $x = 1$ : $\frac{20 5}{5 ^{3/2}} = \frac{20 5}{5 5} = 4 > 0$ , so $T = 30$ minutes is a minimum.

Answer: a) $T = 55 4 + x^{2} + 5 (2 - x)$ ; b) verified; c) (i) $x = 1$ ; (ii) $T = 30$ minutes; (iii) verified, $\frac{d ^{2} T}{d x ^{2}} (1) = 4 > 0$ , so it is a minimum

Exercise 192 HardIB · P2 TZ2 May 2013

A straight street $20$ metres wide is bounded on both sides by two vertical parallel walls, one $13$ metres high and the other $8$ metres high. The intensity of light reaching a point $P$ on the street, at ground level, is proportional to the angle $θ$ , where $θ = APB$ , as shown in the figure.

Straight street 20 m wide between two vertical walls, one 8 m high and the other 13 m high; a point P on the ground of the street and the angle θ between the lines of sight to the upper ends A and B of the walls

a) Find an expression for $θ$ in terms of $x$ , where $x$ is the distance between $P$ and the base of the $8$ metre high wall.

b) i) Calculate the value of $θ$ when $x = 0$ .

(ii) Calculate the value of $θ$ when $x = 20$ .

c) Sketch the graph of $θ$ for $0 \leq x \leq 20$ .

d) Verify that $\frac{d θ}{d x} = \frac{5 ( 744 - 64 x - x ^{2} )}{( x ^{2} + 64 ) ( x ^{2} - 40 x + 569 )}$ .

e) Using the result of part (d), or otherwise, determine the value of $x$ for which the intensity of light at $P$ is a maximum. Give your answer correct to four significant figures.

f) Point $P$ moves across the street at a speed of $0.5 m s^{- 1}$ . Determine the rate of change of $θ$ with respect to time, at the instant when $P$ is at the midpoint of the street.

View solution

a) 1. Let $α$ be the angle that $PA$ makes with the ground (towards the 8 m wall) and $β$ the angle that $PB$ makes with the ground (towards the 13 m wall). Then $θ = π - α - β$ , with $tan α = \frac{8}{x}$ and $tan β = \frac{13}{20 - x}$ . 2. $tan (α + β) = \frac{\frac{8}{x} + \frac{13}{20 - x}}{1 - \frac{104}{x ( 20 - x )}} = \frac{8 ( 20 - x ) + 13 x}{x ( 20 - x ) - 104} = \frac{5 x + 160}{- x ^{2} + 20 x - 104}$ . 3. As $θ = π - (α + β)$ , $tan θ = - tan (α + β) = \frac{5 x + 160}{x ^{2} - 20 x + 104}$ . The denominator is always positive (its discriminant is $400 - 416 < 0$ ), so $θ$ is acute. 4. Therefore $θ = arctan (\frac{5 x + 160}{x ^{2} - 20 x + 104})$ .

b) i) With $x = 0$ : $θ = arctan \frac{160}{104} = 0.994$ rad.

(ii) With $x = 20$ : $θ = arctan \frac{260}{104} = arctan 2.5 = 1.19$ rad.

c) 1. Using the GDC, we plot $θ (x)$ for $0 \leq x \leq 20$ . 2. The curve starts at $(0; 0.994)$ , rises gently to a single maximum near $x \approx 10$ and ends at $(20; 1.19)$ .

Graph of the angle θ against x for x from 0 to 20: smooth curve rising to a maximum near the middle of the street and falling afterwards

d) 1. Let $u = \frac{5 x + 160}{x ^{2} - 20 x + 104}$ . The derivative of $arctan u$ is $\frac{u ^{'}}{1 + u ^{2}}$ (chain rule). 2. By the quotient rule: $u^{'} = \frac{5 ( x ^{2} - 20 x + 104 ) - ( 5 x + 160 ) ( 2 x - 20 )}{( x ^{2} - 20 x + 104 ) ^{2}} = \frac{- 5 x ^{2} - 320 x + 3720}{( x ^{2} - 20 x + 104 ) ^{2}} = \frac{5 ( 744 - 64 x - x ^{2} )}{( x ^{2} - 20 x + 104 ) ^{2}}$ . 3. Also $1 + u^{2} = \frac{( x ^{2} - 20 x + 104 ) ^{2} + ( 5 x + 160 ) ^{2}}{( x ^{2} - 20 x + 104 ) ^{2}}$ , and the numerator is $x^{4} - 40 x^{3} + 633 x^{2} - 2560 x + 36416 = (x^{2} + 64) (x^{2} - 40 x + 569)$ . 4. Dividing, $\frac{d θ}{d x} = \frac{5 ( 744 - 64 x - x ^{2} )}{( x ^{2} + 64 ) ( x ^{2} - 40 x + 569 )}$ .

e) 1. The intensity is maximum when $θ$ is maximum, that is, when $\frac{d θ}{d x} = 0$ : $744 - 64 x - x^{2} = 0$ . 2. $x^{2} + 64 x - 744 = 0 \Rightarrow x = - 32 \pm 1768$ ; the solution in $[0, 20]$ is $x = 10.05$ m. 3. It is a maximum: the numerator $744 - 64 x - x^{2}$ is positive before $10.05$ and negative after it, and the denominator is positive, so $θ^{'}$ changes from $+$ to $-$ .

f) 1. $\frac{d θ}{d t} = \frac{d θ}{d x} \cdot \frac{d x}{d t}$ with $\frac{d x}{d t} = 0.5$ m s $^{- 1}$ (P moves towards the 13 m wall, so $x$ increases). 2. The midpoint is $x = 10$ : $\frac{d θ}{d x} = \frac{5 ( 744 - 640 - 100 )}{164 \cdot 269} = \frac{20}{44116} = 4.53 \cdot 1 0^{- 4}$ . 3. Therefore $\frac{d θ}{d t} = 4.53 \cdot 1 0^{- 4} \cdot 0.5 = 2.27 \cdot 1 0^{- 4}$ rad s $^{- 1}$ (positive: $θ$ increases).

Answer: a) $θ = arctan (\frac{5 x + 160}{x ^{2} - 20 x + 104})$ ; b) i) $0.994$ rad; ii) $1.19$ rad; e) $x = 10.05$ m; f) $2.27 \cdot 1 0^{- 4} rad s^{- 1}$

Exercise 193 HardIB · P2 TZ0 November 2014

Consider the triangle $PQR$ , where $QPR = 3 0^{\circ}$ , $PQ = (x + 2)$ cm and $PR = (5 - x)^{2}$ cm, where $- 2 < x < 5$ .

a) Show that the area of the triangle, $A cm^{2}$ , is given by $A = \frac{1}{4} (x^{3} - 8 x^{2} + 5 x + 50)$ .

b) i) State $\frac{d A}{d x}$ .

(ii) Verify that $\frac{d A}{d x} = 0$ for $x = \frac{1}{3}$ .

c) i) Find $\frac{d ^{2} A}{d x ^{2}}$ and, hence, justify that the maximum area of the triangle $PQR$ occurs when $x = \frac{1}{3}$ .

(ii) State the maximum area of the triangle $PQR$ .

(iii) Find $QR$ when the area of the triangle $PQR$ reaches its maximum value.

Show answer

Answer: b) i) $\frac{d A}{d x} = \frac{1}{4} (3 x^{2} - 16 x + 5)$ ; c) i) $\frac{d ^{2} A}{d x ^{2}} = \frac{1}{4} (6 x - 16)$ ; ii) $12.7 cm^{2}$ ; iii) $QR = 19.8$ cm

Show working and answer

a) 1. Area using $A = \frac{1}{2} PQ \cdot PR sin 3 0^{\circ} = \frac{1}{2} (x + 2) (5 - x)^{2} \cdot \frac{1}{2} = \frac{1}{4} (x + 2) (5 - x)^{2}$ . 2. Expanding: $(x + 2) (x^{2} - 10 x + 25) = x^{3} - 10 x^{2} + 25 x + 2 x^{2} - 20 x + 50 = x^{3} - 8 x^{2} + 5 x + 50$ . 3. Hence $A = \frac{1}{4} (x^{3} - 8 x^{2} + 5 x + 50)$ .

b) i) $\frac{d A}{d x} = \frac{1}{4} (3 x^{2} - 16 x + 5)$ .

(ii) For $x = \frac{1}{3}$ : $3 \cdot \frac{1}{9} - \frac{16}{3} + 5 = \frac{1}{3} - \frac{16}{3} + 5 = - 5 + 5 = 0$ . Hence $\frac{d A}{d x} = 0$ .

c) i) 1. $\frac{d ^{2} A}{d x ^{2}} = \frac{1}{4} (6 x - 16)$ . 2. At $x = \frac{1}{3}$ : $\frac{1}{4} (2 - 16) = - \frac{7}{2} < 0$ , so it is a local maximum. 3. Moreover, $\frac{d A}{d x} = 0$ is also solved at $x = 5$ , which is outside the interval $] - 2, 5 [$ , so on the interval the maximum is at $x = \frac{1}{3}$ .

(ii) $A (\frac{1}{3}) = \frac{1}{4} (\frac{1}{27} - \frac{8}{9} + \frac{5}{3} + 50) = \frac{1}{4} \cdot \frac{1372}{27} = \frac{343}{27} = 12.7 cm^{2}$ .

(iii) 1. For $x = \frac{1}{3}$ : $PQ = \frac{7}{3}$ cm and $PR = (\frac{14}{3})^{2} = \frac{196}{9}$ cm. 2. Cosine rule: $QR^{2} = PQ^{2} + PR^{2} - 2 PQ \cdot PR cos 3 0^{\circ} = \frac{49}{9} + \frac{38416}{81} - 2 \cdot \frac{7}{3} \cdot \frac{196}{9} \cdot \frac{3}{2}$ . 3. Numerically, $QR^{2} \approx 391.7$ , so $QR \approx 19.8$ cm.

Answer: b) i) $\frac{d A}{d x} = \frac{1}{4} (3 x^{2} - 16 x + 5)$ ; c) i) $\frac{d ^{2} A}{d x ^{2}} = \frac{1}{4} (6 x - 16)$ ; ii) $12.7 cm^{2}$ ; iii) $QR = 19.8$ cm

Exercise 194 HardIB · P2 TZ2 May 2014

A group of engineers needs to install pipes to connect two cities $A$ and $B$ that are separated by a river 450 metres wide, as shown in the following diagram.

Diagram of a river 450 m wide between two cities A and B: A on the lower bank, B on the upper bank (ground) 1,000 m away horizontally; the point X is on the upper bank at a distance x from E, the point on that bank opposite A; the pipes run from A to X through the river and from X to B along the ground

They plan to lay the pipes under the river between $A$ and $X$ , and underground between $X$ and $B$ . The cost of laying the pipes under the river is five times greater than the cost of laying the pipes underground. Let $EX = x$ . Let $k$ be the cost, in dollars per metre, of laying the pipes underground.

a) Show that the total cost $C$ , in dollars, of laying the pipes between $A$ and $B$ is given by $C = 5 k 202500 + x^{2} + (1000 - x) k$ .

b) i) Find $\frac{d C}{d x}$ .

(ii) Hence, find the value of $x$ for which the total cost is a minimum, and justify why this value gives a minimum.

c) Find the minimum total cost in terms of $k$ .

The angle formed by the pipes at the point where they meet is $AXB = θ$ .

d) Find $θ$ for the value of $x$ calculated in part (b).

For safety reasons, $θ$ must be at least $12 0^{\circ}$ . Given this new requirement,

e) i) find the new value of $x$ that minimises the total cost.

(ii) Find the percentage by which the minimum total cost has increased.

Show answer

Answer: b) i) $\frac{d C}{d x} = \frac{5 k x}{202500 + x ^{2}} - k$ ; ii) $x = 91.9$ m; c) $C = 3204.5 k$ ; d) $θ = 101. 5^{\circ}$ ; e) i) $x = 259.8$ m; ii) $4.17 %$

Show working and answer

a) 1. By Pythagoras in triangle $AEX$ : $AX = 45 0^{2} + x^{2} = 202500 + x^{2}$ (under the river, cost $5 k$ per metre). 2. The section $XB = 1000 - x$ lies underground (cost $k$ per metre). 3. Total cost: $C = 5 k 202500 + x^{2} + (1000 - x) k$ .

b) i) By the chain rule: $\frac{d C}{d x} = 5 k \cdot \frac{2 x}{2 202500 + x ^{2}} - k = \frac{5 k x}{202500 + x ^{2}} - k$ .

(ii) 1. $\frac{d C}{d x} = 0 \Rightarrow 5 x = 202500 + x^{2} \Rightarrow 25 x^{2} = 202500 + x^{2} \Rightarrow x^{2} = 8437.5$ . 2. Hence $x = 91.9$ m. 3. It is a minimum: $\frac{d ^{2} C}{d x ^{2}} = \frac{5 k \cdot 202500}{( 202500 + x ^{2} ) ^{3/2}} > 0$ (convex), or equivalently $C^{'} < 0$ for $x < 91.9$ and $C^{'} > 0$ for $x > 91.9$ .

c) $C_{m i n} = 5 k 202500 + 8437.5 + (1000 - 91.856) k = 2296.4 k + 908.1 k = 3204.5 k$ .

d) 1. Let $AXE$ be the angle at $X$ in triangle $AEX$ : $tan AXE = \frac{450}{91.856}$ , so $AXE = 78.4 6^{\circ}$ . 2. $E$ , $X$ and $B$ are collinear, so $θ = 18 0^{\circ} - 78.4 6^{\circ} = 101. 5^{\circ}$ .

e) i) 1. $θ \geq 12 0^{\circ}$ is equivalent to $AXE \leq 6 0^{\circ}$ , that is, $tan AXE = \frac{450}{x} \leq 3$ , or $x \geq \frac{450}{3} = 259.8$ . 2. Since $C$ is convex with its minimum at $91.9$ , for $x \geq 259.8$ the cost increases with $x$ , so under the new constraint the minimum is at $x = 259.8$ m.

(ii) 1. $C (259.8) = 5 k 202500 + 67500 + (1000 - 259.8) k = 2598.1 k + 740.2 k = 3338.3 k$ . 2. Increase: $\frac{3338.27 - 3204.54}{3204.54} = 0.0417$ , that is, $4.17 %$ .

Answer: b) i) $\frac{d C}{d x} = \frac{5 k x}{202500 + x ^{2}} - k$ ; ii) $x = 91.9$ m; c) $C = 3204.5 k$ ; d) $θ = 101. 5^{\circ}$ ; e) i) $x = 259.8$ m; ii) $4.17 %$

Exercise 195 HardIB · P1 TZ2 May 2019

A right circular cone of radius $r$ is inscribed in a sphere of centre O and radius $R$ , as shown in the following diagram. The perpendicular height of the cone is $h$ , X denotes the centre of the base and B is a point where the cone touches the sphere.

Cross-section through the axis of a cone inscribed in a sphere: circle of centre O and radius R, triangle with the vertex above and the base below O, height h, base radius r and segment OB equal to R

a) Show that the volume of the cone can be expressed as $V = \frac{π}{3} (2 R h^{2} - h^{3})$ .

b) Given that there exists an inscribed cone with maximum volume, show that the volume of this cone is $\frac{32 π R ^{3}}{81}$ .

Show answer

Answer: a) $V = \frac{π}{3} (2 R h^{2} - h^{3})$ ; b) maximum for $h = \frac{4 R}{3}$ , with $V = \frac{32 π R ^{3}}{81}$ .

Show working and answer

a) 1. $O$ lies on the height. The vertex of the cone is at distance $R$ from $O$ (the radius of the sphere), so $OX = h - R$ (in absolute value, $∣ h - R ∣$ ). 2. In the right-angled triangle $OXB$ (right angle at $X$ ): $r^{2} + (h - R)^{2} = R^{2}$ . 3. Rearranging: $r^{2} = R^{2} - (h - R)^{2} = 2 R h - h^{2}$ . 4. Volume of the cone: $V = \frac{1}{3} π r^{2} h = \frac{π}{3} (2 R h - h^{2}) h = \frac{π}{3} (2 R h^{2} - h^{3})$ .

b) 1. Differentiating with respect to $h$ (with $R$ constant): $\frac{d V}{d h} = \frac{π}{3} (4 R h - 3 h^{2}) = \frac{π}{3} h (4 R - 3 h)$ . 2. Setting this equal to zero with $h > 0$ : $h = \frac{4 R}{3}$ . 3. It is a maximum: $\frac{d ^{2} V}{d h ^{2}} = \frac{π}{3} (4 R - 6 h) = \frac{π}{3} (4 R - 8 R) < 0$ at $h = \frac{4 R}{3}$ . 4. Maximum value: $V = \frac{π}{3} (2 R \cdot \frac{16 R ^{2}}{9} - \frac{64 R ^{3}}{27}) = \frac{π}{3} \cdot \frac{96 R ^{3} - 64 R ^{3}}{27} = \frac{π}{3} \cdot \frac{32 R ^{3}}{27} = \frac{32 π R ^{3}}{81}$ .

Answer: a) $V = \frac{π}{3} (2 R h^{2} - h^{3})$ ; b) maximum for $h = \frac{4 R}{3}$ , with $V = \frac{32 π R ^{3}}{81}$ .

Exercise 196 HardIB · P2 May 1996

(i) Line L passes through the point $P (2, 3, 1)$ and has direction $u = i + 2 j - 3 k$ . A second line M passes through the point $Q (4, 2, 0)$ and has direction $v = 3 i - j + k$ .

a) Calculate, in parametric form and using the parameter $t$ , the equation of the line M.

b) Calculate the vector $w = RP$ , where R is any point on the line M.

c) Calculate the vector (cross) product $u \times w$ and, using the above, express $∣ u \times w ∣$ in terms of $t$ .

d) Deduce that $∣ u \times w ∣$ is a minimum when $t = - \frac{3}{5}$ , and calculate the minimum value.

Show answer

Answer: a) $x = 4 + 3 t$ , $y = 2 - t$ , $z = t$ ; b) $w = - 2 - 3 t 1 + t 1 - t$ ; c) $u \times w = 5 + t 5 + 10 t 5 + 7 t$ , $∣ u \times w ∣ = 150 t^{2} + 180 t + 75$ ; d) minimum $21$ for $t = - \frac{3}{5}$

Show working and answer

(i) a) 1. The line M passes through $Q (4, 2, 0)$ with direction $v = (3, - 1, 1)$ . 2. In parametric form: $x = 4 + 3 t$ , $y = 2 - t$ , $z = t$ .

b) 1. A general point on M is $R (4 + 3 t, 2 - t, t)$ . 2. $w = RP = P - R = (2 - 4 - 3 t, 3 - 2 + t, 1 - t) = - 2 - 3 t 1 + t 1 - t$ .

c) 1. Vector (cross) product with $u = (1, 2, - 3)$ : first component $2 (1 - t) - (- 3) (1 + t) = 5 + t$ ; second $(- 3) (- 2 - 3 t) - 1 \cdot (1 - t) = 5 + 10 t$ ; third $1 \cdot (1 + t) - 2 (- 2 - 3 t) = 5 + 7 t$ . 2. Thus $u \times w = 5 + t 5 + 10 t 5 + 7 t$ . 3. Its modulus squared: $(5 + t)^{2} + (5 + 10 t)^{2} + (5 + 7 t)^{2} = 150 t^{2} + 180 t + 75$ , and therefore $∣ u \times w ∣ = 150 t^{2} + 180 t + 75$ .

d) 1. $∣ u \times w ∣$ is a minimum when its square $h (t) = 150 t^{2} + 180 t + 75$ is a minimum. 2. $h^{'} (t) = 300 t + 180 = 0 \Rightarrow t = - \frac{180}{300} = - \frac{3}{5}$ , and $h^{''} = 300 > 0$ , so it is a minimum. 3. $h (- \frac{3}{5}) = 150 \cdot \frac{9}{25} - 108 + 75 = 54 - 108 + 75 = 21$ . 4. The minimum value of $∣ u \times w ∣$ is $21$ .

Exercise 197 HardIB · P2 TZ1 May 2018

Two submarines A and B have their routes planned so that their positions at time $t$ hours, $0 \leq t < 20$ , are given by the position vectors

$r_{A} = 24 - 1 + t - 1 1 - 0.15$ and $r_{B} = 0 3.2 - 2 + t - 0.5 1.2 0.1$

relative to a fixed point on the surface of the ocean (all lengths are in kilometres).

a) Show that the two submarines would collide at a point P, and write down the coordinates of P.

To avoid the collision, submarine B adjusts its speed so that its position vector is now given by $r_{B} = 0 3.2 - 2 + t - 0.45 1.08 0.09$ .

b) (i) Show that submarine B travels in the same direction as originally planned.

(ii) Find the value of $t$ when submarine B passes through P.

c) (i) Find an expression for the distance between the two submarines in terms of $t$ .

(ii) Find the value of $t$ when the two submarines are closest to each other.

(iii) Find the distance between the two submarines at this time.

Show answer

Answer: a) $P (- 2, 8, - 1.6)$ ; b) (i) the direction vectors are proportional (the new one is $0.9$ times the original); (ii) $t = \frac{40}{9}$ h; c) (i) $D = 5.64 - 2.808 t + 0.3665 t^{2}$ ; (ii) $t \approx 3.83$ h; (iii) $D \approx 0.511$ km

Show working and answer

a) 1. We equate the coordinates of $r_{A}$ and $r_{B}$ at the same time $t$ (for them to collide they must coincide at the same moment):

$x$ -coordinate: $2 - t = - 0.5 t \Rightarrow t = 4$ .
$y$ -coordinate: $4 + t = 3.2 + 1.2 t \Rightarrow 0.8 = 0.2 t \Rightarrow t = 4$ .
$z$ -coordinate: $- 1 - 0.15 t = - 2 + 0.1 t \Rightarrow 1 = 0.25 t \Rightarrow t = 4$ .

All three equations give the same value $t = 4$ (which lies in $0 \leq t < 20$ ), so the submarines coincide at that instant: they would collide.
We substitute $t = 4$ into $r_{A}$ : $(2 - 4, 4 + 4, - 1 - 0.6)$ , that is, $P (- 2, 8, - 1.6)$ .

b) (i) 1. The new direction vector is $- 0.45 1.08 0.09$ . 2. Comparing with the original: $- 0.45 1.08 0.09 = 0.9 - 0.5 1.2 0.1$ . 3. It is a scalar multiple of the original vector, so the vectors are parallel and B follows the same direction.

(ii) 1. B passes through P when $r_{B} = (- 2, 8, - 1.6)$ . Using the $x$ -coordinate: $- 0.45 t = - 2$ . 2. Solving: $t = \frac{2}{0.45} = \frac{40}{9} \approx 4.44$ h. 3. Check with $y$ : $3.2 + 1.08 \cdot \frac{40}{9} = 3.2 + 4.8 = 8$ , and with $z$ : $- 2 + 0.09 \cdot \frac{40}{9} = - 1.6$ . Correct.

c) (i) 1. The vector joining A to B is $r_{A} - r_{B} = 2 - 0.55 t 0.8 - 0.08 t 1 - 0.24 t$ (now using the new path of B). 2. The distance is the magnitude of this vector: $D = (2 - 0.55 t)^{2} + (0.8 - 0.08 t)^{2} + (1 - 0.24 t)^{2}$ . 3. Expanding and collecting terms: $D = 5.64 - 2.808 t + 0.3665 t^{2}$ .

(ii) 1. $D$ is minimal when $D^{2} = 5.64 - 2.808 t + 0.3665 t^{2}$ is minimal (the square root is increasing). 2. This is a parabola with positive coefficient of $t^{2}$ , so the minimum is at the vertex: $\frac{d ( D ^{2} )}{d t} = 0.733 t - 2.808 = 0$ . 3. Solving: $t = \frac{2.808}{0.733} \approx 3.83$ h (within the interval $0 \leq t < 20$ ).

(iii) 1. Substitute $t = 3.83$ (using the GDC or a calculator, with the unrounded value): $D^{2} \approx 0.2615$ . 2. Hence $D \approx 0.511$ km.

Exercise 198 HardIB · P1 TZ2 May 2017

A window is constructed in the shape of a rectangle with a semicircle of radius $r$ metres on top, as shown in the figure. The perimeter of the window is constant and equal to $P$ metres.

Window formed by a rectangle taller than it is wide, with a semicircle on top resting on the full width of the rectangle

a) (i) Find the area of the window in terms of $P$ and $r$ .

(ii) Find the width of the window in terms of $P$ when the area reaches a maximum value, and justify why this is a maximum.

b) Show that in this case the height of the rectangle is equal to the radius of the semicircle.

Show answer

Answer: a) (i) $A = P r - 2 r^{2} - \frac{π r ^{2}}{2}$ ; (ii) width $= \frac{2 P}{π + 4}$ (a maximum, since $A^{''} < 0$ ); b) $h = r$

Show working and answer

a) (i) 1. Let $h$ be the height of the rectangle. The width of the rectangle is the diameter of the semicircle, $2 r$ . 2. The perimeter of the window consists of the base ( $2 r$ ), the two sides ( $2 h$ ) and the arc of the semicircle ( $π r$ ); the top side of the rectangle does not count: $P = 2 r + 2 h + π r$ . 3. Solving for $h$ : $h = \frac{P - 2 r - π r}{2}$ . 4. Area: rectangle plus semicircle, $A = 2 r h + \frac{π r ^{2}}{2}$ . 5. Substituting $h$ : $A = r (P - 2 r - π r) + \frac{π r ^{2}}{2} = P r - 2 r^{2} - \frac{π r ^{2}}{2}$ .

(ii) 1. We differentiate with respect to $r$ (with $P$ constant): $\frac{d A}{d r} = P - 4 r - π r$ . 2. We set this equal to zero: $r = \frac{P}{4 + π}$ . 3. The width is $2 r = \frac{2 P}{π + 4}$ . 4. Second derivative: $\frac{d ^{2} A}{d r ^{2}} = - (4 + π) < 0$ , so it is a maximum. (Also: $A$ is a parabola in $r$ with negative coefficient of $r^{2}$ .)

b) 1. Substituting $P = (4 + π) r$ into $h = \frac{P - 2 r - π r}{2}$ : $h = \frac{( 4 + π ) r - 2 r - π r}{2} = \frac{2 r}{2} = r$ . 2. Hence the height of the rectangle is equal to the radius of the semicircle.

Answer: a) (i) $A = P r - 2 r^{2} - \frac{π r ^{2}}{2}$ ; (ii) width $= \frac{2 P}{π + 4}$ (a maximum, since $A^{''} < 0$ ); b) $h = r$

Exercise 199 HardIB · P2 TZ2 May 2016

Points A, B and T lie on a line of a futsal pitch.

The goal, [AB], is $2$ metres wide. A player situated at point P kicks the ball in the direction of the goal. [PT] is perpendicular to (AB) and is $6$ metres from a parallel line that passes through the centre of [AB]. Let PT be $x$ metres and let $α = APB$ , measured in degrees. Assume that the ball travels along the floor.

Top view of the 2 m goal AB, point T on the same line, point P situated x m from T on the perpendicular, the 6 m dimension and the angle alpha subtended by the goal at P

a) Find the value of $α$ when $x = 10$ .

b) Show that $tan α = \frac{2 x}{x ^{2} + 35}$ .

The value of $α$ is a maximum when the value of $tan α$ is a maximum.

c) (i) Find $\frac{d}{d x} (tan α)$ .

(ii) Hence, or otherwise, find the value of $α$ such that $\frac{d}{d x} (tan α) = 0$ .

(iii) Find $\frac{d ^{2}}{d x ^{2}} (tan α)$ and, hence, show that the value of $α$ never exceeds $1 0^{\circ}$ .

d) Find the set of values of $x$ for which $α \geq 7^{\circ}$ .

Show answer

Answer: a) $α \approx 8.4 3^{\circ}$ ; b) proved; c) (i) $\frac{70 - 2 x ^{2}}{( x ^{2} + 35 ) ^{2}}$ ; (ii) $α \approx 9.5 9^{\circ}$ ; (iii) $\frac{4 x ( x ^{2} - 105 )}{( x ^{2} + 35 ) ^{3}}$ , negative at $x = 35$ , so $α$ is a maximum and does not exceed $1 0^{\circ}$ ; d) $x \in [2.55; 13.7]$

Show working and answer

a) 1. With $x = 10$ : $tan α = \frac{2 \cdot 10}{1 0 ^{2} + 35} = \frac{20}{135}$ (formula from part (b)). 2. Using the GDC: $α = arctan \frac{20}{135} \approx 8.4 3^{\circ}$ .

b) 1. Let M be the midpoint of [AB]. P is $6$ m from the line through M, so $TM = 6$ . Since $AB = 2$ , we have $TB = 5$ and $TA = 7$ . 2. In the right-angled triangles PTA and PTB: $tan APT = \frac{7}{x}$ and $tan BPT = \frac{5}{x}$ . 3. The angle is $α = APT - BPT$ , so $tan α = \frac{\frac{7}{x} - \frac{5}{x}}{1 + \frac{7}{x} \cdot \frac{5}{x}} = \frac{\frac{2}{x}}{1 + \frac{35}{x ^{2}}}$ . 4. Multiplying numerator and denominator by $x^{2}$ : $tan α = \frac{2 x}{x ^{2} + 35}$ .

c) (i) 1. Quotient rule with $u = 2 x$ and $v = x^{2} + 35$ : $\frac{d}{d x} (tan α) = \frac{2 ( x ^{2} + 35 ) - 2 x \cdot 2 x}{( x ^{2} + 35 ) ^{2}}$ . 2. Simplifying: $\frac{70 - 2 x ^{2}}{( x ^{2} + 35 ) ^{2}}$ .

(ii) 1. Equating to zero: $70 - 2 x^{2} = 0$ , and with $x > 0$ : $x = 35$ . 2. Then $tan α = \frac{2 35}{35 + 35} = \frac{1}{35}$ . 3. $α = arctan \frac{1}{35} \approx 9.5 9^{\circ}$ .

(iii) 1. Differentiating again (quotient rule): $\frac{d ^{2}}{d x ^{2}} (tan α) = \frac{- 4 x ( x ^{2} + 35 ) ^{2} - ( 70 - 2 x ^{2} ) \cdot 2 ( x ^{2} + 35 ) \cdot 2 x}{( x ^{2} + 35 ) ^{4}}$ . 2. Simplifying by dividing by $(x^{2} + 35)$ : $\frac{- 4 x ( x ^{2} + 35 ) - 4 x ( 70 - 2 x ^{2} )}{( x ^{2} + 35 ) ^{3}} = \frac{4 x ^{3} - 420 x}{( x ^{2} + 35 ) ^{3}} = \frac{4 x ( x ^{2} - 105 )}{( x ^{2} + 35 ) ^{3}}$ . 3. At $x = 35$ : $x^{2} - 105 = - 70 < 0$ , so the second derivative is negative and $tan α$ has a maximum ( $tan α$ changes from increasing to decreasing; it is the only critical point with $x > 0$ ). 4. Since $α$ is acute, $α$ is a maximum when $tan α$ is, and that maximum is $9.5 9^{\circ} < 1 0^{\circ}$ . Therefore $α$ never exceeds $1 0^{\circ}$ .

d) 1. $α \geq 7^{\circ} ⟺ tan α \geq tan 7^{\circ}$ , i.e. $\frac{2 x}{x ^{2} + 35} \geq tan 7^{\circ} \approx 0.1228$ . 2. Using the GDC we solve $\frac{2 x}{x ^{2} + 35} = tan 7^{\circ}$ (equivalent to $0.1228 x^{2} - 2 x + 4.298 = 0$ ): $x \approx 2.55$ and $x \approx 13.7$ . 3. As $tan α$ increases up to $x = 35 \approx 5.92$ and decreases afterwards, the inequality holds between both solutions. 4. $2.55 \leq x \leq 13.7$ .

Exercise 200 HardIB · P2 TZ2 May 2005

The following figure shows a trapezium $OABC$ where $OA$ is parallel to $CB$ . $O$ is the centre of a circle of radius $r$ cm. $A$ , $B$ and $C$ lie on the circumference. The angle $OCB = θ$ .

Circle with centre O and radius r cm, with the trapezium OABC inscribed: OA is parallel to CB (marked with arrows), A, B and C are on the circumference, and the angle OCB is θ

Let $T$ be the area of the trapezium $OABC$ .

a) Show that $T = \frac{r ^{2}}{2} (sin θ + sin 2 θ)$ .

For a fixed value of $r$ , the value of $T$ changes as $θ$ changes.

b) Verify that $T$ takes its maximum value when $θ$ satisfies the equation $4 cos^{2} θ + cos θ - 2 = 0$ , and verify that this value of $T$ is a maximum.

c) Given that the perimeter of the trapezium is 75 cm, find the maximum value of $T$ .

Show answer

Answer: c) $T_{m a x} \approx 296 c m^{2}$ (more exactly $296.2 c m^{2}$ )

Show working and answer

a) 1. Since $OC = OB = r$ (radii), triangle $OCB$ is isosceles and $OBC = θ$ . 2. The height of triangle $OCB$ from $O$ (also that of the trapezium, since $OA ∥ CB$ ) is $h = r sin θ$ , and the base is $CB = 2 r cos θ$ . 3. Area of the trapezium with parallel sides $OA = r$ and $CB = 2 r cos θ$ : $T = \frac{1}{2} (r + 2 r cos θ) r sin θ = \frac{r ^{2}}{2} (sin θ + 2 sin θ cos θ)$ . 4. Using $sin 2 θ = 2 sin θ cos θ$ : $T = \frac{r ^{2}}{2} (sin θ + sin 2 θ)$ .

b) 1. For fixed $r$ , differentiate with respect to $θ$ : $\frac{d T}{d θ} = \frac{r ^{2}}{2} (cos θ + 2 cos 2 θ)$ . 2. Set this equal to $0$ and use $cos 2 θ = 2 cos^{2} θ - 1$ : $cos θ + 4 cos^{2} θ - 2 = 0$ , that is, $4 cos^{2} θ + cos θ - 2 = 0$ . 3. Solve: $cos θ = \frac{- 1 \pm 33}{8}$ . For $0 < θ < \frac{π}{2}$ the cosine is positive, so $cos θ = \frac{- 1 + 33}{8} \approx 0.5931$ and $θ \approx 0.936$ rad ( $53. 6^{\circ}$ ). 4. Second derivative: $\frac{d ^{2} T}{d θ ^{2}} = \frac{r ^{2}}{2} (- sin θ - 4 sin 2 θ)$ . For $0 < θ < \frac{π}{2}$ both $sin θ$ and $sin 2 θ$ are positive, so this is negative: the critical point is a maximum.

c) 1. Sides of the trapezium: $OA = OC = r$ , $CB = 2 r cos θ$ and $AB = 2 r sin \frac{θ}{2}$ (chord subtending the central angle $θ$ at $AB$ , since $AOB = θ$ by alternate interior angles with $OA ∥ CB$ ). 2. Perimeter: $r (2 + 2 cos θ + 2 sin \frac{θ}{2}) = 75$ . 3. With $θ = 0.9359$ rad (the value from part b): $r (2 + 1.1861 + 0.9022) = 75$ , so $r \approx 18.345$ cm. 4. Maximum area: $T = \frac{r ^{2}}{2} (sin θ + sin 2 θ) = \frac{18.34 5 ^{2}}{2} (0.8052 + 0.9551) \approx 296.2 c m^{2}$ .

Answer: c) $T_{m a x} \approx 296 c m^{2}$ (more exactly $296.2 c m^{2}$ )

Exercise 201 HardIB · P2 TZ0 November 2005

A man PF is standing on horizontal ground at F, a distance $x$ from the base of a vertical wall GE. The man observes the picture AB on the wall. The angle BPA is $θ$ .

A man PF standing on the ground at F, at distance x from the vertical wall GE; on the wall, a picture AB above point D (level with P); θ is the angle BPA

Let $DA = a$ and $DB = b$ , where the angle $PDE$ is a right angle. Find the value of $x$ for which $tan θ$ is a maximum; give your answer in terms of $a$ and $b$ . Justify that this value of $x$ gives a maximum value of $tan θ$ .

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Answer: $x = ab$ (maximum of $tan θ$ )

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As $P$ and $D$ are at the same height, $PD = x$ (horizontal) and $PD ⊥ DE$ . With $DA = a$ and $DB = b$ ( $b > a$ ), the angles that the lines of sight make with the horizontal satisfy $tan DPA = \frac{a}{x}$ and $tan DPB = \frac{b}{x}$ .
Since $θ = DPB - DPA$ , we use the tangent of a difference: $tan θ = \frac{\frac{b}{x} - \frac{a}{x}}{1 + \frac{b}{x} \cdot \frac{a}{x}} = \frac{( b - a ) x}{x ^{2} + ab}$ .
Differentiating with the quotient rule: $\frac{d}{d x} (tan θ) = \frac{( b - a ) ( x ^{2} + ab ) - ( b - a ) x \cdot 2 x}{( x ^{2} + ab ) ^{2}} = \frac{( b - a ) ( ab - x ^{2} )}{( x ^{2} + ab ) ^{2}}$ .
Setting this equal to $0$ with $x > 0$ : $ab - x^{2} = 0$ , so $x = ab$ .
Justification that it is a maximum: as $b - a > 0$ and the denominator is positive, the sign of the derivative is that of $ab - x^{2}$ . It is positive for $x < ab$ and negative for $x > ab$ , so $tan θ$ changes from increasing to decreasing: a maximum.

Answer: $x = ab$ (maximum of $tan θ$ )

Exercise 202 HardIB · P2 TZ2 May 2007

A television screen $BC$ of height one metre is mounted on a wall. The bottom of the television screen, $B$ , is situated one metre above the eye level of an observer. The angles of elevation $(AOC, AOB)$ from the observer's eye level, $O$ , to the top and bottom of the television screen are $α$ and $β$ radians respectively. The horizontal distance from the observer's eye to the wall on which the television screen is mounted is equal to $x$ metres.

The observer's angle of view $(BOC)$ is $θ$ radians, as shown below.

Observer at O, vertical wall at x metres, with screen BC of height 1 m and B 1 m above A; sight lines OB and OC (dashed) with the angles α = AOC, β = AOB and θ = BOC

a) (i) Verify that $θ = arctan \frac{2}{x} - arctan \frac{1}{x}$ .

(ii) Hence, or otherwise, find the exact value of $x$ for which $θ$ is a maximum, and justify why this $x$ gives a maximum of $θ$ .

(iii) Find the maximum value of $θ$ .

b) Find where the observer should stand so that the angle of view is equal to $1 5^{\circ}$ .

Show answer

Answer: a) (ii) $x = 2$ m, maximum; (iii) $θ_{m a x} \approx 0.34$ rad; b) $x \approx 0.65$ m or $x \approx 3.08$ m

Show working and answer

a) (i) 1. In the right-angled triangle with vertex $O$ and the wall, $tan α = \frac{2}{x}$ (height of $C$ above eye level: 2 m) and $tan β = \frac{1}{x}$ (height of $B$ : 1 m). 2. The angles are acute, so $α = arctan \frac{2}{x}$ and $β = arctan \frac{1}{x}$ . 3. As $θ = α - β$ : $θ = arctan \frac{2}{x} - arctan \frac{1}{x}$ .

(ii) 1. Differentiate using $\frac{d}{d x} arctan u = \frac{u ^{'}}{1 + u ^{2}}$ : $\frac{d θ}{d x} = \frac{- 2/ x ^{2}}{1 + 4/ x ^{2}} - \frac{- 1/ x ^{2}}{1 + 1/ x ^{2}} = \frac{- 2}{x ^{2} + 4} + \frac{1}{x ^{2} + 1}$ . 2. Common denominator: $\frac{d θ}{d x} = \frac{- 2 ( x ^{2} + 1 ) + ( x ^{2} + 4 )}{( x ^{2} + 4 ) ( x ^{2} + 1 )} = \frac{2 - x ^{2}}{( x ^{2} + 4 ) ( x ^{2} + 1 )}$ . 3. Setting this equal to $0$ with $x > 0$ : $x^{2} = 2$ , so $x = 2$ . 4. Justification: the denominator is positive, so the sign of $\frac{d θ}{d x}$ is that of $2 - x^{2}$ : positive if $x < 2$ and negative if $x > 2$ . $θ$ changes from increasing to decreasing: maximum at $x = 2$ .

(iii) 1. $θ_{m a x} = arctan \frac{2}{2} - arctan \frac{1}{2} = arctan 2 - arctan \frac{1}{2}$ . 2. Using the GDC (in radians): $θ_{m a x} \approx 0.9553 - 0.6155 \approx 0.340$ rad (equal to $arctan \frac{2}{4}$ ).

b) 1. Combining using the tangent of a difference: $tan θ = \frac{\frac{2}{x} - \frac{1}{x}}{1 + \frac{2}{x ^{2}}} = \frac{x}{x ^{2} + 2}$ . 2. We set $θ = 1 5^{\circ}$ (which is less than the maximum of $\approx 19. 5^{\circ}$ , so there will be two solutions): $\frac{x}{x ^{2} + 2} = tan 1 5^{\circ} = 2 - 3 \approx 0.2679$ . 3. Quadratic equation: $0.2679 x^{2} - x + 0.5359 = 0$ . 4. Solving: $x \approx 0.649$ or $x \approx 3.083$ . 5. The observer must stand approximately $0.65$ m or $3.08$ m from the wall.

Answer: a) (ii) $x = 2$ m, maximum; (iii) $θ_{m a x} \approx 0.34$ rad; b) $x \approx 0.65$ m or $x \approx 3.08$ m

Exercise 203 HardIB · P1 TZ0 November 2011

At 12:00 a boat is located exactly south of a cargo ship, 20 km from it. The boat sails due east at 20 $km h^{- 1}$ , and the cargo ship sails due south at 40 $km h^{- 1}$ .

a) Determine the time at which these two vessels are closest to each other, and justify your answer.

b) If the visibility at sea is 9 $km h^{- 1}$ , determine whether at any moment the captain of either vessel is able to see the other vessel.

Show answer

Answer: a) at 12:24; b) the minimum distance is $80 \approx 8.94 km < 9 km$ , so they can see each other.

Show working and answer

a) 1. We take as origin the initial position of the cargo ship, with east ( $x$ ) and north ( $y$ ) axes. At 12:00 the boat is at $(0, - 20)$ . 2. After $t$ hours the boat (eastwards at 20 km/h) is at $(20 t, - 20)$ and the cargo ship (southwards at 40 km/h) is at $(0, - 40 t)$ . 3. Difference of positions: $(20 t, 40 t - 20)$ . Squared distance: $d^{2} = (20 t)^{2} + (40 t - 20)^{2} = 2000 t^{2} - 1600 t + 400$ . 4. Minimising $d$ is equivalent to minimising $d^{2}$ . Differentiating: $\frac{d}{d t} (d^{2}) = 4000 t - 1600 = 0$ , so $t = 0.4$ h $= 24$ min. 5. It is a minimum because the second derivative is $4000 > 0$ (parabola opening upwards). Time: 12:24.

b) 1. Minimum distance: $d^{2} = 2000 (0.16) - 1600 (0.4) + 400 = 320 - 640 + 400 = 80$ , so $d_{m i n} = 80 \approx 8.94$ km. 2. As $8.94 < 9$ , at that instant the distance is less than the visibility: the captains can indeed see each other.

Answer: a) at 12:24; b) the minimum distance is $80 \approx 8.94 km < 9 km$ , so they can see each other.

Exercise 204 HardIB · P2 TZ1 May 2010

Points $A$ , $B$ and $C$ lie on the circumference of a circle, with centre $O$ and radius $r$ .

The trapezium $OABC$ is formed such that $AB$ is parallel to $OC$ , and the angle $AOC$ is $θ$ , $\frac{π}{2} \leq θ < π$ .

Circle with centre O and radius r, with three points A, B and C on it; trapezium OABC has AB parallel to OC and angle θ at O between OA and OC

a) Show that the angle $BOC$ is $π - θ$ .

b) Show that the area $T$ of the trapezium can be expressed as $T = \frac{1}{2} r^{2} sin θ - \frac{1}{2} r^{2} sin 2 θ$ .

c) (i) Show that when the area is a maximum, the value of $θ$ satisfies $cos θ = 2 cos 2 θ$ .

(ii) Hence, determine the maximum area of the trapezium when $r = 1$ . (Note: It is not necessary to prove that it is a maximum.)

Show answer

Answer: a) and b) proofs; c) (i) proof; (ii) $T = 0.88$

Show working and answer

a) 1. $OA = OB = OC = r$ (they are radii), so triangle $OAB$ is isosceles with $OA = OB$ . 2. $AB ∥ OC$ and $OA$ is a transversal: the angles $OAB$ and $AOC$ are supplementary (co-interior angles), so $OAB = π - θ$ . 3. Since the triangle is isosceles, $OBA = OAB = π - θ$ . 4. $AB ∥ OC$ and $OB$ is a transversal: the alternate interior angles $ABO$ and $BOC$ are equal. Therefore $BOC = π - θ$ .

b) 1. The diagonal $OB$ divides the trapezium into the triangles $OAB$ and $OBC$ . 2. Triangle $OBC$ : two sides $r$ and included angle $π - θ$ : $Area = \frac{1}{2} r^{2} sin (π - θ) = \frac{1}{2} r^{2} sin θ .$ 3. Triangle $OAB$ : the angle at $O$ is $AOB = θ - (π - θ) = 2 θ - π$ , and $sin (2 θ - π) = - sin 2 θ$ : $Area = \frac{1}{2} r^{2} sin (2 θ - π) = - \frac{1}{2} r^{2} sin 2 θ .$ 4. Adding: $T = \frac{1}{2} r^{2} sin θ - \frac{1}{2} r^{2} sin 2 θ .$

c) (i) 1. The area is a maximum where $\frac{d T}{d θ} = 0$ . 2. We differentiate: $\frac{d T}{d θ} = \frac{1}{2} r^{2} cos θ - \frac{1}{2} r^{2} \cdot 2 cos 2 θ = \frac{1}{2} r^{2} (cos θ - 2 cos 2 θ)$ . 3. Setting this equal to $0$ (with $r > 0$ ): $cos θ = 2 cos 2 θ$ .

(ii) 1. With $cos 2 θ = 2 cos^{2} θ - 1$ the condition becomes $cos θ = 4 cos^{2} θ - 2$ , i.e. $4 cos^{2} θ - cos θ - 2 = 0$ . 2. We solve the quadratic equation in $cos θ$ : $cos θ = \frac{1 \pm 33}{8}$ . 3. Since $\frac{π}{2} \leq θ < π$ , $cos θ \leq 0$ , so $cos θ = \frac{1 - 33}{8} \approx - 0.5931$ , and $θ \approx 2.2057$ . 4. With $r = 1$ and the GDC (or calculator): $T = \frac{1}{2} sin θ - \frac{1}{2} sin 2 θ \approx 0.880$ .

Answer: a) and b) proofs; c) (i) proof; (ii) $T = 0.88$

Exercise 205 HardIB · P1 TZ0 November 2011

The following diagram shows a circular lake with centre $O$ , diameter $AB$ and radius 2 km.

Circular lake with centre O and diameter AB: rowing path from A to a point P on the shore and then walking along the arc from P to B, with angle θ at A and angle 2θ at O

Jorg needs to travel from $A$ to $B$ as quickly as possible. He plans to row to the point $P$ and then walk from there to the point $B$ . Jorg can row at 3 $km h^{- 1}$ and walk at 6 $km h^{- 1}$ . Let $PAB = θ$ radians, and let $t$ be the time, in hours, taken by Jorg to travel from $A$ to $B$ .

a) Verify that $t = \frac{2}{3} (2 cos θ + θ)$ .

b) Find the value of $θ$ for which $\frac{d t}{d θ} = 0$ .

c) Which route should Jorg take to reach $B$ from $A$ in the least possible time? Justify your answer.

Show answer

Answer: a) proof; b) $θ = \frac{π}{6}$ ; c) walk from $A$ towards $B$ along the shore ( $t = \frac{π}{3} \approx 1.05$ h).

Show working and answer

a) 1. The angle $APB$ is inscribed in a semicircle (the diameter $AB = 4$ km), so it equals $9 0^{\circ}$ . In the right-angled triangle $APB$ : $AP = AB cos θ = 4 cos θ$ . 2. Time rowing (speed $3$ km/h): $t_{1} = \frac{4 cos θ}{3}$ . 3. The central angle $POB$ is twice the inscribed angle $θ$ : $2 θ$ . The arc $PB$ measures $ℓ = r \cdot 2 θ = 2 \cdot 2 θ = 4 θ$ km. 4. Time walking (speed $6$ km/h): $t_{2} = \frac{4 θ}{6} = \frac{2 θ}{3}$ . 5. Adding: $t = \frac{4 cos θ}{3} + \frac{2 θ}{3} = \frac{2}{3} (2 cos θ + θ) .$

b) 1. We differentiate with respect to $θ$ : $\frac{d t}{d θ} = \frac{2}{3} (- 2 sin θ + 1)$ . 2. Setting this equal to $0$ : $sin θ = \frac{1}{2}$ . Since $0 \leq θ \leq \frac{π}{2}$ , $θ = \frac{π}{6}$ .

c) 1. $θ$ varies over $[0, \frac{π}{2}]$ . We compare the value of $t$ at the critical point and at the ends of the interval:

$θ = 0$ ( $P = B$ , all rowing): $t = \frac{2}{3} \cdot 2 = \frac{4}{3} \approx 1.33$ h.
$θ = \frac{π}{6}$ : $t = \frac{2}{3} (3 + \frac{π}{6}) \approx 1.50$ h.
$θ = \frac{π}{2}$ ( $P = A$ , all walking along the shore): $t = \frac{2}{3} \cdot \frac{π}{2} = \frac{π}{3} \approx 1.05$ h.

Also $\frac{d ^{2} t}{d θ ^{2}} = - \frac{4}{3} cos θ < 0$ at $θ = \frac{π}{6}$ : this is a maximum (the worst route), not a minimum.
The least time is obtained at the end point $θ = \frac{π}{2}$ : Jorg should walk from $A$ towards $B$ along the shore, without rowing.

Answer: a) proof; b) $θ = \frac{π}{6}$ ; c) walk from $A$ towards $B$ along the shore ( $t = \frac{π}{3} \approx 1.05$ h).

Exercise 206 HardIB · P2 TZ0 November 2012

The diagram shows the plan of an art gallery of width $a$ metres. $[AB]$ represents a door, which leads to an exit corridor of width $b$ metres. In order to remove a painting from the art gallery, $CD$ (denoted by $L$ ) is measured for various values of $α$ , as shown in the diagram.

Plan of a rectangular gallery of width a with a door AB leading to a corridor of width b; the segment CD passes through the corner A and forms an angle α with the gallery wall

a) If $α$ is the angle between $[CD]$ and the wall, show that $L = \frac{a}{sin α} + \frac{b}{cos α}$ , $0 < α < \frac{π}{2}$ .

b) If $a = 5$ and $b = 1$ , find the maximum length of a painting that can be removed through this door.

Let $a = 3 k$ and $b = k$ .

c) Calculate $\frac{d L}{d α}$ .

d) Find, in terms of $k$ , the maximum length of a painting that can be removed from the gallery through this door.

e) Find the minimum value of $k$ if a painting 8 metres long is to be removed through this door.

Show answer

Answer: a) proof; b) $L = 7.77$ m; c) $\frac{d L}{d α} = \frac{- 3 k cos α}{sin ^{2} α} + \frac{k sin α}{cos ^{2} α}$ ; d) $L = 5.41 k$ m; e) $k = 1.48$ m

Show working and answer

a) 1. The line $CD$ passes through $A$ and makes the angle $α$ with the gallery wall (horizontal through $A$ ). We split $L = CA + AD$ . 2. In the right-angled triangle with hypotenuse $CA$ , the vertical leg is the width of the gallery $a$ : $sin α = \frac{a}{CA}$ , so $CA = \frac{a}{sin α}$ . 3. In the right-angled triangle with hypotenuse $AD$ , the horizontal leg is the width of the corridor $b$ : $cos α = \frac{b}{AD}$ , so $AD = \frac{b}{cos α}$ . 4. Adding: $L = \frac{a}{sin α} + \frac{b}{cos α}$ , $0 < α < \frac{π}{2}$ .

b) 1. A painting of length $ℓ$ can pass only if it fits in every position of $CD$ , i.e. if $ℓ \leq L (α)$ for all $α$ . The maximum length is therefore the minimum of $L (α)$ . 2. With $a = 5$ , $b = 1$ : $L (α) = \frac{5}{sin α} + \frac{1}{cos α}$ . Differentiating (chain rule): $L^{'} (α) = - \frac{5 cos α}{sin ^{2} α} + \frac{sin α}{cos ^{2} α} .$ 3. $L^{'} (α) = 0$ implies $sin^{3} α = 5 cos^{3} α$ , i.e. $tan α = 35 \approx 1.710$ . 4. It is a minimum: $L \to + \infty$ as $α \to 0^{+}$ and as $α \to \frac{π}{2}^{-}$ , and this is the only critical point. 5. Using the GDC (or calculator) for that $α$ : $L_{m i n} = (5^{2/3} + 1)^{3/2} \approx 7.77$ m.

c) 1. With $a = 3 k$ , $b = k$ : $L = \frac{3 k}{sin α} + \frac{k}{cos α}$ . 2. Differentiating: $\frac{d L}{d α} = - \frac{3 k cos α}{sin ^{2} α} + \frac{k sin α}{cos ^{2} α} .$

d) 1. $\frac{d L}{d α} = 0$ implies $k sin^{3} α = 3 k cos^{3} α$ , so $tan α = 33 \approx 1.442$ , i.e. $α \approx 0.965$ rad. 2. As in b), this is the minimum of $L$ (it tends to infinity at the ends). Its value: $L_{m i n} = k (3^{2/3} + 1)^{3/2} \approx 5.41 k m .$

e) 1. A painting of $8$ m passes through the door if $8 \leq L_{m i n}$ , i.e. $5.41 k \geq 8$ . 2. More precisely, $k \geq \frac{8}{5.4056} \approx 1.48$ . The minimum value is $k = 1.48$ m.

Answer: a) proof; b) $L = 7.77$ m; c) $\frac{d L}{d α} = \frac{- 3 k cos α}{sin ^{2} α} + \frac{k sin α}{cos ^{2} α}$ ; d) $L = 5.41 k$ m; e) $k = 1.48$ m

Exercise 207 HardIB · P2 TZ2 May 2009

Two circles that do not intersect: $C_{1}$ , which contains the points $M$ and $S$ , and $C_{2}$ , which contains the points $N$ and $R$ , with centres $P$ and $Q$ , where $PQ = 50$ . The line segments $[MN]$ and $[SR]$ are both tangent to the two circles.

The reflex angle $MPS$ measures $α$ , the obtuse angle $NQR$ measures $β$ , and the angle $MPQ$ measures $θ$ . The length of arc $MS$ is $l_{1}$ and the length of arc $NR$ is $l_{2}$ . This information is shown in the following diagram.

Two non-intersecting circles, with centres P and Q, joined by two common tangent segments MN and SR; the angles θ, α and β and the arcs l1 and l2 are marked

The radius of $C_{1}$ is $x$ , where $x \geq 0$ , and the radius of $C_{2}$ is equal to 10.

a) Explain why $x < 40$ .

b) Show that $cos θ = \frac{x - 10}{50}$ .

c) (i) Find an expression for $MN$ in terms of $x$ .

(ii) Find the value of $x$ for which $MN$ is a maximum.

d) Find an expression, in terms of $x$ , for

(i) $α$

(ii) $β$

e) The perimeter is given by $l_{1} + l_{2} + MN + SR$ .

(i) Find an expression, $b (x)$ , for the perimeter in terms of $x$ .

(ii) Find the maximum value of the perimeter.

(iii) Find the value of $x$ for which the perimeter is equal to 200.

Show answer

Answer: a) and b) justifications; c) (i) $MN = - x^{2} + 20 x + 2400$ ; (ii) $x = 10$ ; d) (i) $α = 2 π - 2 arccos (\frac{x - 10}{50})$ ; (ii) $β = 2 arccos (\frac{x - 10}{50})$ ; e) (i) $b (x) = 2 π x + (20 - 2 x) arccos (\frac{x - 10}{50}) + 2 - x^{2} + 20 x + 2400$ ; (ii) $275.69$ ; (iii) $x = 21.05$

Show working and answer

a) 1. The circles do not intersect and lie outside one another (the common tangents $MN$ and $SR$ are external). Then the distance between the centres is greater than the sum of the radii: $PQ > x + 10$ . 2. That is $50 > x + 10$ , so $x < 40$ .

b) 1. $PM ⊥ MN$ and $QN ⊥ MN$ , so $PM ∥ QN$ (they point to the same side). 2. We project the segment $PN$ onto the line $PM$ . As $MN ⊥ PM$ , the projection of $PN$ is $PM = x$ . 3. On the other hand $PN = PQ + QN$ : the projection of $PQ$ onto $PM$ is $50 cos θ$ , and that of $QN$ is $10$ (parallel and in the same direction). 4. Thus $x = 50 cos θ + 10$ , that is, $cos θ = \frac{x - 10}{50}$ .

c) (i) 1. Using the same projection, but onto the direction perpendicular to $PM$ : $MN = PQ sin θ = 50 sin θ$ . 2. As $sin θ = 1 - cos^{2} θ$ (with $0 < θ < π$ ): $MN = 50 1 - \frac{( x - 10 ) ^{2}}{2500} = 2500 - (x - 10)^{2} = - x^{2} + 20 x + 2400 .$

(ii) 1. $MN$ is a maximum when the radicand $- x^{2} + 20 x + 2400$ is, a parabola opening downwards. 2. Its vertex is at $x = - \frac{20}{2 ( - 1 )} = 10$ . (Differentiating: $- 2 x + 20 = 0$ implies $x = 10$ , and the second derivative $- 2 < 0$ confirms the maximum.) Then $MN = 50$ .

d) 1. The diagram is symmetric about the line $PQ$ , so $MPQ = SPQ = θ$ and the convex angle $MPS$ measures $2 θ$ . 2. (i) The reflex angle (the one on the left, where arc $l_{1}$ is) is the angle that completes $2 π$ : $α = 2 π - 2 θ = 2 π - 2 arccos \frac{x - 10}{50}$ . 3. (ii) As $QN ∥ PM$ , $QN$ makes the angle $θ$ with the direction of $PQ$ (to the right); likewise $QR$ by symmetry. Then $β = 2 θ = 2 arccos \frac{x - 10}{50}$ .

e) (i) 1. Arc length $=$ radius $\times$ angle: $l_{1} = xα$ and $l_{2} = 10 β$ . Also $SR = MN$ by symmetry. 2. Perimeter: $b (x) = x (2 π - 2 θ) + 20 θ + 2 - x^{2} + 20 x + 2400 = 2 π x + (20 - 2 x) arccos \frac{x - 10}{50} + 2 - x^{2} + 20 x + 2400 .$

(ii) 1. Using the GDC we plot $b (x)$ for $0 \leq x < 40$ : the graph is increasing, so the maximum is reached at the endpoint $x = 40$ (tangent circles). 2. $b (40) = 80 π - 60 arccos \frac{30}{50} + 2 - 1600 + 800 + 2400 = 80 π - 60 arccos 0.6 + 80 \approx 275.69$ .

(iii) 1. Using the GDC, find the intersection of $y = b (x)$ with $y = 200$ (as $b (0) \approx 133.4$ and $b$ is increasing, there is only one solution): $x \approx 21.05$ .

Rate of change

Exercise 208 BasicIB · P1 November 1996

A spherical snowball is melting and its volume is decreasing at a rate of $40 c m^{3}$ per minute. Calculate, in terms of $π$ , the rate of change of the radius when the radius is $5$ cm.

Show answer

Answer: $\frac{d r}{d t} = - \frac{2}{5 π} cm mi n^{- 1}$

Show working and answer

Volume of the sphere: $V = \frac{4}{3} π r^{3}$ , so $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ (chain rule).
The volume decreases: $\frac{d V}{d t} = - 40$ cm $^{3}$ min $^{- 1}$ .
With $r = 5$ : $- 40 = 4 π \cdot 25 \cdot \frac{d r}{d t} = 100 π \frac{d r}{d t}$ .
Solving for the rate: $\frac{d r}{d t} = - \frac{40}{100 π} = - \frac{2}{5 π}$ (the radius decreases).

Answer: $\frac{d r}{d t} = - \frac{2}{5 π} cm mi n^{- 1}$

Exercise 209 MediumIB · P2 TZ2 May 2009

A helicopter $H$ moves vertically, with a speed of $10 m s^{- 1}$ . The helicopter is at a height of $h$ m, directly above the point $Q$ , which is at ground level. The helicopter is observed from the point $P$ , which is also at ground level, where $PQ = 40$ m. This information is shown in the following diagram.

Right-angled triangle PQH with the right angle at Q, PQ = 40 m and QH = h m; the diagram is not to scale

For $h = 30$ ,

a) verify that the rate of change of $HPQ$ is equal to $0.16$ radians per second.

b) find the rate of change of $PH$ .

Show answer

Answer: a) verified: $0.16 rad s^{- 1}$ ; b) $6 m s^{- 1}$

Show working and answer

Let $θ = HPQ$ . In the right-angled triangle $PQH$ : $tan θ = \frac{h}{40}$ , and $\frac{d h}{d t} = 10 m s^{- 1}$ (the helicopter is ascending).

a) 1. Differentiating with respect to $t$ : $sec^{2} θ \frac{d θ}{d t} = \frac{1}{40} \frac{d h}{d t}$ . 2. For $h = 30$ : $tan θ = \frac{3}{4}$ , so $sec^{2} θ = 1 + \frac{9}{16} = \frac{25}{16}$ . 3. $\frac{d θ}{d t} = \frac{16}{25} \cdot \frac{10}{40} = \frac{16}{25} \cdot \frac{1}{4} = \frac{4}{25} = 0.16 rad s^{- 1}$ .

b) 1. By Pythagoras, $PH^{2} = 4 0^{2} + h^{2}$ . If $h = 30$ , $PH = 50$ m. 2. Differentiating with respect to $t$ : $2 PH \frac{d ( PH )}{d t} = 2 h \frac{d h}{d t}$ . 3. $\frac{d ( PH )}{d t} = \frac{h}{PH} \frac{d h}{d t} = \frac{30}{50} \cdot 10 = 6 m s^{- 1}$ (the distance is increasing).

Answer: a) verified: $0.16 rad s^{- 1}$ ; b) $6 m s^{- 1}$

Exercise 210 MediumIB · P2 TZ2 May 2011

A rocket is ascending vertically at a speed of $300 m s^{- 1}$ when it is at a height of $800$ m directly above the launch site.

Diagram of a rocket above the launch site at a height of 800 m, and an observer on the ground 600 m from the launch site, joined to the rocket by a line segment

Calculate the rate of change of the distance between the rocket and an observer situated $600$ m from the launch site, at the same horizontal level as the launch site.

Show answer

Answer: $240 m s^{- 1}$

Show working and answer

Let $h$ be the height of the rocket and $D$ its distance from the observer. The observer is $600$ m from the vertical through the launch site, so (Pythagoras) $D^{2} = 60 0^{2} + h^{2}$ , with $\frac{d h}{d t} = 300 m s^{- 1}$ (the rocket is rising).

Differentiating with respect to $t$ : $2 D \frac{d D}{d t} = 2 h \frac{d h}{d t}$ , so $\frac{d D}{d t} = \frac{h}{D} \frac{d h}{d t}$ .
With $h = 800$ : $D = 60 0^{2} + 80 0^{2} = 1000$ m.
$\frac{d D}{d t} = \frac{800}{1000} \cdot 300 = 240 m s^{- 1}$ (the distance is increasing).

Answer: $240 m s^{- 1}$

Exercise 211 MediumIB · P2 TZ2 May 2014

Sand is poured to form a cone of height $h$ cm and base radius $r$ cm. At all times, the height is equal to the base radius. The height of the cone is increasing at a rate of $0.5 cm mi n^{- 1}$ . Find the rate at which the sand is being poured, in $c m^{3} mi n^{- 1}$ , when the height is equal to $4$ cm.

Show answer

Answer: $8 π c m^{3} mi n^{- 1} \approx 25.1 c m^{3} mi n^{- 1}$

Show working and answer

Since $r = h$ , the volume of the cone is $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π h^{3}$ .
We differentiate with respect to time (chain rule): $\frac{d V}{d t} = π h^{2} \frac{d h}{d t}$ .
With $h = 4$ cm and $\frac{d h}{d t} = 0.5$ cm min $^{- 1}$ (positive: the height is increasing): $\frac{d V}{d t} = π \cdot 16 \cdot 0.5 = 8 π$ .

Answer: $8 π c m^{3} mi n^{- 1} \approx 25.1 c m^{3} mi n^{- 1}$

Exercise 212 MediumIB · P2 TZ0 November 2014Proof

Two cyclists are at the same road junction. One of the cyclists travels north at $20 km h^{- 1}$ . The other cyclist travels west at $15 km h^{- 1}$ . Use analytical calculus to show that the rate at which the distance between the two cyclists changes is independent of time.

Show answer

Answer: shown: the distance increases at $25 km h^{- 1}$ at every instant.

Show working and answer

Let $t$ be the time in hours since they were at the junction. The position of the cyclist travelling north is $y = 20 t$ km; that of the one travelling west is $x = 15 t$ km. They form a right angle.
The distance between them is $D = x^{2} + y^{2}$ , so $D^{2} = x^{2} + y^{2}$ .
We differentiate with respect to $t$ : $2 D \frac{d D}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$ , with $\frac{d x}{d t} = 15$ and $\frac{d y}{d t} = 20$ (both positive: they are moving apart).
We substitute $x = 15 t$ , $y = 20 t$ and $D = 225 t^{2} + 400 t^{2} = 25 t$ : $2 (25 t) \frac{d D}{d t} = 2 (15 t) (15) + 2 (20 t) (20) = 1250 t$ .
Dividing by $50 t$ ( $t > 0$ ): $\frac{d D}{d t} = 25$ km h $^{- 1}$ , which does not depend on $t$ .

Answer: shown: the distance increases at $25 km h^{- 1}$ at every instant.

Exercise 213 MediumIB · P1 TZ0 November 2006

The radius and the height of a cylinder are both equal to $x$ cm. The lateral surface area of the cylinder increases at a constant rate of $10 c m^{2} s^{- 1}$ . When $x = 2$ , calculate the rate of change of:

a) The radius of the cylinder.

b) The volume of the cylinder.

Show answer

Answer: a) $\frac{5}{4 π} cm s^{- 1}$ ; b) $15 c m^{3} s^{- 1}$

Show working and answer

Cylinder with $r = h = x$ . Lateral area: $S = 2 π r h = 2 π x^{2}$ . Volume: $V = π r^{2} h = π x^{3}$ .
Given: $\frac{d S}{d t} = 10$ cm $^{2}$ s $^{- 1}$ (positive: it is increasing).

a) 1. We differentiate $S = 2 π x^{2}$ with respect to $t$ (chain rule): $\frac{d S}{d t} = 4 π x \frac{d x}{d t}$ . 2. With $x = 2$ : $10 = 8 π \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = \frac{5}{4 π}$ cm s $^{- 1}$ .

b) 1. $\frac{d V}{d t} = 3 π x^{2} \frac{d x}{d t}$ . 2. With $x = 2$ : $\frac{d V}{d t} = 3 π \cdot 4 \cdot \frac{5}{4 π} = 15$ cm $^{3}$ s $^{- 1}$ .

Answer: a) $\frac{5}{4 π} cm s^{- 1}$ ; b) $15 c m^{3} s^{- 1}$

Exercise 214 MediumIB · P1 November 2002

A ball is inflated with air, and the spherical ball expands at $8 c m^{3} s^{- 1}$ . Find the exact rate of increase of the radius of the ball when the radius is $2$ cm.

Show answer

Answer: $\frac{1}{2 π} cm s^{- 1}$

Show working and answer

Volume of the sphere: $V = \frac{4}{3} π r^{3}$ .
We differentiate with respect to time (chain rule): $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ .
Given: $\frac{d V}{d t} = 8$ cm $^{3}$ s $^{- 1}$ (positive: it is inflating). With $r = 2$ : $8 = 16 π \frac{d r}{d t}$ .
Solving for the rate: $\frac{d r}{d t} = \frac{8}{16 π} = \frac{1}{2 π}$ .

Answer: $\frac{1}{2 π} cm s^{- 1}$

Exercise 215 MediumIB · P1 TZ1 May 2019

A camera at point C is 3 m from the edge of a straight section of a road, as shown in the diagram below. The camera detects a car travelling along the road at $t = 0$ . It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.

Diagram showing the camera C 3 m from the point O on the edge of the road, the line of the road and the car at X below O; the angle at C between CO and CX is theta

A car travels along the road at a speed of $24 m s^{- 1}$ . Let X be the position of the car and let $OCX = θ$

Calculate $\frac{d θ}{d t}$ , the rate of rotation of the camera in radians per second, at the instant the car passes the point O.

Show answer

Answer: $\frac{d θ}{d t} = - 8 rad s^{- 1}$ (the camera rotates at $8 rad s^{- 1}$ )

Show working and answer

Let $x$ be the distance $OX$ from the car to the point $O$ . In the right-angled triangle $COX$ (right angle at $O$ ): $tan θ = \frac{x}{3}$ .
The car approaches $O$ at $24$ m s $^{- 1}$ , so $\frac{d x}{d t} = - 24$ (the distance is decreasing).
We differentiate with respect to $t$ (chain rule): $sec^{2} θ \frac{d θ}{d t} = \frac{1}{3} \frac{d x}{d t}$ .
At the instant the car passes $O$ : $x = 0$ , $θ = 0$ and $sec^{2} 0 = 1$ . Therefore $\frac{d θ}{d t} = \frac{- 24}{3} = - 8$ .
The negative sign shows that $θ$ is decreasing: the camera rotates at $8$ rad s $^{- 1}$ .

Answer: $\frac{d θ}{d t} = - 8 rad s^{- 1}$ (the camera rotates at $8 rad s^{- 1}$ )

Exercise 216 MediumIB · P2 TZ0 November 2016

A satellite moves along a trajectory defined by the curve $72.5 x^{2} + 71.5 y^{2} = 1$ , where $x = x (t)$ and $y = y (t)$ are expressed in thousands of kilometres and $t$ is the time in seconds. Given that $\frac{d x}{d t} = 7.75 \times 1 0^{- 5}$ when $x = 3.2 \times 1 0^{- 3}$ , find the possible values of $\frac{d y}{d t}$ . Give your answers in standard form.

Show answer

Answer: $\frac{d y}{d t} \approx \pm 2.13 \times 1 0^{- 6}$ (thousands of km per second), i.e. $\pm 2.1 \times 1 0^{- 6}$

Show working and answer

We differentiate $72.5 x^{2} + 71.5 y^{2} = 1$ implicitly with respect to $t$ (chain rule): $145 x \frac{d x}{d t} + 143 y \frac{d y}{d t} = 0$ .
Rearranging: $\frac{d y}{d t} = - \frac{145 x}{143 y} \cdot \frac{d x}{d t}$ .
With $x = 3.2 \times 1 0^{- 3}$ , from the curve: $71.5 y^{2} = 1 - 72.5 (3.2 \times 1 0^{- 3})^{2} \approx 0.99926$ , so $y^{2} \approx 0.013975$ and $y \approx \pm 0.1182$ (there are two values).
Substituting (with $\frac{d x}{d t} = 7.75 \times 1 0^{- 5}$ ): $\frac{d y}{d t} = - \frac{145 \cdot 3.2 \times 1 0 ^{- 3}}{143 \cdot ( \pm 0.1182 )} \cdot 7.75 \times 1 0^{- 5} \approx \mp 0.0275 \cdot 7.75 \times 1 0^{- 5}$ .
Result (in thousands of km per second): $\frac{d y}{d t} \approx \pm 2.13 \times 1 0^{- 6}$ , with a negative sign if $y > 0$ and a positive sign if $y < 0$ .

Answer: $\frac{d y}{d t} \approx \pm 2.13 \times 1 0^{- 6}$ (thousands of km per second), i.e. $\pm 2.1 \times 1 0^{- 6}$

Exercise 217 MediumIB · P1 May 1997

An aeroplane flies horizontally at an altitude of $10000$ metres and at a speed of $800$ km per hour when it passes directly over a radar station. Calculate the exact value of the rate at which the distance from the aeroplane to the radar station is increasing when the aeroplane is $2$ km from the point vertically above the radar station.

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Answer: $\frac{800}{26} km h^{- 1}$

Show working and answer

Let $x$ be the horizontal distance (in km) between the aeroplane and the point above the station, and $D$ the distance to the radar. The altitude is constant, $10$ km: $D^{2} = x^{2} + 1 0^{2}$ .
Differentiating with respect to $t$ (chain rule): $2 D \frac{d D}{d t} = 2 x \frac{d x}{d t}$ , so $\frac{d D}{d t} = \frac{x}{D} \cdot \frac{d x}{d t}$ . Both rates are positive (the aeroplane is moving away).
Data: $\frac{d x}{d t} = 800 km h^{- 1}$ and, at the required instant, $x = 2$ , $D = 4 + 100 = 104 = 226$ .
Substituting: $\frac{d D}{d t} = \frac{2}{2 26} \cdot 800 = \frac{800}{26}$ .

Answer: $\frac{800}{26} km h^{- 1}$

Exercise 218 MediumIB · P1 November 2003

An aeroplane flies in a straight line at constant speed, maintaining a constant altitude of 3 km on a course that takes it directly over the vertical line through an observer on the ground. At a given instant, the angle $θ$ of the observer's line of sight to the aeroplane is $\frac{1}{3} π$ radians and it is increasing at $\frac{1}{60}$ radians per second. Find the speed, in kilometres per hour, at which the aeroplane is moving towards the observer.

Diagram: an observer on the ground, an aeroplane 3 km high, and the angle θ that the line of sight makes with the horizontal; x is the horizontal distance from the aeroplane to the observer's vertical line

Show answer

Answer: the aeroplane moves towards the observer at $240$ km/h.

Show working and answer

Let $x$ be the horizontal distance of the aeroplane from the observer's vertical line (in km). With the height of $3$ km: $tan θ = \frac{3}{x}$ , that is, $x = 3 cot θ$ .
Differentiating with respect to $t$ (chain rule): $\frac{d x}{d t} = - 3 csc^{2} θ \frac{d θ}{d t}$ .
Data: $θ = \frac{π}{3}$ , $\frac{d θ}{d t} = \frac{1}{60}$ rad/s (positive: the angle increases). $csc^{2} \frac{π}{3} = \frac{1}{sin ^{2} ( π /3 )} = \frac{4}{3}$ .
$\frac{d x}{d t} = - 3 \cdot \frac{4}{3} \cdot \frac{1}{60} = - \frac{1}{15}$ km/s. The negative sign shows that $x$ decreases: the aeroplane is moving towards the observer.
Converting to km/h: $\frac{1}{15} \cdot 3600 = 240$ .

Answer: the aeroplane moves towards the observer at $240$ km/h.

Exercise 219 MediumIB · P1 TZ2 May 2004

The following figure shows an isosceles triangle $ABC$ with $AB = 10$ cm and $AC = BC$ . The vertex $C$ moves in a direction perpendicular to $(AB)$ with a speed of 2 cm per second.

Isosceles triangle ABC with horizontal base AB, vertex C at the top, the altitude drawn dashed from C to AB, and a vertical arrow showing the upward movement of C

Calculate the rate of change of the angle $CAB$ at the instant when the triangle is equilateral.

Show answer

Answer: $\frac{d α}{d t} = \frac{1}{10}$ rad/s

Show working and answer

Let $h$ be the height of C above AB and $α = CAB$ . The altitude falls at the midpoint of AB, $5$ cm from A, so $tan α = \frac{h}{5}$ .
Differentiating with respect to $t$ : $sec^{2} α \frac{d α}{d t} = \frac{1}{5} \frac{d h}{d t}$ , with $\frac{d h}{d t} = 2$ cm/s (C moves away from AB).
Rearranging: $\frac{d α}{d t} = \frac{2}{5} cos^{2} α$ .
Equilateral triangle: $α = \frac{π}{3}$ , $cos^{2} α = \frac{1}{4}$ .
$\frac{d α}{d t} = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$ rad/s (the angle increases).

Answer: $\frac{d α}{d t} = \frac{1}{10}$ rad/s

Exercise 220 MediumIB · P1 TZ1 May 2004

The figure shows a sector $OPQ$ of a circle of radius $r$ cm and centre $O$ , where $POQ = θ$ .

Circular sector OPQ with centre O and radius r cm, with the angle θ = POQ at the centre and the radius OP marked with a double arrow

The value of $r$ is increasing at a rate of 2 cm per second, and the value of $θ$ is increasing at a rate of 0.1 rad per second. Find the rate of increase of the area of the sector when $r = 3$ and $θ = \frac{π}{4}$ .

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Answer: $\frac{d S}{d t} = \frac{1}{2} (0.9 + 3 π) \approx 5.16$ cm $^{2}$ /s

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The area of the sector is $S = \frac{1}{2} r^{2} θ$ , with $r$ and $θ$ functions of time.
Product and chain rules: $\frac{d S}{d t} = r \frac{d r}{d t} θ + \frac{1}{2} r^{2} \frac{d θ}{d t}$ .
Data (both increase, so the rates are positive): $\frac{d r}{d t} = 2$ cm/s, $\frac{d θ}{d t} = 0.1$ rad/s, with $r = 3$ and $θ = \frac{π}{4}$ .
$\frac{d S}{d t} = 3 \cdot 2 \cdot \frac{π}{4} + \frac{1}{2} \cdot 9 \cdot 0.1 = \frac{3 π}{2} + 0.45 = \frac{1}{2} (0.9 + 3 π) \approx 5.16$ cm $^{2}$ /s.

Answer: $\frac{d S}{d t} = \frac{1}{2} (0.9 + 3 π) \approx 5.16$ cm $^{2}$ /s

Exercise 221 MediumIB · P1 TZ1 May 2005

A conical tank with its vertex at the bottom is 8 metres in diameter and 12 metres deep. Water flows into the tank at $10 m^{3}$ per minute. Find the rate of change of the depth of the water at the instant when the water is 6 metres deep.

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Answer: $\frac{d h}{d t} = \frac{5}{2 π} m mi n^{- 1}$

Show working and answer

The tank is a cone of radius $4$ m (diameter 8) and height $12$ m. If the water has depth $h$ and surface radius $r$ , similar triangles give $\frac{r}{h} = \frac{4}{12} = \frac{1}{3}$ , that is, $r = \frac{h}{3}$ .
Volume of water: $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \frac{h ^{2}}{9} h = \frac{π h ^{3}}{27}$ .
Differentiating with respect to time (chain rule): $\frac{d V}{d t} = \frac{π h ^{2}}{9} \frac{d h}{d t}$ .
Water enters at $10 m^{3} /min$ , so $\frac{d V}{d t} = 10$ (positive, since the volume increases). With $h = 6$ : $10 = \frac{π \cdot 36}{9} \frac{d h}{d t} = 4 π \frac{d h}{d t}$ .
Rearranging: $\frac{d h}{d t} = \frac{10}{4 π} = \frac{5}{2 π} \approx 0.796$ (positive: the depth increases).

Answer: $\frac{d h}{d t} = \frac{5}{2 π} m mi n^{- 1}$

Exercise 222 MediumIB · P1 TZ2 May 2005

An experiment is carried out in which the number $n$ of bacteria in a liquid is given by $n = 650 \cdot e^{k t}$ , where $t$ is the time in minutes elapsed since the start of the experiment and $k$ is a constant. The number of bacteria doubles every 20 minutes. Find:

a) the exact value of $k$ .

b) the rate of growth of the number of bacteria when $t = 90$ .

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Answer: a) $k = \frac{ln 2}{20}$ ; b) $\frac{d n}{d t} \approx 509.7$ bacteria $mi n^{- 1}$

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a) 1. If the number doubles every 20 minutes, $n (20) = 2 n (0)$ : $650 e^{20 k} = 2 \cdot 650$ . 2. Simplifying: $e^{20 k} = 2$ . 3. Taking natural logarithms: $20 k = ln 2$ , so $k = \frac{ln 2}{20}$ .

b) 1. The rate of growth is the derivative: $\frac{d n}{d t} = 650 k e^{k t}$ (chain rule for $e^{k t}$ ). 2. For $t = 90$ : $\frac{d n}{d t} = 650 \cdot \frac{ln 2}{20} e^{90 l n 2/20} = 650 \cdot \frac{ln 2}{20} \cdot 2^{4.5}$ . 3. Using a calculator, $\frac{d n}{d t} \approx 509.7$ bacteria per minute.

Answer: a) $k = \frac{ln 2}{20}$ ; b) $\frac{d n}{d t} \approx 509.7$ bacteria $mi n^{- 1}$

Exercise 223 MediumIB · P1 TZ1 May 2007

The volume of a solid is given by $V = \frac{4}{3} π r^{3} + π r^{2} h$ . At the instant when the radius is 3 cm, the volume is $81 π c m^{3}$ , the radius is changing at a rate of 2 cm/min and the volume is changing at a rate of $204 π c m^{3}$ /min. Find the rate of change of the height at this instant.

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Answer: $\frac{d h}{d t} = 8 cm mi n^{- 1}$

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Relationship between the quantities: $V = \frac{4}{3} π r^{3} + π r^{2} h$ , with $r$ , $h$ and $V$ functions of time.
Differentiate with respect to $t$ (chain and product rules): $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} + 2 π r h \frac{d r}{d t} + π r^{2} \frac{d h}{d t}$ .
Find $h$ at that instant: with $r = 3$ and $V = 81 π$ , $81 π = 36 π + 9 π h$ , so $h = 5$ cm.
Substitute $r = 3$ , $h = 5$ , $\frac{d r}{d t} = 2$ , $\frac{d V}{d t} = 204 π$ : $204 π = 4 π \cdot 9 \cdot 2 + 2 π \cdot 3 \cdot 5 \cdot 2 + 9 π \frac{d h}{d t} = 72 π + 60 π + 9 π \frac{d h}{d t}$ .
Rearrange: $9 π \frac{d h}{d t} = 72 π$ , so $\frac{d h}{d t} = 8$ (the height is increasing).

Answer: $\frac{d h}{d t} = 8 cm mi n^{- 1}$

Exercise 224 MediumIB · P1 TZ2 May 2007

Car A travels along a straight east-west road, heading west at $60 km h^{- 1}$ . Car B travels along a straight north-south road, heading north at $70 km h^{- 1}$ . The roads intersect at the point $O$ . When car A is $x$ km east of $O$ , and car B is $y$ km south of $O$ , the distance between the cars is equal to $z$ km.

Diagram of two roads intersecting at O: car A on the east-west road at x km east of O, car B on the north-south road at y km south of O, and the distance z km between both cars; compass with the four cardinal points

Find the rate of change of $z$ when car A is 0.8 km east of $O$ and car B is 0.6 km south of $O$ .

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Answer: $\frac{d z}{d t} = - 90 km h^{- 1}$

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Relationship between the quantities (Pythagoras, since the roads are perpendicular): $z^{2} = x^{2} + y^{2}$ .
Differentiate with respect to $t$ : $2 z \frac{d z}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$ , i.e. $z \frac{d z}{d t} = x \frac{d x}{d t} + y \frac{d y}{d t}$ .
Signs of the rates: car A is heading west, towards $O$ , so $x$ decreases: $\frac{d x}{d t} = - 60$ . Car B is heading north, towards $O$ (it is south of it), so $y$ decreases: $\frac{d y}{d t} = - 70$ .
At the instant asked, $x = 0.8$ , $y = 0.6$ and $z = 0.64 + 0.36 = 1$ .
Substitute: $1 \cdot \frac{d z}{d t} = 0.8 (- 60) + 0.6 (- 70) = - 48 - 42 = - 90$ .
The distance between the cars is decreasing at $90$ km/h.

Answer: $\frac{d z}{d t} = - 90 km h^{- 1}$

Exercise 225 MediumIB · P1 TZ1 May 2013

Paint is poured into a tray, where it forms a circular pool of uniform thickness 0.5 cm. If the paint is poured at a constant rate of 4 $cm^{3} s^{- 1}$ , find the rate of increase of the radius of the circle when the radius is 20 cm.

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Answer: $\frac{1}{5 π} cm s^{- 1}$

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The paint forms a cylinder of radius $r$ and height $0.5$ cm: $V = π r^{2} \cdot 0.5 = \frac{π r ^{2}}{2}$ .
We differentiate with respect to time (chain rule): $\frac{d V}{d t} = π r \frac{d r}{d t}$ .
We know that $\frac{d V}{d t} = 4 cm^{3} s^{- 1}$ (the volume is increasing, so the sign is positive) and $r = 20$ cm: $4 = π \cdot 20 \cdot \frac{d r}{d t}$ .
We rearrange: $\frac{d r}{d t} = \frac{4}{20 π} = \frac{1}{5 π} cm s^{- 1}$ .

Answer: $\frac{1}{5 π} cm s^{- 1}$

Exercise 226 MediumIB · P2 TZ1 May 2015

An inner tube of a bicycle can be considered as a cylinder with fixed height 200 cm and radius $r$ cm. The radius $r$ increases as air is pumped into the tube. The air is pumped in so that the volume of air in the tube increases at a constant rate of 30 $cm^{3} s^{- 1}$ . Calculate the rate at which the radius of the inner tube is increasing when $r = 2$ cm.

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Answer: $\frac{3}{80 π} cm s^{- 1}$

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The volume of the cylinder is $V = π r^{2} h = 200 π r^{2}$ (fixed height $h = 200$ cm).
We differentiate with respect to $t$ (chain rule): $\frac{d V}{d t} = 400 π r \frac{d r}{d t}$ .
With $\frac{d V}{d t} = 30$ (the volume is increasing, so the sign is positive) and $r = 2$ : $30 = 400 π \cdot 2 \cdot \frac{d r}{d t}$ .
We rearrange: $\frac{d r}{d t} = \frac{30}{800 π} = \frac{3}{80 π} cm s^{- 1} \approx 0.0119 cm s^{- 1}$ .

Answer: $\frac{3}{80 π} cm s^{- 1}$

Exercise 227 MediumIB · P2 TZ1 May 2021

Two boats $A$ and $B$ are travelling north. Initially, boat $B$ is located 50 metres east of boat $A$ . The distances travelled by boat $A$ and boat $B$ after $t$ seconds are $x$ metres and $y$ metres respectively. The angle $θ$ is the measure in radians of the bearing of boat $B$ from boat $A$ . This information is shown in the following diagram.

Two boats travelling north along two vertical lines 50 metres apart: boat A at height x above the base, boat B at a greater height y, and the angle θ at A between north and the segment AB

a) Show that $y = x + 50 cot θ$ .

At the instant $T$ , the following conditions are true:

Boat $B$ has travelled 10 metres further than boat $A$
Boat $B$ is travelling at twice the speed of boat $A$
The rate of change of the angle $θ$ is $- 0.1$ radians per second

b) Calculate the speed of boat $A$ at the instant $T$ .

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Answer: a) proof; b) $\frac{d x}{d t} = 5.2 m s^{- 1}$

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a) 1. The boats are travelling north. Boat $B$ is $y - x$ metres further north than $A$ , and $50$ m east of it. 2. In the right-angled triangle formed, the angle $θ$ (measured from north, at $A$ ) has adjacent side $y - x$ (north direction) and opposite side $50$ (east direction): $tan θ = \frac{50}{y - x}$ . 3. Solving: $y - x = \frac{50}{tan θ} = 50 cot θ$ , that is $y = x + 50 cot θ$ .

b) 1. We differentiate $y = x + 50 cot θ$ with respect to $t$ (chain rule, with $\frac{d}{d θ} cot θ = - \frac{1}{sin ^{2} θ}$ ): $\frac{d y}{d t} = \frac{d x}{d t} - \frac{50}{sin ^{2} θ} \frac{d θ}{d t} .$ 2. At the instant $T$ : $y = x + 10$ , so $50 cot θ = 10$ and $cot θ = \frac{1}{5}$ . Then $\frac{1}{sin ^{2} θ} = 1 + cot^{2} θ = 1 + \frac{1}{25} = \frac{26}{25}$ . 3. Data: $\frac{d y}{d t} = 2 \frac{d x}{d t}$ and $\frac{d θ}{d t} = - 0.1$ rad/s (the angle is decreasing). 4. Let $v = \frac{d x}{d t}$ : $2 v = v - 50 \cdot \frac{26}{25} \cdot (- 0.1) = v + 5.2$ . 5. Hence $v = 5.2$ .

Answer: a) proof; b) $\frac{d x}{d t} = 5.2 m s^{- 1}$

Exercise 228 MediumIB · P2 TZ0 November 2022

The following diagram shows a liquid contained in a glass flask with a rounded base, formed by a sphere and a narrow cylindrical neck.

Glass flask formed by a sphere and a narrow cylindrical neck, with liquid in the lower part of the sphere

Initially the flask is empty. The liquid is poured into the flask at a rate of 2 $cm^{3} s^{- 1}$ .

You may assume that the liquid does not reach the cylindrical neck.

The volume ( $V cm^{3}$ ) and the height ( $h$ cm) of the liquid in the flask satisfy the equation $V = 5 π h^{2} - \frac{1}{3} π h^{3}$ . Find the rate of change of the height of the liquid in the flask at the instant when the volume of the liquid is equal to 200 $cm^{3}$ .

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Answer: $\frac{d h}{d t} = 0.0261 cm s^{- 1}$

Show working and answer

Relationship between the quantities: $V = 5 π h^{2} - \frac{1}{3} π h^{3}$ . Both depend on $t$ , and we know that $\frac{d V}{d t} = 2 cm^{3} s^{- 1}$ (the volume increases).
Differentiating with respect to $t$ (chain rule): $\frac{d V}{d t} = (10 π h - π h^{2}) \frac{d h}{d t}$ .
We need $h$ when $V = 200$ . Using the GDC, we solve $5 π h^{2} - \frac{1}{3} π h^{3} = 200$ : $h \approx 4.206$ cm (the other solutions, $h < 0$ and $h > 10$ , are not physical; the liquid reaches a height less than the radius of 5 cm, as the figure shows).
Substituting: $2 = π h (10 - h) \frac{d h}{d t}$ , that is $\frac{d h}{d t} = \frac{2}{π \cdot 4.206 \cdot ( 10 - 4.206 )} \approx 0.0261 cm s^{- 1} .$

Answer: $\frac{d h}{d t} = 0.0261 cm s^{- 1}$

Exercise 229 HardIB · P2 TZ2 May 2010

A lighthouse $L$ is situated at sea, $500$ metres from $P$ , the nearest point on a straight coastline. The narrow beam of light emitted by the lighthouse rotates at a constant rate of $8 π$ radians per minute, producing a point of light $S$ that moves along the coastline. You may assume that the height of the lighthouse can be ignored and that the beam of light lies in the horizontal plane defined by sea level. When $S$ is $2000$ metres from $P$ ,

a) Verify that the speed of $S$ , correct to three significant figures, is $214000$ metres per minute.

b) Find the acceleration of $S$ .

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Answer: a) $214000 m mi n^{- 1}$ ; b) $42952518 m mi n^{- 2}$

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Let $θ$ be the angle $PLS$ and $x = PS$ the distance (in metres) from $S$ to $P$ . In the right-angled triangle $LPS$ , with $LP = 500$ : $tan θ = \frac{x}{500}$ , that is, $x = 500 tan θ$ . Also $\frac{d θ}{d t} = 8 π$ rad/min (constant).

a) 1. Differentiating with respect to $t$ using the chain rule: $\frac{d x}{d t} = 500 sec^{2} θ \frac{d θ}{d t}$ . 2. When $x = 2000$ : $tan θ = 4$ , so $sec^{2} θ = 1 + tan^{2} θ = 17$ . 3. $\frac{d x}{d t} = 500 \cdot 17 \cdot 8 π = 68000 π \approx 213628 m mi n^{- 1}$ . 4. To three significant figures, $214000$ metres per minute.

b) 1. Differentiating the speed again (with $\frac{d θ}{d t}$ constant): $\frac{d ^{2} x}{d t ^{2}} = 500 \cdot 2 sec θ \cdot sec θ tan θ \cdot (\frac{d θ}{d t})^{2} = 1000 sec^{2} θ tan θ (\frac{d θ}{d t})^{2}$ . 2. Substituting: $1000 \cdot 17 \cdot 4 \cdot (8 π)^{2} = 4352000 π^{2}$ . 3. Numerically, $\approx 42952518 m mi n^{- 2}$ .

Answer: a) $214000 m mi n^{- 1}$ ; b) $42952518 m mi n^{- 2}$

Exercise 230 HardIB · P2 TZ0 November 2009

The diagram below shows a Ferris wheel, a fairground ride that carries passengers on the outside of the wheel.

Circular Ferris wheel resting on the ground, with a passenger at point A on the upper right and an operator below the centre, at the bottom of the wheel

a) The Ferris wheel, of radius 10 metres, is turning at a rate of $3$ radians per minute. Find the rate at which a passenger on the wheel is moving vertically when they are rising and pass through point $A$ , which is $6$ metres above the centre of the wheel.

b) The operator of the wheel is positioned directly below the centre, so that the bottom of the wheel is at the height of their eyes. When observing the passenger, their line of sight makes an angle $θ$ with the horizontal. Find the rate of change of $θ$ at point $A$ .

Show answer

Answer: a) $24 m mi n^{- 1}$ ; b) $1.5 rad mi n^{- 1}$

Show working and answer

Let $φ$ be the angle that the radius to the passenger makes with the horizontal. Data: $R = 10$ m and $\frac{d φ}{d t} = 3$ rad/min. At $A$ the passenger is $6$ m above the centre, so $sin φ = \frac{6}{10}$ and $cos φ = \frac{8}{10}$ .

a) 1. The height of the passenger relative to the centre is $y = 10 sin φ$ . 2. Differentiating with respect to $t$ (chain rule): $\frac{d y}{d t} = 10 cos φ \frac{d φ}{d t}$ . 3. At $A$ : $\frac{d y}{d t} = 10 \cdot 0.8 \cdot 3 = 24 m mi n^{- 1}$ (positive: the passenger is rising).

b) 1. The operator is on the vertical through the centre, at the level of the lowest point of the wheel, that is, $10$ m below the centre. The height of the passenger above the operator's eyes is $10 + 10 sin φ$ and the horizontal distance is $10 cos φ$ . 2. Relationship: $tan θ = \frac{10 + 10 sin φ}{10 cos φ} = \frac{1 + sin φ}{cos φ}$ . 3. Differentiating with respect to $t$ (quotient rule): $sec^{2} θ \frac{d θ}{d t} = \frac{cos φ \cdot cos φ + ( 1 + sin φ ) sin φ}{cos ^{2} φ} \frac{d φ}{d t} = \frac{1 + sin φ}{cos ^{2} φ} \frac{d φ}{d t}$ . 4. At $A$ : $tan θ = \frac{1.6}{0.8} = 2$ , so $sec^{2} θ = 1 + 4 = 5$ . And $\frac{1 + 0.6}{0.64} \cdot 3 = 7.5$ . 5. Therefore $5 \frac{d θ}{d t} = 7.5$ and $\frac{d θ}{d t} = 1.5$ rad/min (the angle is increasing).

Answer: a) $24 m mi n^{- 1}$ ; b) $1.5 rad mi n^{- 1}$

Exercise 231 HardIB · P2 TZ0 November 2011

A stalactite has the shape of a right circular cone. Its height is $200$ mm and it is increasing at a rate of $3$ mm per century. The radius of the base is $40$ mm and it is decreasing at a rate of $0.5$ mm per century. Determine whether its volume is increasing or decreasing and the rate at which the volume is changing.

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Answer: decreases at a rate of $\frac{3200 π}{3} \approx 3351 m m^{3}$ per century

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Volume of the cone: $V = \frac{1}{3} π r^{2} h$ . Data: $h = 200$ mm, $\frac{d h}{d t} = + 3$ mm/century; $r = 40$ mm, $\frac{d r}{d t} = - 0.5$ mm/century (the radius decreases).
We differentiate with respect to time (product and chain rules): $\frac{d V}{d t} = \frac{1}{3} π (2 r \frac{d r}{d t} h + r^{2} \frac{d h}{d t})$ .
Substituting: $\frac{d V}{d t} = \frac{π}{3} (2 \cdot 40 \cdot (- 0.5) \cdot 200 + 1600 \cdot 3) = \frac{π}{3} (- 8000 + 4800) = - \frac{3200 π}{3}$ .
The result is negative: the volume decreases, at a rate of $\frac{3200 π}{3} \approx 3351$ mm $^{3}$ per century.

Answer: decreases at a rate of $\frac{3200 π}{3} \approx 3351 m m^{3}$ per century

Exercise 232 HardIB · P1 TZ0 November 2005

The diagram shows two points $A$ and $B$ on the circumference of a circle with centre $O$ and radius $4$ cm, where $AOB = θ$ .

Circle with centre O and radius 4 cm, with two points A and B on the circumference, the angle AOB = θ, and the circular segment bounded by the chord AB and the minor arc AB shaded

Points $A$ and $B$ move on the circumference and $θ$ increases at a constant rate. Given that the rate of change of the minor arc $AB$ is equal to the rate of change of the area of the shaded segment, calculate the value of the acute angle $θ$ .

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Answer: $θ = \frac{π}{3}$ rad

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With radius $4$ cm and angle $θ$ (in radians): the arc length is $s = 4 θ$ and the area of the segment is $S = \frac{1}{2} r^{2} (θ - sin θ) = 8 (θ - sin θ)$ .
We differentiate with respect to $t$ (chain rule, with $\frac{d θ}{d t}$ constant and positive): $\frac{d s}{d t} = 4 \frac{d θ}{d t}$ and $\frac{d S}{d t} = 8 (1 - cos θ) \frac{d θ}{d t}$ .
We equate the two rates: $4 \frac{d θ}{d t} = 8 (1 - cos θ) \frac{d θ}{d t}$ .
Since $\frac{d θ}{d t} \neq = 0$ , we simplify: $1 - cos θ = \frac{1}{2} \Rightarrow cos θ = \frac{1}{2}$ .
The acute angle is $θ = \frac{π}{3}$ .

Answer: $θ = \frac{π}{3}$ rad

Exercise 233 HardIB · P2 TZ1 May 2017

A water channel 10 m long has a uniform cross-section in the shape of a semicircle with radius $0.5$ m. It is partially filled with water, as shown in the diagram of the cross-section. The centre of the circle is O and the angle KOL is $θ$ radians.

Semicircular cross-section of a channel with centre O on the upper edge; the water fills the circular segment below the chord KL, the angle KOL is theta, and the radius is 0.5

a) Find an expression for the volume of water $V$ ( $m^{3}$ ) in the channel in terms of $θ$ .

The volume of water increases at a constant rate of $0.0008 m^{3} s^{- 1}$ .

b) Calculate $\frac{d θ}{d t}$ when $θ = \frac{π}{3}$ .

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Answer: a) $V = \frac{5}{4} (θ - sin θ) m^{3}$ ; b) $\frac{d θ}{d t} = 0.00128 rad s^{- 1}$

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a) 1. The water occupies a circular segment of radius $0.5$ m and central angle $θ$ . Its area is the area of the sector minus that of the triangle KOL: $\frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ = \frac{1}{2} r^{2} (θ - sin θ)$ . 2. With $r = 0.5$ : area $= \frac{1}{8} (θ - sin θ)$ . 3. The volume is the area times the length, $10$ m: $V = \frac{10}{8} (θ - sin θ) = \frac{5}{4} (θ - sin θ)$ .

b) 1. Relationship between the rates: $\frac{d V}{d t} = \frac{d V}{d θ} \cdot \frac{d θ}{d t}$ (chain rule), with $\frac{d V}{d t} = 0.0008 > 0$ (the volume is increasing). 2. $\frac{d V}{d θ} = \frac{5}{4} (1 - cos θ)$ . 3. At $θ = \frac{π}{3}$ : $\frac{d V}{d θ} = \frac{5}{4} \cdot \frac{1}{2} = 0.625$ . 4. Therefore $\frac{d θ}{d t} = \frac{0.0008}{0.625} = 0.00128 rad s^{- 1}$ .

Answer: a) $V = \frac{5}{4} (θ - sin θ) m^{3}$ ; b) $\frac{d θ}{d t} = 0.00128 rad s^{- 1}$

Exercise 234 HardIB · P1 November 1997

In the diagram, PT is a fixed tangent to the circle with centre O and radius $12$ cm, and PQ is a chord of the circle.

Circle with centre O and radius 12 cm, with a chord PQ and the tangent PT at P; the circular segment between the chord and the arc on the right is shaded and the angle TPQ is marked

If the angle TPQ is increasing at a rate of $3^{\circ}$ per second, calculate the rate, in $c m^{2} s^{- 1}$ , at which the area of the shaded region is changing when the angle TPQ is $3 0^{\circ}$ .

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Answer: $\frac{d A}{d t} = 1.2 π c m^{2} s^{- 1}$ (the area is increasing)

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The angle between the tangent $PT$ and the chord $PQ$ is half the central angle subtended by the arc it encloses. Let $θ = TPQ$ : the central angle $POQ$ of the shaded arc is $2 θ$ .
The area of the shaded segment is the area of the sector minus the area of triangle $OPQ$ : $A = \frac{1}{2} r^{2} (2 θ) - \frac{1}{2} r^{2} sin 2 θ = 72 (2 θ - sin 2 θ)$ , with $r = 12$ .
Relating the rates (chain rule): $\frac{d A}{d t} = \frac{d A}{d θ} \cdot \frac{d θ}{d t}$ , with $\frac{d A}{d θ} = 72 (2 - 2 cos 2 θ) = 144 (1 - cos 2 θ)$ .
The angle is increasing (positive rate): $\frac{d θ}{d t} = 3^{\circ} s^{- 1} = \frac{3 π}{180} = \frac{π}{60} rad s^{- 1}$ (we must convert to radians before differentiating).
When $θ = 3 0^{\circ}$ : $cos 6 0^{\circ} = \frac{1}{2}$ , so $\frac{d A}{d θ} = 144 \cdot \frac{1}{2} = 72$ .
$\frac{d A}{d t} = 72 \cdot \frac{π}{60} = \frac{6 π}{5} = 1.2 π$ .

Answer: $\frac{d A}{d t} = 1.2 π c m^{2} s^{- 1}$ (the area is increasing)

Exercise 235 HardIB · P1 May 1998

A drop of ink is placed on a piece of absorbent paper.

The ink forms a circular mark which begins to increase in size.

The radius of the circular mark is given by the formula $r = \frac{4 ( 1 + t ^{4} )}{8 + t ^{4}}$ , where $r$ is the radius of the circular mark in centimetres and $t$ is the time in minutes after the ink is placed on the paper.

a) Find $t$ when $r = \frac{17}{6}$ .

b) Write down a simplified expression, in terms of $t$ , for the rate of change of the radius.

c) Find the rate of change of the area of the circular mark when $r = \frac{17}{6}$ .

d) Find the value of $t$ when the rate of change of the radius starts to decrease, i.e. find the value of $t$ , $t > 0$ , at the point of inflexion of the curve $r = \frac{4 ( 1 + t ^{4} )}{8 + t ^{4}}$ .

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Answer: a) $t = 2$ minutes; b) $\frac{d r}{d t} = \frac{112 t ^{3}}{( 8 + t ^{4} ) ^{2}}$ cm/min; c) $\frac{d A}{d t} \approx 27.7$ cm $^{2}$ /min; d) $t = 4 \frac{24}{5} \approx 1.48$ minutes

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a) 1. We equate $\frac{4 ( 1 + t ^{4} )}{8 + t ^{4}} = \frac{17}{6}$ and clear the denominators: $24 (1 + t^{4}) = 17 (8 + t^{4})$ . 2. Expanding: $24 + 24 t^{4} = 136 + 17 t^{4}$ , so $7 t^{4} = 112$ and $t^{4} = 16$ . 3. Since $t > 0$ , $t = 2$ minutes.

b) 1. The rate of change of the radius is $\frac{d r}{d t}$ . We apply the quotient rule: $\frac{d r}{d t} = \frac{16 t ^{3} ( 8 + t ^{4} ) - 4 ( 1 + t ^{4} ) \cdot 4 t ^{3}}{( 8 + t ^{4} ) ^{2}} .$ 2. We factor out $16 t^{3}$ from the numerator: $16 t^{3} [8 + t^{4} - (1 + t^{4})] = 16 t^{3} \cdot 7 = 112 t^{3}$ . 3. Therefore $\frac{d r}{d t} = \frac{112 t ^{3}}{( 8 + t ^{4} ) ^{2}}$ cm/min.

c) 1. The area is $A = π r^{2}$ . We differentiate with respect to $t$ (chain rule): $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ . 2. When $r = \frac{17}{6}$ , $t = 2$ (part a), and $\frac{d r}{d t} = \frac{112 \cdot 8}{2 4 ^{2}} = \frac{896}{576} = \frac{14}{9}$ cm/min. 3. We substitute: $\frac{d A}{d t} = 2 π \cdot \frac{17}{6} \cdot \frac{14}{9} = \frac{238 π}{27} \approx 27.7$ cm $^{2}$ /min.

d) 1. The rate of change of the radius, $r^{'}$ , starts to decrease when $r^{'}$ has a maximum, i.e. when $r^{''} (t) = 0$ with a change of sign (a point of inflexion of $r$ ). 2. We differentiate $r^{'} = 112 t^{3} (8 + t^{4})^{- 2}$ using the product and chain rules: $r^{''} = 112 [3 t^{2} (8 + t^{4})^{- 2} - 2 t^{3} \cdot 4 t^{3} (8 + t^{4})^{- 3}] = \frac{112 t ^{2} [ 3 ( 8 + t ^{4} ) - 8 t ^{4} ]}{( 8 + t ^{4} ) ^{3}} = \frac{112 t ^{2} ( 24 - 5 t ^{4} )}{( 8 + t ^{4} ) ^{3}} .$ 3. For $t > 0$ , $r^{''} = 0$ when $24 - 5 t^{4} = 0$ , i.e. $t^{4} = \frac{24}{5}$ . 4. The sign of $r^{''}$ changes from positive to negative at this value (the factor $24 - 5 t^{4}$ changes from $+$ to $-$ ), so $r^{'}$ changes from increasing to decreasing. 5. $t = 4 \frac{24}{5} \approx 1.48$ minutes.

Answer: a) $t = 2$ minutes; b) $\frac{d r}{d t} = \frac{112 t ^{3}}{( 8 + t ^{4} ) ^{2}}$ cm/min; c) $\frac{d A}{d t} \approx 27.7$ cm $^{2}$ /min; d) $t = 4 \frac{24}{5} \approx 1.48$ minutes

Exercise 236 HardIB · P2 TZ1 May 2009

The diagram below shows a circle with centre at the origin $O$ and radius $r > 0$ .

Circle centred at the origin O with radius r, with the x and y axes; a point P in the first quadrant on the circumference, joined to O by a segment extended beyond P, and a point marked where the circle meets the x-axis

A point $P (x, y)$ , $(x > 0, y > 0)$ moves around the circumference of the circle.

Let $m = tan (arcsin \frac{y}{r})$

a) Given that $\frac{d y}{d t} = 0.001 r$ , show that $\frac{d m}{d t} = (\frac{r}{10 r ^{2} - y ^{2}})^{3}$ .

b) State the geometrical meaning of $\frac{d m}{d t}$ .

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Answer: a) proven; b) the rate of change of the gradient of the line $OP$ .

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a) 1. Let $φ = arcsin \frac{y}{r}$ , so that $sin φ = \frac{y}{r}$ and $cos φ = 1 - \frac{y ^{2}}{r ^{2}} = \frac{r ^{2} - y ^{2}}{r}$ (an acute angle, since $y > 0$ and $x > 0$ ). 2. Then $m = tan φ = \frac{y / r}{r ^{2} - y ^{2} / r} = \frac{y}{r ^{2} - y ^{2}}$ . 3. Differentiate with respect to $y$ (quotient rule): $\frac{d m}{d y} = \frac{r ^{2} - y ^{2} + \frac{y ^{2}}{r ^{2} - y ^{2}}}{r ^{2} - y ^{2}} = \frac{( r ^{2} - y ^{2} ) + y ^{2}}{( r ^{2} - y ^{2} ) ^{3/2}} = \frac{r ^{2}}{( r ^{2} - y ^{2} ) ^{3/2}}$ . 4. Chain rule: $\frac{d m}{d t} = \frac{d m}{d y} \cdot \frac{d y}{d t} = \frac{r ^{2}}{( r ^{2} - y ^{2} ) ^{3/2}} \cdot 0.001 r = \frac{r ^{3}}{1000 ( r ^{2} - y ^{2} ) ^{3/2}}$ . 5. Since $1000 = 1 0^{3}$ : $\frac{d m}{d t} = (\frac{r}{10 r ^{2} - y ^{2}})^{3}$ .

b) 1. $φ$ is the angle that $OP$ makes with the $x$ -axis (its sine is $\frac{y}{r}$ ), so $m = tan φ$ is the gradient of the line $OP$ . 2. Hence $\frac{d m}{d t}$ is the rate of change of the gradient of the line $OP$ with respect to time.

Answer: a) proven; b) the rate of change of the gradient of the line $OP$ .

Rectilinear motion

Exercise 237 BasicIB · P1 TZ1 May 2014

A body moves in a straight line. When it is $s$ metres from a fixed point $O$ on the line, its velocity, $v$ , is given by $v = - \frac{1}{s ^{2}}$ , $s > 0$ . Find the acceleration of the body when it is 50 cm from $O$ .

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Answer: $a = - 64 m s^{- 2}$

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By the chain rule, $a = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = \frac{d v}{d s} \cdot v$ .
With $v = - s^{- 2}$ : $\frac{d v}{d s} = 2 s^{- 3}$ .
Then $a = 2 s^{- 3} \cdot (- s^{- 2}) = - \frac{2}{s ^{5}}$ .
At $50$ cm $= 0.5$ m from $O$ (the units of $v$ are metres): $a = - \frac{2}{0. 5 ^{5}} = - 2 \cdot 32 = - 64$ .

Answer: $a = - 64 m s^{- 2}$

Exercise 238 MediumIB · P2 TZ2 May 2018

A point P moves along a straight line. Its velocity $v m s^{- 1}$ at time $t$ seconds is given by $v (t) = e^{- t} - 8 t^{2} e^{- 2 t}$ , where $t \geq 0$ .

a) Find the first time $t_{1}$ at which P has zero velocity.

b) (i) Find an expression for the acceleration of P at time $t$ .

(ii) Find the value of the acceleration of P at time $t_{1}$ .

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Answer: a) $t_{1} \approx 0.441$ s; b) (i) $a (t) = - e^{- t} - 16 t e^{- 2 t} (1 - t)$ ; (ii) $a (t_{1}) \approx - 2.28 m s^{- 2}$

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a) 1. We set $v (t) = 0$ : $e^{- t} = 8 t^{2} e^{- 2 t}$ . 2. Dividing by $e^{- 2 t} > 0$ : $e^{t} = 8 t^{2}$ . 3. Using the GDC, we solve the equation (or plot $y = e^{t}$ and $y = 8 t^{2}$ ) and take the first positive solution: $t_{1} \approx 0.441$ s.

b) (i) 1. The acceleration is $a (t) = v^{'} (t)$ . We differentiate each term. 2. For $e^{- t}$ : $- e^{- t}$ . 3. For $8 t^{2} e^{- 2 t}$ we apply the product rule: $16 t e^{- 2 t} + 8 t^{2} \cdot (- 2) e^{- 2 t} = 16 t e^{- 2 t} - 16 t^{2} e^{- 2 t}$ . 4. Therefore $a (t) = - e^{- t} - (16 t e^{- 2 t} - 16 t^{2} e^{- 2 t}) = - e^{- t} - 16 t e^{- 2 t} (1 - t)$ .

(ii) We substitute $t_{1} \approx 0.441$ (using the GDC): $a (t_{1}) \approx - 2.28 m s^{- 2}$ .

Answer: a) $t_{1} \approx 0.441$ s; b) (i) $a (t) = - e^{- t} - 16 t e^{- 2 t} (1 - t)$ ; (ii) $a (t_{1}) \approx - 2.28 m s^{- 2}$

Exercise 239 MediumIB · P1 TZ2 May 2017

A particle moves along a straight line.

The displacement, $s$ metres, at time $t$ seconds, is given by $s = t + cos 2 t$ , $t \geq 0$ .

Let $t_{1}$ and $t_{2}$ be the first two times at which the particle is at rest, where $t_{1} < t_{2}$ .

a) Find $t_{1}$ and $t_{2}$ .

b) Find the displacement of the particle when $t = t_{1}$ .

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Answer: a) $t_{1} = \frac{π}{12}$ s and $t_{2} = \frac{5 π}{12}$ s; b) $s = \frac{π}{12} + \frac{3}{2}$ m

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a) 1. The particle is at rest when the velocity is zero: $v = \frac{d s}{d t} = 1 - 2 sin 2 t$ (chain rule). 2. We solve $1 - 2 sin 2 t = 0$ , that is, $sin 2 t = \frac{1}{2}$ . 3. With $t \geq 0$ (so $2 t \geq 0$ ), the first solutions are $2 t = \frac{π}{6}$ and $2 t = \frac{5 π}{6}$ . 4. Therefore $t_{1} = \frac{π}{12}$ and $t_{2} = \frac{5 π}{12}$ .

b) 1. $s (t_{1}) = \frac{π}{12} + cos \frac{π}{6}$ . 2. As $cos \frac{π}{6} = \frac{3}{2}$ : $s (t_{1}) = \frac{π}{12} + \frac{3}{2}$ m.

Answer: a) $t_{1} = \frac{π}{12}$ s and $t_{2} = \frac{5 π}{12}$ s; b) $s = \frac{π}{12} + \frac{3}{2}$ m

Exercise 240 MediumIB · P1 May 1999

A particle moves along a straight line. When it is a distance $s$ from a fixed point, where $s > 1$ , its velocity $v$ is given by $v = \frac{3 s + 2}{2 s - 1}$ . Find the acceleration when $s = 2$ .

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Answer: $a = - \frac{56}{27}$ when $s = 2$

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Acceleration is $a = \frac{d v}{d t}$ . Since $v$ is given as a function of $s$ , we use the chain rule: $a = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ .
We differentiate $v = \frac{3 s + 2}{2 s - 1}$ using the quotient rule: $\frac{d v}{d s} = \frac{3 ( 2 s - 1 ) - 2 ( 3 s + 2 )}{( 2 s - 1 ) ^{2}} = \frac{- 7}{( 2 s - 1 ) ^{2}} .$
When $s = 2$ : $v = \frac{8}{3}$ and $\frac{d v}{d s} = - \frac{7}{9}$ .
Therefore $a = \frac{8}{3} \cdot (- \frac{7}{9}) = - \frac{56}{27}$ .

Answer: $a = - \frac{56}{27}$ when $s = 2$

Exercise 241 MediumIB · P2 TZ0 November 2012

A particle moves along a straight line so that after $t$ seconds its displacement $s$ , in metres, satisfies the equation $s^{2} + s - 2 t = 0$ .

Find, in terms of $s$ , expressions for its velocity and its acceleration.

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Answer: $v = \frac{2}{2 s + 1}$ ; $a = \frac{- 8}{( 2 s + 1 ) ^{3}}$

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We differentiate the equation $s^{2} + s - 2 t = 0$ with respect to $t$ (with $s = s (t)$ , chain rule on $s^{2}$ ): $2 s \frac{d s}{d t} + \frac{d s}{d t} - 2 = 0$ .
We solve for the velocity: $\frac{d s}{d t} (2 s + 1) = 2$ , so $v = \frac{d s}{d t} = \frac{2}{2 s + 1} .$
The acceleration is $a = \frac{d v}{d t}$ . We differentiate $v = 2 (2 s + 1)^{- 1}$ with respect to $t$ (chain rule): $a = - 2 (2 s + 1)^{- 2} \cdot 2 \frac{d s}{d t} = - \frac{4}{( 2 s + 1 ) ^{2}} \cdot v .$
We substitute $v = \frac{2}{2 s + 1}$ : $a = - \frac{8}{( 2 s + 1 ) ^{3}} .$

Answer: $v = \frac{2}{2 s + 1}$ ; $a = \frac{- 8}{( 2 s + 1 ) ^{3}}$

Exercise 242 HardIB · P1 TZ0 November 2010

a) A particle $P$ moves in a straight line. The displacement of $P$ from the origin is given by: $s = 2 sin (π t) + sin (2 π t)$ , $t \geq 0$ , where $t$ represents the time, in seconds, and the displacement is measured in centimetres.

(i) Write down the period of the function $s$ .

(ii) Find the expression for the velocity, $v$ , and the acceleration, $a$ , of $P$ .

(iii) Determine all the solutions of the equation $v = 0$ for $0 \leq t \leq 4$ .

b) Consider the function $f (x) = A sin (a x) + B sin (b x)$ , $A, a, B, b, x \in R$ .

Use mathematical induction to prove that the $(2 n)$ th derivative of $f$ is given by $f^{(2 n)} (x) = (- 1)^{n} (A a^{2 n} sin (a x) + B b^{2 n} sin (b x))$ , for all $n \in Z^{+}$ .

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Answer: a) (i) $2$ s; (ii) $v = 2 π [cos (π t) + cos (2 π t)]$ , $a = - 2 π^{2} [sin (π t) + 2 sin (2 π t)]$ ; (iii) $t = \frac{1}{3}, 1, \frac{5}{3}, \frac{7}{3}, 3, \frac{11}{3}$ ; b) proved.

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a) (i) 1. $sin (π t)$ has period $\frac{2 π}{π} = 2$ and $sin (2 π t)$ has period $\frac{2 π}{2 π} = 1$ . 2. The sum repeats when both terms repeat at the same time: the smallest common time is $2$ s.

(ii) 1. Differentiating (chain rule): $v = \frac{d s}{d t} = 2 π cos (π t) + 2 π cos (2 π t) = 2 π [cos (π t) + cos (2 π t)]$ . 2. Differentiating again: $a = \frac{d v}{d t} = - 2 π^{2} sin (π t) - 4 π^{2} sin (2 π t) = - 2 π^{2} [sin (π t) + 2 sin (2 π t)]$ .

(iii) 1. $v = 0$ is equivalent to $cos (π t) + cos (2 π t) = 0$ . 2. Using $cos (2 π t) = 2 cos^{2} (π t) - 1$ : $2 cos^{2} (π t) + cos (π t) - 1 = 0$ , that is, $(2 cos (π t) - 1) (cos (π t) + 1) = 0$ . 3. If $cos (π t) = \frac{1}{2}$ : $π t = \frac{π}{3} + 2 π n$ or $π t = \frac{5 π}{3} + 2 π n$ , that is, $t = \frac{1}{3} + 2 n$ or $t = \frac{5}{3} + 2 n$ . In $[0, 4]$ : $t = \frac{1}{3}, \frac{5}{3}, \frac{7}{3}, \frac{11}{3}$ . 4. If $cos (π t) = - 1$ : $π t = π + 2 π n$ , that is, $t = 1 + 2 n$ . In $[0, 4]$ : $t = 1, 3$ . 5. Listing the values in order: $t = \frac{1}{3}, 1, \frac{5}{3}, \frac{7}{3}, 3, \frac{11}{3}$ .

b) We prove by induction that $f^{(2 n)} (x) = (- 1)^{n} (A a^{2 n} sin (a x) + B b^{2 n} sin (b x))$ .

Base case $n = 1$ : $f^{'} (x) = A a cos (a x) + B b cos (b x)$ and $f^{''} (x) = - A a^{2} sin (a x) - B b^{2} sin (b x) = (- 1)^{1} (A a^{2} sin (a x) + B b^{2} sin (b x))$ . It holds.
Hypothesis: $f^{(2 k)} (x) = (- 1)^{k} (A a^{2 k} sin (a x) + B b^{2 k} sin (b x))$ .
Step: differentiating once: $f^{(2 k + 1)} (x) = (- 1)^{k} (A a^{2 k + 1} cos (a x) + B b^{2 k + 1} cos (b x))$ .
Differentiating once more: $f^{(2 k + 2)} (x) = (- 1)^{k} (- A a^{2 k + 2} sin (a x) - B b^{2 k + 2} sin (b x)) = (- 1)^{k + 1} (A a^{2 (k + 1)} sin (a x) + B b^{2 (k + 1)} sin (b x))$ .
This is the formula for $n = k + 1$ . By induction, it holds for all $n \in Z^{+}$ .

Exercise 243 HardIB · P2 TZ2 May 2016

A particle moves in such a way that its velocity $v m s^{- 1}$ and its displacement $s$ m are related by the equation $v (s) = arctan (sin s)$ , $0 \leq s \leq 1$ .

The acceleration of the particle is $a m s^{- 2}$ .

a) Find the acceleration of the particle as a function of $s$ .

b) Using an appropriate approximate graph, find the displacement of the particle when its acceleration is equal to $0.25 m s^{- 2}$ .

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Answer: a) $a = \frac{cos s \cdot arctan ( sin s )}{1 + sin ^{2} s}$ ; b) $s \approx 0.296$ m and $s \approx 0.918$ m

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a) 1. The acceleration is $a = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ (chain rule). 2. $\frac{d v}{d s} = \frac{cos s}{1 + sin ^{2} s}$ (derivative of the arctangent with the chain rule). 3. $a = arctan (sin s) \cdot \frac{cos s}{1 + sin ^{2} s}$ .

b) 1. Using the GDC we sketch $y = \frac{cos s \cdot arctan ( sin s )}{1 + sin ^{2} s}$ on $0 \leq s \leq 1$ and the line $y = 0.25$ . 2. They intersect at two points: $s \approx 0.296$ and $s \approx 0.918$ (the acceleration rises to a maximum near $s \approx 0.56$ , of value approximately $0.32$ , and then falls).

Graph of acceleration against displacement between zero and one, with the horizontal line a = 0.25 and its two points of intersection with the curve

Answer: a) $a = \frac{cos s \cdot arctan ( sin s )}{1 + sin ^{2} s}$ ; b) $s \approx 0.296$ m and $s \approx 0.918$ m

Exercise 244 HardIB · P2 TZ1 May 2009

The position vector at time $t$ of a point $P$ is given by $OP = (1 + t) i + (2 - 2 t) j + (3 t - 1) k$ , $t \geq 0$ .

a) Find the coordinates of $P$ when $t = 0$ .

b) Show that $P$ moves along the line $L$ with Cartesian equations $x - 1 = \frac{y - 2}{- 2} = \frac{z + 1}{3}$

c) (i) Find the value of $t$ when $P$ lies on the plane with equation $2 x + y + z = 6$ .

(ii) State the coordinates of $P$ at this instant.

(iii) Hence, find the total distance travelled by $P$ before it meets the plane.

The position vector at time $t$ of another point, $Q$ , is given by $OQ = t^{2} 1 - t 1 - t^{2}$ , $t \geq 0$ .

d) (i) Find the value of $t$ for which the distance of $Q$ from the origin is a minimum.

(ii) Find the coordinates of $Q$ at that instant.

e) Let $a$ , $b$ and $c$ be the position vectors of $Q$ at the instants $t = 0$ , $t = 1$ and $t = 2$ respectively.

(i) Show that the equation $a - b = k (b - c)$ has no solution for $k$ .

(ii) Hence, show that the path of $Q$ is not a straight line.

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Answer: a) $P (1, 2, - 1)$ ; c) (i) $t = 1$ ; (ii) $P (2, 0, 2)$ ; (iii) $14$ ; d) (i) $t \approx 0.76$ ; (ii) $Q (0.578; 0.24; 0.422)$ ; e) (i) no solution for $k$ (gives $\frac{1}{3}$ and $1$ ); (ii) the path of $Q$ is not a line.

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a) 1. For $t = 0$ : $OP = i + 2 j - k$ , i.e. $P (1, 2, - 1)$ .

b) 1. The coordinates are $x = 1 + t$ , $y = 2 - 2 t$ , $z = 3 t - 1$ . 2. We solve for $t$ in each equation: $t = x - 1$ , $t = \frac{y - 2}{- 2}$ and $t = \frac{z + 1}{3}$ . 3. Since the three expressions give the same $t$ , $x - 1 = \frac{y - 2}{- 2} = \frac{z + 1}{3}$ : $P$ moves along the line $L$ .

c) (i) 1. We substitute into the plane $2 x + y + z = 6$ : $2 (1 + t) + (2 - 2 t) + (3 t - 1) = 6$ . 2. We simplify: $3 + 3 t = 6$ , so $t = 1$ .

(ii) 1. With $t = 1$ : $P (2, 0, 2)$ .

(iii) 1. $P$ travels from $(1, 2, - 1)$ ( $t = 0$ ) to $(2, 0, 2)$ ( $t = 1$ ) in a straight line, without turning back. 2. Distance travelled: $(2 - 1)^{2} + (0 - 2)^{2} + (2 + 1)^{2} = 1 + 4 + 9 = 14$ .

d) (i) 1. Squared distance of $Q$ from the origin: $D (t) = OQ^{2} = t^{4} + (1 - t)^{2} + (1 - t^{2})^{2} = 2 t^{4} - t^{2} - 2 t + 2$ . 2. $D^{'} (t) = 8 t^{3} - 2 t - 2 = 0$ , i.e. $4 t^{3} - t - 1 = 0$ . 3. Using the GDC, the only real root is $t \approx 0.76$ . It is a minimum: for $t \geq 0$ , $D^{'} (t)$ is negative at $t = 0$ ( $D^{'} (0) = - 2$ ) and changes to positive at this root.

(ii) 1. With $t \approx 0.7607$ : $t^{2} \approx 0.5786$ , $1 - t \approx 0.2393$ and $1 - t^{2} \approx 0.4214$ . 2. $Q (0.578; 0.24; 0.422)$ .

e) 1. Positions: $a = OQ (0) = (0, 1, 1)$ , $b = OQ (1) = (1, 0, 0)$ and $c = OQ (2) = (4, - 1, - 3)$ .

(i) 1. $a - b = (- 1, 1, 1)$ and $b - c = (- 3, 1, 3)$ . The equation $a - b = k (b - c)$ gives three equations: $- 1 = - 3 k$ , $1 = k$ and $1 = 3 k$ . 2. From the first, $k = \frac{1}{3}$ ; from the second, $k = 1$ . This is a contradiction: the equation has no solution.

(ii) 1. If $Q$ moved along a line, the points at $t = 0, 1, 2$ would be collinear, and then $a - b$ and $b - c$ would be parallel, i.e. there would exist $k$ with $a - b = k (b - c)$ . 2. By the previous part no such $k$ exists, so the path of $Q$ is not a straight line.

Exercise 245 HardIB · P2 TZ2 May 2022

A particle moves in a straight line so that its velocity ( $m s^{- 1}$ ) at time $t$ seconds is given by $v = \frac{( t ^{2} + 1 ) cos t}{4}$ , $0 \leq t \leq 3$

a) Determine when the particle changes direction.

b) Find the times at which the acceleration of the particle is equal to $- 1.9 m s^{- 2}$ .

c) Find the acceleration of the particle when its speed is at a maximum.

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Answer: a) $t = 1.57$ s; b) $t = 2.26$ s and $t = 2.96$ s; c) $a = - 1.84 m s^{- 2}$

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a) 1. The particle changes direction when $v$ becomes zero and changes sign. 2. $v = \frac{( t ^{2} + 1 ) cos t}{4} = 0$ implies $cos t = 0$ (since $t^{2} + 1 > 0$ ). For $0 \leq t \leq 3$ the only solution is $t = \frac{π}{2} \approx 1.57$ ( $\frac{3 π}{2} \approx 4.71$ lies outside the interval). 3. At this instant $v$ changes from positive to negative, so the particle changes direction at $t \approx 1.57$ s.

b) 1. The acceleration is $a = \frac{d v}{d t}$ . By the product rule: $a = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4} .$ 2. Using the GDC, we solve $a (t) = - 1.9$ on $[0, 3]$ (intersection of $y = a (t)$ with $y = - 1.9$ ): there are two solutions, $t \approx 2.26$ s and $t \approx 2.96$ s.

c) 1. The speed is $∣ v ∣$ . On $[0, \frac{π}{2}]$ , $v \geq 0$ and its maximum value is approximately $0.29$ (using the GDC). 2. On $[\frac{π}{2}, 3]$ , $v < 0$ and $a = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4} < 0$ (since $cos t < 0$ and $sin t > 0$ ): $v$ keeps decreasing, so $∣ v ∣$ increases up to $t = 3$ , where $∣ v (3) ∣ = \frac{10∣ cos 3∣}{4} \approx 2.47$ m s $^{- 1}$ . 3. The maximum speed therefore occurs at $t = 3$ s. The acceleration there is $a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2} .$

Answer: a) $t = 1.57$ s; b) $t = 2.26$ s and $t = 2.96$ s; c) $a = - 1.84 m s^{- 2}$

Rolle's theorem, mean value theorem and L'Hôpital's rule

Exercise 246 BasicIB · P3 TZ0 May 2017

Use L'Hôpital's rule to determine the value of $x \to 0 lim \frac{sin ^{2} x}{x ln ( 1 + x )}$ .

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Answer: $x \to 0 lim \frac{sin ^{2} x}{x ln ( 1 + x )} = 1$

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Substituting $x = 0$ gives $\frac{0}{0}$ , an indeterminate form, so we can apply L'Hôpital's rule (differentiating numerator and denominator).
Numerator: $\frac{d}{d x} sin^{2} x = 2 sin x cos x = sin 2 x$ . Denominator (product): $\frac{d}{d x} [x ln (1 + x)] = ln (1 + x) + \frac{x}{1 + x}$ .
At $x = 0$ we still get $\frac{0}{0}$ , so we apply the rule again.
Numerator: $\frac{d}{d x} sin 2 x = 2 cos 2 x$ . Denominator: $\frac{1}{1 + x} + \frac{( 1 + x ) - x}{( 1 + x ) ^{2}} = \frac{1}{1 + x} + \frac{1}{( 1 + x ) ^{2}}$ .
At $x = 0$ : numerator $= 2$ and denominator $= 1 + 1 = 2$ .
The limit is $\frac{2}{2} = 1$ .

Answer: $x \to 0 lim \frac{sin ^{2} x}{x ln ( 1 + x )} = 1$

Exercise 247 BasicIB · P1 TZ1 May 2021

Use L'Hôpital's rule to calculate $x \to 0 lim (\frac{arctan 2 x}{tan 3 x})$ .

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Answer: $\frac{2}{3}$

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Substituting $x = 0$ : the numerator is $arctan 0 = 0$ and the denominator is $tan 0 = 0$ . This is an indeterminate form $\frac{0}{0}$ , so we apply L'Hôpital's rule.
Derivatives (chain rule): $\frac{d}{d x} arctan 2 x = \frac{2}{1 + 4 x ^{2}}$ and $\frac{d}{d x} tan 3 x = \frac{3}{cos ^{2} 3 x}$ .
The limit is $x \to 0 lim \frac{\frac{2}{1 + 4 x ^{2}}}{\frac{3}{c o s ^{2} 3 x}} = \frac{2/1}{3/1} = \frac{2}{3} .$

Answer: $\frac{2}{3}$

Exercise 248 MediumIB · P3 TZ0 November 2015Proof

The function $f : R \to R$ is given by $f : x \to {1, 1 - x, x < 0 x \geq 0$

Prove, using limits, that $f$

a) is continuous at $x = 0$ ;

b) is not differentiable at $x = 0$ .

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Answer: proved.

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a) A function is continuous at $x = 0$ if $x \to 0^{-} lim f (x) = x \to 0^{+} lim f (x) = f (0)$ .

From the left, $f (x) = 1$ , so $x \to 0^{-} lim f (x) = 1$ .
From the right, $f (x) = 1 - x$ , so $x \to 0^{+} lim f (x) = 1 - 0 = 1$ .
Also $f (0) = 1 - 0 = 1$ . The three values coincide, so $f$ is continuous at $x = 0$ .

b) We use the definition of the one-sided derivative, $f^{'} (0^{\pm}) = h \to 0^{\pm} lim \frac{f ( h ) - f ( 0 )}{h}$ .

From the left ( $h < 0$ ): $f^{'} (0^{-}) = h \to 0^{-} lim \frac{1 - 1}{h} = 0$ .
From the right ( $h > 0$ ): $f^{'} (0^{+}) = h \to 0^{+} lim \frac{( 1 - h ) - 1}{h} = h \to 0^{+} lim (- 1) = - 1$ .
The one-sided derivatives are different ( $0 \neq = - 1$ ), so the limit defining $f^{'} (0)$ does not exist and $f$ is not differentiable at $x = 0$ .

Answer: proved.

Exercise 249 MediumIB · P3 TZ2 May 2008

a) Find the value of $x \to 1 lim (\frac{ln x}{sin 2 π x})$ .

b) Using the series expansion of $e^{x^{2}}$ and that of $cos x$ , evaluate $x \to 0 lim (\frac{1 - e ^{x^{2}}}{1 - cos x})$ .

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Answer: a) $\frac{1}{2 π}$ ; b) $- 2$

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a) 1. Substituting $x = 1$ gives $\frac{ln 1}{sin 2 π} = \frac{0}{0}$ : we apply L'Hôpital's rule. 2. $x \to 1 lim \frac{ln x}{sin 2 π x} = x \to 1 lim \frac{1/ x}{2 π cos 2 π x} = \frac{1}{2 π \cdot 1} = \frac{1}{2 π}$ .

b) 1. $e^{x^{2}} = 1 + x^{2} + \frac{x ^{4}}{2} + \dots$ , so $1 - e^{x^{2}} = - x^{2} - \frac{x ^{4}}{2} - \dots$ . 2. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , so $1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ . 3. Dividing numerator and denominator by $x^{2}$ : $\frac{1 - e ^{x^{2}}}{1 - cos x} = \frac{- 1 - \frac{x ^{2}}{2} - \dots}{\frac{1}{2} - \frac{x ^{2}}{24} + \dots}$ . 4. As $x \to 0$ the limit is $\frac{- 1}{1/2} = - 2$ .

Answer: a) $\frac{1}{2 π}$ ; b) $- 2$

Exercise 250 MediumIB · P3 TZ0 November 2007

a) Determine $x \to 0 lim (\frac{ln ( a ^{2} + x ^{2} )}{ln ( a - x ^{3} )})$ , where $a$ is a positive constant different from $1$ .

b) Calculate $x \to 0 lim (\frac{ln ( 1 + x ^{2} )}{ln ( 1 - x ^{2} )})$

c) Calculate $x \to 0 lim (\frac{2 + x ^{2} - 2 cos x}{e ^{x} + e ^{- x} - 2 cos x})$

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Answer: a) $2$ ; b) $- 1$ ; c) $1$

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a) 1. Substituting $x = 0$ : $\frac{ln ( a ^{2} )}{ln a}$ , which is not indeterminate because $a > 0$ and $a \neq = 1$ imply $ln a \neq = 0$ , and the functions are continuous at $x = 0$ . 2. $\frac{ln ( a ^{2} )}{ln a} = \frac{2 ln a}{ln a} = 2$ .

b) 1. Substituting gives $\frac{0}{0}$ : L'Hôpital's rule. 2. $x \to 0 lim \frac{\frac{2 x}{1 + x ^{2}}}{\frac{- 2 x}{1 - x ^{2}}} = x \to 0 lim - \frac{1 - x ^{2}}{1 + x ^{2}} = - 1$ .

c) 1. Substituting gives $\frac{2 + 0 - 2}{1 + 1 - 2} = \frac{0}{0}$ . We use the Maclaurin expansions: $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , $e^{x} + e^{- x} = 2 + x^{2} + \frac{x ^{4}}{12} + \dots$ . 2. Numerator: $2 + x^{2} - 2 + x^{2} - \frac{x ^{4}}{12} + \dots = 2 x^{2} - \frac{x ^{4}}{12} + \dots$ . 3. Denominator: $2 + x^{2} + \frac{x ^{4}}{12} - 2 + x^{2} - \frac{x ^{4}}{12} + \dots = 2 x^{2} + \dots$ . 4. Dividing by $x^{2}$ : $\frac{2 - \frac{x ^{2}}{12} + \dots}{2 + \dots} \to \frac{2}{2} = 1$ .

Answer: a) $2$ ; b) $- 1$ ; c) $1$

Exercise 251 MediumIB · P3 TZ2 May 2007

Find: a) $s \to 4 lim (\frac{s - 3 s + 4}{4 - s})$

b) $x \to 0 lim (\frac{cos x - 1}{x sin x})$

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Answer: a) $- \frac{5}{8}$ ; b) $- \frac{1}{2}$

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a) 1. Substituting $s = 4$ : $\frac{4 - 4}{0} = \frac{0}{0}$ . We multiply numerator and denominator by the conjugate $s + 3 s + 4$ . 2. The numerator becomes $s^{2} - (3 s + 4) = s^{2} - 3 s - 4 = (s - 4) (s + 1)$ . 3. Therefore $\frac{s - 3 s + 4}{4 - s} = \frac{( s - 4 ) ( s + 1 )}{( 4 - s ) ( s + 3 s + 4 )} = - \frac{s + 1}{s + 3 s + 4}$ . 4. As $s \to 4$ : $- \frac{5}{4 + 4} = - \frac{5}{8}$ .

b) 1. Substituting $x = 0$ gives $\frac{0}{0}$ . We use the Maclaurin expansions: $cos x - 1 = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ and $x sin x = x^{2} - \frac{x ^{4}}{6} + \dots$ . 2. Dividing by $x^{2}$ : $\frac{- \frac{1}{2} + \frac{x ^{2}}{24} - \dots}{1 - \frac{x ^{2}}{6} + \dots} \to - \frac{1}{2}$ .

Answer: a) $- \frac{5}{8}$ ; b) $- \frac{1}{2}$

Exercise 252 MediumIB · P3 TZ0 May 2009

Find:

a) $x \to 0 lim \frac{tan x}{x + x ^{2}}$

b) $x \to 1 lim \frac{1 - x ^{2} + 2 x ^{2} ln x}{1 - sin \frac{π x}{2}}$

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Answer: a) $1$ ; b) $\frac{16}{π ^{2}}$

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a) 1. $tan x = x + \frac{x ^{3}}{3} + \dots$ , so $\frac{tan x}{x + x ^{2}} = \frac{x ( 1 + \frac{x ^{2}}{3} + \dots )}{x ( 1 + x )} = \frac{1 + \frac{x ^{2}}{3} + \dots}{1 + x}$ . 2. As $x \to 0$ the limit is $1$ .

b) 1. At $x = 1$ : numerator $1 - 1 + 0 = 0$ and denominator $1 - sin \frac{π}{2} = 0$ . We apply L'Hôpital's rule. 2. Derivatives: numerator $- 2 x + 4 x ln x + 2 x = 4 x ln x$ ; denominator $- \frac{π}{2} cos \frac{π x}{2}$ . At $x = 1$ we again get $\frac{0}{0}$ . 3. Second application: numerator $4 ln x + 4$ , which at $x = 1$ is $4$ ; denominator $\frac{π ^{2}}{4} sin \frac{π x}{2}$ , which at $x = 1$ is $\frac{π ^{2}}{4}$ . 4. The limit is $\frac{4}{π ^{2} /4} = \frac{16}{π ^{2}}$ .

Answer: a) $1$ ; b) $\frac{16}{π ^{2}}$

Exercise 253 MediumIB · P3 TZ0 May 2019

Using L'Hôpital's rule, find $x \to 0 lim (\frac{tan 3 x - 3 tan x}{sin 3 x - 3 sin x})$ .

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Answer: $- 2$

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Substituting $x = 0$ gives $\frac{0}{0}$ (the numerator and the denominator both equal $0$ ), so we apply L'Hôpital's rule.
First application: $\frac{3 sec ^{2} 3 x - 3 sec ^{2} x}{3 cos 3 x - 3 cos x}$ , which at $x = 0$ is again $\frac{0}{0}$ .
Second application: numerator $18 sec^{2} 3 x tan 3 x - 6 sec^{2} x tan x$ and denominator $- 9 sin 3 x + 3 sin x$ , again $\frac{0}{0}$ at $x = 0$ .
Third application: the numerator differentiates to $18 \cdot 3 (2 sec^{2} 3 x tan^{2} 3 x + sec^{4} 3 x) - 6 (2 sec^{2} x tan^{2} x + sec^{4} x)$ , which at $x = 0$ equals $54 - 6 = 48$ ; the denominator, $- 27 cos 3 x + 3 cos x$ , equals $- 27 + 3 = - 24$ .
The limit is $\frac{48}{- 24} = - 2$ .

Answer: $- 2$

Exercise 254 MediumIB · P3 TZ0 November 2017

The function $f$ is defined by $f (x) = {x^{2} - 2, a x + b, x < 1 x \geq 1$ where $a$ and $b$ are real constants.

Given that $f$ is differentiable and continuous at $x = 1$ , calculate the values of $a$ and $b$ .

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Answer: $a = 2$ , $b = - 3$

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Continuity at $x = 1$ : the left-hand limit is $1^{2} - 2 = - 1$ and the right-hand limit is $a + b$ . So $a + b = - 1$ .
Differentiability at $x = 1$ : from the left $f^{'} (x) = 2 x$ , which at $1$ is $2$ ; from the right $f^{'} (x) = a$ . So $a = 2$ .
From $a + b = - 1$ : $b = - 3$ .

Answer: $a = 2$ , $b = - 3$

Exercise 255 MediumIB · P3 TZ0 May 2016

a) Given that $f (x) = ln x$ , use the mean value theorem to show that, for $0 < a < b$ , $\frac{b - a}{b} < ln \frac{b}{a} < \frac{b - a}{a}$ .

b) Hence, show that $ln (1.2)$ lies between $\frac{1}{m}$ and $\frac{1}{n}$ , where $m$ , $n$ are consecutive positive integers to be determined.

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Answer: a) proved; b) $\frac{1}{6} < ln 1.2 < \frac{1}{5}$ , i.e. $m = 6$ and $n = 5$

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a) 1. We apply the mean value theorem to $f (x) = ln x$ on $[a, b]$ with $0 < a < b$ ( $f$ is continuous and differentiable there). There exists $c \in] a, b [$ with $f^{'} (c) = \frac{ln b - ln a}{b - a}$ , i.e. $\frac{1}{c} = \frac{ln \frac{b}{a}}{b - a}$ . 2. As $a < c < b$ and all are positive, $\frac{1}{b} < \frac{1}{c} < \frac{1}{a}$ . 3. Substituting for $\frac{1}{c}$ and multiplying by $b - a > 0$ : $\frac{b - a}{b} < ln \frac{b}{a} < \frac{b - a}{a}$ .

b) 1. Take $a = 1$ and $b = 1.2$ : $\frac{0.2}{1.2} < ln 1.2 < \frac{0.2}{1}$ . 2. $\frac{0.2}{1.2} = \frac{1}{6}$ and $0.2 = \frac{1}{5}$ . 3. Therefore $\frac{1}{6} < ln 1.2 < \frac{1}{5}$ , with $m = 6$ and $n = 5$ .

Answer: a) proved; b) $\frac{1}{6} < ln 1.2 < \frac{1}{5}$ , i.e. $m = 6$ and $n = 5$

Exercise 256 MediumIB · P3 TZ0 November 2016

Using L'Hôpital's rule, find $x \to \infty lim \frac{arcsin ( \frac{1}{x + 1} )}{\frac{1}{x}}$ .

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Answer: $x \to \infty lim \frac{arcsin ( \frac{1}{x + 1} )}{\frac{1}{x}} = 1$

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As $x \to \infty$ , $arcsin \frac{1}{x + 1} \to arcsin 0 = 0$ and $\frac{1}{x} \to 0$ : an indeterminate form $\frac{0}{0}$ , so L'Hôpital's rule can be applied.
Derivative of the numerator (chain rule): $\frac{1}{1 - \frac{1}{x + 1}} \cdot (- \frac{1}{2}) (x + 1)^{- 3/2} = \frac{x + 1}{x} \cdot (- \frac{1}{2}) (x + 1)^{- 3/2}$ .
Derivative of the denominator: $\frac{d}{d x} x^{- 1/2} = - \frac{1}{2} x^{- 3/2}$ .
The quotient of the derivatives is $\frac{x + 1}{x} \cdot \frac{( x + 1 ) ^{- 3/2}}{x ^{- 3/2}} = \frac{x + 1}{x} \cdot (\frac{x}{x + 1})^{3/2} = \frac{x}{x + 1} .$
$x \to \infty lim \frac{x}{x + 1} = 1$ , and by L'Hôpital's rule the required limit is $1$ .

Answer: $x \to \infty lim \frac{arcsin ( \frac{1}{x + 1} )}{\frac{1}{x}} = 1$

Exercise 257 MediumIB · P3 TZ0 May 2006

Find the value of

a) $x \to 0 lim \frac{1 + x - 1}{x}$ ;

b) $x \to 0 lim x ln x$ .

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Answer: a) $\frac{1}{2}$ ; b) $0$

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a) 1. Substituting $x = 0$ gives $\frac{0}{0}$ . Multiply the numerator and denominator by the conjugate $1 + x + 1$ : $\frac{1 + x - 1}{x} \cdot \frac{1 + x + 1}{1 + x + 1} = \frac{( 1 + x ) - 1}{x ( 1 + x + 1 )}$ . 2. Cancel $x$ : $\frac{1}{1 + x + 1}$ , and as $x \to 0$ this tends to $\frac{1}{1 + 1} = \frac{1}{2}$ .

b) 1. As $x \to 0^{+}$ (the logarithm requires $x > 0$ ), $x ln x$ is of the form $0 \cdot (- \infty)$ . Rewrite it: $x ln x = \frac{ln x}{1/ x}$ , of the form $\frac{- \infty}{\infty}$ . 2. Apply L'Hôpital's rule: $x \to 0^{+} lim \frac{1/ x}{- 1/ x ^{2}} = x \to 0^{+} lim (- x) = 0$ .

Answer: a) $\frac{1}{2}$ ; b) $0$

Exercise 258 MediumIB · P3 TZ1 May 2007

a) Use l'Hôpital's rule to calculate

(i) $x \to 1 lim \frac{ln x ^{2}}{x - 1}$

(ii) $x \to 0 lim \frac{tan ^{2} x}{1 - cos x}$ .

b) Giving a reason, state whether the following argument is correct or incorrect.

“Using l'Hôpital's rule, $x \to 3 lim \frac{x - 3}{x ^{2} - 3} = x \to 3 lim \frac{1}{2 x} = \frac{1}{6}$ ”.

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Answer: a) (i) $2$ ; (ii) $2$ ; b) incorrect: it is not of the form $\frac{0}{0}$ , the limit is $0$ .

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a) (i) 1. For $x \to 1$ , $ln x^{2} \to 0$ and $x - 1 \to 0$ : form $\frac{0}{0}$ , so we apply l'Hôpital's rule. 2. $x \to 1 lim \frac{ln x ^{2}}{x - 1} = x \to 1 lim \frac{2/ x}{1} = 2$ .

(ii) 1. For $x \to 0$ , $tan^{2} x \to 0$ and $1 - cos x \to 0$ : form $\frac{0}{0}$ . 2. Differentiate numerator and denominator: $x \to 0 lim \frac{2 tan x sec ^{2} x}{sin x}$ . 3. We still have $\frac{0}{0}$ ; simplify using $tan x = \frac{sin x}{cos x}$ : $\frac{2 sec ^{2} x}{cos x} = \frac{2}{cos ^{3} x}$ . 4. As $x \to 0$ : $\frac{2}{1} = 2$ .

b) 1. The argument is incorrect. 2. L'Hôpital's rule applies only to indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Here, as $x \to 3$ , the numerator tends to $0$ but the denominator tends to $3^{2} - 3 = 6 \neq = 0$ : it is not indeterminate. 3. The limit is directly $\frac{0}{6} = 0$ , not $\frac{1}{6}$ .

Answer: a) (i) $2$ ; (ii) $2$ ; b) incorrect: it is not of the form $\frac{0}{0}$ , the limit is $0$ .

Exercise 259 MediumIB · P3 TZ0 November 2010

Find $x \to 0 lim (\frac{1 - cos x ^{6}}{x ^{12}})$ .

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Answer: $x \to 0 lim \frac{1 - cos x ^{6}}{x ^{12}} = \frac{1}{2}$

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Let $u = x^{6}$ . As $x \to 0$ , $u \to 0$ , and $x^{12} = u^{2}$ . The limit becomes $u \to 0 lim \frac{1 - cos u}{u ^{2}}$ , which is of the form $\frac{0}{0}$ .
Apply l'Hôpital's rule: $u \to 0 lim \frac{sin u}{2 u}$ , again $\frac{0}{0}$ .
Apply l'Hôpital's rule again: $u \to 0 lim \frac{cos u}{2} = \frac{1}{2}$ .

Answer: $x \to 0 lim \frac{1 - cos x ^{6}}{x ^{12}} = \frac{1}{2}$

Exercise 260 MediumIB · P3 TZ0 November 2011

Calculate $x \to \frac{1}{2} lim \frac{\frac{1}{4} - x ^{2}}{cot π x}$ .

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Answer: $\frac{1}{π}$

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As $x \to \frac{1}{2}$ : the numerator $\frac{1}{4} - x^{2} \to 0$ and $cot π x = \frac{cos π x}{sin π x} \to \frac{0}{1} = 0$ . Form $\frac{0}{0}$ .
Apply l'Hôpital's rule: $\frac{d}{d x} (\frac{1}{4} - x^{2}) = - 2 x$ and $\frac{d}{d x} cot π x = - π csc^{2} π x$ .
The limit becomes $x \to \frac{1}{2} lim \frac{- 2 x}{- π csc ^{2} π x} = \frac{- 1}{- π \cdot 1} = \frac{1}{π}$ , because $csc \frac{π}{2} = 1$ .

Answer: $\frac{1}{π}$

Exercise 261 MediumIB · P3 TZ0 May 2012

Use L'Hôpital's rule to find: $x \to 0 lim \frac{e ^{x} - 1 - x cos x}{sin ^{2} x}$

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Answer: $\frac{1}{2}$

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Substituting $x = 0$ , the numerator equals $e^{0} - 1 - 0 \cdot cos 0 = 0$ and the denominator $sin^{2} 0 = 0$ : this is an indeterminate form $\frac{0}{0}$ , so we can apply L'Hôpital's rule.
We differentiate the numerator (product rule on $x cos x$ ) and the denominator (chain rule): $N^{'} (x) = e^{x} - cos x + x sin x, D^{'} (x) = 2 sin x cos x = sin 2 x .$
At $x = 0$ both again equal $0$ ( $1 - 1 + 0 = 0$ and $sin 0 = 0$ ), so we apply the rule once more: $N^{''} (x) = e^{x} + sin x + sin x + x cos x = e^{x} + 2 sin x + x cos x, D^{''} (x) = 2 cos 2 x .$
Now $N^{''} (0) = 1$ and $D^{''} (0) = 2$ , so $x \to 0 lim \frac{e ^{x} - 1 - x cos x}{sin ^{2} x} = \frac{1}{2} .$

Answer: $\frac{1}{2}$

Exercise 262 HardIB · P3 TZ0 May 2014

The function $f$ is given by $f (x) = {e^{- x^{2}} \cdot (- x^{3} + 2 x^{2} + x) a x + b x \leq 1 x > 1$ , where $a$ and $b$ are constants.

a) Find the exact values of $a$ and $b$ , given that $f$ is continuous and differentiable at $x = 1$ .

b) i) Use Rolle's Theorem applied to $f$ to show that $2 x^{4} - 4 x^{3} - 5 x^{2} + 4 x + 1 = 0$ has a root in the interval $] - 1, 1 [$ .

(ii) Hence, show that $2 x^{4} - 4 x^{3} - 5 x^{2} + 4 x + 1 = 0$ has at least two roots or a double root in the interval $] - 1, 1 [$ .

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Answer: a) $a = - \frac{2}{e}$ , $b = \frac{4}{e}$ ; b) shown.

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a) 1. Continuity at $x = 1$ : $f (1) = e^{- 1} (- 1 + 2 + 1) = \frac{2}{e}$ , and the limit from the right is $a + b$ . Therefore $a + b = \frac{2}{e}$ . 2. Derivative for $x < 1$ (product and chain rules): $f^{'} (x) = - 2 x e^{- x^{2}} (- x^{3} + 2 x^{2} + x) + e^{- x^{2}} (- 3 x^{2} + 4 x + 1) = e^{- x^{2}} (2 x^{4} - 4 x^{3} - 5 x^{2} + 4 x + 1)$ . 3. At $x = 1$ : $f^{'} (1) = e^{- 1} (2 - 4 - 5 + 4 + 1) = - \frac{2}{e}$ . For $x > 1$ , $f^{'} (x) = a$ . Being differentiable, $a = - \frac{2}{e}$ . 4. From $a + b = \frac{2}{e}$ : $b = \frac{2}{e} + \frac{2}{e} = \frac{4}{e}$ .

b) i) 1. On $[- 1, 1]$ , $f (x) = e^{- x^{2}} (- x^{3} + 2 x^{2} + x)$ is continuous and differentiable. 2. $f (- 1) = e^{- 1} (1 + 2 - 1) = \frac{2}{e}$ and $f (1) = \frac{2}{e}$ , so $f (- 1) = f (1)$ . 3. By Rolle's Theorem, there exists $c \in] - 1, 1 [$ with $f^{'} (c) = 0$ : $e^{- c^{2}} (2 c^{4} - 4 c^{3} - 5 c^{2} + 4 c + 1) = 0$ . 4. Since $e^{- c^{2}} \neq = 0$ , we have $2 c^{4} - 4 c^{3} - 5 c^{2} + 4 c + 1 = 0$ : the equation has a root in $] - 1, 1 [$ .

(ii) 1. Let $g (x) = 2 x^{4} - 4 x^{3} - 5 x^{2} + 4 x + 1$ , which is continuous as it is a polynomial. 2. $g (- 1) = 2 + 4 - 5 - 4 + 1 = - 2 < 0$ , $g (0) = 1 > 0$ and $g (1) = 2 - 4 - 5 + 4 + 1 = - 2 < 0$ . 3. By the Intermediate Value Theorem there is a root in $] - 1, 0 [$ and another in $] 0, 1 [$ , and they are distinct because $g (0) \neq = 0$ . 4. Hence $g$ has at least two distinct roots in $] - 1, 1 [$ .

Answer: a) $a = - \frac{2}{e}$ , $b = \frac{4}{e}$ ; b) shown.

Exercise 263 HardIB · P3 TZ0 May 2018

The function $f$ is defined by $f (x) = {∣ x - 2∣ + 1 a x^{2} + b x if x < 2 if x \geq 2$ where $a$ and $b$ are real constants.

Given that both $f$ and its derivative are continuous at $x = 2$ , find the value of $a$ and the value of $b$ .

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Answer: $a = - \frac{3}{4}$ , $b = 2$

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If $f$ is continuous at $x = 2$ , the one-sided limits coincide with $f (2)$ . From the left ( $x < 2$ ): $∣ x - 2∣ = 2 - x$ , so $f (x) = 3 - x$ and its limit at $2$ is $1$ . From the right: $4 a + 2 b$ . Therefore $4 a + 2 b = 1$ .
If $f$ is differentiable at $x = 2$ , the one-sided derivatives coincide. From the left: $f^{'} (x) = - 1$ . From the right: $f^{'} (x) = 2 a x + b$ , which at $x = 2$ equals $4 a + b$ . Therefore $4 a + b = - 1$ .
We subtract the second equation from the first: $(4 a + 2 b) - (4 a + b) = 1 - (- 1)$ , that is, $b = 2$ .
We substitute into $4 a + b = - 1$ : $4 a = - 3$ , so $a = - \frac{3}{4}$ .

Answer: $a = - \frac{3}{4}$ , $b = 2$

Exercise 264 HardIB · P3 TZ0 November 2017

The mean value theorem states that if $f$ is a continuous function on $[a, b]$ and differentiable on $] a, b [$ , then $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for some $c \in] a, b [$ .

The function $g$ , defined by $g (x) = x cos (x)$ , satisfies the conditions of the mean value theorem on the interval $[0, 5 π]$ .

a) For $a = 0$ and $b = 5 π$ , use the mean value theorem to calculate the possible values of $c$ for the function $g$ .

b) Sketch the graph of $y = g (x)$ on the interval $[0, 5 π]$ and hence illustrate the mean value theorem for the function $g$ .

View solution

a) 1. We find the average gradient on $[0, 5 π]$ : $g (0) = 0$ and $g (5 π) = 5 π cos 5 π$ , so $\frac{g ( 5 π ) - g ( 0 )}{5 π - 0} = cos 5 π \approx - 0.681$ . 2. We differentiate (product and chain rules): $g^{'} (x) = cos x - x sin x \cdot \frac{1}{2 x} = cos x - \frac{x}{2} sin x$ . 3. The mean value theorem guarantees that there exists $c \in] 0, 5 π [$ with $g^{'} (c) = cos 5 π$ . Using the GDC we solve $g^{'} (x) = - 0.681$ in $] 0, 5 π [$ (for example, by intersecting $y = g^{'} (x)$ with the line $y = - 0.681$ ). 4. The solutions are $c \approx 2.26$ and $c \approx 11.05$ .

b) 1. Using the GDC we sketch $y = g (x)$ on $[0, 5 π]$ : it starts at the origin, rises to a maximum near $(2.26; 0.154)$ , falls to a minimum near $(11.05; - 10.87)$ and rises slightly to $(5 π; - 10.7)$ . 2. The secant joining $(0, 0)$ and $(5 π; - 10.7)$ has gradient $- 0.681$ ; the tangents at $x = 2.26$ and $x = 11.05$ are parallel to it, which illustrates the theorem.

Graph of the function on the interval zero to five pi, with the line joining the endpoints and two tangents parallel to it at the marked points

Answer: a) $c \approx 2.26$ and $c \approx 11.05$ ; b) at the points $(2.26; 0.15)$ and $(11.05; - 10.87)$ the tangents are parallel to the line through $(0, 0)$ and $(5 π, - 10.7)$ .

Exercise 265 HardIB · P3 TZ0 November 2016

a) State the mean value theorem for a function that is continuous on the closed interval $[a, b]$ and differentiable on the open interval $] a, b [$ .

Let $f (x)$ be a function whose first and second derivatives exist on the closed interval $[0, h]$ .

Let $g (x) = f (h) - f (x) - (h - x) f^{'} (x) - \frac{( h - x ) ^{2}}{h ^{2}} (f (h) - f (0) - h f^{'} (0))$ .

b) (i) Find $g (0)$ .

(ii) Find $g (h)$ .

(iii) Apply the mean value theorem to the function $g (x)$ on the closed interval $[0, h]$ to show that there exists $c$ belonging to the open interval $] 0, h [$ such that $g^{'} (c) = 0$ .

(iv) Find $g^{'} (x)$ .

(v) Hence, show that $- (h - c) f^{''} (c) + \frac{2 ( h - c )}{h ^{2}} (f (h) - f (0) - h f^{'} (0)) = 0$ .

(vi) Deduce that $f (h) = f (0) + h f^{'} (0) + \frac{h ^{2}}{2} f^{''} (c)$ .

c) Hence, show that, for $h > 0$ , $1 - cos h \leq \frac{h ^{2}}{2}$ .

Show answer

Answer: b) (i) $g (0) = 0$ ; (ii) $g (h) = 0$ ; (iv) $g^{'} (x) = - (h - x) f^{''} (x) + \frac{2 ( h - x )}{h ^{2}} (f (h) - f (0) - h f^{'} (0))$ ; (v) and (vi) are proved: $f (h) = f (0) + h f^{'} (0) + \frac{h ^{2}}{2} f^{''} (c)$ ; c) proved: $1 - cos h \leq \frac{h ^{2}}{2}$ .

Show working and answer

a) If $f$ is continuous on $[a, b]$ and differentiable on $] a, b [$ , there exists at least one $c \in] a, b [$ such that $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ .

We write $K = f (h) - f (0) - h f^{'} (0)$ , so that $g (x) = f (h) - f (x) - (h - x) f^{'} (x) - \frac{( h - x ) ^{2}}{h ^{2}} K$ .

b) (i) 1. $g (0) = f (h) - f (0) - h f^{'} (0) - \frac{h ^{2}}{h ^{2}} K = K - K$ . 2. $g (0) = 0$ .

(ii) 1. $g (h) = f (h) - f (h) - 0 \cdot f^{'} (h) - 0 \cdot K$ . 2. $g (h) = 0$ .

(iii) 1. $g$ is continuous on $[0, h]$ and differentiable on $] 0, h [$ , because $f$ and $f^{'}$ are (the existence of $f^{'}$ and $f^{''}$ guarantees this). 2. By the mean value theorem there exists $c \in] 0, h [$ with $g^{'} (c) = \frac{g ( h ) - g ( 0 )}{h - 0} = \frac{0 - 0}{h} = 0$ .

(iv) 1. Differentiate each term: $- f^{'} (x)$ ; the product $- (h - x) f^{'} (x)$ gives $+ f^{'} (x) - (h - x) f^{''} (x)$ ; the last term, $- \frac{( h - x ) ^{2}}{h ^{2}} K$ , gives $+ \frac{2 ( h - x )}{h ^{2}} K$ (chain rule). 2. Adding: the terms $- f^{'} (x) + f^{'} (x)$ cancel, leaving $g^{'} (x) = - (h - x) f^{''} (x) + \frac{2 ( h - x )}{h ^{2}} K$ .

(v) 1. Evaluating at $x = c$ and using $g^{'} (c) = 0$ : $- (h - c) f^{''} (c) + \frac{2 ( h - c )}{h ^{2}} (f (h) - f (0) - h f^{'} (0)) = 0$ .

(vi) 1. Since $c < h$ , $h - c \neq = 0$ . Dividing by $h - c$ : $f^{''} (c) = \frac{2}{h ^{2}} (f (h) - f (0) - h f^{'} (0))$ . 2. Rearranging: $f (h) - f (0) - h f^{'} (0) = \frac{h ^{2}}{2} f^{''} (c)$ . 3. Therefore $f (h) = f (0) + h f^{'} (0) + \frac{h ^{2}}{2} f^{''} (c)$ .

c) 1. Take $f (x) = cos x$ : $f (0) = 1$ , $f^{'} (0) = - sin 0 = 0$ , $f^{''} (c) = - cos c$ . 2. By (vi), for $h > 0$ there exists $c \in] 0, h [$ with $cos h = 1 + 0 - \frac{h ^{2}}{2} cos c$ . 3. Then $1 - cos h = \frac{h ^{2}}{2} cos c$ . 4. As $cos c \leq 1$ and $\frac{h ^{2}}{2} > 0$ : $1 - cos h \leq \frac{h ^{2}}{2}$ .

Exercise 266 HardIB · P3 TZ0 May 2015

a) The mean value theorem states that if $f$ is a continuous function on $[a, b]$ and differentiable on $] a, b [$ , then $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for some $c \in] a, b [$ .

(i) Find the two possible values of $c$ for the function defined by $f (x) = x^{3} + 3 x^{2} - 2$ on the interval $[- 3, 1]$ .

(ii) Illustrate this result graphically.

b) (i) The function $f$ is continuous on $[a, b]$ , differentiable on $] a, b [$ and $f^{'} (x) = 0$ for all $x \in] a, b [$ . Show that $f (x)$ is constant on $[a, b]$ .

(ii) Hence, prove that for $x \in [0, 1]$ , $2 arccos x + arccos (1 - 2 x^{2}) = π$ .

View solution

a) (i) 1. $f$ is a polynomial, so it is continuous and differentiable. We calculate the gradient of the chord: $f (- 3) = - 27 + 27 - 2 = - 2$ and $f (1) = 1 + 3 - 2 = 2$ , so $\frac{f ( 1 ) - f ( - 3 )}{1 - ( - 3 )} = \frac{4}{4} = 1$ . 2. Differentiating: $f^{'} (x) = 3 x^{2} + 6 x$ . Setting this equal to $1$ : $3 x^{2} + 6 x - 1 = 0$ . 3. Solving: $x = \frac{- 6 \pm 36 + 12}{6} = - 1 \pm \frac{2 3}{3}$ , that is $c \approx 0.155$ and $c \approx - 2.155$ . Both lie in $] - 3, 1 [$ .

(ii) The chord joins $(- 3, - 2)$ and $(1, 2)$ (gradient $1$ ); at $x = - 2.155$ and at $x = 0.155$ the tangent to the curve is parallel to that chord.

Graph of a cubic function on the interval from -3 to 1, with the chord joining its endpoints and two tangent lines parallel to that chord

b) (i) 1. Take any two points $x_{1} < x_{2}$ in $[a, b]$ . $f$ is continuous on $[x_{1}, x_{2}]$ and differentiable on $] x_{1}, x_{2} [$ . 2. By the mean value theorem there exists $c \in] x_{1}, x_{2} [$ with $f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})$ . 3. As $f^{'} (c) = 0$ , $f (x_{2}) = f (x_{1})$ . So $f$ takes the same value at every point: it is constant on $[a, b]$ .

(ii) 1. Let $g (x) = 2 arccos x + arccos (1 - 2 x^{2})$ for $x \in [0, 1]$ . It is continuous on $[0, 1]$ and differentiable on $] 0, 1 [$ (if $x \in [0, 1]$ , then $1 - 2 x^{2} \in [- 1, 1]$ ). 2. We differentiate using the chain rule: $g^{'} (x) = - \frac{2}{1 - x ^{2}} - \frac{- 4 x}{1 - ( 1 - 2 x ^{2} ) ^{2}} .$ 3. We calculate $1 - (1 - 2 x^{2})^{2} = 4 x^{2} - 4 x^{4} = 4 x^{2} (1 - x^{2})$ and, as $x \geq 0$ , $\cdot = 2 x 1 - x^{2}$ . Then $g^{'} (x) = - \frac{2}{1 - x ^{2}} + \frac{4 x}{2 x 1 - x ^{2}} = - \frac{2}{1 - x ^{2}} + \frac{2}{1 - x ^{2}} = 0.$ 4. By (i), $g$ is constant on $[0, 1]$ . To find the constant we evaluate at $x = 0$ : $g (0) = 2 arccos 0 + arccos 1 = 2 \cdot \frac{π}{2} + 0 = π$ . 5. Therefore $2 arccos x + arccos (1 - 2 x^{2}) = π$ for all $x \in [0, 1]$ .

Answer: a) (i) $c = 0.155$ and $c = - 2.155$ ; (ii) graph; b) proof.

Exercise 267 HardIB · P2 TZ2 May 2022

Consider $x \to 0 lim \frac{arctan ( cos x ) - k}{x ^{2}}$ , where $k \in R$ .

a) Show that a finite limit exists only for $k = \frac{π}{4}$ .

b) Using L'Hôpital's rule, show by algebraic methods that the value of the limit is $- \frac{1}{4}$ .

Show answer

Answer: a) proof; b) the limit is $- \frac{1}{4}$ .

Show working and answer

a) 1. As $x \to 0$ the denominator $x^{2} \to 0$ and the numerator tends to $arctan (cos 0) - k = \frac{π}{4} - k$ . 2. If $\frac{π}{4} - k \neq = 0$ , the quotient would tend to $\pm \infty$ and the limit would not be finite. For the limit to be finite, the numerator must tend to $0$ , that is $k = \frac{π}{4}$ . 3. With $k = \frac{π}{4}$ we have the indeterminate form $\frac{0}{0}$ , which can be studied with L'Hôpital's rule (part b). A finite limit exists only for $k = \frac{π}{4}$ .

b) 1. With $k = \frac{π}{4}$ , we apply L'Hôpital's rule. Derivative of the numerator (chain rule): $\frac{- sin x}{1 + cos ^{2} x}$ ; of the denominator: $2 x$ . Therefore $x \to 0 lim \frac{arctan ( cos x ) - \frac{π}{4}}{x ^{2}} = x \to 0 lim \frac{- sin x}{2 x ( 1 + cos ^{2} x )} .$ 2. It is still $\frac{0}{0}$ , so we apply L'Hôpital's rule again. Derivatives: numerator $- cos x$ ; denominator $2 (1 + cos^{2} x) + 2 x (- 2 cos x sin x)$ . 3. At $x = 0$ : the numerator is $- 1$ and the denominator is $2 \cdot 2 + 0 = 4$ . 4. Therefore the limit is $- \frac{1}{4}$ .

Answer: a) proof; b) the limit is $- \frac{1}{4}$ .

Taylor and Maclaurin series

Exercise 268 BasicIB · P3 TZ1 May 2007

Given that the Maclaurin series for $e^{s i n x}$ is $a + b x + c x^{2} + d x^{3} + \dots$ , find the values of $a$ , $b$ , $c$ and $d$ .

Show answer

Answer: $a = 1$ , $b = 1$ , $c = \frac{1}{2}$ , $d = 0$

Show working and answer

Let $f (x) = e^{s i n x}$ . The Maclaurin coefficients are $f^{(n)} (0) / n!$ .
$f (0) = e^{0} = 1$ .
$f^{'} (x) = cos x e^{s i n x}$ (chain rule), so $f^{'} (0) = 1$ .
$f^{''} (x) = - sin x e^{s i n x} + cos^{2} x e^{s i n x}$ (product rule), so $f^{''} (0) = 0 + 1 = 1$ .
$f^{'''} (x) = - cos x e^{s i n x} - sin x cos x e^{s i n x} - 2 cos x sin x e^{s i n x} + cos^{3} x e^{s i n x}$ , so $f^{'''} (0) = - 1 - 0 - 0 + 1 = 0$ .
Coefficients: $a = 1$ , $b = \frac{1}{1 !} = 1$ , $c = \frac{1}{2 !} = \frac{1}{2}$ , $d = \frac{0}{3 !} = 0$ .

Answer: $a = 1$ , $b = 1$ , $c = \frac{1}{2}$ , $d = 0$

Exercise 269 MediumIB · P3 TZ2 May 2008

a) Given that $y = ln (cos x)$ , verify that the first two non-zero terms of the Maclaurin series for $y$ are $- \frac{x ^{2}}{2} - \frac{x ^{4}}{12}$ .

b) Use this series to find an approximation, in terms of $π$ , for $ln 2$ .

Show answer

Answer: a) verified; b) $ln 2 \approx \frac{π ^{2}}{18} + \frac{π ^{4}}{972}$

Show working and answer

a) 1. $y = ln (cos x)$ . At $x = 0$ : $y (0) = ln 1 = 0$ . 2. $y^{'} = \frac{- sin x}{cos x} = - tan x$ , and $y^{'} (0) = 0$ . 3. $y^{''} = - sec^{2} x$ , and $y^{''} (0) = - 1$ . 4. $y^{'''} = - 2 sec^{2} x tan x$ , and $y^{'''} (0) = 0$ . 5. $y^{(4)} = - 2 (2 sec^{2} x tan^{2} x + sec^{4} x)$ , and $y^{(4)} (0) = - 2$ . 6. Maclaurin series: $y = 0 + 0 \cdot x - \frac{1}{2 !} x^{2} + 0 \cdot x^{3} - \frac{2}{4 !} x^{4} + \dots = - \frac{x ^{2}}{2} - \frac{x ^{4}}{12} + \dots$ .

b) 1. Take $x = \frac{π}{3}$ : $ln (cos \frac{π}{3}) = ln \frac{1}{2} = - ln 2$ . 2. Using the series: $- ln 2 \approx - \frac{1}{2} (\frac{π}{3})^{2} - \frac{1}{12} (\frac{π}{3})^{4} = - \frac{π ^{2}}{18} - \frac{π ^{4}}{972}$ . 3. Therefore $ln 2 \approx \frac{π ^{2}}{18} + \frac{π ^{4}}{972}$ .

Answer: a) verified; b) $ln 2 \approx \frac{π ^{2}}{18} + \frac{π ^{4}}{972}$

Exercise 270 MediumIB · P3 TZ0 May 2011

a) Find the first three terms of the Maclaurin series for $ln (1 + e^{x})$ .

b) Hence, or otherwise, determine the value of $x \to 0 lim \frac{2 ln ( 1 + e ^{x} ) - x - ln 4}{x ^{2}}$ .

Show answer

Answer: a) $ln 2 + \frac{x}{2} + \frac{x ^{2}}{8}$ ; b) $\frac{1}{4}$

Show working and answer

a) 1. $g (x) = ln (1 + e^{x})$ and $g (0) = ln 2$ . 2. $g^{'} (x) = \frac{e ^{x}}{1 + e ^{x}}$ (chain rule), and $g^{'} (0) = \frac{1}{2}$ . 3. By the quotient rule, $g^{''} (x) = \frac{e ^{x} ( 1 + e ^{x} ) - e ^{x} \cdot e ^{x}}{( 1 + e ^{x} ) ^{2}} = \frac{e ^{x}}{( 1 + e ^{x} ) ^{2}}$ , and $g^{''} (0) = \frac{1}{4}$ . 4. Series: $g (x) \approx ln 2 + \frac{1}{2} x + \frac{1/4}{2 !} x^{2} = ln 2 + \frac{x}{2} + \frac{x ^{2}}{8}$ .

b) 1. Substituting the series: $2 ln (1 + e^{x}) \approx 2 ln 2 + x + \frac{x ^{2}}{4}$ . 2. The numerator: $2 ln (1 + e^{x}) - x - ln 4 \approx 2 ln 2 + x + \frac{x ^{2}}{4} - x - 2 ln 2 = \frac{x ^{2}}{4}$ (since $ln 4 = 2 ln 2$ ). 3. Dividing by $x^{2}$ and letting $x \to 0$ : the limit is $\frac{1}{4}$ .

Answer: a) $ln 2 + \frac{x}{2} + \frac{x ^{2}}{8}$ ; b) $\frac{1}{4}$

Exercise 271 MediumIB · P3 TZ0 May 2013

The Taylor series of $x$ about $x = 1$ is given by: $a_{0} + a_{1} (x - 1) + a_{2} (x - 1)^{2} + a_{3} (x - 1)^{3} + \dots$

a) Find the values of $a_{0}$ , $a_{1}$ , $a_{2}$ and $a_{3}$ .

b) Hence, or otherwise, find the value of $x \to 1 lim \frac{x - 1}{x - 1}$ .

Show answer

Answer: a) $a_{0} = 1$ , $a_{1} = \frac{1}{2}$ , $a_{2} = - \frac{1}{8}$ , $a_{3} = \frac{1}{16}$ ; b) $\frac{1}{2}$

Show working and answer

a) 1. The Taylor coefficients are $a_{k} = \frac{f ^{(k)} ( 1 )}{k !}$ with $f (x) = x^{1/2}$ . 2. $f^{'} (x) = \frac{1}{2} x^{- 1/2}$ , $f^{''} (x) = - \frac{1}{4} x^{- 3/2}$ , $f^{'''} (x) = \frac{3}{8} x^{- 5/2}$ . 3. At $x = 1$ : $f (1) = 1$ , $f^{'} (1) = \frac{1}{2}$ , $f^{''} (1) = - \frac{1}{4}$ , $f^{'''} (1) = \frac{3}{8}$ . 4. Therefore $a_{0} = 1$ , $a_{1} = \frac{1}{2}$ , $a_{2} = - \frac{1}{4 \cdot 2} = - \frac{1}{8}$ , $a_{3} = \frac{3}{8 \cdot 6} = \frac{1}{16}$ .

b) 1. Substituting the series: $x - 1 = \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} + \dots$ 2. Dividing by $x - 1$ : $\frac{x - 1}{x - 1} = \frac{1}{2} - \frac{1}{8} (x - 1) + \frac{1}{16} (x - 1)^{2} + \dots$ 3. As $x \to 1$ , all terms containing $(x - 1)$ tend to $0$ , and the limit is $\frac{1}{2}$ .

Answer: a) $a_{0} = 1$ , $a_{1} = \frac{1}{2}$ , $a_{2} = - \frac{1}{8}$ , $a_{3} = \frac{1}{16}$ ; b) $\frac{1}{2}$

Exercise 272 MediumIB · P3 TZ0 May 2015

The function $f$ is given by $f (x) = e^{- x} cos x + x - 1$ .

By finding an appropriate number of derivatives of $f$ , determine the first non-zero term of the corresponding Maclaurin series.

Show answer

Answer: $\frac{x ^{3}}{3}$

Show working and answer

We calculate the first derivatives of $f (x) = e^{- x} cos x + x - 1$ and evaluate them at $0$ .

$f (0) = 1 \cdot 1 + 0 - 1 = 0$ .
$f^{'} (x) = - e^{- x} cos x - e^{- x} sin x + 1$ (product rule), so $f^{'} (0) = - 1 - 0 + 1 = 0$ .
$f^{''} (x) = e^{- x} cos x + e^{- x} sin x + e^{- x} sin x - e^{- x} cos x = 2 e^{- x} sin x$ , so $f^{''} (0) = 0$ .
$f^{'''} (x) = 2 e^{- x} (cos x - sin x)$ , so $f^{'''} (0) = 2$ .
The first non-zero term of $\sum \frac{f ^{(n)} ( 0 )}{n !} x^{n}$ is the one for $n = 3$ : $\frac{2}{3 !} x^{3} = \frac{x ^{3}}{3}$ .

Answer: $\frac{x ^{3}}{3}$

Exercise 273 MediumIB · P3 TZ0 November 2015

Let $f (x) = e^{x} sin x$ .

a) Show that $f^{''} (x) = 2 (f^{'} (x) - f (x))$ .

b) Continue differentiating the result of part (a) and hence find the Maclaurin series expansion of $f (x)$ , up to and including the term in $x^{5}$ .

Show answer

Answer: a) proof; b) $f (x) = x + x^{2} + \frac{x ^{3}}{3} - \frac{x ^{5}}{30} + \dots$

Show working and answer

a) 1. We differentiate using the product rule: $f^{'} (x) = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x)$ . 2. We differentiate again: $f^{''} (x) = e^{x} (sin x + cos x) + e^{x} (cos x - sin x) = 2 e^{x} cos x$ . 3. On the other hand $f^{'} (x) - f (x) = e^{x} sin x + e^{x} cos x - e^{x} sin x = e^{x} cos x$ . 4. Comparing: $f^{''} (x) = 2 (f^{'} (x) - f (x))$ .

b) 1. Initial values: $f (0) = 0$ , $f^{'} (0) = e^{0} (0 + 1) = 1$ and, by a), $f^{''} (0) = 2 (1 - 0) = 2$ . 2. Differentiating the relation $f^{''} = 2 (f^{'} - f)$ successively: $f^{'''} = 2 (f^{''} - f^{'}), f^{(4)} = 2 (f^{'''} - f^{''}), f^{(5)} = 2 (f^{(4)} - f^{'''}) .$ 3. At $x = 0$ : $f^{'''} (0) = 2 (2 - 1) = 2$ ; $f^{(4)} (0) = 2 (2 - 2) = 0$ ; $f^{(5)} (0) = 2 (0 - 2) = - 4$ . 4. The Maclaurin series $\sum \frac{f ^{(n)} ( 0 )}{n !} x^{n}$ is $f (x) = 0 + x + \frac{2}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{- 4}{5 !} x^{5} + \dots = x + x^{2} + \frac{x ^{3}}{3} - \frac{x ^{5}}{30} + \dots$

Answer: a) proof; b) $f (x) = x + x^{2} + \frac{x ^{3}}{3} - \frac{x ^{5}}{30} + \dots$

Exercise 274 HardIB · P3 TZ0 November 2007

a) State the domain and range of the function $f (x) = arcsin (x)$ .

b) Determine the first two non-zero terms of the Maclaurin series for $f (x)$ .

c) Use the following approximation, valid for small angles, $cos (y) \approx 1 - \frac{y ^{2}}{2} + \frac{y ^{4}}{24}$ , to find a series for $cos (arcsin (x))$ up to and including the term in $x^{4}$ .

d) Find the Maclaurin series for $(p + q x^{2})^{r}$ up to and including the term $x^{4}$ , where $p, q, r \in R$ .

e) Find the values of $p$ , $q$ and $r$ so that the series in (d) is identical to the series in your answer to (c).

Show answer

Answer: a) $D_{f} = [- 1, 1]$ , $R_{f} = [- \frac{π}{2}, \frac{π}{2}]$ ; b) $x + \frac{x ^{3}}{6}$ ; c) $1 - \frac{x ^{2}}{2} - \frac{x ^{4}}{8}$ ; d) $p^{r} + r p^{r - 1} q x^{2} + \frac{r q ^{2} ( r - 1 ) p ^{r - 2}}{2} x^{4}$ ; e) $p = 1$ , $q = - 1$ , $r = \frac{1}{2}$

Show working and answer

a) $f (x) = arcsin x$ is defined for $- 1 \leq x \leq 1$ and takes values between $- \frac{π}{2}$ and $\frac{π}{2}$ : $D_{f} = [- 1, 1]$ and $R_{f} = [- \frac{π}{2}, \frac{π}{2}]$ .

b) 1. $f (0) = 0$ . 2. $f^{'} (x) = (1 - x^{2})^{- 1/2}$ , $f^{'} (0) = 1$ . 3. $f^{''} (x) = x (1 - x^{2})^{- 3/2}$ , $f^{''} (0) = 0$ . 4. $f^{'''} (x) = (1 - x^{2})^{- 3/2} + 3 x^{2} (1 - x^{2})^{- 5/2}$ , $f^{'''} (0) = 1$ . 5. Series: $arcsin x \approx x + \frac{1}{3 !} x^{3} = x + \frac{x ^{3}}{6}$ .

c) 1. Let $y = arcsin x \approx x + \frac{x ^{3}}{6}$ . Then $y^{2} \approx x^{2} + \frac{x ^{4}}{3}$ (up to $x^{4}$ ) and $y^{4} \approx x^{4}$ . 2. $cos y \approx 1 - \frac{1}{2} (x^{2} + \frac{x ^{4}}{3}) + \frac{x ^{4}}{24} = 1 - \frac{x ^{2}}{2} - \frac{x ^{4}}{6} + \frac{x ^{4}}{24}$ . 3. Simplifying: $cos (arcsin x) \approx 1 - \frac{x ^{2}}{2} - \frac{x ^{4}}{8}$ .

d) 1. Let $g (x) = (p + q x^{2})^{r}$ . $g (0) = p^{r}$ . 2. $g^{'} (x) = 2 q r x (p + q x^{2})^{r - 1}$ , $g^{'} (0) = 0$ . 3. $g^{''} (x) = 2 q r (p + q x^{2})^{r - 1} + 4 q^{2} r (r - 1) x^{2} (p + q x^{2})^{r - 2}$ , $g^{''} (0) = 2 q r p^{r - 1}$ . 4. $g^{'''} (x) = 12 q^{2} r (r - 1) x (p + q x^{2})^{r - 2} + 8 q^{3} r (r - 1) (r - 2) x^{3} (p + q x^{2})^{r - 3}$ , and $g^{'''} (0) = 0$ ; the fourth derivative at $0$ is $g^{(4)} (0) = 12 q^{2} r (r - 1) p^{r - 2}$ . 5. Series: $g (x) \approx p^{r} + \frac{2 q r p ^{r - 1}}{2} x^{2} + \frac{12 q ^{2} r ( r - 1 ) p ^{r - 2}}{24} x^{4} = p^{r} + r p^{r - 1} q x^{2} + \frac{r ( r - 1 ) q ^{2} p ^{r - 2}}{2} x^{4}$ .

e) 1. Equating with $1 - \frac{x ^{2}}{2} - \frac{x ^{4}}{8}$ : $p^{r} = 1$ , $r p^{r - 1} q = - \frac{1}{2}$ and $\frac{r ( r - 1 ) q ^{2} p ^{r - 2}}{2} = - \frac{1}{8}$ . 2. From $p^{r} = 1$ with $p > 0$ and $r \neq = 0$ we have $p = 1$ . This leaves $r q = - \frac{1}{2}$ and $r (r - 1) q^{2} = - \frac{1}{4}$ . 3. In the second: $(r q) (r - 1) q = - \frac{1}{4}$ , so $- \frac{1}{2} (r - 1) q = - \frac{1}{4}$ and $(r - 1) q = \frac{1}{2}$ , that is, $r q - q = \frac{1}{2}$ . 4. Substituting $r q = - \frac{1}{2}$ : $- \frac{1}{2} - q = \frac{1}{2}$ , so $q = - 1$ and $r = \frac{1}{2}$ . 5. This gives $(1 - x^{2})^{1/2}$ , which is consistent: $cos (arcsin x) = 1 - x^{2}$ .

Exercise 275 HardIB · P3 TZ0 November 2009

The function $f$ is defined as follows: $f (x) = e^{(e^{x} - 1)}$ .

a) Verify that the Maclaurin series for $f (x)$ is $1 + x + x^{2} + \frac{5}{6} x^{3} + \dots$

b) Hence, or otherwise, find the value of $x \to 0 lim \frac{f ( x ) - 1}{f ^{'} ( x ) - 1}$ .

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Answer: a) verified; b) $\frac{1}{2}$

Show working and answer

a) 1. $f (x) = e^{e^{x} - 1}$ , $f (0) = e^{0} = 1$ . 2. $f^{'} (x) = e^{x} f (x)$ (chain rule), $f^{'} (0) = 1$ . 3. $f^{''} (x) = e^{x} f (x) + e^{x} f^{'} (x) = (e^{x} + e^{2 x}) f (x)$ , and $f^{''} (0) = 2$ . 4. $f^{'''} (x) = (e^{x} + 2 e^{2 x}) f (x) + (e^{x} + e^{2 x}) e^{x} f (x)$ , and $f^{'''} (0) = 3 + 2 = 5$ . 5. Series: $f (x) \approx 1 + x + \frac{2}{2} x^{2} + \frac{5}{6} x^{3} = 1 + x + x^{2} + \frac{5}{6} x^{3} + \dots$ .

b) 1. $f (x) - 1 \approx x + x^{2} + \dots$ . 2. Differentiating the series, $f^{'} (x) \approx 1 + 2 x + \frac{5}{2} x^{2} + \dots$ , so $f^{'} (x) - 1 \approx 2 x + \frac{5}{2} x^{2} + \dots$ . 3. Dividing by $x$ : $\frac{f ( x ) - 1}{f ^{'} ( x ) - 1} \approx \frac{1 + x + \dots}{2 + \frac{5}{2} x + \dots} \to \frac{1}{2}$ .

Answer: a) verified; b) $\frac{1}{2}$

Exercise 276 HardIB · P3 TZ0 May 2009

The variables $x$ and $y$ are related by $\frac{d y}{d x} - y \cdot tan x = cos x$ .

Find the Maclaurin series for $y$ up to and including the term in $x^{2}$ , given that when $x = 0$ , $y = - \frac{π}{2}$ .

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Answer: $y \approx - \frac{π}{2} + x - \frac{π}{4} x^{2}$

Show working and answer

Rearranging for the derivative: $\frac{d y}{d x} = cos x + y tan x$ , with $y (0) = - \frac{π}{2}$ .
At $x = 0$ : $y^{'} (0) = cos 0 + y (0) tan 0 = 1 + 0 = 1$ .
Differentiating again (product rule): $\frac{d ^{2} y}{d x ^{2}} = - sin x + y^{'} tan x + y sec^{2} x$ .
At $x = 0$ : $y^{''} (0) = 0 + 0 + y (0) \cdot 1 = - \frac{π}{2}$ .
Maclaurin series up to $x^{2}$ : $y \approx y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2} x^{2} = - \frac{π}{2} + x - \frac{π}{4} x^{2}$ .

Answer: $y \approx - \frac{π}{2} + x - \frac{π}{4} x^{2}$

Exercise 277 HardIB · P3 TZ2 May 2007

Consider the equation $\frac{d y}{d x} + (2 x - 1) y = 0$ , given that $y = 2$ when $x = 0$ .

a) Verify that $\frac{d ^{4} y}{d x ^{4}} = (1 - 2 x) \frac{d ^{3} y}{d x ^{3}} - 6 \frac{d ^{2} y}{d x ^{2}}$ .

b) Obtain a Maclaurin series for $y$ , up to and including the term in $x^{4}$ .

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Answer: a) verified; b) $y \approx 2 + 2 x - x^{2} - \frac{5}{3} x^{3} + \frac{1}{12} x^{4}$

Show working and answer

a) 1. Starting from $\frac{d y}{d x} + (2 x - 1) y = 0$ , we differentiate (product rule): $\frac{d ^{2} y}{d x ^{2}} + 2 y + (2 x - 1) \frac{d y}{d x} = 0$ . 2. Differentiating again: $\frac{d ^{3} y}{d x ^{3}} + 2 \frac{d y}{d x} + 2 \frac{d y}{d x} + (2 x - 1) \frac{d ^{2} y}{d x ^{2}} = 0$ , that is, $\frac{d ^{3} y}{d x ^{3}} + 4 \frac{d y}{d x} + (2 x - 1) \frac{d ^{2} y}{d x ^{2}} = 0$ . 3. Differentiating once more: $\frac{d ^{4} y}{d x ^{4}} + 4 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d ^{2} y}{d x ^{2}} + (2 x - 1) \frac{d ^{3} y}{d x ^{3}} = 0$ . 4. Rearranging: $\frac{d ^{4} y}{d x ^{4}} = (1 - 2 x) \frac{d ^{3} y}{d x ^{3}} - 6 \frac{d ^{2} y}{d x ^{2}}$ .

b) 1. $y (0) = 2$ . 2. From the equation, $y^{'} = (1 - 2 x) y$ : $y^{'} (0) = 2$ . 3. From $y^{''} = - 2 y + (1 - 2 x) y^{'}$ : $y^{''} (0) = - 4 + 2 = - 2$ . 4. From $y^{'''} = - 4 y^{'} + (1 - 2 x) y^{''}$ : $y^{'''} (0) = - 8 - 2 = - 10$ . 5. From (a): $y^{(4)} (0) = y^{'''} (0) - 6 y^{''} (0) = - 10 + 12 = 2$ . 6. Series: $y \approx 2 + 2 x + \frac{- 2}{2} x^{2} + \frac{- 10}{6} x^{3} + \frac{2}{24} x^{4} = 2 + 2 x - x^{2} - \frac{5}{3} x^{3} + \frac{1}{12} x^{4}$ .

Answer: a) verified; b) $y \approx 2 + 2 x - x^{2} - \frac{5}{3} x^{3} + \frac{1}{12} x^{4}$

Exercise 278 HardIB · P3 TZ0 May 2018

The function $f$ is given by $f (x) = (arcsin x)^{2}$ , $- 1 \leq x \leq 1$ .

a) Show that $f^{'} (0) = 0$ .

The function $f$ satisfies the equation $(1 - x^{2}) f^{''} (x) - x f^{'} (x) - 2 = 0$ .

b) Differentiate the above equation twice to show that $(1 - x^{2}) f^{(4)} (x) - 5 x f^{(3)} (x) - 4 f^{''} (x) = 0$ , where $f^{(3)} (x)$ and $f^{(4)} (x)$ are respectively the 3rd and 4th derivatives of $f (x)$ .

c) Hence, show that the Maclaurin series of $f (x)$ up to and including the $x^{4}$ term is $x^{2} + \frac{1}{3} x^{4}$ .

d) Use this series approximation for $f (x)$ , with $x = \frac{1}{2}$ , to find an approximate value of $π^{2}$ .

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Answer: a) $f^{'} (0) = 0$ ; b) proved; c) $x^{2} + \frac{1}{3} x^{4}$ ; d) $π^{2} \approx \frac{39}{4} = 9.75$

Show working and answer

a) 1. We differentiate using the chain rule: $f^{'} (x) = 2 arcsin x \cdot \frac{1}{1 - x ^{2}}$ . 2. At $x = 0$ : $arcsin 0 = 0$ , so $f^{'} (0) = 0$ .

b) 1. We start from $(1 - x^{2}) f^{''} - x f^{'} - 2 = 0$ and differentiate (product rule on the first two terms): $(1 - x^{2}) f^{'''} - 2 x f^{''} - x f^{''} - f^{'} = 0$ , that is, $(1 - x^{2}) f^{(3)} - 3 x f^{''} - f^{'} = 0$ . 2. We differentiate again: $(1 - x^{2}) f^{(4)} - 2 x f^{(3)} - 3 x f^{(3)} - 3 f^{''} - f^{''} = 0$ . 3. We group terms: $(1 - x^{2}) f^{(4)} - 5 x f^{(3)} - 4 f^{''} = 0$ . This proves the result.

c) 1. We need $f (0), f^{'} (0), f^{''} (0), f^{(3)} (0), f^{(4)} (0)$ . We have $f (0) = 0$ and, from (a), $f^{'} (0) = 0$ . 2. In the given equation with $x = 0$ : $f^{''} (0) - 0 - 2 = 0$ , so $f^{''} (0) = 2$ . 3. In the first differentiated equation with $x = 0$ : $f^{(3)} (0) - 0 - f^{'} (0) = 0$ , so $f^{(3)} (0) = 0$ . 4. In the equation from part (b) with $x = 0$ : $f^{(4)} (0) - 0 - 4 f^{''} (0) = 0$ , so $f^{(4)} (0) = 8$ . 5. The Maclaurin series is $f (0) + f^{'} (0) x + \frac{f ^{''} ( 0 )}{2 !} x^{2} + \frac{f ^{(3)} ( 0 )}{3 !} x^{3} + \frac{f ^{(4)} ( 0 )}{4 !} x^{4} = x^{2} + \frac{8}{24} x^{4} = x^{2} + \frac{1}{3} x^{4}$ .

d) 1. With $x = \frac{1}{2}$ : $f (\frac{1}{2}) = (arcsin \frac{1}{2})^{2} = (\frac{π}{6})^{2} = \frac{π ^{2}}{36}$ . 2. Using the series: $f (\frac{1}{2}) \approx \frac{1}{4} + \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{4} + \frac{1}{48} = \frac{13}{48}$ . 3. Equating: $\frac{π ^{2}}{36} \approx \frac{13}{48}$ , so $π^{2} \approx \frac{36 \cdot 13}{48} = \frac{39}{4} = 9.75$ .

Answer: a) $f^{'} (0) = 0$ ; b) proved; c) $x^{2} + \frac{1}{3} x^{4}$ ; d) $π^{2} \approx \frac{39}{4} = 9.75$

Exercise 279 HardIB · P3 TZ0 November 2018

Consider the differential equation $(x + 2)^{2} \frac{d y}{d x} = (x + 1) y$ , where $x \neq = - 2$ , with the initial condition $y = 2$ when $x = 1$ .

a) Show that $\frac{d ^{3} y}{d x ^{3}} = - \frac{3 x + 7}{( x + 2 ) ^{2}} \frac{d ^{2} y}{d x ^{2}}$ .

Taylor polynomials about $x = 1$ are used to approximate $y (x)$ .

b) Find the Taylor polynomial of

(i) degree 2;

(ii) degree 3.

c) Find the difference between the approximate values of $y (1.05)$ obtained using each of the two answers given in part (b).

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Answer: b) (i) $y \approx 2 + \frac{4}{9} (x - 1) + \frac{1}{81} (x - 1)^{2}$ ; (ii) $y \approx 2 + \frac{4}{9} (x - 1) + \frac{1}{81} (x - 1)^{2} - \frac{10}{2187} (x - 1)^{3}$ ; c) the difference is $\frac{10}{2187} (0.05)^{3} \approx 5.7 \cdot 1 0^{- 7}$

Show working and answer

a) 1. We start from $(x + 2)^{2} y^{'} = (x + 1) y$ and differentiate with respect to $x$ (product rule on both sides): $2 (x + 2) y^{'} + (x + 2)^{2} y^{''} = y + (x + 1) y^{'}$ . 2. We rearrange: $(x + 2)^{2} y^{''} = y + (x + 1) y^{'} - 2 (x + 2) y^{'} = y - (x + 3) y^{'}$ . 3. We differentiate again: $2 (x + 2) y^{''} + (x + 2)^{2} y^{'''} = y^{'} - y^{'} - (x + 3) y^{''}$ , that is, $(x + 2)^{2} y^{'''} = - (x + 3) y^{''} - 2 (x + 2) y^{''} = - (3 x + 7) y^{''}$ . 4. We divide by $(x + 2)^{2}$ (with $x \neq = - 2$ ): $\frac{d ^{3} y}{d x ^{3}} = - \frac{3 x + 7}{( x + 2 ) ^{2}} \frac{d ^{2} y}{d x ^{2}}$ . This proves the result.

b) 1. Values at $x = 1$ , with $y (1) = 2$ :

From $(x + 2)^{2} y^{'} = (x + 1) y$ : $9 y^{'} (1) = 2 \cdot 2 = 4$ , so $y^{'} (1) = \frac{4}{9}$ .
From $(x + 2)^{2} y^{''} = y - (x + 3) y^{'}$ : $9 y^{''} (1) = 2 - 4 \cdot \frac{4}{9} = \frac{2}{9}$ , so $y^{''} (1) = \frac{2}{81}$ .
From $(x + 2)^{2} y^{'''} = - (3 x + 7) y^{''}$ : $9 y^{'''} (1) = - 10 \cdot \frac{2}{81}$ , so $y^{'''} (1) = - \frac{20}{729}$ .

The Taylor polynomial of degree $n$ is $k = 0 \sum n \frac{y ^{(k)} ( 1 )}{k !} (x - 1)^{k}$ .

(i) Degree 2: $y (x) \approx 2 + \frac{4}{9} (x - 1) + \frac{1}{2} \cdot \frac{2}{81} (x - 1)^{2} = 2 + \frac{4}{9} (x - 1) + \frac{1}{81} (x - 1)^{2}$ .

(ii) Degree 3: we add $\frac{1}{6} \cdot (- \frac{20}{729}) (x - 1)^{3} = - \frac{10}{2187} (x - 1)^{3}$ , so $y (x) \approx 2 + \frac{4}{9} (x - 1) + \frac{1}{81} (x - 1)^{2} - \frac{10}{2187} (x - 1)^{3}$ .

c) 1. The difference between the two approximations at $x = 1.05$ is exactly the cubic term. 2. $- \frac{10}{2187} (0.05)^{3} = \frac{10}{2187} \cdot 0.000125 \approx 5.7 \cdot 1 0^{- 7}$ (the degree-3 term is slightly smaller).

Exercise 280 HardIB · P3 TZ0 May 2017

Let the Maclaurin series for $tan x$ be $tan x = a_{1} x + a_{3} x^{3} + a_{5} x^{5} + \dots$ , where $a_{1}$ , $a_{3}$ and $a_{5}$ are constants.

a) Find the series for $sec^{2} x$ in terms of $a_{1}$ , $a_{3}$ and $a_{5}$ , up to and including the $x^{4}$ term,

(i) by differentiating the series for $tan x$ given above;

(ii) by using the relation $sec^{2} x = 1 + tan^{2} x$ .

b) Hence, by comparing the two series obtained, determine the values of $a_{1}$ , $a_{3}$ and $a_{5}$ .

Show answer

Answer: a) (i) $sec^{2} x = a_{1} + 3 a_{3} x^{2} + 5 a_{5} x^{4} + \dots$ ; (ii) $sec^{2} x = 1 + a_{1}^{2} x^{2} + 2 a_{1} a_{3} x^{4} + \dots$ ; b) $a_{1} = 1$ , $a_{3} = \frac{1}{3}$ , $a_{5} = \frac{2}{15}$

Show working and answer

a) (i) 1. We differentiate the series $tan x = a_{1} x + a_{3} x^{3} + a_{5} x^{5} + \dots$ term by term. 2. Since $\frac{d}{d x} tan x = sec^{2} x$ : $sec^{2} x = a_{1} + 3 a_{3} x^{2} + 5 a_{5} x^{4} + \dots$ .

(ii) 1. We use $sec^{2} x = 1 + tan^{2} x$ and square the series, keeping terms up to $x^{4}$ : $tan^{2} x = (a_{1} x + a_{3} x^{3} + \dots)^{2} = a_{1}^{2} x^{2} + 2 a_{1} a_{3} x^{4} + \dots$ . 2. Therefore $sec^{2} x = 1 + a_{1}^{2} x^{2} + 2 a_{1} a_{3} x^{4} + \dots$ .

b) 1. We equate the coefficients of the two series (a function has a unique Maclaurin series). 2. Constant term: $a_{1} = 1$ . 3. Coefficient of $x^{2}$ : $3 a_{3} = a_{1}^{2} = 1$ , so $a_{3} = \frac{1}{3}$ . 4. Coefficient of $x^{4}$ : $5 a_{5} = 2 a_{1} a_{3} = \frac{2}{3}$ , so $a_{5} = \frac{2}{15}$ .

Exercise 281 HardIB · P3 TZ0 November 2017

Consider the function $f (x) = sin (p arcsin x)$ , $- 1 < x < 1$ and $p \in R$ .

a) Show that $f^{'} (0) = p$ .

The function $f$ and its derivatives satisfy $(1 - x^{2}) f^{(n + 2)} (x) - (2 n + 1) x f^{(n + 1)} (x) + (p^{2} - n^{2}) f^{(n)} (x) = 0$ , $n \in N$ where $f^{(n)} (x)$ denotes the $n$ th derivative of $f (x)$ and $f^{(0)} (x)$ is $f (x)$ .

b) Show that $f^{(n + 2)} (0) = (n^{2} - p^{2}) f^{(n)} (0)$ .

c) For $p \in R - {\pm 1, \pm 3}$ , show that the Maclaurin series for $f (x)$ , up to and including the term in $x^{5}$ , is $p x + \frac{p ( 1 - p ^{2} )}{3 !} x^{3} + \frac{p ( 9 - p ^{2} ) ( 1 - p ^{2} )}{5 !} x^{5}$ .

d) Using the above or otherwise, calculate $x \to 0 lim \frac{sin ( p arcsin x )}{x}$ .

e) If $p$ is an odd integer, prove that the Maclaurin series for $f (x)$ is a polynomial of degree $p$ .

Show answer

Answer: a) $f^{'} (0) = p$ ; b) $f^{(n + 2)} (0) = (n^{2} - p^{2}) f^{(n)} (0)$ ; c) proved; d) $x \to 0 lim \frac{sin ( p arcsin x )}{x} = p$ ; e) the Maclaurin series is a polynomial of degree $p$ .

Show working and answer

a) 1. We differentiate using the chain rule: $f^{'} (x) = cos (p arcsin x) \cdot \frac{p}{1 - x ^{2}}$ . 2. At $x = 0$ : $cos 0 \cdot \frac{p}{1} = p$ . Hence $f^{'} (0) = p$ .

b) 1. We substitute $x = 0$ into $(1 - x^{2}) f^{(n + 2)} (x) - (2 n + 1) x f^{(n + 1)} (x) + (p^{2} - n^{2}) f^{(n)} (x) = 0$ . 2. This gives $f^{(n + 2)} (0) + (p^{2} - n^{2}) f^{(n)} (0) = 0$ . 3. Rearranging: $f^{(n + 2)} (0) = (n^{2} - p^{2}) f^{(n)} (0)$ .

c) 1. $f (0) = sin 0 = 0$ and, by (a), $f^{'} (0) = p$ . 2. Using the formula from (b) with $n = 0$ : $f^{''} (0) = - p^{2} f (0) = 0$ . 3. $n = 1$ : $f^{(3)} (0) = (1 - p^{2}) f^{'} (0) = p (1 - p^{2})$ . 4. $n = 2$ : $f^{(4)} (0) = (4 - p^{2}) f^{''} (0) = 0$ . 5. $n = 3$ : $f^{(5)} (0) = (9 - p^{2}) f^{(3)} (0) = p (9 - p^{2}) (1 - p^{2})$ . 6. Maclaurin series: $f (0) + f^{'} (0) x + \frac{f ^{''} ( 0 )}{2 !} x^{2} + \frac{f ^{(3)} ( 0 )}{3 !} x^{3} + \frac{f ^{(4)} ( 0 )}{4 !} x^{4} + \frac{f ^{(5)} ( 0 )}{5 !} x^{5} = p x + \frac{p ( 1 - p ^{2} )}{3 !} x^{3} + \frac{p ( 9 - p ^{2} ) ( 1 - p ^{2} )}{5 !} x^{5}$ .

d) 1. From the series, $\frac{f ( x )}{x} = p + \frac{p ( 1 - p ^{2} )}{3 !} x^{2} + \dots$ . 2. As $x \to 0$ only the first term remains: the limit is $p$ . (Also using L'Hôpital: $\frac{f ^{'} ( x )}{1} \to f^{'} (0) = p$ .)

e) 1. Since $f (0) = 0$ , the formula from (b) gives $f^{(n)} (0) = 0$ for every even $n$ (each even derivative is obtained from the previous even one). The coefficients of the even powers are zero. 2. Take $p$ a positive odd integer (if $p < 0$ , $f = - sin (∣ p ∣ arcsin x)$ and the same argument applies with $∣ p ∣$ ). For odd $n$ less than $p$ , the factor $n^{2} - p^{2} \neq = 0$ , so $f^{(n)} (0) \neq = 0$ up to and including the derivative of order $p$ (starting with $f^{'} (0) = p \neq = 0$ ). 3. For $n = p$ : $f^{(p + 2)} (0) = (p^{2} - p^{2}) f^{(p)} (0) = 0$ , and from there all subsequent odd derivatives are also $0$ . 4. All coefficients of degree greater than $p$ are zero, and the coefficient of $x^{p}$ is not: the series is a polynomial of degree $p$ .

Exercise 282 HardIB · P3 TZ0 May 2016

The function $f$ is defined by $f (x) = e^{x} sin x$ , $x \in R$ .

a) By finding a suitable number of derivatives of $f$ , determine the Maclaurin series for $f (x)$ up to and including the term in $x^{3}$ .

b) Using the above or otherwise, determine the exact value of $x \to 0 lim \frac{e ^{x} sin x - x - x ^{2}}{x ^{3}}$ .

c) The Maclaurin series is to be used to find an approximate value of $f (0.5)$ .

(i) Use the Lagrange form of the remainder term to find an upper bound for the absolute value of the error in this approximation.

(ii) From the Lagrange remainder term, deduce whether the approximation will be greater or less than the true value of $f (0.5)$ .

Show answer

Answer: a) $f (x) \approx x + x^{2} + \frac{x ^{3}}{3}$ ; b) $\frac{1}{3}$ ; c) (i) $∣ R ∣ \leq 0.0082$ ; (ii) the approximation is greater than the true value, because the Lagrange remainder is negative.

Show working and answer

a) 1. $f (x) = e^{x} sin x$ . Differentiating with the product rule: $f^{'} (x) = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x)$ . 2. $f^{''} (x) = e^{x} (sin x + cos x) + e^{x} (cos x - sin x) = 2 e^{x} cos x$ . 3. $f^{'''} (x) = 2 e^{x} cos x - 2 e^{x} sin x = 2 e^{x} (cos x - sin x)$ . 4. At $x = 0$ : $f (0) = 0$ , $f^{'} (0) = 1$ , $f^{''} (0) = 2$ , $f^{'''} (0) = 2$ . 5. The series up to $x^{3}$ is $x + \frac{2}{2 !} x^{2} + \frac{2}{3 !} x^{3} = x + x^{2} + \frac{x ^{3}}{3}$ .

b) 1. Substituting $e^{x} sin x = x + x^{2} + \frac{x ^{3}}{3} + O (x^{4})$ : $\frac{e ^{x} sin x - x - x ^{2}}{x ^{3}} = \frac{\frac{x ^{3}}{3} + O ( x ^{4} )}{x ^{3}} = \frac{1}{3} + O (x)$ . 2. The limit is $\frac{1}{3}$ .

c) (i) 1. The Lagrange remainder after the term of degree 3 is $R = \frac{f ^{(4)} ( c )}{4 !} x^{4}$ with $c$ between $0$ and $x$ . For $x = 0.5$ , $0 < c < 0.5$ . 2. $f^{(4)} (x) = \frac{d}{d x} [2 e^{x} (cos x - sin x)] = 2 e^{x} (cos x - sin x) + 2 e^{x} (- sin x - cos x) = - 4 e^{x} sin x$ . 3. Then $∣ R ∣ = \frac{4 e ^{c} sin c}{24} \cdot 0. 5^{4}$ . The function $e^{c} sin c$ is increasing on $] 0, 0.5 [$ (its derivative $e^{c} (sin c + cos c) > 0$ ), so its maximum value is at $c = 0.5$ : $e^{0.5} sin 0.5 \approx 0.7904$ . 4. Bound: $∣ R ∣ \leq \frac{4 \cdot 0.7904}{24} \cdot 0.0625 \approx 0.0082$ .

(ii) 1. For $0 < c < 0.5$ , $e^{c} > 0$ and $sin c > 0$ , so $f^{(4)} (c) < 0$ and $R < 0$ . 2. Since $f (0.5) = P_{3} (0.5) + R$ with $R < 0$ , the true value is less than the approximation: the approximation is greater than $f (0.5)$ .

Exercise 283 HardIB · P3 TZ1 May 2008

The function $f$ is defined by $f (x) = ln (\frac{1}{1 - x})$ .

a) Write down the constant term in the Maclaurin series for $f (x)$ .

b) Calculate the first three derivatives of $f (x)$ and then show that the Maclaurin series for $f (x)$ up to and including the term in $x^{3}$ is $x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}$ .

c) Use this series to find an approximate value for $ln 2$ .

d) Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.

e) How good is this upper bound as an estimate of the actual error?

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Answer: a) $0$ ; b) $f^{'} (x) = (1 - x)^{- 1}$ , $f^{''} (x) = (1 - x)^{- 2}$ , $f^{'''} (x) = 2 (1 - x)^{- 3}$ and series $x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}$ ; c) $ln 2 \approx \frac{2}{3}$ ; d) $R_{3} (\frac{1}{2}) \leq 0.25$ ; e) the actual error is $\approx 0.0265$ : the bound is much larger, so it is not a good estimate.

Show working and answer

a) 1. $f (0) = ln \frac{1}{1 - 0} = ln 1 = 0$ : the constant term is $0$ .

b) 1. We rewrite $f (x) = - ln (1 - x)$ . 2. First derivative (chain rule): $f^{'} (x) = \frac{1}{1 - x} = (1 - x)^{- 1}$ . 3. Second derivative: $f^{''} (x) = (1 - x)^{- 2}$ . 4. Third derivative: $f^{'''} (x) = 2 (1 - x)^{- 3}$ . 5. At $0$ : $f (0) = 0$ , $f^{'} (0) = 1$ , $f^{''} (0) = 1$ , $f^{'''} (0) = 2$ . 6. Maclaurin series: $f (x) \approx 0 + 1 \cdot x + \frac{1}{2 !} x^{2} + \frac{2}{3 !} x^{3} = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}$ .

c) 1. With $x = \frac{1}{2}$ : $f (\frac{1}{2}) = ln \frac{1}{1/2} = ln 2$ . 2. Approximation: $ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{24} = \frac{12 + 3 + 1}{24} = \frac{16}{24} = \frac{2}{3}$ .

d) 1. Lagrange form: $R_{3} (x) = \frac{f ^{(4)} ( c )}{4 !} x^{4}$ with $c$ between $0$ and $x$ . 2. Fourth derivative: $f^{(4)} (x) = 6 (1 - x)^{- 4}$ . 3. With $x = \frac{1}{2}$ and $0 < c < \frac{1}{2}$ : $R_{3} (\frac{1}{2}) = \frac{6}{24 ( 1 - c ) ^{4}} \cdot \frac{1}{16} = \frac{1}{64 ( 1 - c ) ^{4}}$ . 4. Since $1 - c > \frac{1}{2}$ , $(1 - c)^{4} > \frac{1}{16}$ , so $\frac{1}{( 1 - c ) ^{4}} < 16$ . 5. Hence $R_{3} (\frac{1}{2}) < \frac{16}{64} = \frac{1}{4} = 0.25$ .

e) 1. The actual error is $ln 2 - \frac{2}{3} \approx 0.6931 - 0.6667 = 0.0265$ . 2. The bound $0.25$ is almost ten times larger than the actual error: it is a valid bound but a very poor one, and it does not give a good estimate of the actual error.

Exercise 284 HardIB · P3 TZ0 November 2011

a) Given that $y = ln (\frac{1 + e ^{- x}}{2})$ , show that $\frac{d y}{d x} = \frac{e ^{- y}}{2} - 1$ .

b) Using the above, by repeatedly differentiating the differential equation, find the Maclaurin series for $y$ up to and including the term in $x^{3}$ , showing that two of the terms are zero.

Show answer

Answer: a) proven; b) $y \approx 0 - \frac{1}{2} x + \frac{1}{8} x^{2} + 0 x^{3} + \dots$

Show working and answer

a) 1. Differentiate using the chain rule: $\frac{d y}{d x} = \frac{- e ^{- x}}{1 + e ^{- x}}$ . 2. By the definition of $y$ : $e^{y} = \frac{1 + e ^{- x}}{2}$ , so $1 + e^{- x} = 2 e^{y}$ and $e^{- x} = 2 e^{y} - 1$ . 3. Substitute: $\frac{d y}{d x} = - \frac{2 e ^{y} - 1}{2 e ^{y}} = - 1 + \frac{1}{2 e ^{y}} = \frac{e ^{- y}}{2} - 1$ .

b) 1. Differentiate $y^{'} = \frac{e ^{- y}}{2} - 1$ with respect to $x$ : $y^{''} = - \frac{1}{2} e^{- y} y^{'}$ . 2. Differentiate again (product rule): $y^{'''} = \frac{1}{2} e^{- y} (y^{'})^{2} - \frac{1}{2} e^{- y} y^{''}$ . 3. At $x = 0$ : $y (0) = ln \frac{1 + 1}{2} = 0$ . 4. $y^{'} (0) = \frac{e ^{0}}{2} - 1 = - \frac{1}{2}$ . 5. $y^{''} (0) = - \frac{1}{2} \cdot 1 \cdot (- \frac{1}{2}) = \frac{1}{4}$ . 6. $y^{'''} (0) = \frac{1}{2} \cdot \frac{1}{4} - \frac{1}{2} \cdot \frac{1}{4} = 0$ . 7. Maclaurin series: $y = 0 + (- \frac{1}{2}) x + \frac{1/4}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \dots = - \frac{1}{2} x + \frac{1}{8} x^{2} + 0 \cdot x^{3} + \dots$ . The constant term and the $x^{3}$ term are both zero.

Answer: a) proven; b) $y \approx 0 - \frac{1}{2} x + \frac{1}{8} x^{2} + 0 x^{3} + \dots$

Exercise 285 HardIB · P2 TZ2 May 2021

a) Write down the first three terms of the expansion of $(1 + t)^{- 1}$ in ascending powers of $t$ .

b) Using the Maclaurin series for $cos x$ and the result obtained in part (a), show that the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24}$ .

c) Using the Maclaurin series for $arctan x$ and the result obtained in part (b), find $x \to 0 lim \frac{x arctan 2 x}{sec x - 1}$ .

Show answer

Answer: a) $1 - t + t^{2}$ ; b) proof; c) $4$

Show working and answer

a) 1. Binomial expansion for a negative exponent: $(1 + t)^{- 1} = 1 + (- 1) t + \frac{( - 1 ) ( - 2 )}{2 !} t^{2} + \dots = 1 - t + t^{2} - \dots$

b) 1. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , and $sec x = \frac{1}{cos x} = (1 + t)^{- 1}$ with $t = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . 2. By part a): $sec x \approx 1 - t + t^{2}$ . We substitute, keeping terms up to $x^{4}$ : $1 - (- \frac{x ^{2}}{2} + \frac{x ^{4}}{24}) + (- \frac{x ^{2}}{2})^{2} = 1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \frac{x ^{4}}{4} .$ 3. As $- \frac{1}{24} + \frac{1}{4} = \frac{5}{24}$ : $sec x \approx 1 + \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24}$ .

c) 1. $arctan u = u - \frac{u ^{3}}{3} + \dots$ , so $arctan 2 x = 2 x - \frac{8 x ^{3}}{3} + \dots$ and $x arctan 2 x = 2 x^{2} - \frac{8 x ^{4}}{3} + \dots$ . 2. By b): $sec x - 1 = \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24} + \dots$ . 3. We divide numerator and denominator by $x^{2}$ : $x \to 0 lim \frac{2 x ^{2} - \frac{8}{3} x ^{4} + \dots}{\frac{x ^{2}}{2} + \frac{5}{24} x ^{4} + \dots} = x \to 0 lim \frac{2 - \frac{8}{3} x ^{2} + \dots}{\frac{1}{2} + \frac{5}{24} x ^{2} + \dots} = \frac{2}{1/2} = 4.$

Answer: a) $1 - t + t^{2}$ ; b) proof; c) $4$

Exercise 286 HardIB · P1 TZ1 May 2021

Let $f (x) = 1 + x$ for $x > - 1$ .

a) Show that $f^{''} (x) = - \frac{1}{4 ( 1 + x ) ^{3}}$ .

b) Use mathematical induction to prove that $f^{(n)} (x) = (- \frac{1}{4})^{n - 1} \cdot \frac{( 2 n - 3 )!}{( n - 2 )!} \cdot (1 + x)^{\frac{1}{2} - n}$ , for $n \in Z$ , $n \geq 2$ .

Let $g (x) = e^{m x}$ , $m \in Q$ .

Consider the function $h$ defined by $h (x) = f (x) \times g (x)$ for $x > - 1$ .

c) Given that the term in $x^{2}$ in the Maclaurin series for $h (x)$ has coefficient $\frac{7}{4}$ , calculate the possible values of $m$ .

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Answer: a) proof; b) proof by induction; c) $m = \frac{3}{2}$ and $m = - \frac{5}{2}$

Show working and answer

a) 1. We write $f (x) = (1 + x)^{1/2}$ . Then $f^{'} (x) = \frac{1}{2} (1 + x)^{- 1/2}$ . 2. $f^{''} (x) = \frac{1}{2} \cdot (- \frac{1}{2}) (1 + x)^{- 3/2} = - \frac{1}{4} (1 + x)^{- 3/2} = - \frac{1}{4 ( 1 + x ) ^{3}}$ .

b) Proof by induction of $P (n) : f^{(n)} (x) = (- \frac{1}{4})^{n - 1} \frac{( 2 n - 3 )!}{( n - 2 )!} (1 + x)^{\frac{1}{2} - n}$ , for $n \geq 2$ .

Base case $n = 2$ : $(- \frac{1}{4})^{1} \frac{1 !}{0 !} (1 + x)^{- 3/2} = - \frac{1}{4} (1 + x)^{- 3/2}$ , which agrees with part a). So $P (2)$ holds.
Hypothesis: suppose $P (k)$ is true for some $k \geq 2$ : $f^{(k)} (x) = (- \frac{1}{4})^{k - 1} \frac{( 2 k - 3 )!}{( k - 2 )!} (1 + x)^{\frac{1}{2} - k}$ .
Differentiating (power rule): $f^{(k + 1)} (x) = (- \frac{1}{4})^{k - 1} \frac{( 2 k - 3 )!}{( k - 2 )!} (\frac{1}{2} - k) (1 + x)^{- \frac{1}{2} - k} .$
We rewrite the coefficient: $(- \frac{1}{4})^{k - 1} (\frac{1}{2} - k) = (- \frac{1}{4})^{k} (- 4) \cdot \frac{1 - 2 k}{2} = (- \frac{1}{4})^{k} \cdot 2 (2 k - 1)$ . Then $f^{(k + 1)} (x) = (- \frac{1}{4})^{k} \frac{2 ( 2 k - 1 ) ( 2 k - 3 )!}{( k - 2 )!} (1 + x)^{\frac{1}{2} - (k + 1)} .$
We use $(2 k - 1)! = (2 k - 1) (2 k - 2) (2 k - 3)! = 2 (k - 1) (2 k - 1) (2 k - 3)!$ and $(k - 1)! = (k - 1) (k - 2)!$ . Multiplying and dividing by $(k - 1)$ : $\frac{2 ( 2 k - 1 ) ( 2 k - 3 )!}{( k - 2 )!} = \frac{2 ( k - 1 ) ( 2 k - 1 ) ( 2 k - 3 )!}{( k - 1 ) ( k - 2 )!} = \frac{( 2 k - 1 )!}{( k - 1 )!} .$
Therefore $f^{(k + 1)} (x) = (- \frac{1}{4})^{k} \frac{( 2 k - 1 )!}{( k - 1 )!} (1 + x)^{\frac{1}{2} - (k + 1)}$ , which is $P (k + 1)$ .
As $P (2)$ is true and $P (k) \Rightarrow P (k + 1)$ , it is proved for all integers $n \geq 2$ .

c) 1. Maclaurin expansions: $f (0) = 1$ , $f^{'} (0) = \frac{1}{2}$ , $f^{''} (0) = - \frac{1}{4}$ , so $f (x) = 1 + \frac{x}{2} - \frac{x ^{2}}{8} + \dots$ . And $g (x) = e^{m x} = 1 + m x + \frac{m ^{2} x ^{2}}{2} + \dots$ . 2. We multiply; the coefficient of $x^{2}$ in $h = f \cdot g$ is $\frac{m ^{2}}{2} + \frac{1}{2} \cdot m - \frac{1}{8} .$ 3. Setting this equal to $\frac{7}{4}$ : $\frac{m ^{2}}{2} + \frac{m}{2} = \frac{15}{8}$ , that is $4 m^{2} + 4 m - 15 = 0$ . 4. $m = \frac{- 4 \pm 16 + 240}{8} = \frac{- 4 \pm 16}{8}$ : $m = \frac{3}{2}$ or $m = - \frac{5}{2}$ (both rational).

Answer: a) proof; b) proof by induction; c) $m = \frac{3}{2}$ and $m = - \frac{5}{2}$

Exercise 287 HardIB · P1 TZ0 November 2021

a) Use mathematical induction to prove that $\frac{d ^{n}}{d x ^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ for $n \in Z^{+}$ .

b) Using the above, or otherwise, determine the Maclaurin series of $f (x) = x^{2} e^{x}$ in ascending powers of $x$ up to and including the term in $x^{4}$ .

c) Using the above, or otherwise, calculate the value of $x \to 0 lim [\frac{( x ^{2} e ^{x} - x ^{2} ) ^{3}}{x ^{9}}]$ .

Show answer

Answer: a) proof by induction; b) $f (x) \approx x^{2} + x^{3} + \frac{x ^{4}}{2}$ ; c) $1$

Show working and answer

a) Proof by induction of $P (n) : \frac{d ^{n}}{d x ^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ .

Base case $n = 1$ : by the product rule, $\frac{d}{d x} (x^{2} e^{x}) = 2 x e^{x} + x^{2} e^{x} = [x^{2} + 2 x + 0] e^{x}$ , which is $P (1)$ .
Inductive hypothesis: suppose $P (k)$ is true for some $k \in Z^{+}$ : $\frac{d ^{k}}{d x ^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x}$ .
We differentiate using the product rule: $\frac{d ^{k + 1}}{d x ^{k + 1}} (x^{2} e^{x}) = (2 x + 2 k) e^{x} + [x^{2} + 2 k x + k (k - 1)] e^{x} = [x^{2} + (2 k + 2) x + 2 k + k (k - 1)] e^{x} .$
Since $2 k + k (k - 1) = k (k + 1)$ and $2 k + 2 = 2 (k + 1)$ , this becomes $[x^{2} + 2 (k + 1) x + (k + 1) k] e^{x}$ , which is $P (k + 1)$ .
Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , $P (n)$ is true for all $n \in Z^{+}$ .

b) 1. From a), $f^{(n)} (0) = n (n - 1)$ , so $f (0) = 0$ , $f^{'} (0) = 0$ , $f^{''} (0) = 2$ , $f^{'''} (0) = 6$ , $f^{(4)} (0) = 12$ . 2. The Maclaurin series $\sum \frac{f ^{(n)} ( 0 )}{n !} x^{n}$ up to $x^{4}$ is: $f (x) \approx \frac{2}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{12}{4 !} x^{4} = x^{2} + x^{3} + \frac{x ^{4}}{2} .$

c) 1. From b): $x^{2} e^{x} - x^{2} = x^{3} + \frac{x ^{4}}{2} + \dots = x^{3} (1 + \frac{x}{2} + \dots)$ . 2. Cubing: $(x^{2} e^{x} - x^{2})^{3} = x^{9} (1 + \frac{x}{2} + \dots)^{3}$ . 3. Dividing by $x^{9}$ and letting $x \to 0$ : the limit is $1^{3} = 1$ .

Answer: a) proof by induction; b) $f (x) \approx x^{2} + x^{3} + \frac{x ^{4}}{2}$ ; c) $1$

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