CAIE IGCSE Additional Mathematics Calculus Question Bank

Practise differentiation and integration with standard and composite functions, then apply derivatives, areas and kinematics to optimisation and rate problems.

[Maximum number: 2]

It is given that $y=3\tan^2x$ for $0^\circ<x<360^\circ$ .

Show that $\frac{dy}{dx}=m\tan x\sec^2x$ , where m is an integer to be found.

$\frac{dy}{dx}=6\tan x\sec^2x$
B2; B1 for $\frac{d}{dx}(\tan^2x)=2\tan x\sec^2x$

[Maximum number: 10]

The equation of a curve is $y=kxe^{-2x}$ , where k is a constant.

(a)

Find $\frac{dy}{dx}$ .

[ 2 ]

$\frac{d}{dx}(e^{-2x})=-2e^{-2x}$
B1
$\frac{dy}{dx}=ke^{-2x}-2kxe^{-2x}$
B1
FT for product rule $k e^{-2x}+kx\cdot\frac{d}{dx}(e^{-2x})$ .

(b)

Find the coordinates of the stationary point on the curve $y=10xe^{-2x}$ .

[ 3 ]

Equate $\frac{dy}{dx}=0$ and find
10-20x=0
M1
$\left(\frac{1}{2},\frac{5}{e}\right)$
A2; A1 for $x=\frac{1}{2}$ only. Equivalent $y=5e^{-1}$ or decimal 1.84 accepted.

(c)

Use your answer to part (a) to find $\int 4xe^{-2x}\,dx$ .

[ 3 ]

$-2xe^{-2x}-e^{-2x}+c$
B3
B2 for $-2xe^{-2x}-e^{-2x}$ , or $\int4xe^{-2x}\,dx=-2xe^{-2x}+\int2e^{-2x}\,dx$ .
B1 for using $kxe^{-2x}=\int(ke^{-2x}-2kxe^{-2x})\,dx$ or better.

(d)

Find the exact value of $\int_0^1 4xe^{-2x}\,dx$ .

[ 2 ]

$-2e^{-2}-e^{-2}-(0-e^0)$
M1
$1-\frac{3}{e^2}$
or $1-3e^{-2}$
A1

[Maximum number: 7]

A curve has equation $y=\ln(5-3x)$ , where $x<\frac{5}{3}$ . The normal to the curve at the point where x=-5 cuts the x-axis at the point P.
Find the equation of the normal and the x-coordinate of P.

$\frac{dy}{dx}=\frac{-3}{5-3x}, \qquad \left.\frac{dy}{dx}\right|_{x=-5}=-\frac{3}{20}.$
At (x=-5), (y=\ln20), so the normal gradient is $\frac{20}{3}$ . Hence
$y-\ln20=\frac{20}{3}(x+5).$
At the (x)-axis, (y=0), giving
$x=-5-\frac{3\ln20}{20}=-5.449\ldots.$

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Covered syllabus points

Calculus question 1

Calculus question 2

Question (a)

Question (b)

Question (c)

Question (d)

Calculus question 3