CAIE IGCSE Additional Math Calculus Questions

Use this calculus unit hub to connect differentiation and integration rules with curve applications, rates, optimisation, areas and particle motion.

Exam points

Differentiate standard, product, quotient and composite forms for tangents, rates and optimisation.
Integrate powers and composite forms, evaluate definite integrals and use conditions to recover functions or areas.
Apply derivatives, integrals and graphs to model areas, rates and straight-line particle motion.

Covered syllabus points

14. Calculus

[Maximum number: 3]

Given that $y=\frac{4 x^{3}-5}{x^{2}}$ , show that $\frac{\mathrm{d} y}{\mathrm{~d} x}$ can be written as $\frac{2\left(2 x^{3}+5\right)}{x^{3}}$ .

Attempts to find $y=4 x-5 x^{-2}$
at least one term correct

$\frac{\mathrm{d} y}{\mathrm{~d} x}=4+10 x^{-3}$
Correct completion to given answer $\frac{2\left(2 x^{3}+5\right)}{x^{3}}$
Alternative version

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{12 x^{4}-8 x^{4}+10 x}{x^{4}}$ or equivalent
unsimplified form

M1 for an attempt at the quotient rule with correct structure
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x^{2}\left(\text { their } 12 x^{2}\right)-\left(4 x^{3}-5\right)(\text { their } 2 x)}{\left(x^{2}\right)^{2}}$

Correct completion to given answer $\frac{2\left(2 x^{3}+5\right)}{x^{3}}$

[Maximum number: 6]

(a)

Given that $y=x \cos 2 x$ , find $\frac{\mathrm{d} y}{\mathrm{~d} x}$ .

[ 2 ]

$\cos2x+x(-2\sin2x)$ oe, isw, nfww
M1 for $\frac{d}{dx}(\cos2x)=-2\sin2x$ soi, or M1 FT for correct product-rule structure.
A1

(b)

Hence find $\int x \sin 2 x \mathrm{~d} x$ .

[ 4 ]

$-\frac12x\cos2x+\frac14\sin2x+c$ oe, isw, nfww
M3 for $-\frac12x\cos2x+\frac14\sin2x$ or equivalent integration-by-parts working;
M2/M1 for partial correct working as in markscheme.
A1

[Maximum number: 9]

The point P lies on the curve y=(5 x+2)^{2/3}.
The x-coordinate of P is 5.
The normal to the curve at P intersects the line x+y=11 at the point Q.
The point R is the reflection of Q in the tangent to the curve at P.
Find the coordinates of R.

[When x=5 ] y=9
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2}{3}(5 x+2)^{-\frac{1}{3}} \times 5$ oe
B1 for $\frac{\mathrm{d} y}{\mathrm{~d} x}=k(5 x+2)^{-\frac{1}{3}}, k \neq \frac{10}{3}$ $\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}=\frac{10}{9}$
FT their $\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}$ providing previous B1
awarded

Equation of normal e.g.
$y-\text { their } 9=-\frac{1}{\text { their } \frac{10}{9}}(x-5) \text { oe, soi }$
FT $\frac{-1}{\text { their }\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}}$ and their y-coordinate

Eliminates one variable e.g.
$11-x-\text { their } 9=-\frac{1}{\text { their } \frac{10}{9}}(x-5)$
dep on previous M mark

For Q: x=-25, y=36
A1 for each

Coordinates of R:(35,-18)
or ((10-their(-25)), (18-their36))
FT their coordinates of Q

[Maximum number: 6]

Show that the curve $y=x-\ln \left(x^{2}+2 x\right)$ has exactly one stationary point.
Find the x-coordinate of this point.

$\frac{dy}{dx}=1-\frac{2x+2}{x^2+2x}$
B1 for $\frac{d(-\ln(x^2+2x))}{dx}=\frac{1}{x^2+2x}\times f(x)$ soi.
Equates their first derivative to 0 and simplifies as far as 2(x+1)=x(x+2) oe
M1 FT their $\frac{dy}{dx}$ .
$x^2-2=0$ or $x^2=2$ $x=\sqrt{2}$ [ $x=-\sqrt{2}$ ]
Justification that the negative solution should be rejected, e.g. $x=-\sqrt{2}$ gives $y=2-\ln(2-2\sqrt{2})$ , which is impossible.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question 4