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Learn/ A Level Mathematics/ Past Papers/ 2023/ May/June 2023 /Paper 3 (Variant 3)

Pure Mathematics 3 — Variant 3

A Level Mathematics · 9709/33 · May/June 2023

Question Paper

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Mark Scheme

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Questions

10 of 11 questions shown here, checked against the paper. The others are in the PDF above — they contain diagrams or maths we could not yet verify, and we would rather show you the original than a transcription we are not sure of.

1 [4]
Solve the equation ln⁡(x+5)=5+ln⁡x\ln(x + 5) = 5 + \ln x. Give your answer correct to 3 decimal places.
2 [3]
Find the quotient and remainder when 2x4−272x^{4} - 27 is divided by x2+x+3x^{2} + x + 3.
3 [4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z−3−i∣⩽3|z - 3 - i| \leqslant 3 and ∣z∣⩾∣z−4i∣|z| \geqslant |z - 4i|.
4 [5]
The parametric equations of a curve are x=cos⁡θ2−sin⁡θx = \frac{\cos\theta}{2 - \sin\theta}, y=θ+2cos⁡θy = \theta + 2\cos\theta. Show that dydx=(2−sin⁡θ)2\frac{\mathrm{d}y}{\mathrm{d}x} = (2 - \sin\theta)^{2}.
6
6(a) [4]
Express 3cos⁡x+2cos⁡(x−60∘)3\cos x + 2\cos(x - 60^\circ) in the form Rcos⁡(x−α)R\cos(x - \alpha), where R>0R > 0 and 0∘<α<90∘0^\circ < \alpha < 90^\circ. State the exact value of RR and give α\alpha correct to 2 decimal places.
6(b) [4]
Hence solve the equation 3cos⁡2θ+2cos⁡(2θ−60∘)=2.53\cos 2\theta + 2\cos(2\theta - 60^\circ) = 2.5 for 0∘<θ<180∘0^\circ < \theta < 180^\circ.
7
7(a) [4]
Use the substitution u=cos⁡xu = \cos x to show that ∫0πsin⁡2x e2cos⁡x dx=∫−112ue2u du\int_{0}^{\pi} \sin 2x \, e^{2\cos x} \, \mathrm{d}x = \int_{-1}^{1} 2u e^{2u} \, \mathrm{d}u.
7(b) [4]
Hence find the exact value of ∫0πsin⁡2x e2cos⁡x dx\int_{0}^{\pi} \sin 2x \, e^{2\cos x} \, \mathrm{d}x.
8 [8]
The variables xx and yy satisfy the differential equation dydx=y2+4x(y+4)\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^{2} + 4}{x(y + 4)} for x>0x > 0. It is given that x=4x = 4 when y=23y = 2\sqrt{3}. Solve the differential equation to obtain the value of xx when y=2y = 2.
9
The lines ll and mm have equations l:r=ai+3j+bk+λ(ci−2j+4k)l: \mathbf{r} = a\mathbf{i} + 3\mathbf{j} + b\mathbf{k} + \lambda(c\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}), m:r=i+2j+3k+μ(2i−3j+k)m: \mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \mu(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}). Relative to the origin OO, the position vector of the point PP is 4i+7j−2k4\mathbf{i} + 7\mathbf{j} - 2\mathbf{k}.
9(a) [4]
Given that ll is perpendicular to mm and that PP lies on ll, find the values of the constants aa, bb and cc.
9(b) [6]
The perpendicular from PP meets line mm at QQ. The point RR lies on PQPQ extended, with PQ:QR=2:3PQ : QR = 2 : 3. Find the position vector of RR.
10
Let f(x)=21−8x−2x2(1+2x)(3−x)2f(x) = \frac{21 - 8x - 2x^{2}}{(1 + 2x)(3 - x)^{2}}.
10(a) [5]
Express f(x)f(x) in partial fractions.
10(b) [5]
Hence obtain the expansion of f(x)f(x) in ascending powers of xx, up to and including the term in x2x^{2}.
11
The complex number zz is defined by z=5a−2i3+aiz = \frac{5a - 2i}{3 + ai}, where aa is an integer. It is given that arg⁡z=−14π\arg z = -\frac{1}{4}\pi.
11(a) [6]
Find the value of aa and hence express zz in the form x+iyx + iy, where xx and yy are real.
11(b) [3]
Express z3z^{3} in the form reiθr\mathrm{e}^{i\theta}, where r>0r > 0 and −π<θ⩽π-\pi < \theta \leqslant \pi. Give the simplified exact values of rr and θ\theta.

Published by Cambridge Assessment International Education. Reproduced for study use.