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Learn/ A Level Mathematics/ Past Papers/ 2023/ May/June 2023 /Paper 2 (Variant 2)

Pure Mathematics 2 — Variant 2

A Level Mathematics · 9709/22 · May/June 2023

Question Paper

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Mark Scheme

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Questions

3 of 7 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 [5]
Solve the equation sec⁡2θ+5tan⁡2θ=9+17sec⁡θ\sec^{2}\theta + 5\tan^{2}\theta = 9 + 17\sec\theta for 0∘<θ<360∘0^\circ < \theta < 360^\circ.
6
6(a) [4]
Show that 4sin⁡(θ+13π)cos⁡(θ−13π)≡3+2sin⁡2θ4\sin\left(\theta + \frac{1}{3}\pi\right)\cos\left(\theta - \frac{1}{3}\pi\right) \equiv \sqrt{3} + 2\sin 2\theta.
6(b) [2]
Find the exact value of 4sin⁡1724πcos⁡124π4\sin\frac{17}{24}\pi \cos\frac{1}{24}\pi.
6(c) [4]
Find the exact value of ∫018π4sin⁡(2x+13π)cos⁡(2x−13π) dx\int_{0}^{\frac{1}{8}\pi} 4\sin\left(2x + \frac{1}{3}\pi\right)\cos\left(2x - \frac{1}{3}\pi\right) \, \mathrm{d}x.
7
A curve has parametric equations x=2t+3t+2x = \frac{2t + 3}{t + 2}, y=t2+at+1y = t^{2} + at + 1, where aa is a constant. It is given that, at the point PP on the curve, the gradient is 11.
7(a) [4]
Show that the value of tt at PP satisfies the equation 2t3+(a+8)t2+(4a+8)t+4a−1=02t^{3} + (a + 8)t^{2} + (4a + 8)t + 4a - 1 = 0.
7(b) [2]
It is given that (t+1)(t + 1) is a factor of 2t3+(a+8)t2+(4a+8)t+4a−12t^{3} + (a + 8)t^{2} + (4a + 8)t + 4a - 1. Find the value of aa.
7(c) [3]
Hence show that PP is the only point on the curve at which the gradient is 11.

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