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Learn/ A Level Mathematics/ Past Papers/ 2023/ May/June 2023 /Paper 2 (Variant 1)

Pure Mathematics 2 — Variant 1

A Level Mathematics · 9709/21 · May/June 2023

Question Paper

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Mark Scheme

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Questions

6 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 [4]
Use logarithms to solve the equation 12x=32x+112^{x} = 3^{2x+1}. Give your answer correct to 3 significant figures.
2 [5]
A curve has equation y=2+3ln⁡x1+2xy = \frac{2 + 3\ln x}{1 + 2x}. Find the equation of the tangent to the curve at the point (1,23)(1, \frac{2}{3}). Give your answer in the form ax+by+c=0ax + by + c = 0, where aa, bb and cc are integers.
3
It is given that ∫0a(3e2x−1) dx=12\int_{0}^{a} (3\mathrm{e}^{2x} - 1) \, \mathrm{d}x = 12, where aa is a positive constant.
3(a) [4]
Show that a=12ln⁡(9+23a)a = \frac{1}{2}\ln(9 + \frac{2}{3}a).
3(b) [3]
Use an iterative formula, based on the equation in (a), to find the value of aa correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures.
4
The polynomial p(x)p(x) is defined by p(x)=2x3+3x2+kx−30,p(x) = 2x^{3} + 3x^{2} + kx - 30, where kk is a constant. It is given that (x−3)(x - 3) is a factor of p(x)p(x).
4(a) [2]
Find the value of kk.
4(b) [3]
Hence find the quotient when p(x)\mathrm{p}(x) is divided by (x−3)(x - 3) and factorise p(x)\mathrm{p}(x) completely.
4(c) [3]
It is given that aa is one of the roots of the equation p(x)=0\mathrm{p}(x) = 0. Given also that the equation ∣4y−5∣=a|4y - 5| = a is satisfied by two real values of yy, find these two values of yy.
6 [7]
Show that ∫14π13π(4cos⁡22x+1cos⁡2x) dx=343+16π−1\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \left( 4\cos^{2}2x + \frac{1}{\cos^{2}x} \right) \, \mathrm{d}x = \frac{3}{4}\sqrt{3} + \frac{1}{6}\pi - 1.
7
7(a) [3]
Express 7cos⁡θ+24sin⁡θ7\cos\theta + 24\sin\theta in the form Rcos⁡(θ−α)R\cos(\theta - \alpha), where R>0R > 0 and 0∘<α<90∘0^\circ < \alpha < 90^\circ. Give the value of α\alpha correct to 2 decimal places.
7(b) [4]
Solve the equation 7cos⁡θ+24sin⁡θ=187\cos\theta + 24\sin\theta = 18 for 0∘<θ<360∘0^\circ < \theta < 360^\circ.
7(c) [4]
As β\beta varies, the greatest possible value of 1507cos⁡12β+24sin⁡12β+50\frac{150}{7\cos\frac{1}{2}\beta + 24\sin\frac{1}{2}\beta + 50} is denoted by VV. Find the value of VV and determine the smallest positive value of β\beta (in degrees) for which the value of VV occurs.

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