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Learn/ A Level Mathematics/ Past Papers/ 2022/ Oct/Nov 2022 /Paper 2 (Variant 2)

Pure Mathematics 2 — Variant 2

A Level Mathematics · 9709/22 · Oct/Nov 2022

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]
Solve the equation sec⁡θ=5cosec⁡θ\sec \theta = 5 \operatorname{cosec} \theta for 0∘<θ<360∘0^\circ < \theta < 360^\circ.
2 [5]
The solutions of the equation ∣4x−1∣=∣x+3∣|4x - 1| = |x + 3| are x=px = p and x=qx = q, where p<qp < q. Find the exact values of pp and qq, and hence determine the exact value of ∣p−2∣−∣q−1∣|p - 2| - |q - 1|.
3 [5]
The variables xx and yy satisfy the equation y=Axky = Ax^{k}, where AA and kk are constants. The graph of ln⁡y\ln y against ln⁡x\ln x is a straight line passing through the points (0.56,2.87)(0.56, 2.87) and (0.81,3.47)(0.81, 3.47), as shown in the diagram. Find the value of kk, and the value of AA correct to 2 significant figures.
4
The polynomial p(x)p(x) is defined by p(x)=ax3+23x2−ax−8,p(x) = ax^{3} + 23x^{2} - ax - 8, where aa is a constant. It is given that (2x+1)(2x + 1) is a factor of p(x)p(x).
4(a) [5]
Find the value of aa and hence factorise p(x)p(x) completely.
4(b) [2]
Hence solve the equation p(e4y)=0p(\mathrm{e}^{4y}) = 0, giving your answer correct to 3 significant figures.
5
The curve with equation y=xln⁡(4x+1)−3xy = x\ln(4x + 1) - 3x has one stationary point PP.
5(a) [4]
Show that the xx-coordinate of PP satisfies the equation x=2x+0.75ln⁡(4x+1)−0.25.x = \dfrac{2x + 0.75}{\ln(4x + 1)} - 0.25.
5(b) [2]
Show by calculation that the xx-coordinate of PP lies between 1.8 and 1.9.
5(c) [3]
Use an iterative formula, based on the equation in part (a), to find the xx-coordinate of PP correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
6 [9]
The diagram shows the curves y=63x+2y = \dfrac{6}{3x + 2} and y=3e−x−3y = 3\mathrm{e}^{-x} - 3 for values of xx between 0 and 4. The shaded region is bounded by the two curves and the lines x=0x = 0 and x=4x = 4. Find the exact area of the shaded region, giving your answer in the form ln⁡a+b+ced\ln a + b + c\mathrm{e}^{d}.

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