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Learn/ A Level Mathematics/ Past Papers/ 2020/ May/June 2020 /Paper 3 (Variant 3)

Pure Mathematics 3 — Variant 3

A Level Mathematics · 9709/33 · May/June 2020

Question Paper

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Mark Scheme

PDF · 374 KB · 15 pages

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Questions

9 of 10 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 inequality ∣2x−1∣>3∣x+2∣|2x - 1| > 3|x + 2|.
2 [5]
Find the exact value of ∫01(2−x)e−2x dx\int_{0}^{1} (2 - x)e^{-2x} \, \mathrm{d}x.
3
3(a) [2]
Show that the equation ln⁡(1+e−x)+2x=0\ln(1 + e^{-x}) + 2x = 0 can be expressed as a quadratic equation in exe^{x}.
3(b) [4]
Hence solve the equation ln⁡(1+e−x)+2x=0\ln(1 + e^{-x}) + 2x = 0, giving your answer correct to 3 decimal places.
4
The equation of a curve is y=xtan⁡−1(12x)y = x \tan^{-1}(\frac{1}{2}x).
4(a) [3]
Find dydx\frac{\mathrm{d}y}{\mathrm{d}x}.
4(b) [3]
The tangent to the curve at the point where x=2x = 2 meets the yy-axis at the point with coordinates (0,p)(0, p). Find pp.
5 [6]
By first expressing the equation tan⁡θtan⁡(θ+45∘)=2cot⁡2θ\tan\theta \tan(\theta + 45^\circ) = 2\cot 2\theta as a quadratic equation in tan⁡θ\tan\theta, solve the equation for 0∘<θ<90∘0^\circ < \theta < 90^\circ.
6
6(a) [2]
By sketching a suitable pair of graphs, show that the equation x5=2+xx^{5} = 2 + x has exactly one real root.
6(b) [2]
Show that if a sequence of values given by the iterative formula xn+1=4xn5+25xn4−1x_{n+1} = \frac{4x_{n}^{5} + 2}{5x_{n}^{4} - 1} converges, then it converges to the root of the equation in part (a).
6(c) [3]
Use the iterative formula with initial value x1=1.5x_{1} = 1.5 to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
7
Let f(x)=2(2x−1)(2x+1)f(x) = \frac{2}{(2x - 1)(2x + 1)}.
7(a) [2]
Express f(x)f(x) in partial fractions.
7(b) [2]
Using your answer to part (a), show that (f(x))2=1(2x−1)2−12x−1+12x+1+1(2x+1)2(f(x))^{2} = \frac{1}{(2x - 1)^{2}} - \frac{1}{2x - 1} + \frac{1}{2x + 1} + \frac{1}{(2x + 1)^{2}}.
7(c) [5]
Hence show that ∫12(f(x))2 dx=25+12ln⁡(59)\int_{1}^{2} (f(x))^{2} \, \mathrm{d}x = \frac{2}{5} + \frac{1}{2}\ln(\frac{5}{9}).
8
Relative to the origin OO, the points AA, BB and DD have position vectors given by OA→=i+2j+k\overrightarrow{OA} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}, OB→=2i+5j+3k\overrightarrow{OB} = 2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} and OD→=3i+2k\overrightarrow{OD} = 3\mathbf{i} + 2\mathbf{k}. A fourth point CC is such that ABCDABCD is a parallelogram.
8(a) [5]
Find the position vector of CC and verify that the parallelogram is not a rhombus.
8(b) [3]
Find angle BADBAD, giving your answer in degrees.
8(c) [2]
Find the area of the parallelogram correct to 3 significant figures.
9
9(a) [5]
The complex numbers uu and ww are such that u−w=2iu - w = 2i and uw=6uw = 6. Find uu and ww, giving your answers in the form x+iyx + iy, where xx and yy are real and exact.
9(b) [5]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z−2−2i∣⩽2|z - 2 - 2i| \leqslant 2, 0⩽arg⁡z⩽14π0 \leqslant \arg z \leqslant \frac{1}{4}\pi and Re⁡z⩽3\operatorname{Re} z \leqslant 3.

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