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Learn/ A Level Mathematics/ Past Papers/ 2021/ Oct/Nov 2021 /Paper 3 (Variant 3)

Pure Mathematics 3 — Variant 3

A Level Mathematics · 9709/33 · Oct/Nov 2021

Question Paper

PDF · 120 KB · 20 pages

Mark Scheme

PDF · 319 KB · 14 pages

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Questions

9 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 [3]
Find the quotient and remainder when 2x4+12x^{4} + 1 is divided by x2−x+2x^{2} - x + 2.
2
2(a) [1]
Sketch the graph of y=∣2x−3∣y = |2x - 3|.
2(b) [3]
Solve the inequality ∣2x−3∣<3x+2|2x - 3| < 3x + 2.
3 [4]
Solve the equation 4x−2=4x−424^{x-2} = 4^{x} - 4^{2}, giving your answer correct to 3 decimal places.
4 [5]
Find the exact value of ∫13ππxsin⁡12x dx\int_{\frac{1}{3}\pi}^{\pi} x \sin \frac{1}{2}x \, \mathrm{d}x.
5 [5]
Solve the equation sin⁡θ=3cos⁡2θ+2\sin \theta = 3\cos 2\theta + 2, for 0∘⩽θ⩽360∘0^\circ \leqslant \theta \leqslant 360^\circ.
6
6(a) [5]
By first expanding cos⁡(x−60∘)\cos(x - 60^\circ), show that the expression 2cos⁡(x−60∘)+cos⁡x2\cos(x - 60^\circ) + \cos x can be written in the form Rcos⁡(x−α)R\cos(x - \alpha), where R>0R > 0 and 0∘<α<90∘0^\circ < \alpha < 90^\circ. Give the exact value of RR and the value of α\alpha correct to 2 decimal places.
6(b) [2]
Hence find the value of xx in the interval 0∘<x<360∘0^\circ < x < 360^\circ for which 2cos⁡(x−60∘)+cos⁡x2\cos(x - 60^\circ) + \cos x takes its least possible value.
7
The equation of a curve is ln⁡(x+y)=x−2y\ln(x + y) = x - 2y.
7(a) [4]
Show that dydx=x+y−12(x+y)+1\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y - 1}{2(x + y) + 1}.
7(b) [3]
Find the coordinates of the point on the curve where the tangent is parallel to the xx-axis.
9
Let f(x)=1(9−x)xf(x) = \frac{1}{(9 - x)\sqrt{x}}.
9(a) [4]
Find the xx-coordinate of the stationary point of the curve with equation y=f(x)y = f(x).
9(b) [6]
Using the substitution u=xu = \sqrt{x}, show that ∫04f(x) dx=13ln⁡5\int_{0}^{4} f(x) \, \mathrm{d}x = \frac{1}{3}\ln 5.
10
A large plantation of area 2020 km2\text{km}^{2} is becoming infected with a plant disease. At time tt years the area infected is xx km2\text{km}^{2} and the rate of increase of xx is proportional to the ratio of the area infected to the area not yet infected. When t=0t = 0, x=1x = 1 and dxdt=1\frac{\mathrm{d}x}{\mathrm{d}t} = 1.
10(a) [2]
Show that xx and tt satisfy the differential equation dxdt=19x20−x\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{19x}{20 - x}.
10(b) [5]
Solve the differential equation and show that when t=1t = 1 the value of xx satisfies the equation x=e0.9+0.05xx = \mathrm{e}^{0.9 + 0.05x}.
10(c) [3]
Use an iterative formula based on the equation in part (b), with an initial value of 22, to determine xx correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
10(d) [1]
Calculate the value of tt at which the entire plantation becomes infected.

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