Pure Mathematics 3 — Variant 3

A Level Mathematics · 9709/33 · Oct/Nov 2022

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Questions

11 of 11 questions shown here, checked against the paper.

1 [4]
Solve the equation ln⁡(2x−1)=2ln⁡(x+1)−ln⁡x\ln(2x - 1) = 2\ln(x + 1) - \ln x. Give your answer correct to 3 decimal places.
2 [5]
Expand 1+2x1−2x\sqrt{\frac{1 + 2x}{1 - 2x}} in ascending powers of xx, up to and including the term in x2x^{2}, simplifying the coefficients.
3 [5]
Find the exact value of ∫014πxsec⁡2x dx\int_{0}^{\frac{1}{4}\pi} x\sec^{2}x \, \mathrm{d}x.
4 [5]
The parametric equations of a curve are x=2t−tan⁡t,y=ln⁡(sin⁡2t),x = 2t - \tan t, \quad y = \ln(\sin 2t), for 0<t<12π0 < t < \frac{1}{2}\pi. Show that dydx=cot⁡t\frac{\mathrm{d}y}{\mathrm{d}x} = \cot t.
5
5(a) [4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z+2∣⩽2|z + 2| \leqslant 2 and Im⁡z⩾1\operatorname{Im} z \geqslant 1.
5(b) [2]
Find the greatest value of arg⁡z\arg z for points in the shaded region.
6 [6]
Solve the quadratic equation (1−3i)z2−(2+i)z+i=0(1 - 3i)z^{2} - (2 + i)z + i = 0, giving your answers in the form x+iyx + iy, where xx and yy are real.
7
7(a) [4]
Show that the equation 5sec⁡x+tan⁡x=4\sqrt{5}\sec x + \tan x = 4 can be expressed as Rcos⁡(x+α)=5R\cos(x + \alpha) = \sqrt{5}, 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.
7(b) [4]
Hence solve the equation 5sec⁡2x+tan⁡2x=4\sqrt{5}\sec 2x + \tan 2x = 4, for 0∘<x<180∘0^\circ < x < 180^\circ.
8
The curve with equation y=x3ex−1y = \frac{x^{3}}{e^{x} - 1} has a stationary point at x=px = p, where p>0p > 0.
8(a) [3]
Show that p=3(1−e−p)p = 3(1 - e^{-p}).
8(b) [2]
Verify by calculation that pp lies between 2.52.5 and 33.
8(c) [3]
Use an iterative formula based on the equation in part (a) to determine pp correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
9
With respect to the origin OO, the position vectors of the points AA, BB and CC are given by OA→=(052)\overrightarrow{OA} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix}, OB→=(101)\overrightarrow{OB} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} and OC→=(4−3−2)\overrightarrow{OC} = \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix}. The midpoint of ACAC is MM and the point NN lies on BCBC, between BB and CC, and is such that BN=2NCBN = 2NC.
9(a) [3]
Find the position vectors of MM and NN.
9(b) [2]
Find a vector equation for the line through MM and NN.
9(c) [4]
Find the position vector of the point QQ where the line through MM and NN intersects the line through AA and BB.
10
A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time tt minutes after filling begins the volume of water in the pool is VV litres. The pool has a small leak and loses water at a rate of 0.01V0.01V litres per minute. The differential equation satisfied by VV and tt is of the form dVdt=a−bV\frac{\mathrm{d}V}{\mathrm{d}t} = a - bV.
10(a) [1]
Write down the values of the constants aa and bb.
10(b) [6]
Solve the differential equation and find the value of tt when V=1000V = 1000.
10(c) [2]
Obtain an expression for VV in terms of tt and hence state what happens to VV as tt becomes large.
11
Let f(x)=5−x+6x2(3−x)(1+3x2)\mathrm{f}(x) = \frac{5 - x + 6x^{2}}{(3 - x)(1 + 3x^{2})}.
11(a) [5]
Express f(x)\mathrm{f}(x) in partial fractions.
11(b) [5]
Find the exact value of ∫01f(x) dx\int_{0}^{1} \mathrm{f}(x) \, \mathrm{d}x, simplifying your answer.

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