CAIE A-Level Mathematics 3. Pure Mathematics 3 Question Bank

CAIE A-Level Mathematics 3. Pure Mathematics 3 Question Bank
Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2028, 2029 and 20302028–2030

Practise Pure Mathematics 3 techniques for algebra, functions, calculus, vectors and equations through exact and numerical work.

Question 3(a)

[Maximum number: 4]

Solve the inequality 3x42x+5|3 x-4| \leqslant|2 x+5|.

Question 3

Question 3(a)

(a)

Show that the equation log3(2x+1)=1+2log3(x1)\log _{3}(2 x+1)=1+2 \log _{3}(x-1) can be written as a quadratic equation in x.

[ 3 ]

Question 3(b)

(b)

Hence solve the equation log3(4y+1)=1+2log3(2y1)\log _{3}(4 y+1)=1+2 \log _{3}(2 y-1), giving your answer correct to 2 decimal places.

[ 2 ]

Question 4

[Maximum number: 4]

The complex number u is given by u=1i3u=-1-\mathrm{i} \sqrt{3}.

Question 4(a)

(a)

Express u in the form r(cosθ+isinθ)r(\cos \theta+\mathrm{i} \sin \theta), where r>0 and π<θπ-\pi<\theta \leqslant \pi. Give the exact values of r and θ\theta.
The complex number v is given by v=5(cos16π+isin16π)v=5\left(\cos \frac{1}{6} \pi+\mathrm{i} \sin \frac{1}{6} \pi\right).

[ 2 ]

Question 4(b)

(b)

Express the complex number vu\frac{v}{u} in the form reiθr \mathrm{e}^{\mathrm{i} \theta} where r>0 and π<θπ-\pi<\theta \leqslant \pi.

[ 2 ]

Question 4(a)

[Maximum number: 3]

The polynomial p(x) is defined by

p(x)=x410x3+20x230x+40.\mathrm{p}(x)=x^{4}-10 x^{3}+20 x^{2}-30 x+40 .

Find the quotient when p(x) is divided by (x2+3)\left(x^{2}+3\right) and show that the remainder is -11.

Question 5

[Maximum number: 7]
Figure for Question 5 — CAIE A-Level Mathematics

The diagram shows the curve with equation y=8e12x1y=8 \mathrm{e}^{-\frac{1}{2} x}-1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB.

Question 5(a)

(a)

Show that the x-coordinate of B is 6ln26 \ln 2.

[ 2 ]

Question 5(b)

(b)

Find the area of the shaded region. Give your answer in the form pln2qp \ln 2-q, where p and q are positive integers.

[ 5 ]

Question 4

[Maximum number: 5]

Given that aa+1413x dx=ln2\int_{a}^{a+14} \frac{1}{3 x} \mathrm{~d} x=\ln 2, find the value of the positive constant a.

Question 7

[Maximum number: 8]

The equation of a curve is y=tan1(4x)y=\tan ^{-1}(4 x).

Question 7(a)

(a)

Find the exact values of x when the gradient of the curve is 14\frac{1}{4}.

[ 3 ]

Question 7(b)

(b)

Find the exact value of 00.25ydx\int_0^{0.25}y\,dx.

[ 5 ]

Question 7

[Maximum number: 11]

The polynomial p(x) is defined by

p(x)=2x4+kx3+kx2+17x+18,\mathrm{p}(x)=2 x^{4}+k x^{3}+k x^{2}+17 x+18,

where k is a constant. It is given that (x+2) is a factor of p(x).

Question 7(a)

(a)

Find the value of k.

It is given that the equation p(x)=0 has exactly two real roots, denoted by α\alpha and β\beta, where α\alpha is an integer and β\beta is not an integer.

[ 2 ]

Question 7(b)

(b)

State the value of α\alpha and show that β\beta satisfies the equation x=2x4.53x=\sqrt[3]{-2 x-4.5}.

[ 4 ]

Question 7(c)

(c)

Show by calculation that 1.4<β<1.0-1.4<\beta<-1.0.

[ 2 ]

Question 7(d)

(d)

Use an iterative formula, based on the equation in part (b), to find the value of β\beta correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

[ 3 ]

Question 6

[Maximum number: 7]

The polynomials f(x) and g(x) are defined by

f(x)=4x3+ax2+8x+15 and g(x)=x2+bx+18\mathrm{f}(x)=4 x^{3}+a x^{2}+8 x+15 \quad \text { and } \quad \mathrm{g}(x)=x^{2}+b x+18

where a and b are constants.

Question 6(a)

(a)

Given that (x+3) is a factor of f(x), find the value of a.

[ 2 ]

Question 6(b)

(b)

Given that the remainder is 40 when g(x) is divided by (x-2), find the value of b.

[ 2 ]

Question 6(d)

(c)

Hence solve the equation f(cosecθ)g(cosecθ)=0\mathrm{f}(\operatorname{cosec} \theta)-\mathrm{g}(\operatorname{cosec} \theta)=0 for 0<θ<2π0<\theta<2 \pi.

[ 3 ]

Question 10

[Maximum number: 9]

The diagram shows a tank for holding water. The tank is in the shape of a cube of side 50 cm. At time t seconds, the depth of water in the tank is h cmh \mathrm{~cm}. Water is poured into the tank at a rate of 5000 cm3 s15000 \mathrm{~cm}^{3} \mathrm{~s}^{-1}. Water pours out of the tank through a hole in the bottom at a rate proportional to h2h^{2}.

When h=20, the depth of the water is increasing at a rate of 0.4cms10.4 \mathrm{cms}^{-1}.

Question 10(a)

(a)

Show that dh dt=500h2250\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{500-h^{2}}{250}.

[ 4 ]

Question 10(b)

(b)

Given that h=0 when t=0, find the time taken for the depth of the water in the tank to reach 20 cm.

[ 5 ]