CIE AS/A level Pure Mathematics 3 Paper 3 Prediction - 2025

CIE AS/A level Pure Mathematics 3 Paper 3 Prediction- 2025

To excel in A level Math Exam, consistent practice with CIE AS & A Level Revision resources is key. CIE AS/A level Pure Mathematics 3 Paper 3 Prediction will guide you for exam pattern.

IITian Academy offers a vast collection of questions that can aid your understanding of specific topics and solidify your concepts. By practicing regularly and focusing on these key areas, you’ll be well-prepared for the A level Math exam

Question 1

(i) Expand$\frac{1}{\sqrt{(1-4x)}}$ in ascending powers of x, up to and including the term in $x^{2}$, simplifying the
coefficients.

(ii) Hence find the coefficient of$ x^{2}$ in the expansion of$ \frac{1+2x}{\sqrt{(4-16x)}}$

Answer/Explanation

(i) Either Obtain correct (unsimplified) version of x or$ x^{2}$ term from$ (1-4x)^{\frac{1}{2}}$

Obtain 1 + 2x

Obtain +$ 6x^{2}$

Or Differentiate and evaluate f(0) and f′(0) where f′(x) =$ k(1-4x)^{-\frac{3}{2}}$

Obtain 1 + 2x

Obtain + 6x^{2}

(ii) Combine both x^{2}
terms from product of 1 + 2x and answer from part (i)
Obtain 5

Question 2

Let f(x) $\frac{6x^{2}+8x+9}{(2-x)(3+2x)^{2}}$

(i) Express f(x) in partial fractions.

(ii) Hence, showing all necessary working, show that $\int_{-1}^{0}f(x)dx=1+\frac{1}{2}In (\frac{3}{4}).$

Answer/Explanation

(i) State or imply the form

Use a correct method to find a constant

Obtain one of A = 1, B = – 1, C = 3

Obtain a second value

Obtain the third value

[Mark the form $\frac{A}{2-x}+\frac{Dx+E}{(3+2x)^{2}} $, where A = 1, D = – 2 and E = 0, B1M1A1A1A1 as above.]

(ii) Integrate and obtain terms$ -In(2-x)-\frac{1}{2} In (3+2x)-\frac{3}{2(3+2x)^{2}}$

Substitute correctly in an integral with terms a ln (2 – x),

b ln (3 + 2x) and c / (3 + 2x) where abc ≠ 0

Obtain the given answer after full and correct working
[Correct integration of the

A, D, E form gives an extra constant term if integration by

parts is used for the second partial fraction.]

Question 3

The diagram shows the curve$ y = 5sin^{2}xcox^{3}x $ for$ 0 ≤ x ≤\frac{1}{2}\pi $ , and its maximum point M. The shaded region R is bounded by the curve and the x-axis.
(i) Find the x-coordinate of M, giving your answer correct to 3 decimal places.

(ii) Using the substitution u = sin x and showing all necessary working, find the exact area of R.

Answer/Explanation

(i) Use product rule

Obtain correct derivative in any form Equate derivative to zero and obtain an equation in a single trig function

Obtain a correct equation, e.g.

Obtain answer x = 0.685

(ii) Use the given substitution and reach$ a\int (u^{2}-u^{4})du$ Obtain correct integral with a = 5 and limits 0 and 1

Use correct limits in an integral of the form $a\left ( \frac{1}{3} u^{3}-\frac{1}{5}u^{5}\right )$

Obtain answer $\frac{2}{3}$

Question 4

(i) Show that the equation $log_{10}(x-4)=2-log_{10}x$ can be written as a quadratic equation in x.

(ii) Hence solve the equation $log_{10}(x-4)=2-log_{10} $ x giving your answer correct to 3 significant figures.

Answer/Explanation

1(i) Use law for the logarithm of a product or quotient

$Use log_{10}100=2 or 10^{2}=100$
obtain$ x^{2}-4x-100=0$ , or equivalent

1(ii) Solve a 3-term quadratic equation
Obtain answer 12.2 only

Question 5

The sequence of values given by the iterative formula

$X_{n+1}=\frac{2{x_{n}}^{6}+12x_{n}}{3x_{n}^{5}+8}$

with initial value$ x_{1}$ = $2, converges to \alpha$

(i) Use the formula to calculate ! correct to 4 decimal places. Give the result of each iteration to
6 decimal places.

(ii) State an equation satisfied by ! and hence find the exact value of $\alpha $

Answer/Explanation

(i) Use the iterative formula correctly at least once

Obtain answer 1.3195

Show sufficient iterations to 6 d.p. to justify 1.3195 to 4 d.p., or show there is a sign change in (1.31945, 1.31955)

(ii) State x = $\frac{2x^{6}+12x}{3x^{5}+8}$ , or equivalent

State answer$ \sqrt[5]{4}$

Question 6

The parametric equations of a curve are

x = 2 sin 1 + sin 21, y = 2 cos 1 + cos 21, where 0 < 1 < 0.

(i) Obtain an expression for$\frac{dy}{dx} $ in terms of $\Theta $

(ii) Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the
y-axis.

Answer/Explanation

4(i) $obtain \frac{dx}{d\Theta }=2cos\Theta +2cos2\Theta or \frac{dy}{d\Theta }=-2sin\Theta -2sin2\Theta$
use$ dy/dx=dy/d\Theta \div dx/d\Theta$

Obtain correct$ \frac{dy}{dx}e.g.-\frac{2sin\Theta +2sin2\Theta }{2cos\Theta +2cos2\Theta }$

4(ii) Equate denominator to zero and use any correct double angle formula

Obtain correct 3-term quadratic in cosθ in any form Solve forθ

Obtain$ x = \sqrt[3]{3} and y=\frac{1}{2} y = 1$ 2 , or exact equivalents

Question 7

When $\left ( a+bx \right )\sqrt{1 + 4x}$ , where a and b are constants, is expanded in ascending powers of x, the coefficients of x and x² are 3 and −6 respectively.

Find the values of a and b.

Answer/Explanation

Ans:

State or imply 1 + 2x as first terms of the expansion of $\sqrt{1 +4x}$

State or imply − 2x² as third term of the expansion of $\sqrt{1 +4x}$

Form an expression for the coefficient of x or coefficient of x² in the expansion of (a + bx) $\sqrt{1 +4x}$ and equate to given coefficient

Obtain 2a + b = 3, or equivalent

Obtain– 2a + 2b = – 6 or equivalent

Obtain answer a = 2 and b = – 1

Question 8

With respect to the origin O, the points A, B, C, D have position vectors given by

$\vec{OA}=i+3j+2k, \vec{OB}=2i+4j+k,\vec{OD}=-3i+j+2k$

(i) Find the equation of the plane containing A, B and C, giving your answer in the form ax + by + cz = d.

(ii) The line through D parallel to OA meets the plane with equation x + 2y − z = 7 at the point P.
Find the position vector of P and show that the length of DP is$ 2\sqrt{14}$

Answer/Explanation

(ii) Correctly form an equation for the line through D parallel to OA

Obtain a correct equation e.g $r=-3i+j+2k+\lambda (i+3j+2k)$

Substitute components in the equation of the plane and solve for λ

Obtain λ = 2 and position vector -i+7j+6k for p

Obtain the given answer correctly

Question 9

The points A and B have position vectors 2i + j + k and i − 2j + 2k respectively. The line l has vector equation r = i + 2j − 3k + μ(i – 3j – 2k).

(a)Find a vector equation for the line through A and B.

(b) Find the acute angle between the directions of AB and l, giving your answer in degrees.

Answer/Explanation

Ans:

(a)Obtain direction vector −i− 3j+k

Use a correct method to form a vector equation

Obtain answer r = 2i + j + k + λ (-i-3j+k)

or r = i 2j + 2k + λ (-i-3j+k)

(b) Carry out the correct process for evaluating the scalar product of the direction vectors.

Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors

Obtain answer 61.1 °

(c)

Express general point of AB or l in component form, e.g. (2 –λ, 1 – 3λ, 1 + λ) or (1 + µ, 2 – 3 µ, – 3 – 2µ)

Equate at least two pairs of components and solve for λ or for µ

Obtain a correct answer for λ or µ, e.g. λ = $= 6, \frac{1}{6}, or -\frac{14}{9}; \mu = -5, \frac{2}{3} or – \frac{11}{9}$

Verify that all three equations are not satisfied, and the lines do not intersect

Express general point of AB or l in component form,
e.g. (1 –λ*,− 2 – 3λ*, 2 +λ*) or (1 + µ*, 2 – 3µ*, – 3 – 2µ*)

Question 10

(a) Showing all necessary working, express the complex number$ \frac{2+3i}{1-2i}$in the form $re^{i\Theta }$where r > 0 and −0 < 1 ≤ 0. Give the values of r and 1 correct to 3 significant figures.

(b) On an Argand diagram sketch the locus of points representing complex numbers z satisfying the
equation| z − 3 + 2i |= 1. Find the least value of z for points on this locus, giving your answer
in an exact form.

▶️Answer/Explanation

(a)Multiplying by the conjugate:
$ \frac{2+3i}{1-2i} $
$ \frac{(2+3i)(1+2i)}{(1-2i)(1+2i)} $
$ \frac{4+7i}{5} $
So, $r = \frac{4}{5}$ and $7 = 5\sin\theta$, from which we can solve for $\theta$.
$ r = \sqrt{\left(\frac{4}{5}\right)^2 + \left(\frac{7}{5}\right)^2} $
$ = \sqrt{\frac{16}{25} + \frac{49}{25}} $
$ = \sqrt{\frac{65}{25}} $
$ = \frac{\sqrt{65}}{5} $
$ \approx 1.61 $
$ \sin\theta = \frac{7}{5} $
$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$,
$ \theta = \arcsin\left(\frac{7}{5}\right) $
Since both the real and imaginary parts are positive, $\theta$ lies in the first quadrant.
$ \theta \approx 2.09 $
(b) Given the equation of the circle with center $3-2i$ and radius $1$:
$ |z-3+2i| = 1 $
$ |x+iy – (3-2i)| = 1 $
$ |x+iy – 3 + 2i| = 1 $
$ |(x-3) + (y+2)i| = 1 $
$ \sqrt{(x-3)^2 + (y+2)^2} = 1 $
The center of the circle is $3-2i$, so its distance from the origin is:
$ \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} $
Therefore, the least value of $|z|$ is this distance minus the radius of the circle, which is $1$:
$ \sqrt{13} – 1 $
Hence, the least value of $|z|$ for points on the locus represented by the circle is $\sqrt{13} – 1$.

CIE AS/A level Pure Mathematics 3 Paper 3 Prediction

CIE AS/A level Pure Mathematics 3 Paper 3 Prediction - 2025