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

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

Covered syllabus points

3.1 Algebra 3.2 Logarithmic and exponential functions 3.3 Trigonometry 3.4 Differentiation 3.5 Integration 3.6 Numerical solution of equations 3.7 Vectors 3.8 Differential equations 3.9 Complex numbers

[Maximum number: 4]

Solve the inequality |2 x+3|>3|x+2|.

Square both sides:
$(2x+3)^2>9(x+2)^2.$
Then
$4x^2+12x+9>9x^2+36x+36,$
so
$5x^2+24x+27<0.$
Factorising,
(5x+9)(x+3)<0.
The critical values are x=-3 and $x=-\frac95$ . The product is negative between them, hence
$-3<x<-\frac95.$

B1 for the non-modular squared inequality or equivalent linear equations. M1 for solving the quadratic/equations. A1 for critical values -3 and $-\frac95$ . A1 for $-3<x<-\frac95$ , with strict inequalities.

Figure for Answer to Question 3. Pure Mathematics 3 question 1 — CAIE A-Level Mathematics A2

[Maximum number: 5]

(a)

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

[ 3 ]

Use law of logarithm of a power

M1

$\log _{3}(2 x+1)=1+\log _{3}(x-1)^{2}$

Use $\log _{3} 3=1$

B1

$\begin{aligned} & \log _{3}(2 x+1)=\log _{3} 3+2 \log _{3}(x-1) & {\left[\log 3\left(\frac{2 x+1}{(x-1)^{2}}\right)=\log _{3} 3 \text { or }\left(\frac{2 x+1}{(x-1)^{2}}\right)=3\right]} \end{aligned}$

SC For candidates scoring M0 B0 due to combining logs before dealing with coefficient 2, and confusing coefficients, allow $\log _{3}(\ldots)=c$ leading to $(\ldots)=3^{c} \quad \mathbf{B1}$ .

Obtain $3 x^{2}-8 x+2=0$ or $1.5 x^{2}-4 x+1=0$

A1

OE 3 terms only and =0 required.

(b)

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

[ 2 ]

Solve 3-term quadratic equation from part 3(a) or restart to find y

M1

$\begin{aligned} & y=\frac{4 \pm \sqrt{10}}{6} \text { or } y=1.1937 \ldots \text { or } y=0.1396 \ldots & (x=2.3874 \text { or } x=0.2792) \end{aligned}$

May solve for x but must find $y=\frac{x}{2}$ to gain M1.

Obtain answer 1.19

A1

CAO. 2 dp required.

[Maximum number: 8]

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

(a)

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

[ 3 ]

Differentiate to obtain $\frac{k_1}{1+Bx^2}$ or $\frac{k_2}{x^2+\frac1B}$ . M1

Obtain
$\frac{dy}{dx}=\frac4{1+16x^2}.$
A1

Set $\frac{dy}{dx}=\frac14$ and obtain
$x=\pm\frac{\sqrt{15}}4.$
A1
Accept exact equivalents, including $\pm\frac{\sqrt{15}}{\sqrt{16}}$ .

Alternative method: rewrite as $\tan y=4x$ , use $\sec^2y=k\frac{dx}{dy}$ , and obtain the same exact values.

(b)

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

[ 5 ]

Commence integration by parts and reach
$Ax\tan^{-1}(4x)\pm\int x\frac{B}{1+Cx^2}\,dx.$
*M1

Obtain
$x\tan^{-1}(4x)-\int\frac{4x}{1+16x^2}\,dx.$
A1

Complete the integration and obtain
$x\tan^{-1}(4x)-\frac18\ln(1+16x^2).$
A1FT

Substitute the limits correctly. DM1

Obtain
$\frac1{16}\pi-\frac18\ln 2.$
A1
Accept exact equivalents such as $\frac1{16}\pi+\frac18\ln\frac12$ .

[Maximum number: 6]

Find the exact value of $\int_{0}^{6} \frac{x(x+1)}{x^{2}+4} \mathrm{~d} x$ .

Split fraction to obtain $1+\frac{x-4}{x^{2}+4}$

B1

Attempt integration and obtain $p \ln \left(x^{2}+4\right)$ or $q \tan ^{-1}\left(\frac{x}{2}\right)$ from correct
working

M1

Marking guidance:

Allow for $p \ln \left(x^{2}+4\right)$ from $\int \frac{x}{x^{2}+4} \mathrm{~d} x$ but only if a
correct method for splitting has been used.

Obtain $\frac{1}{2} \ln \left(x^{2}+4\right)$

A1 FT

Follow through is on their coefficients in the partial fraction.

Allow from $\frac{x^{2}}{x^{2}+4}+\frac{x}{x^{2}+4}$ even if the split of the fraction is not complete. If $1-\frac{4}{x^{2}+4}+\frac{x}{x^{2}+4}$ later seen or implied, award the B1.

Only available from a correct split, not from an approach using parts that is incomplete.

Obtain $-2 \tan ^{-1}\left(\frac{x}{2}\right)$

A1 FT

Only available from a correct split, not from an approach using parts that is incomplete.

Correct use of correct limits 0 and 6 in an expression involving $p \ln \left(x^{2}+4\right)$ ,
$q \tan ^{-1}\left(\frac{x}{2}\right)$ and no incorrect terms.

M1

p and q should be constants.

The x term is not required at this stage.

Obtain $6+\frac{1}{2} \ln 10-2 \tan ^{-1} 3$

A1

ISW

Or three term equivalent. (Must combine the ln terms.) Accept with $\frac{1}{2} \ln |10|$ .

5

Alternative method for question 5

Use the substitution $x=2 \tan \theta$ to obtain $\int 2 \tan ^{2} \theta+\tan \theta \mathrm{d} \theta$

B1

Attempt integration and obtain $p \tan \theta$ or $r \ln (\sec \theta)$ from correct working

M1

Obtain $2 \tan \theta(-2 \theta)$ and

A1 FT

Follow through on their coefficients after the substitution.

Obtain ln $\sec \theta$

A1 FT

Follow through on their coefficients after the substitution.

Use correct limits 0 and $\tan ^{-1} 3$ in an expression involving $u \tan \theta, v \ln \sec \theta$
and no incorrect terms

M1

u and v should be constants. The $\theta$ term is not required at
this stage.

Obtain $6+\ln \left|\sec \left(\tan ^{-1} 3\right)\right|-2 \tan ^{-1} 3$

A1

ISW

Or three term equivalent.

Not required to simplify $\ln \left|\sec \left(\tan ^{-1} 3\right)\right|$ .

All question bank results loaded

Covered syllabus points

3. Pure Mathematics 3 question 1

3. Pure Mathematics 3 question 2

Question (a)

Question (b)

3. Pure Mathematics 3 question 3

Question (a)

Question (b)

3. Pure Mathematics 3 question 4