CAIE A-Level Mathematics 1 Pure Mathematics 1 Question Bank

Practise Pure Mathematics 1 through quadratics, functions, coordinate geometry, circular measure, trigonometry, series, differentiation and integration with exact working.

Covered syllabus points

1.1 Quadratics 1.2 Functions 1.3 Coordinate geometry 1.4 Circular measure 1.5 Trigonometry 1.6 Series 1.7 Differentiation 1.8 Integration

[Maximum number: 5]

(a)

Express $3 y^{2}-12 y-15$ in the form $3(y+a)^{2}+b$ , where a and b are constants.

[ 2 ]

$3(y-2)^{2}-27$ or a=-2, b=-27

B1 B1

(b)

Hence find the exact solutions of the equation $3 x^{4}-12 x^{2}-15=0$ .

[ 3 ]

$\left(x^{2}-2\right)^{2}=9$ leading to $x^{2}-2= \pm 3$

M1

Must be $x^{2}$ unless substitution is clear.

$x^{2}=-1$ or $x^{2}=5$

M1

Marking guidance:

Allow omission of -1 if $\pm 3$ seen.

$x= \pm \sqrt{5}$

A1

B1 SC if M1M1 not awarded. Ignore $\pm \mathrm{i}, \mathrm{i},-\mathrm{i}, \sqrt{ }-1$ .
Use of calculator with no working scores 0/3.

Alternative method for Question 1(b)

$3 x^{4}-12 x^{2}-15=0$ leading to $3\left(x^{2}-5\right)\left(x^{2}+1\right)[=0]$

(M1)

$x^{2}=-1$ or $x^{2}=5$

(M1)

Allow omission of -1 if factors seen. Factorising or other
valid method.

$x= \pm \sqrt{5}$

(A1)

B1 SC if M1M1 not scored. Ignore $\pm \mathrm{i}, \mathrm{i},-\mathrm{i}, \sqrt{ }-1$ .
Use of calculator with no working scores 0/3.

[Maximum number: 8]

Functions f and g are defined by

\begin{aligned} & \mathrm{f}(x)=(x+a)^{2}-a \text { for } x \leqslant-a, \\ & \mathrm{~g}(x)=2 x-1 \text { for } x \in \mathbb{R}, \end{aligned}

where a is a positive constant.

(a)

Find an expression for $\mathrm{f}^{-1}(x)$ .

[ 3 ]

$y=(x+a)^{2}-a$ leading to $(x+a)^{2}=y \pm a$

*M1

x and y may be interchanged initially.

Marking guidance:

Allow ± errors for these method marks.

$x=[ \pm] \sqrt{y \pm a} \pm a$

DM1

Alternative method for first 2 marks of Question 8(a)

$x=(y+a)^{2}-a$ leading to $y^{2}+2 a y+a^{2}-a-x[=0]$

*M1

Allow ± errors for this method mark.

$y=\frac{-2 a \pm \sqrt{4 a^{2}-4\left(a^{2}-a-x\right)}}{2}$

DM1

$\left[y\right.$ or $\left.f^{-1}(x)=\right]-\sqrt{x+a}-a$

A1

OE
Must choose negative root.

(b)

State the domain of the function $\mathrm{f}^{-1}$ .

[ 1 ]

$x \geqslant-a$

B1

Marking guidance:

Ignore infinity limit if included.

(c)

State the range of the function $\mathrm{f}^{-1}$ .

[ 1 ]

y or $f^{-1}[(x)] \leqslant-a$

B1

Marking guidance:

Ignore negative infinity limit if included.

(d)

Given that $a=\frac{7}{2}$ , solve the equation gf(x)=0.

[ 3 ]

$\left[\operatorname{gf}\left(\frac{7}{2}\right)=\right] 2\left(\left(x+\frac{7}{2}\right)^{2}-\frac{7}{2}\right)-1$ or $2 x^{2}+4\left(\frac{7}{2}\right) x+2\left(\frac{7}{2}\right)^{2}-2\left(\frac{7}{2}\right)-1[=0]$

B1

OE

Alternatively, $[\mathrm{gf}(x)=0 \Rightarrow] \mathrm{f}(x)=\frac{1}{2}$ .

$\begin{aligned} {[x=]-\frac{7}{2} \pm 2 \text { or } \frac{-14 \pm \sqrt{14^{2}-4 \times 2 \times \frac{33}{2}}}{4}} {\left[\frac{-14 \pm \sqrt{64}}{4}\right] \text { or factorising }} \end{aligned}$

M1

OE

Solving their three term quadratic equation as far as two solutions or correctly selecting the negative root only.

Alternatively, $\pm \sqrt{\frac{1}{2}+\frac{7}{2}}-\frac{7}{2}$ .

$[x=]-\frac{11}{2}$

A1

If B1M0 scored then award SCB1 for the correct final answer.

[Maximum number: 11]

Figure for Question 1. Pure Mathematics 1 question 3 — CAIE A-Level Mathematics AS

The diagram shows the curve with equation $x=y^{2}+1$ . The points A(5,2) and B(2,-1) lie on the curve.

(a)

Find an equation of the line A B.

[ 2 ]

Gradient of $A B=\frac{2-(-1)}{5-2}$

M1

Expect 1, must be from $\Delta y / \Delta x$ .

Equation of A B is y-2=1(x-5) or y+1=1(x-2)

A1

OE. Expect y=x-3.

(b)

Find the volume of revolution when the region between the curve and the line A B is rotated through $360^{\circ}$ about the y-axis.

[ 9 ]

$[\pi] \int x^{2} \mathrm{~d} y=[\pi] \int\left(y^{2}+1\right)^{2} \mathrm{~d} y=[\pi] \int\left(y^{4}+2 y^{2}+1\right) \mathrm{d} y$

M1

For curve: Attempt to square $y^{2}+1$ and attempt integration.

Subtracting curve equation from line equation before squaring is M0. Integration before squaring M0.

$[\pi]\left(\frac{y^{5}}{5}+\frac{2 y^{3}}{3}+y\right)$

A2, 1, 0

$[\pi] \int(y+3)^{2} \mathrm{~d} y=[\pi] \int\left(y^{2}+6 y+9\right) \mathrm{d} y$

M1

For line: Attempt to square their y+3 and attempt
integration.

$[\pi]\left(\frac{y^{3}}{3}+3 y^{2}+9 y\right)$ or $[\pi]\left(\frac{(y+3)^{3}}{3}\right)$

A2, 1, 0

Not available for incorrect line equations.

$[\pi]\left\{\frac{8}{3}+12+18-\left(-\frac{1}{3}+3-9\right)\right\}$ or $[\pi]\left\{\frac{32}{5}+\frac{16}{3}+2-\left(-\frac{1}{5}-\frac{2}{3}-1\right)\right\}$

DM1

Apply limits $-1 \rightarrow 2$ to either integral providing
they have been awarded M1. Expect $15 \frac{3}{5}[\pi]$
and/or $39[\pi]$ . Some evidence of substitution of
both -1 and 2 must be seen. Dependent on at least
one of the first 2 M1 marks.

Volume $=[\pi]\left(39-15 \frac{3}{5}\right)$

DM1

Appropriate subtraction. Dependent on at least one of the first 2 M1 marks.

$=23 \frac{2}{5} \pi$ or $\frac{117}{5} \pi$ or awrt 73.5[1327]

A1

[Maximum number: 7]

Figure for Question 1. Pure Mathematics 1 question 4 — CAIE A-Level Mathematics AS

The diagram shows a motif formed by the major arc A B of a circle with radius r and centre O, and the minor $\operatorname{arc} A O B$ of a circle, also with radius r but with centre C. The point C lies on the circle with centre O.

(a)

Given that angle $A C B=k \pi$ radians, state the value of the fraction k.

[ 1 ]

$k=\frac{2}{3}$

B1

Marking guidance:

Allow $\mathrm{ACB}=\frac{2 \pi}{3}$ .

(b)

State the perimeter of the shaded motif in terms of $\pi$ and r.

[ 1 ]

Perimeter of shaded area $=2 \pi r$

B1

(c)

Find the area of the shaded motif, giving your answer in terms of $\pi, r$ and $\sqrt{3}$ .

[ 5 ]

Major sector $O A B=\frac{1}{2} r^{2} \times \frac{4 \pi}{3}$

*M1

Expect $\frac{2}{3} \pi r^{2}$ . Finds area of any relevant sector or
triangle. Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$

*M1

$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$

A1

Shaded area $=\frac{2}{3} \pi r^{2}-2\left(\frac{1}{6} \pi r^{2}-\frac{r^{2} \sqrt{3}}{4}\right)$

DM1

$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$

A1

6(c)

Alternative method for Question 6(c)

Sector CAOB $=[2] \times \frac{1}{2} r^{2}$ their $\frac{1}{3} \pi$

*M1

Expect [2] $\times \frac{1}{6} \pi r^{2}$ .
Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$

*M1

$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$

A1

Shaded area $=\pi r^{2}-\left\{\frac{1}{3} \pi r^{2}+2\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)\right\}$

DM1

$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$

A1

6(c)

Alternative method for Question 6(c)

Area of rhombus AOBC $=[2] \times \frac{1}{2} r^{2} \sin \frac{\pi}{3}$

M1

Expect [2] $\times \frac{\sqrt{3}}{4}$ .
Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$

M1

$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$

A1

Shaded area $=\pi r^{2}-\left\{\frac{\sqrt{3}}{2} r^{2}-4\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)\right\}$

DM1

$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$

A1

All question bank results loaded

Covered syllabus points

1. Pure Mathematics 1 question 1

Question (a)

Question (b)

1. Pure Mathematics 1 question 2

Question (a)

Question (b)

Question (c)

Question (d)

1. Pure Mathematics 1 question 3

Question (a)

Question (b)

1. Pure Mathematics 1 question 4

Question (a)

Question (b)

Question (c)