CAIE A-Level Mathematics AS 1 Pure Mathematics 1 Questions

Express f(x) as: $a-(x-3)^{2}$ or $a-(3-x)^{2}$ where $a= \pm 19$ or $\pm 1$
OE

If the form $-f(x)=\left(x^{2}-6 x-10\right)$ is used the form must be returned to $f(x)=\ldots$

Completed square form must give $-x^{2}$. Answers must come from completion of the square (not calculus or graphs).

$19-(3-x)^{2}$ or $19-(x-3)^{2}$
OE

$f(x) \leqslant 19$ or $y \leqslant 19$ with $\leqslant$, not <
or $-\infty<\mathrm{f}(x) \leqslant 19$ or $-\infty \leqslant \mathrm{f}(x) \leqslant 19$
or $(-\infty, 19]$ or $[-\infty, 19]$

A1 FT

Using their constant following the award of M1.
SC B1 answer only or answer from a method not
involving completion of the square.

$\mathrm{g}^{-1}(x)=\frac{1}{4}(x-k)$$g^{-1} f(x)=\frac{1}{4}\left(10+6 x-x^{2}-k\right)=4 x+k$
OE
May use their completed square form for f(x).

Simplify the quadratic equation obtained from $\mathrm{g}^{-1} \mathrm{f}(x)=\mathrm{g}(x)$ provided k is
present and apply $b^{2}-4 a c=0$ to this quadratic equation

*M1

Expect $x^{2}+10 x-10+5 k=0$.

Obtain 100-4(5 k-10)=0 and hence k=7
Use their k to form and solve a quadratic in x
Marking guidance:

Allow if their quadratic has two solutions.

(-5,-13) only
SC B1 if no method seen.

Alternative Method for first 4 marks

State $\mathrm{f}(x)=\operatorname{gg}(x)$$\operatorname{gg}(x)=16 x+5 k$
Apply $b^{2}-4 a c=0$ to quadratic equation obtained from $\mathrm{f}(x)=\operatorname{gg}(x)$

(*M1)

Provided k is present.

100-4(5 k-10)=0 and hence k=7

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

*M1

x and y may be interchanged initially.

Marking guidance:

Allow ± errors for these method marks.

$x=[ \pm] \sqrt{y \pm a} \pm a$
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}$$\left[y\right.$ or $\left.f^{-1}(x)=\right]-\sqrt{x+a}-a$
OE
Must choose negative root.

$x \geqslant-a$
Marking guidance:

Ignore infinity limit if included.

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

Ignore negative infinity limit if included.

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

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

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

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

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

$[x=]-\frac{11}{2}$
If B1M0 scored then award SCB1 for the correct final answer.

Gradient of $A B=\frac{2-(-1)}{5-2}$
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)
OE. Expect y=x-3.

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

$[\pi] \int(y+3)^{2} \mathrm{~d} y=[\pi] \int\left(y^{2}+6 y+9\right) \mathrm{d} y$
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 $[\pi]\left\{\frac{32}{5}+\frac{16}{3}+2-\left(-\frac{1}{5}-\frac{2}{3}-1\right)\right\}$
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 $=[\pi]\left(39-15 \frac{3}{5}\right)$
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]

$k=\frac{2}{3}$
Marking guidance:

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

Perimeter of shaded area $=2 \pi r$

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

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

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

Alternative method for Question 6(c)

Area of rhombus AOBC $=[2] \times \frac{1}{2} r^{2} \sin \frac{\pi}{3}$
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)$$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$
Shaded area $=\pi r^{2}-\left\{\frac{\sqrt{3}}{2} r^{2}-4\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)\right\}$$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$