CAIE A-Level Mathematics AS 1.2 Functions Questions

$y=(x+a)^{2}-a$ 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.