2017 Paper 2 Question 3

$(\frac{a}{2},0),$ $(0,b).$

$(a+1,0).$

$(\frac{a+1}{2},0).$

$(0,a),$ $(b,0).$

$a=1.$
It must be excluded as $g$ is not defined when $x=1.$

$g^2(x)=x.$
$g^{-1}(x)=1-\frac{1}{1-x}.$

$b=0$ or $b=2.$

We apply a scaling of scale factor $\frac{1}{2}$ parallel to the $x\text{-axis}$ on $y=f(x)$

$$\begin{array}{c}(\frac{a}{2},0)∎\\ (0,b)∎\end{array}$$

We apply a translation of $1$ unit in the positive $x\text{-axis}$ direction on $y=f(x)$

$$(a+1,0)∎$$

We apply a scaling of scale factor $\frac{1}{2}$ parallel to the $x\text{-axis}$ on $y=f(x-1)$ from the previous part

$$(a+1,0)∎$$

We apply a reflection of the curve $y=f(x)$ about the line $y=x.$

$$\begin{array}{c}(0,a)∎\\ (b,0)∎\end{array}$$

$$a=1∎$$

$x=1$ has to be excluded from the domain of $g$ as $g(1)$ is not defined $∎$

$$\begin{array}{rl}{g}^2(x)& =gg(x)\\ & =g(1-\frac{1}{1-x})\\ & =1-\frac{1}{1-(1-\frac{1}{1-x})}\\ & =1-(1-x)\\ & =x∎\end{array}$$$$\begin{array}{rl}gg(x)& =x\\ g^{-1}gg(x)& =g^{-1}(x)\\ g(x)& =g^{-1}(x)\\ g^{-1}(x)& =1-\frac{1}{1-x}∎\end{array}$$

$$\begin{array}{c}{g}^2(b)=g^{-1}(b)\\ b=1-\frac{1}{1-b}\\ 1-b=\frac{1}{1-b}\\ {(1-b)}^2=1\\ 1-b=\pm 1\\ b=0∎\text{or}b=2∎\end{array}$$