2021 Paper 1 Question 6

$gh\left(2\right)=\frac{1}{4}.$

$x=\frac{2}{5}.$

$k=-\frac{b}{2}.$
It must be excluded as $f$ is not defined when $x=-\frac{b}{2}.$

$b=-1.$
$a\in ℝ,a\ne -\frac{1}{2}.$

$f^{-1}(-4)=\frac{4-a}{9}.$

$$\begin{array}{rl}& gh\left(2\right)\\ & =g(\frac{1}{2}\left(2^2\right)+3)\\ & =g\left(5\right)\\ & =\frac{5+1}{5\left(5\right)-1}\\ & =\frac{1}{4}∎\end{array}$$

$$\begin{array}{c}g(x)=1.4\\ \frac{x+1}{5x-1}=\frac{7}{5}\\ 5(x+1)=7(5x-1)\\ 5x+5=35x-7\\ -30x=-12\\ x=\frac{2}{5}∎\end{array}$$

Note that division by zero is undefined $$\begin{array}{c}2x+b=0\\ 2x=-b\\ x=-\frac{b}{2}\end{array}$$

Hence $k=-\frac{b}{2}$ and it must be excluded from the domain of $f$ since $f$ is not defined when $x=-\frac{b}{2}∎$

$$\begin{array}{c}\text{Let }y=\frac{x+a}{2x+b}\\ y(2x+b)=x+a\\ 2yx+yb=x+a\\ 2yx-x=a-yb\\ x(2y-1)=a-yb\\ x=\frac{a-yb}{2y-1}\\ f^{-1}(x)=\frac{a-xb}{2x-1}\end{array}$$$$\begin{array}{rl}f(x)& =f^{-1}(x)\\ \frac{x+a}{2x+b}& =\frac{a-xb}{2x-1}\end{array}$$

Since this is valid for all $x$ in the domain of $f,$ by comparing,

$$b=-1∎$$

Substituting this into $f(x):$

$$f(x)=\frac{x+a}{2x-1}$$

We observe that if $a=-\frac{1}{2},$ then $f(x)=\frac{x-\frac{1}{2}}{2x-1}=\frac{1}{2}$ which is not a one-to-one function and does not have an inverse.

Hence $a$ can take all real values except $-\frac{1}{2}$

$$a\in ℝ,a\ne -\frac{1}{2}∎$$

$$\begin{array}{rl}{f}^{-1}(-4)& =f(-4)\\ & =\frac{-4+a}{2(-4)-1}\\ & =-\frac{-4+a}{9}\\ & =\frac{4-a}{9}∎\end{array}$$