2018 Paper 1 Question 5

$b=-1.$
$f^{-1}(x)=\frac{x+a}{x-1}.$

$$\begin{array}{rl}ff(x)& =f\left(\frac{x+a}{x+b}\right)\\ & =\frac{\frac{x+a}{x+b}+a}{\frac{x+a}{x+b}+b}\\ & =\frac{\frac{x+a+ax+ab}{x+b}}{\frac{x+a+bx+b^2}{x+b}}\\ & =\frac{x+a+ax+ab}{x+a+bx+b^2}\end{array}$$

Since $ff=g,$ $$\begin{array}{c}\frac{x+a+ax+ab}{x+a+bx+b^2}=x\\ x+a+ax+ab=x(x+a+bx+b^2)\\ x+a+ax+ab=x^2+xa+x^{2}b+x{b}^2\\ (1+a)x+ab+a=(1+b)x^2+(a+b^2)x\end{array}$$

Comparing coefficients of $x^2,$ $$\begin{array}{rl}1+b& =0\\ b& =-1∎\end{array}$$

$$\begin{array}{rl}ff(x)& =g(x)\\ ff(x)& =x\\ f(x)& =f^{-1}(x)\\ f^{-1}(x)& =f(x)\\ & =\frac{x+a}{x+b}\\ & =\frac{x+a}{x-1}∎\end{array}$$