2019 Paper 1 Question 5

$f^{-1}(x)=\frac{\ln (x+4)}{2}.$
$D_{f^{-1}}=R_f=(-4,\infty ).$

$x=\ln 3-2.$

$$\begin{array}{c}\text{Let }y={\mathrm{e}}^{2x}-4\\ {\mathrm{e}}^{2x}=y+4\\ 2x=\ln (y+4)\\ x=\frac{\ln (y+4)}{2}\end{array}$$$$f^{-1}(x)=\frac{\ln (x+4)}{2}■$$

$$\begin{array}{rl}{D}_{f^{-1}}& =R_f\\ & =(-4,\infty )■\end{array}$$

$$\begin{array}{c}fg(x)=5\\ f^{-1}fg(x)=f^{-1}(5)\\ g(x)=\frac{\ln (5+4)}{2}\\ x+2=\frac{\ln 9}{2}\\ x+2=\frac{\ln 3^2}{2}\\ x+2=\frac{2\ln 3}{2}\\ x+2=\ln 3\\ x=\ln 3-2∎\end{array}$$