2016 Paper 1 Question 10

$f^{-1}(x)={(x-1)}^2.$
$D_{f^{-1}}=R_f=[1,\infty ).$

$x=2.62.$

$g(4)=6.$
$g(7)=8.$
$g(12)=9.$

$g$ does not have an inverse as $g(7)=g(8)=8$ so $g$ is not a one-to-one function.

$$\begin{array}{c}\text{Let }y=\sqrt{x}+1\\ \sqrt{x}=y-1\\ x={(y-1)}^2\end{array}$$$$f^{-1}(x)={(x-1)}^2■$$

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

$$\begin{array}{c}ff(x)=x\\ f(\sqrt{x}+1)=x\\ \sqrt{\sqrt{x}+1}+1=x\\ \sqrt{\sqrt{x}+1}=x-1\\ \sqrt{x}+1={(x-1)}^2\\ \sqrt{x}+1=x^2-2x+1\\ \sqrt{x}=x^2-2x\\ x={(x^2-2x)}^2\\ x=x^4-4x^3+4x^2\\ x^4-4x^3+4x^2-x=0\\ x^3-4x^2+4x-1=0\end{array}$$Solving with a GC,

$$x=0.382,1\text{or}2.62$$

From $\sqrt{\sqrt{x}+1}+1=x,$ we observe that $x>1.$ Hence

$$x=2.62∎$$

From $ff(x)=x,$ we can apply $f^{-1}$ on both sides

$$\begin{array}{rl}ff(x)& =x\\ f^{-1}ff(x)& =f^{-1}(x)\\ f(x)& =f^{-1}(x)\end{array}$$

Hence the value of $x$ obtained satisfies the equation $f(x)=f^{-1}(x)∎$

$$\begin{array}{rl}g(2)& =2+g\left(1\right)\\ & =2+1+g\left(0\right)\\ & =3+1\\ & =4\end{array}$$

$$\begin{array}{rl}g(4)& =2+g\left(2\right)\\ & =2+4\\ & =6∎\end{array}$$

$$\begin{array}{rl}g(6)& =2+g\left(3\right)\\ & =2+1+g\left(2\right)\\ & =3+4\\ & =7\end{array}$$

$$\begin{array}{rl}g(7)& =1+g\left(6\right)\\ & =1+7\\ & =8∎\end{array}$$

$$\begin{array}{rl}g(12)& =2+g\left(6\right)\\ & =2+2+g\left(3\right)\\ & =2+7\\ & =9∎\end{array}$$

$$\begin{array}{rl}g(8)& =2+g\left(4\right)\\ & =2+6\\ & =8\end{array}$$

Since $g(7)=g(8)=8,$ $g$ is not a one-to-one function and $g$ does not have an inverse $∎$