2015 Paper 2 Question 3

All horizontal lines $y=k,k\in ℝ,$ cuts the graph of $y=f(x)$ at most once. Hence $f$ is one-to-one and has an inverse.

$f^{-1}(x)=\sqrt{\frac{x-1}{x}}.$
$D_{f^{-1}}=R_f=(-\infty ,0).$

$(-\infty ,\frac{2-\sqrt{3}}{2}]\cup [\frac{2+\sqrt{3}}{2},\infty ).$

All horizontal lines $y=k,k\in ℝ,$ cuts the graph of $y=f(x)$ at most once. Hence $f$ is one-to-one and has an inverse $∎$

$$\begin{array}{c}\text{Let }y=\frac{1}{1-x^2}\\ y(1-x^2)=1\\ y-y{x}^2=1\\ y{x}^2=y-1\\ x^2=\frac{y-1}{y}\end{array}$$

Since $x>1,$ $$x=\sqrt{\frac{y-1}{y}}$$

$$f^{-1}(x)=\sqrt{\frac{x-1}{x}}■$$

$$\begin{array}{rl}{D}_{f^{-1}}& =R_f\\ & =(-\infty ,0)■\end{array}$$

$$\begin{array}{c}y=\frac{2+x}{1-x^2}\\ y(1-x^2)=2+x\\ y-y{x}^2=2+x\\ y{x}^2+x-(y-2)=0\end{array}$$

The range of $f$ corresponds to the values of $y$ for which there are at least one real root to the equation above in $x.$ $$\begin{array}{c}{b}^2-4ac\ge 0\\ 1+4y(y-2)\ge 0\\ 1+4y^2-8y\ge 0\\ 4y^2-8y+1\ge 0\end{array}$$

Solving $4y^2-8y+1=0,$ $$\begin{array}{rl}y& =\frac{-(-8)\pm \sqrt{{(-8)}^2-4\left(4\right)\left(1\right)}}{2\left(4\right)}\\ & =\frac{8\pm \sqrt{48}}{8}\\ & =\frac{8\pm 4\sqrt{3}}{8}\\ & =\frac{2-\sqrt{3}}{2}\text{ or }\frac{2+\sqrt{3}}{2}\end{array}$$

Hence the solution to the inequality is

$$y\le \frac{2-\sqrt{3}}{2}\text{ or }y\ge \frac{2+\sqrt{3}}{2}$$

$$R_f=(-\infty ,\frac{2-\sqrt{3}}{2}]\cup [\frac{2+\sqrt{3}}{2},\infty )∎$$