The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{1}{1-{{x}^{2}}}$, $x\in \mathbb{R

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{1}{1-{{x}^{2}}}$, $x\in \mathbb{R}$, $x>1$. Show that $\mathrm{f}$ has an inverse. [2] Find ${{\mathrm{f}}^{-1}}\left( x ight)$ and state the domain of ${{\mathrm{f}}^{-1}}$. [3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{2+x}{1-{{x}^{2}}}$, $x\in \mathbb{R}$, $x
e \pm 1$. Find algebraically the range of $\mathrm{g}$, giving your answer in terms of $\sqrt{3}$ as simply as possible. [5]

Accepted Answer

(a)(i) $f^{-1}(x)=\sqrt{1-\frac1x}$, domain $(- \infty,0)$; (b) $R_g=(-\infty,1- rac{\sqrt3}{2}]\cup[1+ rac{\sqrt3}{2},\infty)$

N2015 P2 Q3

N2015 P2 Q3

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