<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{1}{1-{{x}^{2}}}$, $x\in \mathbb{R}$, $x>1$.</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that $\mathrm{f}$ has an inverse.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.</span></div><p style="text-align: right;">[3]</p></li></ol></li><li><div class="list-item-content"><span class="list-item-text">The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{2+x}{1-{{x}^{2}}}$, $x\in \mathbb{R}$, $x\ne \pm 1$. Find algebraically the range of $\mathrm{g}$, giving your answer in terms of $\sqrt{3}$ as simply as possible.</span></div><p style="text-align: right;">[5]</p></li></ol>

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

Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.

The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{2+x}{1-{{x}^{2}}}$, $x\in \mathbb{R}$, $x\ne \pm 1$. Find algebraically the range of $\mathrm{g}$, giving your answer in terms of $\sqrt{3}$ as simply as possible.

N2015 P2 Q3

N2015 P2 Q3

Video Solution:

Handwritten Solution: