Question

If $F:[1,\infty )\to [2,\infty )$ is given by $f\left(x\right)=x+\frac{1}{x},\mathrm{then}{f}^{-1}\left(x\right)$ equals

(a) $\frac{x+\sqrt{x^2-4}}{2}$

(b) $\frac{x}{1+x^2}$

(c) $\frac{x-\sqrt{x^2-4}}{2}$

(d) $1+\sqrt{x^2-4}$

Answer 1

$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow y+\frac{1}{y}=x \\ \Rightarrow y^2+1=xy \\ \Rightarrow y^2-xy+1=0 \\ \Rightarrow y^2-2\times y\times \frac{x}{2}+{\left(\frac{x}{2}\right)}^2-{\left(\frac{x}{2}\right)}^2+1=0 \\ \Rightarrow y^2-2\times y\times \frac{x}{2}+{\left(\frac{x}{2}\right)}^2=\frac{x^2-1}{4} \\ \Rightarrow {\left(y-\frac{x}{2}\right)}^2=\frac{x^2-1}{4} \\ \Rightarrow y-\frac{x}{2}=\frac{\sqrt{x^2-4}}{2} \\ \Rightarrow y=\frac{x}{2}+\frac{\sqrt{x^2-4}}{2} \\ \Rightarrow y=\frac{x+\sqrt{x^2-4}}{2} \\ \Rightarrow f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2} \end{gathered}$$

So, the answer is (a).