Question

If $f:[1,\infty ]\to [2,\infty ]$ is given by $f\left(x\right)=x+\frac{1}{x}$, 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

The correct option is A $\frac{x+\sqrt{x^2-4}}{2}$

Clearly, for $f:[1,\infty )\to [2,\infty ),f\left(x\right)=x+\frac{1}{x}$ is a bijection.
Let $f\left(x\right)=y$

Then, $x+\frac{1}{x}=y$
$\Rightarrow x^2-xy+1=0$
$\Rightarrow x=\frac{y\pm \sqrt{y^2-4}}{2}$
$\Rightarrow x=\frac{y+\sqrt{y^2-4}}{2}$ $[\because x\ge 1]$
$\Rightarrow f^{-1}\left(y\right)=\frac{y+\sqrt{y^2-4}}{2}$
Hence, $f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2},\forall xϵ[1,\infty )$.