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

Let $y=x+\frac{1}{x}$

$\Rightarrow x^2-xy+1=0\Rightarrow x=\frac{y\pm \sqrt{y^2-4}}{2}$

$\because x\in (2,\infty )$

$\therefore x=\frac{y+\sqrt{y^2-4}}{2}$

Hence, $f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2}$