Question

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

• A. $\frac{x+\sqrt{x^2-4}}{2}$
• B. $\frac{x}{1+x^2}$
• C. $\frac{x-\sqrt{x^2-4}}{2}$
• D. $1+\sqrt{x^{-4}}$

Answer 1

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

$f\left(x\right)=x+\frac{1}{x},$ for $f^{-1}\left(x\right)$ Put f (x) = y
$\therefore y=\frac{x^2+1}{x}$ or $x^2$ - xy + 1 = 0
$\Rightarrow x=\frac{y\pm \sqrt{y^2-4}}{2}$
$\therefore f^{-1}\left(y\right)=\frac{y+\sqrt{y^2-4}}{2}$ (neglect - ve, as x > 1)
$\therefore f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2}$