MCQ
If $f : [1, \infty) \rightarrow [2, \infty)$ is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then $f^{-1}$ equals to:
  • $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
  • B
    $\frac{\text{x}}{1+\text{x}^2}$
  • C
    $\frac{\text{x}-\sqrt{\text{x}^2-4}}{2}$
  • D
    $1+\sqrt{\text{x}^2-4}$

Answer

Correct option: A.
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

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