Q56 / 150 · MHT CET 2024 (11 May Shift 2) · Mathematics

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

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

Solution

$\begin{aligned} & f(x)=x+\frac{1}{x} \\ & \text { let } y=x+\frac{1}{x} \\ \therefore \quad & x y=x^2+1 \\ \therefore \quad & x^2-x y+1=0\end{aligned}$ $\begin{array}{ll}\therefore & x=\frac{y \pm \sqrt{y^2-4}}{2} \\ \therefore & x=\frac{y+\sqrt{y^2-4}}{2} \\ \therefore & \mathrm{f}^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\end{array} \quad \ldots[\because x \in[1, \infty)]$

Hence, the correct answer is option: 'A'.

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