Q42 / 160 · TS EAMCET 2019 (04 May Shift 1) · Mathematics

If $f:[1, \infty) \rightarrow[0, \infty)$ is given by $f(x)=x-\frac{1}{x}$, then $f^{-1}(x)=$

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

Solution

We have, $f(x)=x-\frac{1}{x}$ $$ \begin{aligned} f\left(f^{-1}(x)\right) & =f^{-1}(x)-\frac{1}{f^{-1}(x)} \\ x & =\frac{\left(f^{-1} x\right)^2-1}{f^{-1}(x)} \end{aligned} $$ $$ \begin{aligned} & \Rightarrow\left(f^{-1}(x)\right)^2-x f^{-1}(x)-1=0 \\ & \Rightarrow \quad f^{-1}(x)=\frac{x+\sqrt{x^2+4}}{2} \quad[\because x \in(1, \infty)] \\ & \therefore \quad f^{-1}(x)=\frac{1}{2}\left(x+\sqrt{x^2+4}\right) \end{aligned} $$

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

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