If F : [1, ∞)→[2, ∞) is given by

Question:

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

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

(b) $\frac{x}{1+x^{2}}$

(c) $\frac{x-\sqrt{x^{2}-4}}{2}$

(d) $1+\sqrt{x^{2}-4}$

Solution:

Let $f^{-1}(x)=y$

$\Rightarrow f(y)=x$

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

$\Rightarrow y^{2}+1=x y$

$\Rightarrow y^{2}-x y+1=0$

$\Rightarrow y^{2}-2 \times y \times \frac{x}{2}+\left(\frac{x}{2}\right)^{2}-\left(\frac{x}{2}\right)^{2}+1=0$

$\Rightarrow y^{2}-2 \times y \times \frac{x}{2}+\left(\frac{x}{2}\right)^{2}=\frac{x^{2}-1}{4}$

$\Rightarrow\left(y-\frac{x}{2}\right)^{2}=\frac{x^{2}-1}{4}$

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

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

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

$\Rightarrow f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$

So, the answer is (a).

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