Question

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

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

Answer 1

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

Let $y=x+\frac{1}{x}$
$\Rightarrow x^2-yx+1=0\Rightarrow x=\frac{y+\sqrt{y^2-4}}{2}=f^{-1}\left(y\right)$
Hence $f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2}$
Note: -ve term in step two is discarded using given domain and co-domain.