Question

If $f\left(x\right)+f\left(\frac{1}{x}\right)=0,f\left(e\right)=1;g\left(x\right)=f^{-1}\left(x\right)$ then $g{}^{\prime }\left(x\right)$ equals

• A. $e^x$
• B. $x$
• C. $x^2$
• D. $e^{-x}$

Answer 1

The correct option is A $e^x$

$f\left(x\right)+f\left(\frac{1}{x}\right)=0$ is satisfied by $f\left(x\right)=k\ln x$, where $k$ is a constant.

Given that $f\left(e\right)=1\Rightarrow k=1$

$f\left(x\right)=\ln x$

$\Rightarrow f^{-1}x=e^x$

$\Rightarrow g\left(x\right)=e^x$

$\Rightarrow g^{\prime }\left(x\right)=e^x$