Question

$f\left(x\right)=\frac{e^x+e^{-x}}{2}$ then the inverse of $f\left(x\right)$ is

• A. ${\log }_e\left(x+\sqrt{x^2+1}\right)$
• B. ${\log }_e\sqrt{x^2+1}$
• C. ${\log }_e\left(x+\sqrt{x^2-1}\right)$
• D. ${\log }_e\left(x-\sqrt{x^2-1}\right)$

Answer 1

The correct option is C ${\log }_e\left(x+\sqrt{x^2-1}\right)$

Let $f\left(x\right)=y=\frac{e^x+e^{-x}}{2}$

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

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

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

$\Rightarrow e^x=\frac{2y\pm \sqrt{4y^2-4}}{2}=\frac{2y\pm 2\sqrt{y^2-1}}{2}=y\pm \sqrt{y^2-1}$

$\Rightarrow x{\log }_{e}e={\log }_e(y\pm \sqrt{y^2-1})$

$\Rightarrow x={\log }_e(y+\sqrt{y^2-1})$ since ${\log }_e(y-\sqrt{y^2-1})$ is not defined.

$\Rightarrow x={\log }_e(y+\sqrt{y^2-1})$

$\therefore f^{-1}\left(x\right)={\log }_e(x+\sqrt{x^2-1})$