Question

The inverse of the function $f\left(x\right)={\log }_2(x+\sqrt{x^2+1})$ is

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

Answer 1

The correct option is D $\frac{2^x-2^{-x}}{2}$

We have to find the inverse of $f\left(x\right)$ which means $f$ is one one and onto function
Let $f\left(x\right)=y={\log }_2(x+\sqrt{x^2+1})$
$\therefore 2^y=x+\sqrt{x^2+1}$ ...(i)
$2^{-y}=\frac{1}{x+\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}-x}{1}$ ...(ii)
By subtracting (i) and (ii) we have
$\therefore 2^y-2^{-y}=2x$
or $x=\frac{1}{2}(2^y-2^{-y})$
$\therefore f^{-1}\left(y\right)=\frac{1}{2}(2^y-2^{-y})$
Replace $y\to x$
$\therefore f^{-1}\left(x\right)=\frac{1}{2}(2^x-2^{-x})$