If $f(x)=\log_{e}(x+\sqrt{1+x^{2}})$ t

CUET Preparation Today

Start Free Mocks

Home Questions YFKK4K8PZR

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

If $f(x)=\log_{e}(x+\sqrt{1+x^{2}})$ then $f^{-1}(x)$ is equal to

Options:

$\frac{e^{x}-e^{-x}}{2}$

$\frac{e^{x}+e^{-x}}{2}$

$\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

$\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$

Correct Answer:

$\frac{e^{x}-e^{-x}}{2}$

Explanation:

$y=\log_e(x+\sqrt{1+x^2})$

so $e^y=x+\sqrt{1+x^2}$ ...(1)

as it is an odd function

$e^{-y}-x+\sqrt{1+x^2}$ ...(2)

so eq(1) - eq(2)

$e^y-e^{-y}=2x$

so $\frac{e^y-e^{-y}}{2}$

so $f^{-1}(x)=\frac{e^x-e^{-x}}{2}$