Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Write explicitly function of y defined by the following equation.

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

Options:

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

$\log_e(x-\sqrt{x^2-1})$

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

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

Correct Answer:

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

Explanation:

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

$⇒e^{2y}-2xe^y-1=0$  (Multiplying by $e^y$)

$⇒e^y=\frac{2x±\sqrt{4x^2+4}}{2}=x±\sqrt{x^2+1}$

$⇒e^y=x+\sqrt{x^2+1}$  (as $\sqrt{x^2+1}>x$, then $x-\sqrt{x^2+1}<0$, which is not possible)

$⇒y=\log_e(x+\sqrt{x^2+1})$