Question

if $e^x \, =\, y+ \sqrt{1+y^2},$ then the value of y is

• A. $\frac{1}{2}(e^x+e^{-x})$
• B. $\frac{1}{2}(e^x - e^{-x})$
• C. $e^x \, - \, e ^{\frac{-x}{2}}$
• D. $e^x \, + \, e ^{\frac{-x}{2}}$

Answer 1

Answer: (B)

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

Answer 2

Given $e^x =\, y + \sqrt{1+y^2}$
$\Rightarrow \, \, e^x \, - y = \sqrt{1+y^2}$
Squaring both side, we have
$e^{2x} + y^2 - 2e^xy \, = 1+y^2$
$\Rightarrow 2e^{x}y=e^{2x}-1$
$\Rightarrow \, \, y=\frac{e^{2x}-1}{2e^x}$

$\Rightarrow \, y=\frac{1}{2} \, [e^x - e^{-x}]$

So, the correct answer is (B): $\Rightarrow \, y=\frac{1}{2} \, [e^x - e^{-x}]$