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)

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