Mathematics | VITEEE 2016 | Discussion Page

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

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

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

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

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

Option B

Given $e^{x}=y+\sqrt{1+y^{2}}$

$\Rightarrow e^{x}-y=\sqrt{1+y^{2}}$

Squaring both side, we have

$e^{2x}+y^{2}-2e^{x}y=1+y^{2}$

$\Rightarrow 2e^{x}y=e^{2x}-1$

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

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