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