If f(x)=ex+e−x2, then the inverse function of f(x) is

f(x)=ex+e−x2Let f−1(x)=y⇒ x=f(y)⇒ x=ey+e−y2⇒2x=ey+1ey⇒ (ey)2−(2x)ey+1=0∴ ey= 2x±4x2−42=x±x2−1∴ ey>0⇒ ey=x+x2−1y=loge(x+x2−1)

If f(x)=ex+e−x2, then the inverse function of f(x) is

f(x)=ex+e−x2Let f−1(x)=y⇒ x=f(y)⇒ x=ey+e−y2⇒2x=ey+1ey⇒ (ey)2−(2x)ey+1=0∴ ey= 2x±4x2−42=x±x2−1∴ ey>0⇒ ey=x+x2−1y=loge(x+x2−1)

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$I f f (x) = \frac{e^{x} + e^{- x}}{2}, t h e n t h e i n v e r s e f u n c t i o n o f f (x) i s$

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

$\log_{e} \sqrt{x^{2} + 1}$

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{e^{x} + e^{- x}}{2} \\ L e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = \frac{e^{y} + e^{- y}}{2} \\ \Rightarrow 2 x = e^{y} + \frac{1}{e^{y}} \\ \Rightarrow {(e^{y})}^{2} - (2 x) e^{y} + 1 = 0 \\ ∴ e^{y} = \frac{2 x \pm \sqrt{4 x^{2} - 4}}{2} \\ = x \pm \sqrt{x^{2} - 1} \\ ∴ e^{y} > 0 \Rightarrow e^{y} = x + \sqrt{x^{2} - 1} \\ y = \log_{e} (x + \sqrt{x^{2} - 1}) \end{array}$