The inverse of the function f(x) = e^x - e^ - x e^x + e^ - x + 2 … - Vidyadip

MCQ

The inverse of the function $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} + 2$ is given by

A
${\log _e}{\left( {\frac{{x - 2}}{{x - 1}}} \right)^{1/2}}$
✓
${\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$
C
${\log _e}{\left( {\frac{x}{{2 - x}}} \right)^{1/2}}$
D
${\log _e}{\left( {\frac{{x - 1}}{{x + 1}}} \right)^{ - 2}}$

✓

Answer

Correct option: B.

${\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$

b
(b) $y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} + 2\,\, $

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

$ \Rightarrow \,\,{e^{2x}} = \frac{{1 - y}}{{y - 3}} = \frac{{y - 1}}{{3 - y}}\,\, $

$\Rightarrow \,\,x = \frac{1}{2}{\log _e}\,\left( {\frac{{y - 1}}{{3 - y}}} \right)$

$ \Rightarrow {f^{ - 1}}(y) = {\log _e}\,{\left( {\frac{{y - 1}}{{3 - y}}} \right)^{1/2}} $

$\Rightarrow {f^{ - 1}}(x) = {\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$.

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