MCQ
The inverse of the function $\frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}$ is
  • $\frac{1}{2}{\log _{10}}\left( {\frac{{1 + x}}{{1 - x}}} \right)$
  • B
    $\frac{1}{2}{\log _{10}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$
  • C
    $\frac{1}{4}{\log _{10}}\left( {\frac{{2x}}{{2 - x}}} \right)$
  • D
    None of these

Answer

Correct option: A.
$\frac{1}{2}{\log _{10}}\left( {\frac{{1 + x}}{{1 - x}}} \right)$
a
(a) $y = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}$

$\Rightarrow x = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + y}}{{1 - y}}} \right)$

Let $y = f(x)$ ==> $x = \pi ,\,\,f(\pi ) = - \tan \frac{\pi }{4} = - 1$

==> ${f^{ - 1}}(y) = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + y}}{{1 - y}}} \right)$ 

==>${f^{ - 1}}(x) = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + x}}{{1 - x}}} \right)$

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