Question

chapter :realtion and function
class 12th
answer fast
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Answer 1

Step-by-step explanation:

Let, $\rm\:y=f^{-1}(x)$

$\rm \implies \: f(y) = x$

$\rm \implies \: \frac{ {10}^{y} - {10}^{ - y} }{ {10}^{y} + {10}^{ - y} } = x$

$\rm \implies \: \frac{ ({10}^{y} - {10}^{ - y}) +({10}^{y} + {10}^{ - y}) }{ ({10}^{y} - {10}^{ - y}) - ({10}^{y} + {10}^{ - y}) } = \frac{x + 1}{x - 1}$

$\rm \implies \: \frac{ {10}^{y} - {10}^{ - y}+{10}^{y} + {10}^{ - y} }{ {10}^{y} - {10}^{ - y} - {10}^{y} - {10}^{ - y} } = \frac{x + 1}{x - 1}$

$\rm \implies \: \frac{ 2 \times {10}^{y} }{ - 2 \times {10}^{ - y} } = \frac{x + 1}{x - 1}$

$\rm \implies \: - \frac{ {10}^{y} }{ {10}^{ - y} } = \frac{x + 1}{x - 1}$

$\rm \implies \: - {10}^{2y} = \frac{x + 1}{x - 1}$

$\rm \implies \: {10}^{2y} = \frac{ 1 + x}{ 1 - x}$

$\rm \implies \: log( {10})^{2y} = log \bigg( \frac{ 1 + x}{ 1 - x} \bigg)$

$\rm \implies \: 2y. log( {10}) = log \bigg( \frac{ 1 + x}{ 1 - x} \bigg)$

$\rm \implies \: y = \frac{1}{2}. log \bigg( \frac{ 1 + x}{ 1 - x} \bigg)$

$\rm \implies \: y = log \sqrt{ \frac{ 1 + x}{ 1 - x}}$

$\rm \implies \: f ^{ - 1} (x) = log \sqrt{ \frac{ 1 + x}{ 1 - x}}$