Question

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Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x}}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{y}{1} = \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x}}$

can be further rewritten as

$\rm :\longmapsto\:\dfrac{1}{y} = \dfrac{ {e}^{x} + {e}^{ - x} }{{e}^{x} - {e}^{ - x}}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = \dfrac{( {e}^{x} + {e}^{ - x} ) + ({e}^{x} - {e}^{ - x})}{({e}^{x} + {e}^{ - x}) - ({e}^{x} - {e}^{ - x})}$

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = \dfrac{ {e}^{x} + {e}^{ - x} + {e}^{x} - {e}^{ - x}}{{e}^{x} + {e}^{ - x} - {e}^{x} + {e}^{ - x}}$

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = \dfrac{2 {e}^{x} }{2{e}^{ - x} }$

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = \dfrac{ {e}^{x} }{{e}^{ - x} }$

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = {e}^{x} \times {e}^{x}$

$\rm :\longmapsto\:\dfrac{1 + y}{1 - y} = {e}^{2x}$

On taking log both sides, we get

$\rm :\longmapsto\:log \bigg(\dfrac{1 + y}{1 - y}\bigg) = log{e}^{2x}$

$\rm :\longmapsto\:log \bigg(\dfrac{1 + y}{1 - y}\bigg) = 2x log{e}$

$\rm :\longmapsto\:log \bigg(\dfrac{1 + y}{1 - y}\bigg) = 2x$

$\rm :\longmapsto\:x \: = \: \dfrac{1}{2} log \bigg(\dfrac{1 + y}{1 - y}\bigg)$

Hence,

Inverse of

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x}}$

is

$\rm :\longmapsto \: \: \dfrac{1}{2} log \bigg(\dfrac{1 + x}{1 - x}\bigg)$

Hence, Option (3) is correct .

Formula Used :-

$\boxed{ \rm{ log {x}^{y} = y \: logx}}$

$\boxed{ \rm{ loge = 1}}$

$\boxed{ \rm{ \frac{a}{b} = \frac{c}{d} \implies \: \frac{a + b}{a - b} = \frac{c + d}{c - d} \: is \: componendo \: and \: dividendo}}$

$\boxed{ \rm{ {x}^{ - y} = \frac{1}{ {x}^{y} }}}$

Remark :-

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x}}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} \: \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x}}$

$\rm \:  =  \:  \: \dfrac{({e}^{x} + {e}^{ - x})\dfrac{d}{dx}({e}^{x} - {e}^{ - x}) - ({e}^{x} - {e}^{ - x})\dfrac{d}{dx}({e}^{x} + {e}^{ - x})}{ {({e}^{x} + {e}^{ - x})}^{2} }$

$\rm \:  =  \:  \: \dfrac{({e}^{x} + {e}^{ - x})({e}^{x} + {e}^{ - x}) - ({e}^{x} - {e}^{ - x})({e}^{x} - {e}^{ - x})}{ {({e}^{x} + {e}^{ - x})}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {({e}^{x} + {e}^{ - x})}^{2} - {({e}^{x} - {e}^{ - x})}^{2} }{ {({e}^{x} + {e}^{ - x})}^{2} }$

$\rm \:  =  \:  \: \dfrac{4{e}^{x}{e}^{ - x}}{ {({e}^{x} + {e}^{ - x})}^{2} }$

$\rm \:  =  \:  \: \dfrac{4}{ {({e}^{x} + {e}^{ - x})}^{2} }$

$\bf\implies \:\dfrac{dy}{dx} > 0$

$\bf\implies \:y \: is \: always \: increasing.$

$\bf\implies \:y \: is \: one \: one \: as \: well \: as \: onto$

$\bf\implies \:inverse \: of \: y \: exist.$