Range of the function y = 2 x + 2 − x 2 x − 2 − x is

2 x + 2 − x is always > 0 , i.e., the domain is R . y = 2 x + 2 − x 2 x − 2 − x = 2 2 x + 1 2 2 x − 1 Applying Componendo Dividendo: 1 − y 1 + y = 2 2 ⋅ 2 2 x = 2 2 x > 0 Therefore, 1 − y 1 + y > 0 , which implies 1 − y 2 ( 1 + y ) 2 > 0 . This leads to 1 − y 2 > 0 , so − 1 < y < 1 .

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Range of the function $y = \frac{2 ^{x} - 2 ^{- x}}{2 ^{x} + 2 ^{- x}}$ is

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Solution:

$2^{x} + 2^{- x}$ is always $> 0$ , i.e., the domain is $R$ .

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

Applying Componendo Dividendo:

$\frac{1 + y}{1 - y} = \frac{2 \cdot 2 ^{2 x}}{2} = 2^{2 x} > 0$

Therefore, $\frac{1 + y}{1 - y} > 0$ , which implies $\frac{( 1 + y ) ^{2}}{1 - y ^{2}} > 0$ .

This leads to $1 - y^{2} > 0$ , so $- 1 < y < 1$ .

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

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