Range of the function y -2 x -2- x -2 x -2- x isA-R B- -1-1C- -1-1-D- 0-1

The correct option is B ( 1,1)2^x+2^ x is always gt; 0 i,e., domain is R y=2^x 2^ x/2^x+2^ x=2^2x 1/2^2x+1 ⇒1+y/1 y=2.2^2x/2 (Componendo Dividendo) =2^2x ...

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Standard IX

Mathematics

Expressing X in Terms of Y, to Find the Range of a Function

Range of the ...

Question

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

R

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(- 1, 1)

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[- 1, 1]

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(0, 1)

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Solution

The correct option is B $(- 1, 1)$
$2^{x} + 2^{- x}$ is always > 0 i,e., domain is R
$y = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}} = \frac{2^{2 x} - 1}{2^{2 x} + 1}$
$\Rightarrow \frac{1 + y}{1 - y} = \frac{{2.2}^{2 x}}{2}$ (Componendo Dividendo)
$= 2^{2 x} > 0$
$\Rightarrow \frac{1 + y}{1 - y} > 0$ i.e., $\frac{(1 + y)^{2}}{1 - y^{2}} > 0$
$\Rightarrow 1 - y^{2} > 0$ $\Rightarrow$ $- 1 < y < 1$

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Similar questions

$R a n g e o f t h e f u n c t i o n y = \frac{2^{x} - 2^{- x}}{2^{x} + 2^{- x}} i s$

Q. Range of the function

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Range of the function y=2x−2−x2x+2−x is

The correct option is B (−1,1)2x+2−x is always > 0 i,e., domain is R y=2x−2−x2x+2−x=22x−122x+1 ⇒1+y1−y=2.22x2 (Componendo Dividendo) =22x>0 ⇒1+y1−y>0 i.e., (1+y)21−y2>0 ⇒1−y2>0 ⇒ −1<y<1

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