For x - the range of fx-2x - 12x-1 is

Let y=2^x 12^x+1 ⇒ 2^x y +y = 2^x 1 ⇒ y+1 = 2^x (1 y) ⇒ 2^x = y+11 y As 2^x 0 ∀ x∈ℝ, So, y+11 y gt;0 ⇒y+1y 1 lt;0 ⇒ y ∈ ( 1, 1) Hence, R(f)=( 1,1) ...

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Standard XI

Mathematics

Range

For x ∈ℝ, the...

Question

For $x \in R$ , the range of $f (x) = \frac{2^{x} - 1}{2^{x} + 1}$ is

R

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R - [- 1, 1]

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(- 1, 1)

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(- 1, 1]

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Solution

The correct option is C $(- 1, 1)$
Let $y = \frac{2^{x} - 1}{2^{x} + 1}$
$\Rightarrow 2^{x} y + y = 2^{x} - 1 \Rightarrow y + 1 = 2^{x} (1 - y)$
$\Rightarrow 2^{x} = \frac{y + 1}{1 - y}$

As $2^{x} > 0 \forall x \in R$ ,
So,
$\frac{y + 1}{1 - y} > 0$
$\Rightarrow \frac{y + 1}{y - 1} < 0$
$\Rightarrow y \in (- 1, 1)$

Hence, $R (f) = (- 1, 1)$

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Range

Standard XI Mathematics

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For x∈R, the range of f(x)=2x−12x+1 is

The correct option is C (−1,1)Let y=2x−12x+1 ⇒2xy+y=2x−1⇒y+1=2x(1−y) ⇒2x=y+11−y As 2x>0 ∀ x∈R, So, y+11−y>0 ⇒y+1y−1<0 ⇒y∈(−1,1) Hence, R(f)=(−1,1)

For $x \in R$ , the range of $f (x) = \frac{2^{x} - 1}{2^{x} + 1}$ is