Question

please find the question in the attachment. mathematics class 11​

Answer 1

☆ $\sf{ 0 < {e}^{x} }$

$\sf{\implies f(x) = \frac{ {e}^{x} - 1}{ {e}^{x} + 1} }$

$\sf{\implies \frac{{e}^{x} -1}{ {e}^{x} + 1 } = \frac{ {e}^{x} + 1}{ {e}^{x} + 1} - \frac{2}{ {e}^{x} + 1 }}$

$\sf{ \implies \: f(x) = 1 - \frac{2}{ {e}^{x} + 1 } }$

→ Adding 1 :-

$\sf{ \implies \: {e}^{x} + 1>0 + 1 \rightarrow \: {e}^{x} + 1 > 1 }$

$\sf{ \implies \: 0 < \frac{1}{ {e}^{x} + 1 } < 1}$

→ Multiplying by 2 :-

$\sf{ \implies \: 0 < \frac{2}{ {e}^{x} + 1 } < 2}$

→ Now multiplying by negative :-

$\sf{ \implies \: -2 < - \frac{2}{ {e}^{x} + 1 } < 0}$

→ Now adding 1 :-

$\sf{ \implies \: 1 - 2 < 1 - \frac{2}{ {e}^{x} + 1 } < 1 }$

$\sf{ \implies \: -1 < 1 - \frac{2}{ {e}^{x} + 1 } < 1 }$

$\sf{ \implies \: -1 < \frac{{e}^{x} - 1 }{ {e}^{x} + 1 } < 1 }$

Answer is option D (-1 , 1)

Answer 2

${\underline{\sf{Question}}}$

Find range of

$\sf{f(x) =\dfrac{ {e}^{x} - 1}{ {e}^{x} + 1} }$

${\underline{\sf{Solution}}}$

$\sf{f(x) =\dfrac{ {e}^{x} - 1}{ {e}^{x} + 1} }$

$\sf{\implies\:f(x)=\dfrac{{e}^{x}+1-2}{ {e}^{x} + 1 } }$

$\sf{ \implies\:f(x) = 1-\dfrac{2}{ {e}^{x} + 1 }}$

Now apply the limits .

$\sf\lim_{x\to+\infty}1-\dfrac{2}{\infty}=1$

$\sf\lim_{x\to-\infty}1-\dfrac{2}{0+1}=-1$

Therefore ,Range of $\sf{f(x) =\dfrac{ {e}^{x} - 1}{ {e}^{x} + 1}}$ is (-1,1)

Correct option D) (-1 , 1)

$\rule{200}2$

Graph of Given Function :

$\sf{f(x) = \dfrac{ {e}^{x} - 1}{ {e}^{x} + 1} }$

$\sf{f(-x) =\dfrac{ {e}^{-x} - 1}{ {e}^{-x} + 1}=-f(x)}$

Thus, f(x) is odd function

We know that the graph of odd function will be symmetric about origin.

we have, f (0) = 0

and range =(-1,1)

Note :

Refer to the attachment for graph .