Let f :(0,1) rightarrow R be a function defin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \frac{1}{1-e^{-x}} = \frac{e^x}{e^x-1}$.Then $f(-x) = \frac{1}{1-e^x}$.$g(x) = f(-x) - f(x) = \frac{1}{1-e^x} - \frac{e^x}{e^x-1} =

Vedclass · Accepted Answer

Both $(I)$ and $(II)$ are true

Vedclass · Answer

(D) Given $f(x) = \frac{1}{1-e^{-x}} = \frac{e^x}{e^x-1}$.Then $f(-x) = \frac{1}{1-e^x}$.$g(x) = f(-x) - f(x) = \frac{1}{1-e^x} - \frac{e^x}{e^x-1} = \frac{1}{1-e^x} + \frac{e^x}{1-e^x} = \frac{1+e^x}{1-e^x}$.Now,$g'(x) = \frac{(1-e^x)(e^x) - (1+e^x)(-e^x)}{(1-e^x)^2} = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1-e^x)^2} = \frac{2e^x}{(1-e^x)^2}$.Since $e^x > 0$ and $(1-e^x)^2 > 0$ for all $x \in (0,1)$,$g'(x) > 0$.Therefore,$g(x)$ is an increasing function in $(0,1)$.Since $g(x)$ is strictly increasing,it is also a one-one function in $(0,1)$.Thus,both statements $(I)$ and $(II)$ are true.

Let $f :(0,1) \rightarrow R$ be a function defined by $f(x)=\frac{1}{1-e^{-x}}$,and $g(x)=(f(-x)-f(x))$. Consider two statements:
$(I)$ $g$ is an increasing function in $(0,1)$
$(II)$ $g$ is one-one in $(0,1)$
Then,

Similar Questions

$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\begin{cases} -2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2 \end{cases}$ and $g(x)=|f(x)|+f(|x|)$, then

Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$. Then $f$ is:

Let $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$ and $A = \{x \in \mathbb{R} : f(x) = 1\}$. Then,$A$ has

Let $f: R \rightarrow R$ be defined as $f(x)=3x$. Choose the correct answer.

The number of one-one functions from the set $A = \{1, 2, 3, 4, 5\}$ to itself is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $f :(0,1) \rightarrow R$ be a function defined by $f(x)=\frac{1}{1-e^{-x}}$,and $g(x)=(f(-x)-f(x))$. Consider two statements:$(I)$ $g$ is an increasing function in $(0,1)$$(II)$ $g$ is one-one in $(0,1)$Then,