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(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} =

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

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(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,

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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,