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Let f:(0,1) \rightarrow \mathbb{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,

Option: 1

Both (I) and (II) are true


Option: 2

Neither (I) nor (II) is true


Option: 3

Only (I) is true


Option: 4

Only (II) is true


Answers (1)

best_answer

f(x)=\frac{1}{1-e^{-x}}
\mathrm{~g}(\mathrm{x})=(\mathrm{f}(-\mathrm{x})-\mathrm{f}(\mathrm{x}))
          =\frac{1}{1-\mathrm{e}^{\mathrm{x}}}-\frac{1}{1-\mathrm{e}^{-\mathrm{x}}}
          =\frac{1}{1-\mathrm{e}^{\mathrm{x}}}-\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}-1}

\mathrm{~g}(\mathrm{x})=\frac{1+\mathrm{e}^{\mathrm{x}}}{1-\mathrm{e}^{\mathrm{x}}}
g^{\prime}(x)=\frac{\left(1-e^{x}\right)\left(e^{x}\right)-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}}
=\frac{e^{x}-e^{2 x}+e^{x}+e^{2 x}}{\left(1-e^{x}\right)^{2}}
g^{\prime}(x)=\frac{2 e^{x}}{\left(1-e^{x}\right)^{2}}
\mathrm{~g}^{\prime}(\mathrm{x})>0 \Rightarrow \mathrm{g}(\mathrm{x}) \uparrow

g(x) \text { is one-one }

Posted by

Pankaj

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