JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 15)
Let $$f:(0,1)\to\mathbb{R}$$ be a function defined $$f(x) = {1 \over {1 - {e^{ - x}}}}$$, and $$g(x) = \left( {f( - x) - f(x)} \right)$$. Consider two statements
(I) g is an increasing function in (0, 1)
(II) g is one-one in (0, 1)
Then,
Both (I) and (II) are true
Neither (I) nor (II) is true
Only (II) is true
Only (I) is true
Explanation
$g(x)=f(-x)-f(x)$
$$ \begin{aligned} & =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\ & =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\ & =\frac{1+e^{x}}{1-e^{x}} \\\\ g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\ & =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0 \end{aligned} $$
$$ \begin{aligned} & =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\ & =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\ & =\frac{1+e^{x}}{1-e^{x}} \\\\ g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\ & =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0 \end{aligned} $$
So both statements are correct
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