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,
Solution
$g(x)=f(-x)-f(x)$
<br/><br/>
$$
\begin{aligned}
& =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\
& =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\
& =\frac{1+e^{x}}{1-e^{x}} \\\\
g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\
& =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0
\end{aligned}
$$<br/><br/>
<p>So both statements are correct</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
This question is part of PrepWiser's free JEE Main question bank. 99 more solved questions on Application of Derivatives are available — start with the harder ones if your accuracy is >70%.