Q64 / 125 · BITSAT 2021 · Mathematics

$f(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right)$ is

  • Correct answer
    A
    an increasing function
  • B
    a decreasing function
  • C
    an even function
  • D
    None of these

Solution

$\begin{array}{l} \because \mathrm{f}(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right) \\ \therefore \mathrm{f}(-x)=\frac{e^{-2 x}-1}{e^{-2 x}+1}=\frac{1-e^{2 x}}{1+e^{2 x}} \\ \Rightarrow \mathrm{f}(-x)=\frac{-\left(e^{2 x}-1\right)}{e^{2 x}+1}=-\mathrm{f}(x) \end{array}$ $\therefore \mathrm{f}(x)$ is an odd function. Again, $\mathrm{f}^{\prime}(x)=\frac{4 e^{2 x}}{\left(1+e^{2 x}\right)^{2}}>0, \forall x \in R$ $\Rightarrow \mathrm{f}(x)$ is an increasing function.

Hence, the correct answer is option: 'A'.

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