If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$, such that $f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$, then…

If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$, such that $f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$, then $\mathrm{f}$ is
  1. a periodic function
  2. an even function
  3. an odd function
  4. a neither even nor odd function

Solution

$f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$ $=\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1}$ $f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}$ $=\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}}=\frac{1+e^{2 x}}{-\left(e^{2 x}-1\right)}$ $\therefore f(-x)=-f(x)$ $\therefore \mathrm{f}(\mathrm{x})$ is an odd function.

Asked in: MHT CET 2020 (13 Oct Shift 1)

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