Q39 / 110 · VITEEE 2023 · Mathematics

The function \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{e}^{\mathrm{x}}-1}+\frac{\mathrm{x}}{2}+1\) is -

  • A
    an odd function
  • Correct answer
    B
    an even function
  • C
    neither an odd nor an even function
  • D
    a periodic function

Solution

\(\begin{aligned} & f(x)=\frac{x}{e^x-1}+\frac{x}{2}+1 \\ & =\frac{2 x+x e^x-x}{2\left(e^x-1\right)}+1=\frac{x+x e^x}{2\left(e^x-1\right)}+1 \\ & f(-x)=\frac{-x-x e^x}{2\left(e^x-1\right)}+1=\frac{x+x e^x}{2\left(e^x-1\right)}+1 \\ & \therefore f(-x)=f(x) \text { for all } x \\ & \therefore f(x) \text { is an even function. } \end{aligned}\)

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

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