$f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$ is

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Aicharya · Accepted Answer

Given function $f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$. $\begin{aligned} \quad f(-x) & =\frac{-x}{e^{-x}-1}-\frac{x}{2}+2 \cos ^3\left(-\frac{x}{2}\right) \ & =\frac{-x e^x}{1-e^x}-\frac{x}{2}+2 \cos ^3 \frac{x}{2} \ & =\frac{x e^x}{e^x-1}-\frac{x}{2}+2 \cos ^3 \frac{x}{2} \ & =\frac{x\left(e^x-1+1\right)}{e^x-1}-\frac{x}{2}+2 \cos ^3 \frac{x}{2} \ & =x+\frac{x}{e^x-1}-\frac{x}{2}+2 \cos ^3 \frac{x}{2} \ & =\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}=f(x) \end{aligned}$ $\because f(-x)=f(x), \forall x \in R-\{0\}$ $\therefore f(x)$ is an even function. Hence, option (4) is correct.

\(f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}\) on \(R-\{0\}\) is

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