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