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