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Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}, & \mathrm{x}>0 \\ 0, & \mathrm{x}=0\end{array}\right.$ be a real valued function.
Which one of the following statements is correct?
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Which one of the following statements is correct?
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Verified Answer
The correct answer is:
$\mathrm{f}(\mathrm{x})$ is a strictly increasing function in $(0, \mathrm{x})$,
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}>0$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{xe}^{\mathrm{x}}-\left(\mathrm{e}^{\mathrm{x}}-1\right)}{\mathrm{x}^{2}}=\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+1}{\mathrm{x}^{2}}$
$=\left(\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{2}}\right)$, which is a strictly increasing function.
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{xe}^{\mathrm{x}}-\left(\mathrm{e}^{\mathrm{x}}-1\right)}{\mathrm{x}^{2}}=\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+1}{\mathrm{x}^{2}}$
$=\left(\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{2}}\right)$, which is a strictly increasing function.
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