MHT CET · Maths · Application of Derivatives
The minimum value of the function \(f(x)=x \log x\) is
- A -e
- B e
- C \(\frac{1}{\mathrm{e}}\)
- D \(-\frac{1}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
\(f(x)=x \log x\)
\(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}\)
When \(1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}\)
\(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})\right]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0\)
Thus \(\mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\frac{1}{\mathrm{e}}\) and
\(f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}\)
\(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}\)
When \(1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}\)
\(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})\right]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0\)
Thus \(\mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\frac{1}{\mathrm{e}}\) and
\(f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}\)
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