MHT CET · Maths · Application of Derivatives
The maximum value of \(\frac{\log x}{x}\) is
- A e
- B \(2 \mathrm{e}\)
- C \(\frac{1}{\mathrm{e}}\)
- D \(\frac{2}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{f}(x)=\frac{\log x}{x} \Rightarrow \mathrm{f}^{\prime}(x)=\frac{1}{x^2}-\frac{\log x}{x^2}\)
For maximum or minimum value of \(\mathrm{f}(x)\),
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow \frac{1-\log _e x}{x^2}=0
\end{aligned}\)
\(\therefore \quad \log _{\mathrm{e}} x=1\) or \(x=\mathrm{e}\), which lie in \((0, \infty)\).
For \(x=\mathrm{e}, \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-\frac{1}{\mathrm{e}^3}\), which is -ve .
\(\therefore \quad y\) is maximum at \(x=\mathrm{e}\) and its maximum value \(=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\).
For maximum or minimum value of \(\mathrm{f}(x)\),
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow \frac{1-\log _e x}{x^2}=0
\end{aligned}\)
\(\therefore \quad \log _{\mathrm{e}} x=1\) or \(x=\mathrm{e}\), which lie in \((0, \infty)\).
For \(x=\mathrm{e}, \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-\frac{1}{\mathrm{e}^3}\), which is -ve .
\(\therefore \quad y\) is maximum at \(x=\mathrm{e}\) and its maximum value \(=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\).
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