MHT CET · Maths · Application of Derivatives
The maximum value of the function \(\frac{\log x}{x}, x \neq 0\) is
- A \(e^{2}\)
- B \(\frac{1}{e}\)
- C \(\frac{1}{e^{2}}\)
- D \(e\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{e}\)
Step-by-step Solution
Detailed explanation
Let \(y=\frac{\log x}{x}\)
\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x} \cdot \frac{1}{\mathrm{x}}-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\)
Put \(\frac{d y}{d x}=0\), we get
\(1-\log x=0 \Rightarrow \log x=1 \Rightarrow \log x=\log e \Rightarrow x=e \)
\( \text { Now } \frac{d^{2} y}{d x^{2}} =\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}\)\(=\frac{-x-2 x+2 x \log x}{x^{4}} \)
\( =\frac{2 \log x-3}{x^{3}} \)
\(\text {At } x=e, \frac{d^{2} y}{d x^{2}} < 0\)
from (1), maximum value is \(y=\frac{1}{\mathrm{e}}\)
\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x} \cdot \frac{1}{\mathrm{x}}-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\)
Put \(\frac{d y}{d x}=0\), we get
\(1-\log x=0 \Rightarrow \log x=1 \Rightarrow \log x=\log e \Rightarrow x=e \)
\( \text { Now } \frac{d^{2} y}{d x^{2}} =\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}\)\(=\frac{-x-2 x+2 x \log x}{x^{4}} \)
\( =\frac{2 \log x-3}{x^{3}} \)
\(\text {At } x=e, \frac{d^{2} y}{d x^{2}} < 0\)
from (1), maximum value is \(y=\frac{1}{\mathrm{e}}\)
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