KCET · Maths · Indefinite Integration
The maximum value of \(\frac{\log _{e} x}{x}\), if \(x>0\) is
- A \(e\)
- B 1
- C \(\frac{1}{e}\)
- D \(-\frac{1}{e}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{e}\)
Step-by-step Solution
Detailed explanation
Let \(y=\frac{\log _{e} x}{x}\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log _{e} x \cdot 1}{x^{2}}=\frac{1-\log _{e} x}{x^{2}}\)
For maximum, put \(\frac{d y}{d x}=0\)
\(\Rightarrow \quad 1-\log _{e} x=0\)
\(\Rightarrow \quad \log _{e} x=1\)
\(\Rightarrow \quad x=e\)
Now, \(\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(0-\frac{1}{x}\right)-\left(1-\log _{e} x\right)(2 x)}{x^{4}}\)
\(=\frac{-x-2 x\left(1-\log _{e} x\right)}{x^{4}}\)
\(\therefore \quad\left(\frac{d^{2} y}{d x^{2}}\right)_{\text {at } x=e} < 0\)
\(\therefore \quad y(e)=\frac{\log _{e} e}{e}=\frac{1}{e}\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log _{e} x \cdot 1}{x^{2}}=\frac{1-\log _{e} x}{x^{2}}\)
For maximum, put \(\frac{d y}{d x}=0\)
\(\Rightarrow \quad 1-\log _{e} x=0\)
\(\Rightarrow \quad \log _{e} x=1\)
\(\Rightarrow \quad x=e\)
Now, \(\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(0-\frac{1}{x}\right)-\left(1-\log _{e} x\right)(2 x)}{x^{4}}\)
\(=\frac{-x-2 x\left(1-\log _{e} x\right)}{x^{4}}\)
\(\therefore \quad\left(\frac{d^{2} y}{d x^{2}}\right)_{\text {at } x=e} < 0\)
\(\therefore \quad y(e)=\frac{\log _{e} e}{e}=\frac{1}{e}\)
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