A) e
B) 1
C) 2e
D) 2
Correct Answer: B
Solution :
Let\[y=\frac{\log x}{x}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x.1}{{{x}^{2}}}\] \[=\frac{1-\log x}{{{x}^{2}}}\] For maxima and minima, \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[1-log\text{ }x=0\] \[\Rightarrow \] \[x=e\] Again, differentiating w.r.t.\[x,\]we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( 0-\frac{1}{x} \right)-(1-\log x)2x}{{{x}^{4}}}\] At \[x=e,\frac{{{d}^{2}}y}{d{{x}^{2}}}<0\] \[\therefore \]y is maximum at\[x=e\] and maximum value\[=\frac{{{\log }_{e}}e}{e}=\frac{1}{e}\]
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