The maximum value of {{{{In\;x}}}}{{{x}}} is

Correct Answer is : 1e{1}{{{e}}}e1

Correct Answer is : 1e{1}{{{e}}}e1

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 71 of 120

Marks: +1, -0

\frac{{{\rm{In\;x}}}}{{\rm{x}}}

Solution:

Concept:

To find the maximum of a function, set its first derivative to zero and check that the second derivative is negative at that point.

Explanation:

Let

f(x) = \frac{\ln x}{x}

Differentiate using the quotient rule:

f'(x) = \frac{1 - \ln x}{x^2}

Set

f'(x) = 0

, so

1 - \ln x = 0

. This gives

\ln x = 1

and hence

x = e

Now find the second derivative:

f''(x) = \frac{-(3 - 2\ln x)}{x^3}

x = e

f''(e) = \frac{-(3 - 2\cdot 1)}{e^3} = \frac{-1}{e^3} < 0

, which confirms a maximum.

The maximum value is

f(e) = \frac{\ln e}{e} = \frac{1}{e}

Answer:

Option B:

\frac{1}{e}

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