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UPSC NDA Math Model Paper 5

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Question : 51 of 120

Marks: +1, -0

The maximum value of

f(x)=\frac{\log x}{x}(x \neq O, x \neq 1)

is

e
$\frac{1}{e}$
$e^{2}$
$\frac{1}{e^{2}}$

Solution:

Let

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

Differentiating both sides, we get

f'(x)=\frac{d}{dx}\left(\frac{\ln x}{x}\right)

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

=\frac{1-\ln x}{x^{2}}

For the maximum value of

f(x)

f'(x)=0

\Rightarrow \frac{1-\ln x}{x^{2}}=0

\Rightarrow 1-\ln x=0

\Rightarrow \ln x=1

\Rightarrow x=e^{1}=e

\;\;\;(\ln_a b = c \Rightarrow b = a^{c})

At x = e, we get maximum value of f(x)

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

=\frac{\ln e}{e}

=\frac{1}{e}

(∵ ln e = 1)

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