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MHT CET 2017 Math Solved Paper

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Question : 4 of 50

Marks: +1, -0

The maximum value of

\frac{\log x}{x}

(x ≠ 0, x ≠ 1) is

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

Solution:

(b) since,

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

After differentiating on both sides w.r.t. x, we get

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

=

\frac{1 - \log x}{x^{2}}

For maximum or minimum value of f(x), put

f'(x) = 0

⇒

\frac{1 - \log x}{x^{2}} = 0

⇒

\log x = 1 \Rightarrow x = e

Now,

f''(x) = \frac{3 + 2 \log x}{x^{3}}

∴

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

After substituting x = e in eq. (i), we get

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

Hence, maximum value of

f(x)

is

\frac{1}{e}

at x = e

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