The maximum value of logx-x is

The correct option is D 1/e f'(x)=1x^2 log xx^2 f'(x)=0 at x=6 f'(x) gt;0 when x lt;e f'(x) lt;0 when x gt;e So ...

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Standard XII

Mathematics

Local Maxima

The maximum v...

Question

The maximum value of $\frac{log x}{x}$ is

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1/e

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e^{2}

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Solution

The correct option is D 1/e
$f^{'} (x) = \frac{1}{x^{2}} - \frac{log x}{x^{2}}$
$f^{'} (x) = 0$ at $x = 6$
$f^{'} (x) > 0$ when $x < e$
$f^{'} (x) < 0$ when $x > e$
So $f (x)$ is maxima at $x = e$
$f (e) = \frac{1}{e}$

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The maximum value of logxx is

The correct option is D 1/ef′(x)=1x2−logxx2f′(x)=0 at x=6f′(x)>0 when x<ef′(x)<0 when x>eSo f(x) is maxima at x=ef(e)=1e

The maximum value of $\frac{log x}{x}$ is

The correct option is D 1/e
$f^{'} (x) = \frac{1}{x^{2}} - \frac{log x}{x^{2}}$
$f^{'} (x) = 0$ at $x = 6$
$f^{'} (x) > 0$ when $x < e$
$f^{'} (x) < 0$ when $x > e$
So $f (x)$ is maxima at $x = e$
$f (e) = \frac{1}{e}$