The minimum value of x x in the interval [2, , ) is - Vidyadip

MCQ

The minimum value of ${{\log x} \over x}$ in the interval $[2,\,\infty )$ is

A
${{\log 2} \over 2}$
B
Zero
C
${1 \over e}$
✓
Does not exist

✓

Answer

Correct option: D.

Does not exist

d
(d) Let $y = \frac{{\log x}}{x}$

==> $\frac{{dy}}{{dx}} = \frac{{x.\frac{1}{x} - \log x}}{{{x^2}}}$$ = \frac{{1 - \log x}}{{{x^2}}}$

Put $\frac{{dy}}{{dx}} = 0 \Rightarrow \frac{{1 - \log x}}{{{x^2}}} = 0$

==> $1 - \log x = 0$ ==> $x = e$ and $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 3x + 2x\log x}}{{{x^4}}}$

At $x = e$, $\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{ - {e^3}}} < 0$

$\therefore$ In $[2, \infty ) $ the function ${p^2} = q$ will be maximum and minimum value does not exist.

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