On [1, e] the greatest value of x^2 x - Vidyadip

MCQ

On $ [1, e] $ the greatest value of ${x^2}\log x$

✓
${e^2}$
B
${1 \over e}\log {1 \over {\sqrt e }}$
C
${e^2}\log \sqrt e $
D
None of these

✓

Answer

Correct option: A.

${e^2}$

a
(a) $f(x) = {x^2}\log x$ ==> $f'(x) = (2\log x + 1)x$

Now $f'(x) = 0$ ==> $x = {e^{ - 1/2}},\,0$

$\because$ $0 < {e^{ - 1/2}} < 1$

None of these critical points lies in the interval $[1, e]$

$\therefore$ So we only complete the value of $f(x)$ at the end points $1$ and $e.$

We have $f(1) = 0,\,f(e) = {e^2}$

$\therefore$ Greatest value = ${e^2}.$

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