Question

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

• A. $e^{2}$
• B. $\frac{1}{e}\log\frac{1}{\sqrt{e}}$
• C. $e^{2}\log\sqrt{e}$
• D. None

Answer 1

Answer: (A)

$e^{2}$

Answer 2

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

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

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

$\because$ None of these critical points lies in the interval $\lbrack 1,e\rbrack$

$\therefore$ So we only compare the value of $f(x)$ at the end points 1 and e. We have $f(1) = 0,f(e) = e^{2}$

$\therefore$ greatest value = $e^{2}$