Explanation

(a)
Given, $f(x) = x^{2}\log x$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \left( 2\log x + 1 \right)x$
For a maximum put $f^{'}(x) = 0$
$\Rightarrow \ \ \ (2\log{x + 1})x = 0\ \ \Rightarrow \ \ \ x = e^{- 1/2},\ 0$
$\because\ \ 0 < e^{- 1/2} < 1$
None of these critical points lies in the interval $\lbrack 1,\ e\rbrack$
So, we only compute the value of $f(x)$ at the end points 1 and $e$
We have, $f(1) = 0,\ \ f(e) = e^{2}$
Hence, greatest value of $f(x) = e^{2}$