MCQ
The function ${x^2}\log x$ in the interval $ (1, e)$ has
- AA point of maximum
- BA point of minimum
- CPoints of maximum as well as of minimum
- ✓Neither a point of maximum nor minimum
and $f''(x) = 2(1 + \log x) + 1$
Now $f''(1) = 3 + 2{\log _e}1$ and $f''(e) = 3 + 2{\log _e}e$
$f(x)$ has local minimum at $\frac{1}{{\sqrt e }}$,
but $x$ lies only in interval $(1,e)$ so that ${y_2} = \sqrt x $ has not extremum in $(1,e).$
Hence neither a point of maximum nor minimum.
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