MCQ
Difference between the greatest and the least values of the function$f (x) = x(ln x - 2)$ on $[1, e^2]$ is
- A$2$
- ✓$e$
- C$e^2$
- D$1$
$y' = x\left( {\frac{1}{x}} \right) + (ln x - 2) = ln x - 1$
$\frac{{dy}}{{dx}}= ln x - 1 = 0$==>$x = e$
now$f (1) = - 2$
$f (e) = - e$(least)
$f (e^2) = 0$(greatest)
difference $= 0 - (-e) = e$ Ans. 
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