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
    $1$

Answer

Correct option: B.
$e$
b
$y = x (ln x - 2)$

$y' = x\left( {\frac{1}{x}} \right) + (ln x - 2) = ln x - 1$

$\frac{{dy}}{{dx}}= ln x - 1 = 0$==>$x = e$

now$f (1) = - 2$

        $f (e) = - e$(least)

        $f (e^2) = 0$(greatest)

difference $= 0 - (-e) = e$ Ans. 

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0), f(0)=0$ and $f(1)=\frac{1}{K},$ then the value of $K$ is
If $\int \frac{\mathrm{d} \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{\mathrm{e}}|\mathrm{f}(\theta)|+\mathrm{C}$  where $\mathrm{C}$ is a constant of integration, then the ordered pair $(\lambda, f(\theta))$ is equal to
If $A$ is square matrix such that $A^{2}=A$, then $(1+A)^{3}-7 A$ is equal to
Given the function $f(x) = 2x \sqrt {{x^3}\, - \,\,1}    + 5 \sqrt x  \sqrt {1\,\, - \,\,{x^4}}  + 7x^2 \sqrt {x\,\, - \,\,1} + 3x + 2$ then :
If a right circularcone having maximum volume, is inscribed in a sphere of radius $3\, cm$, then the curved surface area (in $cm^2$) of this cone is
Choose the correct answer from the given four options:he two curves $x^3 - 3xy^2 + 2 = 0$ and $3x^2y - y^3 - 2 = 0$ intersect at an angle of:
The first step in formulating an LP problem is:
$\int_{}^{} {\frac{{{e^{\sqrt x }}\cos {e^{\sqrt x }}}}{{\sqrt x }}dx} = $
Let $f:(0,2) \rightarrow R$ be defined as $f( x )=\log _{2}\left(1+\tan \left(\frac{\pi x }{4}\right)\right)$ Then, $\lim _{n \rightarrow \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f(1)\right)$ is equal to
If $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{J}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix},$ then $B$ equals: