- ✓${e^2}$
- B${1 \over e}\log {1 \over {\sqrt e }}$
- C${e^2}\log \sqrt e $
- DNone of these
Now $f'(x) = 0$ ==> $x = {e^{ - 1/2}},\,0$
$\because$ $0 < {e^{ - 1/2}} < 1$
None of these critical points lies in the interval $[1, e]$
$\therefore$ So we only complete the value of $f(x)$ at the end points $1$ and $e.$
We have $f(1) = 0,\,f(e) = {e^2}$
$\therefore$ Greatest value = ${e^2}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ If $\mathrm{f}(-10 \sqrt{2})=2 \sqrt{2}$, then $\mathrm{f}^{\prime \prime}(-10 \sqrt{2})=$
$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$
$2.$ The area of the region bounded by the curves $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty < \mathrm{a} < \mathrm{b} < -2$, is
$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$3.$ $\int_{-1}^1 g^{\prime}(x) d x=$
$(A)$ $2 g(-1)$ $(B)$ 0 $(C)$ $-2 g(1)$ $(D)$ $2 \mathrm{~g}(1)$
Give the answer question $1,2$ and $3.$
$( S 1):|(\overrightarrow{ a } \times \overrightarrow{ b })+(\overrightarrow{ c } \times \overrightarrow{ b })|-|\overrightarrow{ c }|=6(2 \sqrt{2}-1)$
$( S 2): \angle ABC =\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then