MCQ
$\int_{1/e}^e {|\log x|\,dx = } $
- A$1 - \frac{1}{e}$
- ✓$2\,\left( {1 - \frac{1}{e}} \right)$
- C${e^{ - 1}} - 1$
- DNone of these
$ = [x - x\log x]_{1/e}^1 + [x\log x - x]_1^e$
$ = (1 - 0) - \left\{ {\frac{1}{e} - \frac{1}{e}( - 1)} \right\} + e - e + 1$
$ = 2 - \frac{2}{e} = 2\left( {1 - \frac{1}{e}} \right)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$................$(E)$
$1.$ If the point $P(a, b, c)$, with reference to $( E )$, lies on the plane $2 x+y+z=1$, then the value of $7 a+b+c$ is
$(A)$ $0$ $(B)$ $12$ $(C)$ $7$ $(D)$ $6$
$2.$ Let $\omega$ be a solution of $x^3-1=0$ with $\operatorname{Im}(\omega)>0$. If $a=2$ with $b$ and $c$ satisfying $( E )$, then the value of $\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$ is equal to
$(A)$ $-2$ $(B)$ $2$ $(C)$ $3$ $(D)$ $-3$
$3.$ Let $b=6$, with $a$ and $c$ satisfying (E). If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^2+b x+c=0$, then
$\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n$ is
$(A)$ $6$ $(B)$ $7$ $(C)$ $\frac{6}{7}$ $(D)$ $\infty$
Give the answer question $1,2$ and $3.$