MCQ
$\mathop \smallint \limits_{ - \pi /2}^{\pi /2} \frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx =$ . .. .
- A$\frac{\pi }{2}$
- B$4\pi \;$
- ✓$\frac{\pi }{4}$
- D$\frac{\pi }{8}$
Using, $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x,$ we get :
$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^{ - x}}}}dx} $
Adding $(i)$ and $(ii),$ we get;
$2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}xdx} $
$ \Rightarrow 2I.\int\limits_0^{{\rm{x}}/2} {{{\sin }^2}xdx} $
$2 \mathrm{I}=2 \times \frac{\pi}{4} \Rightarrow \mathrm{I}=\frac{\pi}{4}$
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$A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)$
If $A^7-(\beta-1) A^6-\beta A^5$ is a singular matrix, then the value of $9 \beta$ is
($A$) $\mathrm{M}$ is invertible
($B$) There exists a nonzero column matrix $\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)$ such that $M\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)=\left(\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right)$
($C$) The set $\left\{\mathrm{X} \in \mathbb{R}^3: \mathrm{MX}=\mathbf{0} \neq \neq 0\right\}$, where $\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)$
($D$) The matrix $(M-2 I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix