MCQ
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is
- ✓$\frac{\pi}{4}$
- B$4 \pi$
- C$\frac{\pi}{2}$
- D$2 \pi$
$I =\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x }{1+3^{- x }} dx =\int_{-\pi / 2}^{\pi / 2} \frac{3^{ x } \cos ^{2} x }{1+3^{ x }} dx$
$2 I =\int_{-\pi / 2}^{\pi / 2} \frac{\left(1+3^{ x }\right) \cos ^{2} x }{1+3^{ x }} dx$
$=\int_{-\pi / 2}^{\pi / 2} \cos ^{2} xdx =2 \int_{0}^{\pi / 2} \cos ^{2} x dx$
$\Rightarrow I =\int_{0}^{\pi / 2} \cos ^{2} x dx =\frac{\pi}{4}$
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Statement $-1 :$ If $A \ne I,A \ne - I$ then $\det \left( A \right) = - 1$
Statement $-2 :$ If $A \ne I,A \ne - I$ then ${\rm{tr}}\left( A \right) \ne 0$
$\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$.
Then the ordered pair $( m , M )$ is equal to
$|f( x )-f( y )| \leq\left|( x - y )^{2}\right|, \forall( x , y ) \in R$ If $f(0)=1,$ then
$STATEMENT-1$: $y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$ and
$STATEMENT-2$ : $\mathrm{y}(\mathrm{x})$ is given by $\frac{1}{\mathrm{y}}=\frac{2 \sqrt{3}}{\mathrm{x}}-\sqrt{1-\frac{1}{\mathrm{x}^2}}$