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(A) . The period of $\sin^2 x$ is $\pi$. Thus,$A-V$.$B$. The maximum value of $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2+b^2}$. Here,$\frac{\pi}{3}

Vedclass · Accepted Answer

$A-V, B-I, C-II, D-III$

Vedclass · Answer

(A) . The period of $\sin^2 x$ is $\pi$. Thus,$A-V$.$B$. The maximum value of $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2+b^2}$. Here,$\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2} = \frac{\pi}{3} 	imes 2 = \frac{2\pi}{3}$. Thus,$B-I$.$C$. The period of $\sin \frac{x}{3}$ is $\frac{2\pi}{1/3} = 6\pi$ and the period of $\cos \frac{x}{2}$ is $\frac{2\pi}{1/2} = 4\pi$. The period of the sum is the $LCM(6\pi, 4\pi) = 12\pi$. Thus,$C-II$.$D$. For $y=|\sin x|$ and $y=1$,we have $|\sin x|=1$,so $\sin x = \pm 1$. In the interval $(0, \pi)$,$\sin x = 1$ at $x = \frac{\pi}{2}$. Thus,$D-III$.Therefore,the correct match is $A-V, B-I, C-II, D-III$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

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