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(A) Given $2 \sin ^2 	heta + \sin ^2 2 	heta = 2$. Using $\sin 2 	heta = 2 \sin 	heta \cos 	heta$,we get $2 \sin ^2 	heta + 4 \sin ^2 	heta \co

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$A-q, s; B-p, r, s, t; C-t; D-r$

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(A) Given $2 \sin ^2 	heta + \sin ^2 2 	heta = 2$. Using $\sin 2 	heta = 2 \sin 	heta \cos 	heta$,we get $2 \sin ^2 	heta + 4 \sin ^2 	heta \cos ^2 	heta = 2$. Dividing by $2$,$\sin ^2 	heta + 2 \sin ^2 	heta (1 - \sin ^2 	heta) = 1$. $3 \sin ^2 	heta - 2 \sin ^4 	heta - 1 = 0 \Rightarrow 2 \sin ^4 	heta - 3 \sin ^2 	heta + 1 = 0$. $(2 \sin ^2 	heta - 1)(\sin ^2 	heta - 1) = 0$. So $\sin ^2 	heta = \frac{1}{2}$ or $\sin ^2 	heta = 1$. Thus $	heta = \frac{\pi}{4}, \frac{\pi}{2}$. $(B)$ Let $y = \frac{3x}{\pi}$. Then $f(x) = [2y] \cos [y]$. The function $[2y]$ is discontinuous at $2y = k \in \mathbb{Z}$,i.e.,$y = \frac{k}{2}$. The function $\cos [y]$ is discontinuous at $y = k \in \mathbb{Z}$. For $x \in [0, \pi]$,$y \in [0, 3]$. Discontinuities occur at $y \in \{0.5, 1, 1.5, 2, 2.5, 3\}$. Converting back to $x = \frac{y\pi}{3}$,we get $x \in \{\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}, \pi\}$. Comparing with options,the set of points is $\{\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi\}$. $(C)$ Volume = $|(\hat{i}+\hat{j}) \cdot ((\hat{i}+2\hat{j}) 	imes (\hat{i}+\hat{j}+\pi\hat{k}))| = |\det \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 0 \ 1 & 1 & \pi \end{bmatrix}| = |\pi(2-1)| = \pi$. $(D)$ $\vec{a} + \vec{b} = -\sqrt{3}\vec{c}$. Squaring both sides: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 3|\vec{c}|^2$. $1 + 1 + 2 \cos \alpha = 3(1) \Rightarrow 2 \cos \alpha = 1 \Rightarrow \cos \alpha = \frac{1}{2} \Rightarrow \alpha = \frac{\pi}{3}$.

Column $I$	Column $II$
$(A)$ Root$(s)$ of the equation $2 \sin ^2 \theta + \sin ^2 2 \theta = 2$	$(p)$ $\frac{\pi}{6}$
$(B)$ Points of discontinuity of the function $f(x) = [\frac{6x}{\pi}] \cos [\frac{3x}{\pi}]$,where $[y]$ denotes the largest integer less than or equal to $y$	$(q)$ $\frac{\pi}{4}$
$(C)$ Volume of the parallelepiped with its edges represented by the vectors $\hat{i}+\hat{j}, \hat{i}+2\hat{j}$ and $\hat{i}+\hat{j}+\pi\hat{k}$	$(r)$ $\frac{\pi}{3}$
$(D)$ Angle between vectors $\vec{a}$ and $\vec{b}$ where $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $\vec{a}+\vec{b}+\sqrt{3}\vec{c}=\overrightarrow{0}$	$(s)$ $\frac{\pi}{2}$
	$(t)$ $\pi$

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