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(A) Let $D_1$ be the first die and $D_2$ be the second die. The outcomes for $D_1$ are ${1, 1, 2, 2, 3, 4}$ and for $D_2$ are ${1, 2, 2, 3, 3, 4}$. To

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$\frac{1}{2}$

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(A) Let $D_1$ be the first die and $D_2$ be the second die. The outcomes for $D_1$ are ${1, 1, 2, 2, 3, 4}$ and for $D_2$ are ${1, 2, 2, 3, 3, 4}$. Total outcomes $= 6 	imes 6 = 36$.We want the sum $S = 4$ or $S = 5$.For $S = 4$,the possible pairs $(D_1, D_2)$ are $(1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (4, 	ext{none})$.Counting frequencies: $(1, 3)$ occurs $2 	imes 2 = 4$ times,$(2, 2)$ occurs $2 	imes 2 = 4$ times,$(3, 1)$ occurs $1 	imes 1 = 1$ time. Total for $S=4$ is $4+4+1 = 9$.For $S = 5$,the possible pairs $(D_1, D_2)$ are $(1, 4), (2, 3), (2, 3), (3, 2), (3, 2), (4, 1)$.Counting frequencies: $(1, 4)$ occurs $2 	imes 1 = 2$ times,$(2, 3)$ occurs $2 	imes 2 = 4$ times,$(3, 2)$ occurs $1 	imes 2 = 2$ times,$(4, 1)$ occurs $1 	imes 1 = 1$ time. Total for $S=5$ is $2+4+2+1 = 9$.Total favorable outcomes $= 9 + 9 = 18$.Probability $= \frac{18}{36} = \frac{1}{2}$.

column-$I$	column-$II$
$(A)$ $A$ line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$,then $d^2$ is	$(p)$ $-4$
$(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are	$(q)$ $0$
$(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} \cdot \vec{b}=0$,$(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2\|\vec{b}+\vec{c}\|=\|\vec{b}-\vec{a}\|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$,then the possible values of $\mu$ are	$(r)$ $4$
$(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) dx$ is	$(s)$ $5$
	$(t)$ $6$

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