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Match the Statements / Expressions in Column $I$ with the Statements / Expressions in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ The minimum value of $\frac{x^2+2 x+4}{x+2}$ is | $(p)$ $0$ |
| $(B)$ Let $A$ and $B$ be $3 \times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skewsymmetric, and $(A+B)(A-B)=(A-B)(A+B)$. If $(A B)^t=(-1)^k A B$, where $(A B)^t$ is the transpose of the matrix $A B$, then the possible values of $k$ are | $(q)$ $1$ |
| $(C)$ Let $\mathrm{a}=\log _3 \log _3 2$. An integer $\mathrm{k}$ satisfying $1<2^{\left(-k+3^{-2}\right)}<2$, must be less than | $(r)$ $2$ |
| $(D)$ If $\sin \theta=\cos \phi$, then the possible values of $\frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right)$ are | $(s)$ $3$ |
$(A) \rightarrow(r) ;(B) \rightarrow(q, s) ;(C) \rightarrow(r, s) ;(D) \rightarrow(p, r)$
$(A) \rightarrow(r) ;(B) \rightarrow(p, q) ;(C) \rightarrow(r, p) ;(D) \rightarrow(p, s)$
$(A) \rightarrow(q) ;(B) \rightarrow(q, r) ;(C) \rightarrow(r, s) ;(D) \rightarrow(p, s)$
$(A) \rightarrow(q) ;(B) \rightarrow(q, r) ;(C) \rightarrow(r, s) ;(D) \rightarrow(s, q)$
Solution
$ \text { (A) } y=\frac{x^2+2 x+4}{x+2} $
$ \Rightarrow x^2+(2-y) x+4-2 y=0 $
$ \Rightarrow y^2+4 y-12 \geq 0 $
$ y \leq-6 \text { or } y \geq 2$
minimum value is $2$
$ \text { (B) }(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})=(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B}) $
$ \Rightarrow \mathrm{AB}=\mathrm{BA}$
as $A$ is symmetric and $B$ is skew symmetric
$ \Rightarrow(A B)^t=-A B $
$ \Rightarrow k=1 \text { and } k=3$
$(C)$ $\mathrm{a}=\log _3 \log _3 2 \Rightarrow 3^{-\mathrm{a}}=\log _2 3$
$ \text { Now } 1<2^{-\mathrm{k}+\log _2^3}<2 $
$ \Rightarrow 1<3.2^{-\mathrm{k}}<2 $
$ \Rightarrow \log _2\left(\frac{3}{2}\right)<\mathrm{k}<\log _2(3) $
$ \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}<2 \text { and } \mathrm{k}<3 \text {. }$
$ \text { (D) } \sin \theta=\cos \phi \Rightarrow \cos \left(\frac{\pi}{2}-\theta\right)=\cos \phi $
$ \frac{\pi}{2}-\theta=2 \mathrm{n} \pi \pm \phi $
$ \frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right)=-2 \mathrm{n} $
$ \Rightarrow 0 \text { and } 2 \text { are possible. }$
