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(C) Let $y = \frac{x^2+2x+4}{x+2} = \frac{x(x+2)+4}{x+2} = x + \frac{4}{x+2} = (x+2) + \frac{4}{x+2} - 2$. By $AM$-$GM$ inequality,for $x > -2$,$(x+2)

Vedclass · Accepted Answer

$(A) \rightarrow (r); (B) \rightarrow (q, s); (C) \rightarrow (r, s); (D) \rightarrow (p, s)$

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(C) Let $y = \frac{x^2+2x+4}{x+2} = \frac{x(x+2)+4}{x+2} = x + \frac{4}{x+2} = (x+2) + \frac{4}{x+2} - 2$. By $AM$-$GM$ inequality,for $x > -2$,$(x+2) + \frac{4}{x+2} \geq 2\sqrt{(x+2) \cdot \frac{4}{x+2}} = 4$. Thus,$y \geq 4 - 2 = 2$. The minimum value is $2$ (Option $r$).$(B)$ Given $(A+B)(A-B) = (A-B)(A+B) \Rightarrow A^2 - AB + BA - B^2 = A^2 + AB - BA - B^2 \Rightarrow 2BA = 2AB \Rightarrow AB = BA$. Since $A^t = A$ and $B^t = -B$,$(AB)^t = B^t A^t = -BA = -AB$. Thus,$(-1)^k AB = -AB$,which implies $(-1)^k = -1$. This holds for odd integers $k$. In the set ${1, 2, 3}$,$k$ can be $1$ or $3$ (Options $q, s$).$(C)$ $a = \log_3 \log_3 2$. Note $3^{-a} = 3^{-\log_3 \log_3 2} = (\log_3 2)^{-1} = \log_2 3$. The inequality is $1 $(D)$ $\sin 	heta = \cos \phi \Rightarrow \cos(\frac{\pi}{2} - 	heta) = \cos \phi \Rightarrow \frac{\pi}{2} - 	heta = 2n\pi \pm \phi \Rightarrow 	heta \pm \phi - \frac{\pi}{2} = -2n\pi$. Thus,$\frac{1}{\pi}(	heta \pm \phi - \frac{\pi}{2}) = -2n$,which are even integers. Among options,$0$ is even (Option $p$).

Column $I$	Column $II$
$(A)$ The minimum value of $\frac{x^2+2x+4}{x+2}$ for $x > -2$ is	$(p)$ $0$
$(B)$ Let $A$ and $B$ be $3 \times 3$ matrices of real numbers,where $A$ is symmetric,$B$ is skew-symmetric,and $(A+B)(A-B)=(A-B)(A+B)$. If $(AB)^t=(-1)^k AB$,where $(AB)^t$ is the transpose of the matrix $AB$,then the possible values of $k$ are	$(q)$ $1$
$(C)$ Let $a=\log_3 \log_3 2$. An integer $k$ satisfying $1 < 2^{(-k+3^{-a})} < 2$,must be less than	$(r)$ $2$
$(D)$ If $\sin \theta = \cos \phi$,then the possible values of $\frac{1}{\pi}(\theta \pm \phi - \frac{\pi}{2})$ are	$(s)$ $3$

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