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$$\begin{gathered} \mathrm{To} \mathrm{find} \mathrm{range} \mathrm{of} \frac{\mathrm{Quadratic}}{\mathrm{Quadratic}},\frac{\mathrm{Quadratic}}{\mathrm{Linear}}\mathrm{or} \frac{\mathrm{Linear}}{\mathrm{Quadratic}} \mathrm{always} \mathrm{follow} \mathrm{the} \mathrm{following} \mathrm{procedure}: \\ \left(\mathrm{A}\right) \mathrm{y}=\frac{{\mathrm{x}}^2+2\mathrm{x}+4}{\mathrm{x}+2} \\ \Rightarrow \mathrm{y}\left(\mathrm{x}+2\right)={\mathrm{x}}^2+2\mathrm{x}+4 \\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}+4-\mathrm{yx}-\mathrm{y}=0 \\ \Rightarrow {\mathrm{x}}^2+\left(2-\mathrm{y}\right)+\left(4-2\mathrm{y}\right)=0 \\ \mathrm{For} \mathrm{roots} \mathrm{to} \mathrm{be} \mathrm{real} , \mathrm{Discriminant}\ge 0 \\ {\left(2-\mathrm{y}\right)}^2-4\left(4-2\mathrm{y}\right)\ge 0 \\ \Rightarrow 4+{\mathrm{y}}^2-4\mathrm{y}-16+8\mathrm{y}\ge 0 \\ \Rightarrow {\mathrm{y}}^2+4\mathrm{y}-12\ge 0 \\ \Rightarrow {\mathrm{y}}^2+6\mathrm{y}-2\mathrm{y}-12\ge 0 \\ \Rightarrow \left(\mathrm{y}+6\right)\left(\mathrm{y}-2\right)\ge 0 \\ \mathrm{y}\ge 2 \mathrm{or} \mathrm{y}\le -6 \\ \mathrm{Now} \mathrm{y}\ge 2 \\ \mathbf{\Rightarrow }\mathbf{Minimum}\mathbf{ }\mathbf{value}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{ } \left(\mathrm{Note}: \mathrm{It} \mathrm{is} \mathrm{minimum} \mathrm{positive} \mathrm{value}. \mathrm{Since} \mathrm{all} \mathrm{the} \mathrm{options} \mathrm{are} \mathrm{positive} \mathrm{numbers} \mathrm{hence} \mathrm{we} \mathrm{have} \mathrm{answered} \mathrm{minimum} \mathrm{positive} \mathrm{value}.\right) \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{r}\right)\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{match}\mathbf{.} \\ \left(\mathrm{B}\right)\mathbf{ }\left(\mathrm{A}+\mathrm{B}\right)\left(\mathrm{A}-\mathrm{B}\right)=\left(\mathrm{A}-\mathrm{B}\right)\left(\mathrm{A}+\mathrm{B}\right) \\ \Rightarrow {\mathrm{A}}^2-\mathrm{AB}+\mathrm{BA}-{\mathrm{B}}^2={\mathrm{A}}^2+\mathrm{AB}-\mathrm{BA}+{\mathrm{B}}^2 \\ \Rightarrow 2\mathrm{BA}=2\mathrm{AB} \\ \Rightarrow \mathrm{BA}=\mathrm{AB} \\ \mathrm{Now} {\left(\mathrm{AB}\right)}^{\mathrm{t}}={\left(-1\right)}^{\mathrm{k}}\mathrm{AB} \\ \mathrm{We} \mathrm{kno} \mathrm{that} {\left(\mathrm{AB}\right)}^{\mathrm{t}}={\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}} \\ \Rightarrow {\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}}={\left(-1\right)}^{\mathrm{k}}\mathrm{AB} \\ \mathrm{A} \mathrm{is} \mathrm{symmetric} \mathrm{matrix}, \mathrm{hence} {\mathrm{A}}^{\mathrm{t}}=\mathrm{A} \\ \mathrm{B} \mathrm{is} \mathrm{skew}-\mathrm{symmetric} \mathrm{matrix}, {\mathrm{henceB}}^{\mathrm{t}}=-\mathrm{B} \\ \Rightarrow {\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}}={\left(-1\right)}^{\mathrm{k}}\mathrm{AB} \\ \Rightarrow -\mathrm{BA}={\left(-1\right)}^{\mathrm{k}}\mathrm{AB} \\ \mathrm{Since} \mathrm{BA}=\mathrm{AB} \\ \Rightarrow -1={\left(-1\right)}^{\mathrm{k}} \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} -1 \mathrm{we} \mathrm{get}, \\ {\left(-1\right)}^{\mathrm{k}+1}=1 \\ \Rightarrow \mathrm{k}+1 \mathrm{is} \mathrm{even}. \\ \Rightarrow \mathrm{k} \mathrm{is} \mathrm{odd}. \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{q}\mathbf{,}\mathbf{s}\right)\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{match}\mathbf{.} \\ \left(\mathrm{C}\right) 1<2^{\left(-\mathrm{k}+3^{-\mathrm{a}}\right)}<2 \\ \mathrm{Taking} {\log }_2 \mathrm{on} \mathrm{each} \mathrm{side} \\ {\log }_{2}1<{\log }_2{2}^{\left(-\mathrm{k}+3^{-\mathrm{a}}\right)}<{\log }_{2}2 \\ \mathrm{We} \mathrm{know} {\mathrm{thatlog}}_{\mathrm{a}}1=0 \mathrm{and} {\log }_{\mathrm{a}}\mathrm{a}=1 {\log }_{\mathrm{a}}{\mathrm{m}}^{\mathrm{n}}={\mathrm{nlog}}_{\mathrm{a}}\mathrm{m} \\ \Rightarrow 0<\left(-\mathrm{k}+3^{-\mathrm{a}}\right){\log }_{2}2<1 \\ \Rightarrow 0<\left(-\mathrm{k}+3^{-\mathrm{a}}\right)<1 ; \mathrm{inequality}\left(\mathrm{i}\right) \\ \mathrm{a}={\log }_3{\log }_3\left(2\right) \\ \Rightarrow 3^{\mathrm{a}}={\log }_3\left(2\right) \\ \Rightarrow \frac{1}{3^{\mathrm{a}}}=\frac{1}{{\log }_3\left(2\right)} \\ \mathrm{Now} {\log }_{\mathrm{a}}\mathrm{b}=\frac{1}{{\log }_{\mathrm{b}}\mathrm{a}} \\ \Rightarrow 3^{-\mathrm{a}}={\log }_{2}3 \\ \mathrm{Hence} \mathrm{inequality} \mathrm{becomes}: \\ 0<-\mathrm{k}+{\log }_{2}3<1 \\ \mathrm{Adding} -{\log }_{2}3 \mathrm{to} \mathrm{each} \mathrm{side}, \\ -{\log }_{2}3<-\mathrm{k}<1-{\log }_{2}3 \\ \Rightarrow -{\log }_{2}3<-\mathrm{k}<{\log }_{2}2-{\log }_{2}3 \\ \mathrm{As} \mathrm{loga}-\mathrm{logb}=\log \left(\frac{\mathrm{a}}{\mathrm{b}}\right) \\ \Rightarrow -{\log }_{2}3<-\mathrm{k}<{\log }_2\left(\frac{2}{3}\right) \\ \mathrm{Multiplying} \mathrm{each} \mathrm{term} \mathrm{by} \text{'}-1\text{'} \mathrm{so} \mathrm{inequality} \mathrm{will} \mathrm{reverse} \\ \Rightarrow -{\log }_2\left(\frac{2}{3}\right)<\mathrm{k}<{\log }_{2}3 \\ \Rightarrow {\log }_2{\left(\frac{2}{3}\right)}^{-1}<\mathrm{k}<{\log }_{2}3 \\ \Rightarrow {\log }_2\left(\frac{3}{2}\right)<\mathrm{k}<{\log }_{2}3 \\ \mathrm{But} \mathrm{k} \mathrm{is} \mathrm{an} \mathrm{integer} \mathrm{and} \mathrm{only} \mathrm{integer} \mathrm{between} {\log }_2\left(\frac{3}{2}\right) \mathrm{and} {\log }_{2}3 \mathrm{will} \mathrm{be} {\log }_{2}2=1 \\ \Rightarrow \mathrm{k}=1 \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{r}\right)\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{match}\mathbf{.} \\ \left(\mathrm{D}\right) \mathrm{Given} \mathrm{that} \\ \sin \mathrm{\theta }=\cos \mathrm{\varphi } \\ \Rightarrow \cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)=\cos \mathrm{\varphi } \\ \mathrm{Since} \cos \mathrm{is} \mathrm{a} \mathrm{periodic} \mathrm{function} \mathrm{with} \mathrm{a} \mathrm{period} \mathrm{of} 2\mathrm{\pi }, \mathrm{hence} \mathrm{general} \mathrm{solution} \mathrm{will} \mathrm{be}, \\ \frac{\mathrm{\pi }}{2}-\mathrm{\theta }=2\mathrm{n\pi }\pm \mathrm{\varphi } \mathrm{where} \mathrm{n}\in \mathrm{I} \\ \mathrm{\theta }\pm \mathrm{\varphi }-\frac{\mathrm{\pi }}{2}=-2\mathrm{n\pi } \\ \Rightarrow \frac{1}{\mathrm{\pi }}\left(\mathrm{\theta }\pm \mathrm{\varphi }-\frac{\mathrm{\pi }}{2}\right)=-2\mathrm{n} \\ \Rightarrow \frac{1}{\mathrm{\pi }}\left(\mathrm{\theta }\pm \mathrm{\varphi }-\frac{\mathrm{\pi }}{2}\right)=\mathrm{Even} \mathrm{integer} \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{p}\mathbf{,}\mathbf{r}\right)\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{match}\mathbf{.} \end{gathered}$$

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