Match the following : \

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Professor PC Thomas Classes · Answer

(A) - r; (B) - q, s; (C) - r, s; (D) - p, r (A) $y=\frac{x^{2}+2 x+4}{x+2}$ Put $x+2=t$, then\ $y=\frac{(t-2)^{2}+2(t-2)+4}{t}=\frac{t^{2}-2 t+4}{t}$ $=\left(t+\frac{4}{t} ight)-2$\ for $t>0$ we have\ $\frac{t+\frac{4}{t}}{2} \geq \sqrt{t . \frac{4}{t}}$ (A.M. - G.M. inequality)\ $\Rightarrow \quad t+\frac{4}{t} \geq 4$ Similarly, for $t<0$, we have by applying A.M. - G.M. to\ $-t,-4 / t \Rightarrow t+\frac{4}{t} \leq-4$\ Then the minimum value is 2 .\ (B) $(A+B)(A-B)=(A-B)(A+B)$ gives $A^{2}-A B+B A-B^{2}$\ $=A^{2}+A B-B A-B^{2} \Rightarrow A B=B A$.\ Now, $(A B)^{\mathrm{T}}=B^{T} A^{T}=(-B)(A)=-B A$\ $=-A B=(-1)^{k} A B$\ Thus $k$ has to be odd, so $k$ can take, among the choices, the values 1 and 3 .\ (C) $a=\log _{3}\left(\log _{3} 2 ight) \Rightarrow 3^{-a}=3^{-\log _{3} \log _{3} 2}$\ $=3^{\log _{3} \frac{1}{\log _{3} 2}}=\frac{1}{\log _{3} 2}=\log _{2} 3$\ Now, $\left\{2^{-k+3^{-a}}=2^{-k+\log _{2} 3}=2^{-k} \cdot 2^{\log _{2} 3} ight.$\ Now, $1<3 \cdot 2^{-k}<2 \Rightarrow \frac{1}{3}<2^{-k}<\frac{2}{3}$\ $\Rightarrow 3>2^{k}>\frac{3}{2} \Rightarrow \frac{3}{2}<2^{k}<3$\ The only integer between $3 / 2$ and 3 is 2 , which gives $2^{k}=2 \Rightarrow k=1$\ Then $k$ is less than 2 and 3 .\ (D) $\sin heta=\cos \phi \Rightarrow \cos (\pi / 2- heta)=\cos \phi$\ $\Rightarrow \pi / 2- heta=2 n \pi \pm \phi \Rightarrow-2 n \pi$\ $= heta-\phi-\pi / 2$\ $\Rightarrow \frac{1}{\pi}\left( heta \pm \phi-\frac{\pi}{2} ight)=-2 n=$ an even integer.\ So it can take the values 0 or 2 .\