Question

In the section, each question has some statements $(A,B,C,D,\dots )$ given in $Column–I$ and some statements $(p,q,r,s,t,\dots )$in the$column–II$.

Any given statement is $column–I$ can have correct matching with ONE OR MORE statement(s) in the $column–II$ for example, if for a given questions, statement $B$ matches with the statements given in $q$ and $r$, then for that particular question against statement $B$, darken the bubbles corresponding to $q$ and $r$ in the $ORS$. i.e., answer $r$ will be $q$ and $r$. Match that statements are given in $Column–I$ with the intervals/union of intervals given in $Column–II$

Column 1	Column 2
The minimum value of$\frac{x^2+2x+4}{x+2}=$	$0$
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 ${\left(AB\right)}^t={(-1)}^{k}AB$, where ${\left(AB\right)}^t=$ is the possible values of $k$ are	$1$
Let $a={\log }_3{\log }_{2}2$. An integer $k$ satisfying $1<2^{(-k+3^{-a})}<2$must be less than	$2$
If $\sin \theta =\cos \varphi$then the possible values of $\frac{1}{\mathrm{\pi }}\left(\theta \pm \varphi -\frac{\mathrm{\pi }}{2}\right)$ are	$3$

Answer 1

Explanation for (A)

Finding the minimum value of

$\frac{x^2+2x+4}{x+2}$

Simplifying the function, we get

$$\begin{gathered} y=\frac{x^2+2x+4}{x+2} \\ \Rightarrow yx+2y=x^2+2x+4 \\ \Rightarrow x^2+2x-yx+4-2y=0 \\ \Rightarrow x^2+(2-y)x+(4-2y)=0 \end{gathered}$$

For $x$ to be real, discriminant should be greater than or equal to zero,

$$\begin{gathered} b^2-4ac\ge 0 \\ \Rightarrow {(2-y)}^2-4(4-2y)\ge 0 \\ \Rightarrow 4+y^2-4y-16+8y\ge 0 \\ \Rightarrow y^2+4y-12\ge 0 \\ \Rightarrow y^2+6y-2y-12\ge 0 \\ \Rightarrow y(y+6)-2(y+6)\ge 0 \\ \Rightarrow (y+6)(y-2)\ge 0 \\ \Rightarrow y\le -6,y\ge 2 \end{gathered}$$

Hence, the minimum value is 2.

Therefore, (A) matches to (III).

Explanation for (B)

Given that $A$is symmetric,$B$ is skew-symmetric of order $3\times 3$ and $(A+B)(A-B)=(A-B)(A+B).$

$$\begin{gathered} \Rightarrow A^2-AB+BA-B^2=A^2+AB-BA-B^2 \\ \Rightarrow AB=BA \\ \because {\left(AB\right)}^T={(-1)}^{k}AB \\ \Rightarrow B^T{A}^T={(-1)}^{k}AB \\ \Rightarrow -BA={(-1)}^{k}AB \\ \Rightarrow -AB={(-1)}^{k}AB \\ \therefore {(-1)}^k=-1 \end{gathered}$$

Therefore $k$ is an odd number.

Hence, possible matches are (II) and (IV).

Explanation for (C)

$$\begin{gathered} \Rightarrow a={\log }_3{\log }_{3}2 \\ \Rightarrow 3^{-a}=3^{-\mathrm{lo}{g}_3{\log }_{3}2} \\ =3^{{\log }_3{\left({\log }_{}2\right)}^{-1}} \\ ={\left({\log }_{3}2\right)}^{-1} \\ ={\left(\frac{\log 2}{{\log }_3}\right)}^{-1} \\ ={\log }_{2}3 \\ \Rightarrow 3^{-a}={\log }_{2}3 \\ Now \\ 1<2^{-k+{\log }_{2}3}<2 \\ \Rightarrow 1<2^{-k}.2^{{\log }_{2}3}<2 \\ \Rightarrow 1<2^{-k}.3<2 \\ \Rightarrow \frac{1}{3}<2^{-k}<\frac{2}{3} \\ \Rightarrow {\log }_2\left(\frac{1}{3}\right)<-k<\log \left(\frac{2}{3}\right) \\ \Rightarrow -{\log }_2\left(\frac{1}{3}\right)>k>\log \left(\frac{2}{3}\right) \\ \Rightarrow \log \left(\frac{3}{2}\right)<k<{\log }_{2}3 \\ \frac{3}{2}<2<3 \\ \Rightarrow {\log }_{2}2=1 \\ \Rightarrow k=1 \end{gathered}$$

Therefore, (C) matches to (II).

Explanation for (D)

$$\begin{gathered} \Rightarrow \sin \theta =\cos \varphi \\ \Rightarrow \cos (\frac{\mathrm{\pi }}{2}-\theta )=\cos \varphi \\ \Rightarrow \frac{\mathrm{\pi }}{2}-\theta =2m\pi \pm \varphi \\ \Rightarrow \theta \pm \varphi -\frac{\mathrm{\pi }}{2}=2n\pi \\ \Rightarrow \frac{1}{\mathrm{\pi }}\left(\theta \pm \varphi -\frac{\mathrm{\pi }}{2}\right)=2n \end{gathered}$$

Thus,Possible values are even

Therefore, (D) matches to (I),(III).