\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)

\(m_{1} m_{2}=\frac{a}{b}=\frac{6}{-7}\)

\((p+2 q) x+(p-3 q) y-p+q=0 \)
\(p(x+y-1)+q(2 x-3 y+1)=0 \)
\(x+y=1 ...(1)\)
\(2 x-3 y=-1..(2)\)
\(\begin{array}{r} 3 \times(1) \Rightarrow 3 x+3 y=3 \\ (2) \Rightarrow 2 x-3 y=-1 \\ \hline 5 x=2 \end{array}\)
\(x=\frac{2}{5}, y=\frac{3}{5} \)
\(\therefore\left(\frac{2}{5}, \frac{3}{5}\right) \)

\(e=1+\frac{1}{\lfloor 1}+\frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots \ldots\)
\(e^{-1}=1-\frac{1}{\lfloor 1}+\frac{1}{\lfloor 2}-\frac{1}{\lfloor 3}+\frac{1}{4}-\ldots\)
\(\frac{e+e^{-1}}{2}=1+\frac{1}{\lfloor 2}+\frac{1}{\lfloor 4}+\ldots\)
\(\frac{1}{\lfloor 2}+\frac{1}{\lfloor 4}+\ldots . \quad=\frac{e+e^{-1}}{2}-1\)
\(=\frac{\mathrm{e}-2+\mathrm{e}^{-1}}{2} \)
\(=\frac{(\mathrm{e}-1)^{2}}{2 \mathrm{e}} \)
 

\(\mathrm{e}^{-2 x}=1-\frac{2 x}{1 !}+\frac{(2 x)^{2}}{2 !}-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{4}}{4 !}-\frac{(2 x)^{5}}{5 !}+\ldots\)
\(\text { Coefficient of } x^{5} \text { is } \frac{-2^{5}}{5 !}=\frac{-32}{120}=\frac{-4}{15}\)

\(1+3+5+7+\ldots \ldots+17 \text { is equal to } 9^{2}=81\)

\(\text {Perimeter }=4+2 \sqrt{2}\)
\(a+a+\sqrt{2} =4+2 \sqrt{2} \)
\(2 a+\sqrt{2} a =4+2 \sqrt{2} \)
\(\therefore a =2 \)
\(\text {Equation of line is } \frac{x}{2}+\frac{y}{2}=1\)
\(x+y-2=0\)

\(\text { Slope of } 3 x-y+5=0 \text { is } \frac{-3}{-1}=3=m_{1}\)
Let m₂ be the slope of the second line
\(\text {Given } \tan \theta=\tan 45^{\circ}=1 \Rightarrow \frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\pm 1\)
\(\frac{3-m_{2}}{1+3 m_{2}}=1 \quad \frac{m_{2}-3}{1+3 m_{2}}=1\)
\(3-m_{2}=1+3 m_{2} \quad m_{2}-3=1+3 m_{2}\)
\(2=4 m_{2} \quad-2 m_{2}=4\)
\(\mathrm{m}_{2}=\frac{1}{2} \quad \mathrm{~m}_{2}=-2\)
\(\left(\frac{1}{2},-2\right)\)

\({ }^{52} \mathrm{C}_{5} \text { includes all possibilities (zero king, }1 \text { king, } 2 \text { kings, } 3 \text { kings, } 4 \text { kings }){ }^{48} C_{5} \text { has no kings}\)
\(\text { Réquired possibilities }{ }^{52} \mathrm{C}_{5}-{ }^{48} \mathrm{C}_{5}\)

\({ }^{(\mathrm{n}-1)} \mathrm{C}_{\mathrm{r}}+{ }^{(\mathrm{n}+1)} \mathrm{C}_{(\mathrm{r}-1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\)

\(\frac{{ }^{2 n} C_{3}}{{ }^{n} C_{3}} =\frac{11}{1} \)
\(\frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)} =\frac{11}{1} \)
\(\frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)} =11 \)
\(4(2 n-1) =11(n-2) \)
\(8 n-4 =11 n-22 \)
\(18 =3 n \)
\(n = 6\)

No. of ways of answering = 5³ = 125
Correct answer - 1
.'. Number of incorrect answer = 125 - 1 = 124

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\(\sqrt{2+2 \cos 4 \theta} =\sqrt{2+2\left(2 \cos ^{2} 2 \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} 2 \theta} \)
\(=2 \cos 2 \theta \)
\(\sqrt{2+\sqrt{2+2 \cos 4 \theta}} =\sqrt{2+2 \cos 2 \theta} \)
\(=\sqrt{2+2\left(2 \cos ^{2} \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} \theta} \)
\(=\pm 2 \cos \theta \)
\(\Rightarrow \pi<2 \theta<\frac{3 \pi}{2} \)
\(\Rightarrow \frac{\pi}{2}<\theta<\frac{3 \pi}{4} \text { in II quadrant. } \)
\(\therefore \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=-2 \cos \theta\)

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation

\(\text { If } x \in(-\infty,-2), \text { Let } x=-3\)
\(\text { then } f(x)=|-5|+|1|=6=-2 x\)
\(\text { If } x \in(-2,-2), \text { Let } x=0 \text { then }\)
\(f(x)=|0-2|+|0+2|=4\)
\(\text { If } x \in(2, \infty), \text { Let } x=4\)
\(\text { then } f^{}(x)=|2|+|6|=8=2 x\)
\(\therefore \mathrm{f}(x)=\left\{\begin{array}{rl} -2 x & x \in(-\infty,-2] \\ 4 & x \in(-2,2] \\ 2 x & x \in(2, \infty) \end{array}\right.\)