Let us detine the events as
E₁ : Students reside in a hostel
E₂ : Students are day scholars
A : Students get A grade
Then,
P(E₁) = Probability that student reside in a hostel
\(=60 \%=\frac{60}{100}\)
and P(E₂) = Probability that students are day scholars = 1 - \(\frac{60}{100}=\frac{40}{100}\)
Also, P(A/E₁) = Probability that hostelers get A grade
\(=30 \%=\frac{30}{100}\)
and P(A/E₂)= Probability that students having day scholars get A grade
\(=20 \%=\frac{20}{100}\)
\(\therefore\) The probability that the selecting student is a hosteler having A grade,
\(P\left(E_1 / A\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)}\)
                                 [by Baye's theorem]
\(\begin{aligned} =\frac{\frac{60}{100} \times \frac{30}{100}}{\left(\frac{60}{100} \times \frac{30}{100}\right)+\left(\frac{40}{100} \times \frac{20}{100}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{1800}{1800+800}=\frac{1800}{2600}=\frac{18}{26}=\frac{9}{13} \end{aligned}\)

We have,
\(\overrightarrow{A B}\) = (position vector of B) - (position vector of A)
\(=(\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}-2 \hat{j}-6 \hat{k}\)
\(\begin{aligned} \overrightarrow{B C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=2 \hat{i}-\hat{j}+\hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k}) \end{aligned}\)
\(=-\hat{i}+3 \hat{j}+5 \hat{k}\)
Here, \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=0\)
\(\Rightarrow\) A, B and C are the vertices of a triangle.
Now, \(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)
= - 2 - 3 + 5 = 0
\(\Rightarrow \overrightarrow{B C} \perp \overrightarrow{C A} \Rightarrow \angle C=90^{\circ}\)

Now, area of \(\Delta A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{B C}|\)
\(=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ 2 & -1 & 1 \end{array}\right|=\frac{1}{2}|(8 \hat{i}-11 \hat{j}-5 \hat{k})|\)
\(=\frac{1}{2} \sqrt{64+121+25}\)
\(=\frac{1}{2} \sqrt{210}\) sq units

We have, \(A=\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right]\)
Now, \(A^2=A \cdot A=\left[\begin{array}{lll}9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9\end{array}\right]\)
and \(A^3=A^2 \cdot A=\left[\begin{array}{ccc} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{array}\right]\)
Now, consider LHS \(=A^3-4 A^2-3 A+11 I\)
\(=\left[\begin{array}{ccc} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{array}\right]-4\left[\begin{array}{ccc} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{array}\right] \)
\(-3\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{array}\right]+11\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=O=\mathrm{RHS}\)
Now, \(A^3-4 A^2-3 A+11 I=O\)
Pre-multiplied by \(A^{-1}\), we get \(A^{-1} A^3-4 A^{-1} A^2-3 A^{-1} A+11 A^{-1}=O\)
\(\Rightarrow A^2-4 A-3 I+11 A^{-1}=O \quad\left[\because A^{-1} A=I\right]\)
\(\Rightarrow 11 A^{-1}=3 I+4 A-A^2\)
\(\Rightarrow 11 A^{-1}=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]+4\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{array}\right]-\left[\begin{array}{ccc} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{array}\right] \)
\(\Rightarrow 11 A^{-1}=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]+\left[\begin{array}{ccc} 4 & 12 & 8 \\ 8 & 0 & -4 \\ 4 & 8 & 12 \end{array}\right]-\left[\begin{array}{ccc} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{array}\right] \)
\( \Rightarrow 11 A^{-1}=\left[\begin{array}{ccc} -2 & 5 & 3 \\ 7 & -1 & -5 \\ -4 & -1 & 6 \end{array}\right] \Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc} -2 & 5 & 3 \\ 7 & -1 & -5 \\ -4 & -1 & 6 \end{array}\right]\)

Given, equations of lines are
\(\begin{aligned} \frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}=\lambda \end{aligned}\)  (say)        ....(i)
and \(\begin{aligned} \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu \end{aligned}\) (say) ....(ii)
The coordinates of points on line (i) and (ii) will be \((2 \lambda+1,3 \lambda+b, 4 \lambda+3) \text { and }(5 \mu+4,2 \mu+1, \mu) \text {. }\)
If the lines intersect, then point will be common.
So, \(\begin{aligned} 2 \lambda+1=5 \mu+4 \Rightarrow 2 \lambda-5 \mu=3 \end{aligned}\)     ....(iii)
\(\begin{aligned} 3 \lambda+b=2 \mu+1 \Rightarrow 3 \lambda-2 \mu=1-b \end{aligned}\)    ....(iv)
and \(\begin{aligned} 4 \lambda+3=\mu \Rightarrow 4 \lambda-\mu=-3 \end{aligned}\)            ...(v)
On solving Eqs. (iii) and (v), we get
\(\lambda=-1 \text { and } \mu=-1\)
Since, given lines are intersect to each other.
\(\therefore\) The value of \(\lambda\) and \(\mu\) satisfies the Eq. (iv), we get
3 (-1) - 2(-1) = 1 - b
\(\begin{aligned} \Rightarrow -3+2 & =1-b \end{aligned}\)
\(\begin{aligned} \Rightarrow & b =1+1 \Rightarrow b=2 \end{aligned}\)
Now, putting \(\lambda\) = -1 and b = 2 in coordinates of points on line (i), we get
= (2(-1) + 1, 3(-1) + 2, 4(-1) + 3)
= (-2 + 1, -3 + 2, -4 + 3) = (-1, -1, -1)
Hence, the intersection point is (-1, -1, -1).

We have,
\(A=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]\) 
\(\begin{array}{l} \Rightarrow A^{\prime}=\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \end{array}\) 
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\}\) 
\(=\frac{1}{2}\left[\begin{array}{ccc} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]\) 
which is symmetric matrix and
\(Q =\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]-\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\} \)
\(=\frac{1}{2}\left[\begin{array}{rrr} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{array}\right]=\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right] \) 
which is skew-symmetric matrix
Now,\(P+Q=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\) 
 \(=\left[\begin{array}{rrr} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]+\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right]=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]=A\) 
Hence, A is represented as sum of symmetric and skew-symmetric matrix.

\(f(x)=\tan ^{-1} 1-\tan ^{-1} x-\tan ^{-1} x-\tan ^{-1} 2 \)
\(=\tan ^{-1} 1-\tan ^{-1} 2-2 \tan ^{-1} x \)
\(\Rightarrow f^{\prime}(x)=0-0-\frac{2}{1+x^{2}} \Rightarrow f^{\prime}(x)=\frac{-2}{1+x^{2}} \)

\(\operatorname{LHL}_{x \rightarrow 0}=\lim _{x \rightarrow 0^{-}}\left\{\frac{\sin (a+1) x}{x}+\frac{\sin x}{x}\right\} \)
\(= \lim _{h \rightarrow 0}\left\{\frac{\sin (a+1)(0-h)}{0-h}+\frac{\sin (0-h)}{0-h}\right\} \)
\(=\lim _{h \rightarrow 0}\left\{\frac{\sin (a+1) h}{(a+1) h} \cdot(a+1)+\frac{\sin h}{h}\right\} \)
\(\Rightarrow a+1+1 =a+2 \)
\(\mathrm{RHL} =\lim _{x \rightarrow 0}\left[\frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b \sqrt{x^{3}}}\right] \)
\(=\lim _{h \rightarrow 0} \frac{\sqrt{h+b h^{2}}-\sqrt{h}}{b \sqrt{h^{3}}} \)
\(=\lim _{h \rightarrow 0} \frac{h+b h^{2}-h}{b h \sqrt{h}\left[\sqrt{h+b h^{2}}+\sqrt{h}\right]} \)
\(=\lim _{h \rightarrow 0} \frac{b h^{2}}{b h \sqrt{h} \sqrt{h}[\sqrt{1+b h}+1]}\)
\(=\lim _{h \rightarrow 0} \frac{1}{\sqrt{1+b h}+1}=\frac{1}{2}\)
For continuity at x = 0,
\(\mathrm{LHL}=\underset{x=0}{\mathrm{RHL}}=f(0) \)
\(\Rightarrow a+2=\frac{1}{2}=c \ \Rightarrow \ c=\frac{1}{2}, a=-\frac{3}{2}[\text { from {i}] } \)
\(\therefore \)  Number are \(a=-\frac { 3 }{ 2 } ,b\quad any\quad value\quad (\neq 0),\quad c=\frac { 1 }{ 2 } \)

Given, \(\vec{a}+\vec{b}+\vec{c}=0\)
\((\vec{a}+\vec{b}+\vec{c})^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\)\(+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)
\(\begin{array}{ll} \Rightarrow & 0=9+25+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \end{array}\)
\(\begin{array}{ll} \Rightarrow & -\frac{50}{2}=(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \end{array}\)
\(\begin{array}{ll} \therefore & (\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-25 \end{array}\)

Given, equations of lines are
\(\begin{aligned} \vec{r} & =\lambda(\hat{i}-\hat{j}+\hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r} & =\hat{i}-\hat{j}+\mu(-2 \hat{j}+\hat{k}) \end{aligned}\)
These lines are not parallel as \((\hat{i}-\hat{j}+\hat{k})\) is not parallel to \((-2 \hat{j}+\hat{k})\)

Let AB be the shortest distance between the lines such that AB makes right angle with both the lines.
Let the position vector of the point A lying on the line
\(\vec{r}=\lambda(\hat{i}-\hat{j}+\hat{k})=\lambda(\hat{i}-\hat{j}+\hat{k})\)                            ...(i)
and let the position vector of the point B lying on the line
\(\vec{r}=\hat{i}-\hat{j}+\mu(-2 \hat{j}+\hat{k})=\hat{i}+(-1-2 \mu) \hat{j}+\mu \hat{k}\)   ...(ii)
Now, \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\)
\(=(1-\lambda) \hat{i}+(-1-2 \mu+\lambda) \hat{j}+(\mu-\lambda) \hat{k}\)
Now, \(\overrightarrow{A B}\) is perpendicular to both the lines
\(\hat{i}-\hat{j}+\hat{k} \text { and }-2 \hat{j}+\hat{k}\)
\(\begin{array}{lr} \therefore \quad 1(1-\lambda)-1(-1-2 \mu+\lambda)+1(\mu-\lambda) & =0 \end{array}\)
\(\begin{array}{lr} \Rightarrow & 2+3 \mu-3 \lambda=0 \end{array}\)                                                      ...(iii)
\(\begin{array}{lr} \text { and } 0(1-\lambda)-2(-1-2 \mu+\lambda)+1(\mu-\lambda) & =0 \end{array}\)
\(\begin{array}{lr} \Rightarrow \quad 2+5 \mu-3 \lambda & =0 \end{array}\)                                                 ...(iv)
From Eqs. (iii) and (iv), we get
\(\begin{aligned} 2+3 \mu & =2+5 \mu \Rightarrow 2 \mu=0 \Rightarrow \mu=0 \end{aligned}\)
\(\begin{aligned} \therefore \lambda & =\frac{2}{3} \end{aligned}\)
So, the position vector of the points at which they should be so that the distance between them is the shortest are \(\frac{2}{3}(\hat{i}-\hat{j}+\hat{k}) \text { and } \hat{i}-\hat{j}\)
                                                                          [using Eqs. (i) and (iii)]
\(\begin{gathered} \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\frac{1}{3} \hat{i}-\frac{1}{3} \hat{j}-\frac{2}{3} \hat{k} \end{gathered}\)
\(\begin{gathered} \therefore \quad|\overrightarrow{A B}|=\sqrt{\left(\frac{1}{3}\right)^2+\left(\frac{-1}{3}\right)^2+\left(\frac{-2}{3}\right)^2}=\sqrt{\frac{2}{3}} \end{gathered}\)
Hence, the shortest distance AB = \(\sqrt{\frac{2}{3}}\) units

Let E₁, E₂, E₃ and A be events such that
E₁ = Two red balls are transferred from Bag I to Bag II
E₂ = Two black balls are transferred from Bag I to Bag II
E₃ = Out of two transferred ball one is red and other is black.
A = Drawing a red ball from Bag II
Here, \(P\left(\frac{E_1}{A}\right)+P\left(\frac{E_3}{A}\right)\) is required.
Now, \(\begin{aligned} P\left(E_1\right)=\frac{{ }^3 C_2}{{ }^7 C_2}=\frac{3!}{2!1!} \times \frac{2!5!}{7!}=3 \times \frac{2}{7 \times 6}=\frac{1}{7} \end{aligned}\)
\(\begin{aligned} P\left(E_2\right)=\frac{{ }^4 C_2}{{ }^7 C_2}=\frac{4!}{2!2!} \times \frac{2!5!}{7!}=\frac{4 \times 3}{2} \times \frac{2}{7 \times 6}=\frac{2}{7} \end{aligned}\)
\(\begin{aligned} P\left(E_3\right)=\frac{{ }^3 C_1 \times{ }^4 C_1}{C_2}=\frac{3 \times 4}{1} \times \frac{2!5!}{7!}=\frac{4}{7} \end{aligned}\)
\(\begin{aligned} P\left(\frac{A}{E_1}\right)=\frac{7}{9}, P\left(\frac{A}{E_2}\right)=\frac{5}{9}, P\left(\frac{A}{E_3}\right)=\frac{6}{9} \end{aligned}\)
\(\begin{aligned} \therefore P\left(\frac{E_3}{A}\right)+P\left(\frac{E_3}{A}\right) \end{aligned}\)
\(\begin{aligned} =\frac{P\left(E_1\right) \cdot P\frac{A}{E_1}+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{\frac{1}{7} \times \frac{7}{9}+\frac{4}{7} \times \frac{6}{9}}{\frac{1}{7} \times \frac{7}{9}+\frac{2}{7} \times \frac{5}{9}+\frac{4}{7} \times \frac{6}{9}} \end{aligned}\)
\(\begin{aligned} =\frac{\frac{7}{63}+\frac{24}{63}}{\frac{7}{63}+\frac{10}{63}+\frac{24}{63}}=\frac{\frac{31}{63}}{\frac{41}{63}}=\frac{31}{41}=0.75 \end{aligned}\)

\(f(x)=\left\{\begin{array}{ll} \sin x-\cos x, & \text { if } x \neq 0 \\ -1 & \text { if } x=0 \end{array}\right.\)
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I: \(\text { If } c \neq 0, \text { then } f(c)=\sin c-\cos c \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(\sin x-\cos x)=\sin c-\cos c \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x ≠ 0
Case II: \(\text { If } c=0, \text { then } f(0)=-1 \)
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1 \)
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1 \)
\(\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function

Not reflexive, as (1, 1) \(\notin\) R.
Not symmetric, as (2, 3) \(\in\) R but (3, 2) \(\notin\) R.
Transitive, as relation R in a non empty set containing one element is transitive

(i) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right] \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right] \Rightarrow\left(\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right]=\mathrm{A}\)
Thus \(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\)
(ii) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right], \mathrm{B}=\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right] \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc} 5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{array}\right]\)
Therefore \((A+B)^{\prime}=\left[\begin{array}{cc} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)
Now \(\mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right], \mathrm{B}^{\prime}=\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right] \text {, }\)
So \(A^{\prime}+B^{\prime}=\left[\begin{array}{rr} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)

Thus \((\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \)

(iii) We have
\(k B=k\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]\) 
and 
\((k B)^{\prime} =\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]^{\prime}=\left[\begin{array}{cc} 2 k & k \\ -k & 2 k \\ 2 k & 4 k \end{array}\right].\)
\(=k\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right]=k B^{\prime} \) 
Thus, (kB)' = kB'

R = {(P₁, P₂): P₁ and P₂ have same the number of sides}
R is reflexive since (P₁, P₁) ∈ R as the same polygon has the same number of sides with itself.
Let (P₁, P₂) ∈ R.
⇒ P₁ and P_2­ have the same number of sides.
⇒ P₂ and P₁ have the same number of sides.
⇒ (P₂, P₁) ∈ R
∴ R is symmetric.
Now,
Let (P₁, P₂), (P₂, P₃) ∈ R.
⇒ P₁ and P₂ have the same number of sides. Also, P₂ and P₃ have the same number of sides.
⇒ P₁ and P₃ have the same number of sides.
⇒ (P₁, P₃) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.

By the equation, \(\frac { 1 }{ 2 } \left| \begin{matrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} \right| =\pm 35\)
\({1\over2}[2(4-4)+6(5-k)+(1)920-4k)]=\le35\)
\({1\over2}[0+30-6k+20-4k]=\le35\)
\(50-10k=\le70\)
\(10k=50\mp 70\)
10k = -20, 120.
Hence k = -2, 12

Probability of chosing a bag:
\(P(I)=\frac{1}{3}, P(I I)=\frac{1}{3}, P(I I I)=\frac{1}{3} .\)
E : one white and one red ball is drawn.
\(P(E / I)=\frac{{ }^{1} C_{1} \times{ }^{3} C_{1}}{{ }^{6} C_{2}}=\frac{1 \times 3 \times 2}{6 \times 5}=\frac{1}{5}\)
\(P(E / I I)=\frac{{ }^{2} C_{1} \times{ }^{1} C_{1}}{{ }^{4} C_{2}}=\frac{2 \times 1}{6}=\frac{1}{3} \)
\(P(E / I I)=\frac{{ }^{4} C_{1} \times{ }^{2} C_{1}}{{ }^{9} C_{2}}=\frac{4 \times 2 \times 2}{9 \times 8}=\frac{2}{9} \)
Using Bayes' Theorem,
Probability of drawing one white and one red from bag III is
\(P(I I I E) =\frac{P(I I I) \cdot P(E / I I I)}{P(I) \cdot P(E / I)+P(I I) \cdot P(E / I I)+P(I I I) \cdot P(E / I I I)} \)
\(=\frac{\frac{1}{3} \times \frac{2}{9}}{\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{9}}=\frac{\frac{2}{9}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{9}} \)
\(=\frac{\frac{2}{9}}{\frac{9+15+10}{45}}=\frac{2}{9} \times \frac{45}{34}=\frac{5}{17} \)

\(\because\) Two cards are drawn at random from a pack of 52 cards one-by-one without replacement.
\(\therefore\) The required probability = P {(the first is a red jack card and the second is a jack card) or (the first is a red non-jack card and the second is a jack card)}
\(=\frac{2}{52} \times \frac{3}{51}+\frac{24}{52} \times \frac{4}{51}=\frac{1}{26}\)

Let \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
Now, projection vector of \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}\)
Here, \(\begin{aligned} \vec{a} \cdot \vec{b} & =(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k}) \end{aligned}\)
\(\begin{aligned} =2 \times 1+(-1) \times(-2)+1 \times 1 \end{aligned}\)
= 2 + 2 + 1 = 5
and \(|\vec{b}|=\sqrt{(1)^2+(-2)^2+(1)^2}=\sqrt{1+4+1}=\sqrt{6}\)
\(\therefore\) Projection vector of \(\vec{a} \text { on } \vec{b}=\frac{5}{6}(\hat{i}-2 \hat{j}+\hat{k})\)

Consider a natural number n in codomain N.
Case I : When n  is odd
Therefore, n = 2r + 1 for some \(r \in N\).
Then, there exists \(4 r+1 \in N\) such that
\(f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1\)
Therefore, f is onto.
Case II : When n is even
Therefore, n =2r  for some \(r \in N\)
Then, there exists \(4 r \in N\) such that \(f(4 r)=\frac{4 r}{2}=2 r\)
Hence, f is surjective.
Now, it can be observed that
\(f(1)=\frac{1+1}{2}=\frac{2}{2}=1 \text { and } f(2)=\frac{2}{2}=1\)
Here, f(1) = f(2) but \(1 \neq 2\)
Hence, f is not injective.

Let \(A=\left[\begin{array}{rrr} 1 & \lambda & 0 \\ 3 & -1 & 2 \\ 4 & 1 & 5 \end{array}\right] \)Since the matrix is singular. 
\(\therefore |A|=0 \)
\(\Rightarrow\left|\begin{array}{rrr} 1 & \lambda & 0 \\ 3 & -1 & 2 \\ 4 & 1 & 5 \end{array}\right|=0   \)
\(\Rightarrow 1(-5-2)-\lambda(15-8)+0=0 \)
\(\Rightarrow -7-7 \lambda=0 \)
\(\Rightarrow -7 \lambda=7 \)
\(\therefore\lambda=-1\)

a = -2, b = 0 and c = -3

a₃₂ = -11
Alternative Method:
Given \(A =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|\)
⇒ a₃₂ \(=\left| \begin{matrix} 5 & 8 \\ 2 & 1 \end{matrix} \right|\)
⇒ a₃₂ = 5-16 = -11

(b) Both A and R are correct; R is not the correct explanation of A

(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

Given, P(E₁) = 0.65, P(E/E₁) = 0.35 and \(P\left(\frac{E}{E_2}\right)=0.80\)
(i) P(E₂) = 1 - P(E₁) = 1 - 0.65 = 0.35
(ii) \(\begin{aligned} P(E) & =P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right) \end{aligned}\)
\(\begin{aligned} =0.65 \times 0.35+0.35 \times 0.80 \end{aligned}\)
\(\approx 0.23+0.28=0.51\)
(iii) (a) \(\begin{aligned} P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.65 \times 0.35}{0.65 \times 0.35+0.35 \times 0.80}=0.45 \end{aligned}\)

or (b) \(\begin{aligned} P\left(\frac{E_2}{E}\right) & =\frac{P\left(E_2\right) P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.35 \times 0.80}{0.65 \times 0.35+0.35 \times 0.80} \end{aligned}\)
= 0.55

Given, \(\vec{F}_1=6 \hat{i}+0 \hat{j} k N\)
\(\begin{aligned} \therefore \quad\left|\vec{F}_1\right|=\sqrt{6^2+0^2}=6 \mathrm{kN} \\ \end{aligned}\)
\(\vec{F}_2=-4 \hat{i}+4 \hat{j} k N\)
\(\therefore \quad\left|\vec{F}_2\right|=\sqrt{(-4)^2+4^2}=4 \sqrt{2} k N\)
\(\vec{F}_3=-3 \hat{i}-3 \hat{j} k N\)
\(\Rightarrow\left|\vec{F}_3\right|=\sqrt{(-3)^2+(-3)^2}=3 \sqrt{2} \mathrm{kN}\)
(i) Magnitude of force of team \(A=\left|\vec{F}_1\right|=6 \mathrm{kN}\)
(ii) Since, magnitude of force of team A is greater than other teams, therefore team A will win the game.
(iii) Resultant force,
\( \vec{F}=\vec{F}_1+\vec{F}_2+\vec{F}_3\)
\(=(6 \hat{i}+0 \hat{j})+(-4 \hat{i}+4 \hat{j})+(-3 \hat{i}-3 \hat{j}) \)
\(\vec{F}=(-\hat{i}+\hat{j}) k N \)
\(\Rightarrow |\vec{F}|=\sqrt{(-1)^2+(1)^2}=\sqrt{2} k N \)
Or
Resultant force \(\vec{F}=-\hat{i}+\hat{j}\)
Let \(\vec{F}\) makes an angle \(\theta\) with the X-axis,then its direction cosine along X-axis is cos \(\theta\).
\(\therefore \cos \theta=\frac{f_x}{\sqrt{f_x^2+f_y^2}} \text { (where } f_x \text { and } f_y\) are direction ratios along X-axis and Y-axis, respectively.)
\(\begin{aligned} \cos \theta & =\frac{-1}{\sqrt{(-1)^2+(1)^2}}=\frac{-1}{\sqrt{2}} \end{aligned}\)
\(\begin{aligned} \Rightarrow & & \theta & =\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \end{aligned}\)
\(\Rightarrow \theta =\pi-\frac{\pi}{4}=\frac{3 \pi}{4}\)
where '\(\theta\)' is the angle made by the resultant force with the '+' ve direction of the X-axis.

We have, R = {(l₁,l₂) :l₁ is parallel to l₂}
(i) If l₁ is parallel to l₂, then l₂ is parallel to l_1.
So, if (l₁, l₂) \(\in R,\) then (l₂, l₁) \(\in R\)
\(\therefore\) R is symmetric.
(ii) If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃.
So, if \(\left(l_1, l_2\right) \in R,\left(l_2, l_3\right) \in R\), then (l₁,l₃)\(\in R\)
\(\therefore\) R is transitive.
(iii) Let equation of line parallel to y = 3x + 2 be y = mx + c, where m is the slope of line. Since, y = 3x + 2 and y = mx + c are parallel. Slope of (y =3x + 2) = Slope of(y = mx + c)
\(\Rightarrow\) 3 = m i.e. m = 3
Hence, the required line is
y = 3x + c, where c \(\in R\)
Or
We have, S = {(I₁, I₂) : l₁ is perpendicular to l₂}
For Symmetric If I₁, is perpendicular to I₂, then l₂ is perpendicular to l₁.
So, if (l₁,l₂) \(\in S\), then (l₂, l₁) \(\in S\)
\(\therefore\) S is symmetric.
For Transitive If I₁, is perpendicular to l₂, and l₂, is perpendicular to l₃, then I₁, is not perpendicular to l₃, it is parallel to I₃.
So, if \(\left(l_1, l_2\right) \in S,\left(l_2, l_3\right) \in S \text {, then }\left(l_1, l_3\right) \notin S \text {. }\)
\(\therefore\) S is not transitive.