We have, f(x) = x³
Let x₁, x₂ ∈ R such that
f(x₁) = f(x₂)
⇒ x₁³ = x₂³  
⇒ x₁ = x₂
∴ f(x) is one-one function.

When a die is thrown, then sample space is
\(S=\{1,2,3,4,5,6\} \Rightarrow n(S)=6\) 
Also, A : number is even and B : number is red
\(\therefore A=\{2,4,6\} \text { and } B=\{1,2,3\} \text { and } A \cap B=\{2\}\) 
\(\Rightarrow n(A)=3, n(B)=3 \text { and } n(A \cap B)=1\) 
Now, \(P(A)=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}, P(B)=\frac{n(B)}{n(S)}=\frac{3}{6}=\frac{1}{2}\) 
and \(P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{6}\) 
Now, \(P(A) \times P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6}=P(A \cap B)\) 
\(\therefore P(A \cap B) \neq P(A) \times P(B)\) 
Thus, A and B are not independent events.

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

1

Note that
f '(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0, in every interval of R
Therefore, the function f is increasing on R.

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\)  A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

\(y+2x=7,-x=-2\Rightarrow x=2,y=3\)

\(|A|=\left|\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right|=-4-15=-19 \neq 0\)
Matrix formed by cofactor of each element in IAI
\(A_{11}=-2 \quad A_{12}=-5 \quad A_{21}=-3 \quad A_{22}=2\)
\(\therefore \operatorname{adj} A =\left[\begin{array}{cc} -2 & -5 \\ -3 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{lr} -2 & -3 \\ -5 & 2 \end{array}\right] \)
\(\therefore A^{-1} =\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-19}\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right] \)
\(=\frac{-1}{-19}\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19} A . \)
\(\therefore A^{-1} =k A \)
\(\text { So, } k =\frac{1}{19}\)

The given matrix is \(A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]\)
 It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C₁) for easier calculation. 
\(|A|=1\left|\begin{array}{ll} 1 & 2 \\ 0 & 4 \end{array}\right|-0\left|\begin{array}{ll} 0 & 1 \\ 0 & 4 \end{array}\right|+0\left|\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}\right|=1(4-0)-0+0=4\)
\(\therefore 27|A|=27(4)=108 \\ \text { Now, } 3 A=3\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]=\left[\begin{array}{llc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \)
\(\begin{aligned} \therefore|3 A|=3\left|\begin{array}{lr} 3 & 6 \\ 0 & 12 \end{array}\right|-0\left|\begin{array}{rr} 0 & 3 \\ 0 & 12 \end{array}\right|+0\left|\begin{array}{ll} 0 & 3 \\ 3 & 6 \end{array}\right| \\ \end{aligned}\)
\(=3(36-0)=3(36)=108\)
 From equations (i) and (ii), we have: 
\(|3 A|=27|A|\) 
Hence, the given result is proved. 

R = {(a, b): a ≤ b²}
It can be observed that \(\left(\frac{1}{2}, \frac{1}{2}\right) \in R, \operatorname{Since} \frac{1}{2}>\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)
∴ R is not reflexive.
Now, (1, 4) ∈ R as 1 < 4²
But, 4 is not less than 1².
∴ (4, 1) ∉ R
∴ R is not symmetric.
Now,
(3, 2), (2, 1.5) ∈ R
(as 3 < 2² = 4 and 2 < (1.5)² = 2.25)
But, 3 > (1.5)² = 2.25
∴ (3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(i) We have : \({ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right| \)
\(\therefore \ { a }_{ 11 }=\frac { 1 }{ 2 } \left| -3+1 \right| =\frac { 2 }{ 2 } =1;\quad { a }_{ 12 }=\frac { 1 }{ 2 } \left| -3+2 \right| =\frac { 1 }{ 2 } =1;\)
\({ a }_{ 13 }=\frac { 1 }{ 2 } \left| -3+3 \right| =0;\quad { a }_{ 14 }=\frac { 1 }{ 2 } \left| -3+4 \right| =\frac { 1 }{ 2 } ;\)
\({ a }_{ 21 }=\frac { 1 }{ 2 } \left| -6+1 \right| =\frac { 5 }{ 2 } ;\quad { a }_{ 22 }=\frac { 1 }{ 2 } \left| -6+2 \right| =2;\)
\( { a }_{ 23 }=\frac { 1 }{ 2 } \left| -6+3 \right| =\frac { 3 }{ 2 } ;\quad { a }_{ 24 }=\frac { 1 }{ 2 } \left| -6+4 \right| =1;\)
\({ a }_{ 31 }=\frac { 1 }{ 2 } \left| -9+1 \right| =4;\quad { a }_{ 32 }=\frac { 1 }{ 2 } \left| -9+2 \right| =\frac { 7 }{ 2 } ;\)
\({ a }_{ 33 }=\frac { 1 }{ 2 } \left| -9+3 \right| =3;\quad { a }_{ 32 }=\frac { 1 }{ 2 } \left| -9+4 \right| =\frac { 5 }{ 2 } ;\)
Hence, \(A=\left[ \begin{matrix} 1 & \frac { 1 }{ 2 } & 0 & \frac { 1 }{ 2 } \\ \frac { 5 }{ 2 } & 2 & \frac { 3 }{ 2 } & 1 \\ 4 & \frac { 7 }{ 2 } & 3 & \frac { 5 }{ 2 } \end{matrix} \right] \)
(ii) We have: \({ a }_{ ij }=2i-j.\)
\(​​\therefore {a }_{ 11 }=2-1=1;\quad { a }_{ 12 }=2-2=0;\)
\({ a }_{ 13 }=2-3=-1;\quad { a }_{ 14 }=2-4=-2;\)
\({ a }_{ 21 }=4-1=3;\quad { a }_{ 22 }=4-2=2;\)
\({ a }_{ 23 }=4-3=1;\quad { a }_{ 24 }=4-4=0;\)
\({ a }_{ 31 }=6-1=5;\quad { a }_{ 32 }=6-2=4;\)
\({ a }_{ 33 }=6-3=3;\quad { a }_{ 32 }=6-4=2;\)
Hence, \(A=\left[ \begin{matrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{matrix} \right] .\)

E : Son on one end = {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
No.of exhaustive cases = 3! = 6
where s = son, m = mother and f = father 
F : Father in middle = {(m, s ,f), (s, f, m)},
\(E\cap F\)={(m, s, f), (s, f, m)}
\(P(E\cap F)=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
and \(P(F)=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 3 } }{ \frac { 1 }{ 3 } } =1\)

\(P(\overset { \_ }{ A } \cup \overset { \_ }{ B } )=1-P(A\cap B)\)
\(\frac { 1 }{ 4 } =1-P(A\cap B)\)
\(P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
And P(A)P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
Thus \(P(A\cap B)\neq P(A)P(B)\)
Hence, A and B are not independent.

Let the events be:
E₁ : Item from machine A
E₂ : Item from machine B
And A : Item is defective.
\(P({ E }_{ 1 })=\frac { 60 }{ 100 } =\frac { 3 }{ 5 } \)
\(P({ E }_{ 2 })=\frac { 40 }{ 100 } =\frac { 2 }{ 5 } \)
Also   \(P(A/{ E }_{ 1 })=\frac { 2 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 1 }{ 100 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) }{ \left( \frac { 3 }{ 5 } \right) \left( \frac { 2 }{ 100 } \right) +\left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) } \)
\(=\frac { 2 }{ 6+2 } =\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)

 The equation of the curve is given as: " id="MathJax-Element-566-Frame" role="presentation" style="position: relative;" tabindex="0">The equation of the curve is given as: 
\(6 y=x^{3}+2\)
 The rate of change of the position of the particle with respect to time (t) is given by, " id="MathJax-Element-568-Frame" role="presentation" style="position: relative;" tabindex="0">The rate of change of the position of the particle with respect to time (t) is given by, 
\(6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0 \Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}\)
 When the y -coordinate of the particle changes 8 times as fast as the " id="MathJax-Element-570-Frame" role="presentation" style="position: relative;" tabindex="0">When the y -coordinate of the particle changes 8 times as fast as the 
\(x \text { -coordinate i.e., }\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right), \text { we have: }\)
\(2\left(8 \frac{d x}{d t}\right)=x^{2} \frac{d x}{d t} \)
\(\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t} \)
\(\Rightarrow\left(x^{2}-16\right) \frac{d x}{d t}=0 \)
\(\Rightarrow x^{2}=16 \)
\(\Rightarrow x=\pm 4 \)
\(\text { When } x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11 \text { . }\)
\(\text { When } x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3} \text { . }\)
 Hence, the points required on the curve are (4,11) and (−4,−313) . " id="MathJax-Element-579-Frame" role="presentation" style="position: relative;" tabindex="0">Hence, the points required on the curve are (4,11) and 

Let E₁: Event that the student knows the answer
E₂ : Event that the student guesses the answer
E : Event that the answer is correct
Here, E₁ and E₂ are mutually exclusive and exhaustive events.
\(\therefore \quad P\left(E_1\right)=\frac{3}{5} \text { and } P\left(E_2\right)=\frac{2}{5}\)
Now, \(P\left(\frac{E}{E_1}\right)=P\) (the student answered correctly, given he knows the answer)
= 1
\(\begin{aligned} P\left(\frac{E}{E_2}\right) & =P \end{aligned}\) ( the student answered correctly, given he guesses)
\(=\frac{1}{3}\)
The probability that the student knows the answer given that he answered it correctly is given by \(P\left(\frac{E_1}{E}\right)\)
By using Baye's theorem, we get
\(P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right) \cdot P\left(E_1\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) \cdot P\left(E_2\right)}\)
\(\begin{aligned} =\frac{1 \times \frac{3}{5}}{1 \times \frac{3}{5}+\frac{1}{3} \times \frac{2}{5}}=\frac{\frac{3}{5}}{\frac{3}{5}+\frac{2}{15}} \end{aligned}\)
\(\begin{aligned} =\frac{3 \times 15}{5 \times 11}=\frac{3 \times 3}{11}=\frac{9}{11} \end{aligned}\)

Let E₁, E₂, E₃ and A denote the following events.
E₁ = Bag I is chosen, E₂ = Bag II is chosen,
E₃ = bag III is chosen, A = The balls drawn from the chosen bag are white and red.
Since, one of the bags is chosen at random.
\(\therefore \quad P\left(E_1\right)=\frac{1}{3}=P\left(E_2\right)=P\left(E_3\right)\)
If one white and red balls are chosen from bag I with replacement.
Then, \(P\left(\frac{A}{E_1}\right)=\frac{1}{6} \times \frac{3}{6} \times 2\)
Similarly, \(P\left(\frac{A}{E_2}\right)=\frac{2}{4} \times \frac{1}{4} \times 2 \text { and } P\left(\frac{A}{E_3}\right)=\frac{4}{9} \times \frac{2}{9} \times 2\)
By Baye's theorem,
Required probability \(P\left(\frac{E_3}{A}\right)=\frac{P\left(E_3\right) \times P\left(\frac{A}{E_3}\right)}{\sum_{i=1}^3 P\left(E_i\right) \times P\left(\frac{A}{E_i}\right)}\)
\(=\frac{\frac{1}{3} \times \frac{4}{9} \times \frac{2}{9} \times 2}{\frac{1}{3} \times \frac{1}{6} \times \frac{3}{6} \times 2+\frac{1}{3} \times \frac{2}{4} \times \frac{1}{4} \times 2+\frac{1}{3} \times \frac{4}{9} \times \frac{2}{9} \times 2}=\frac{64}{199}\)

Given, \(f(x)=\frac{x-1}{x-2}\) and \(f: A \rightarrow B\) , where \(A=R-\{2\}\) and \(B=R-\{1\}\)
For one-one Let \(f\left(x_{1}\right)=f\left(x_{3}\right),\) for some \(x_{1}, x_{2} \in A\) \(\Rightarrow \frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}\)
\( \Rightarrow \left(x_{1}-1\right)\left(x_{2}-2\right)=\left(x_{2}-1\right)\left(x_{1}-2\right)\)
\( \Rightarrow x_{1} x_{2}-2 x_{1}-x_{2}+2=x_{1} x_{2}-2 x_{2}-x_{1}+2 \)
\(\Rightarrow -x_{1}=-x_{2} \Rightarrow x_{1}=x_{2}\)
Thus,
\(f\left(x_{1}\right)=f\left(x_{2}\right)\)
\(\Rightarrow\)\(x_{1}=x_{2}\)
So, \(f(x) is one-ont function. For onto Let y=\frac{x-1}{x-2} \Rightarrow x y-2 y=x-1\)
\(\Rightarrow x(y-1)=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1} \)
\(\text { Clearly, } x \in R-\{2\}, \forall y \in R-\{1\} \)
\(\left[\text { if } \frac{2 y-1}{y-1}=2,\right. \text { then } 2 y-1=2 y-2\)
\(\Rightarrow-1=-2,\) which is not true So, range of \(f(x)=R-\{1\}\)
\(\therefore\) Range = Codomain So, f(x) is onto function and hence invertible.
Also, from Equation (i), we get
\(f^{-1}(y)=\frac{2 y-1}{y-1} \quad\left[\because y=f(x) \Rightarrow x=f^{-1}(y)\right]\)
on \(f^{-1}(x)=\frac{2 x-1}{x-1}\)

We have a relation R in set A = {1, 2, 3, 4, 5} defined as
R = {(a, b): la - bl is divisible by 2}
Clearly, R = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)}
Reflexive For any a ∈ A, we have |a - a| = 0, which is divisible by 2.
⇒ (a, a) ∈ R, ∀ a ∈ A
Thus, R is reflexive.
Symmetric Let a, b∈A,such that(a, b)∈R
\( \Rightarrow |a-b| \) Iis divisible by 2
\( \Rightarrow |a-b| \) = \(2\lambda\) for some \( \lambda \epsilon N\) 
\(\Rightarrow |b-a|=2 \lambda \) for some \( \lambda \epsilon N\) \([\because|a-b|=|b-a|] \) 
Thus, R is symmetric
Transitive Let a, b, c ∈  A, such that (a, b) ∈ R and 
⇒ |a - b| is divisible by 2 and |b - c| is divisible by 2.
⇒ |a - b| = 2λand |b - c| = 2, for some λ μ∈N
⇒ (a-b) = ±2A and (b-c) = ±2μ
Now, |a-c| = |(a-b) + (b-c)| = |±2λ + (±2λ)1
= I± 2λ ± 2μ = 2|± λ ± μ| [some positive number]
⇒ |a-c|is divisible by 2 ⇒ (a, c) ∈ R
Thus, R is transitive.
Hence, R is an equivalence relation.
Now, [a] = {xe A : (a, x)e R}
ஃ Equivalence class [1] = {1, 3, 5}, [2] = {2, 4}, [3] = {1, 3, 5},
[4] = {2, 4} and [5] = {1, 3, 5}
Hence, [1] = [3] = [5] = {1, 3, 5} and [2] = [4] = {2, 4}

Given \(A=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \) 
|A| = 1(1+3)-2(-1-1)+1(3-1)
= 4 + 4 + 2 = 10
Co-factor Matrix is :
\(=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] \)
\(\therefore\) adj A = transpose of above matrix
\(=\left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\therefore { A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
\(=\frac { 1 }{ 10 } \left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \)
Given set of equations are:
x + 2y + z = 4
-x + y + z = 0
x - 3y + z = 4
\(\Rightarrow \quad \left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \quad A.X=B\)
Multiplying both sides by A^-1 , we get
\({ A }^{ -1 }AX={ A }^{ -1 }B\)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} \frac { 8 }{ 5 } +\frac { 2 }{ 5 } \\ \frac { 4 }{ 5 } -\frac { 4 }{ 5 } \\ \frac { 4 }{ 5 } +\frac { 6 }{ 5 } \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 0 \\ 2 \end{matrix} \right] \)
\(\therefore \ x=2,y=0,z=2\)

We have: \(A=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \)
\(\therefore \quad \quad { A }^{ 2 }=AA=\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times (-1) & 2\times 1+0\times 3+1\times 0 \\ 2\times 2+1\times 2+3\times 1 & 2\times 0+1\times 1+3\times (-1) & 2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times (-1) & 1\times 1+(-1)\times 3+0\times 0 \end{matrix} \right]\)
\( =\left[ \begin{matrix} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{matrix} \right] .\)
\(\therefore \quad { A }^{ 2 }-5A=6I=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{matrix} \right] -5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] +6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{matrix} \right] +\left[ \begin{matrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{matrix} \right] +\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix} \right]\)
\(=\left[ \begin{matrix} 5-10+6 & -1+0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2+0+6 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{matrix} \right] \)

Given \(\left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right| \Rightarrow x^{2}-36=36-36\) 
\(\Rightarrow \quad x^{2}=36 \Rightarrow x=\pm 6\)

As in skew symmetric matrix, a_ij = -a_ji
⇒ a_ii = – a_ii
⇒ 2a_ii = 0
⇒ a_ii = 0, i.e. diagonal elements are zeroes.

As a₂₃ = \(\frac { { (-2+6) }^{ 2 } }{ 5 } =\frac { 16 }{ 5 } \)

y = f(x)
⇒ 5x + 4
⇒ \(\frac{y-4}{5}\)
∴ f^-1(y) = \(\frac{y-4}{5}\)
or f^-1(x) = \(\frac{x-4}{5}\)

Not reflexive, as l₁ R l₂
⇒ l₁ ⊥ l₁ Not true
Symmetric, true as l₁ R l₂ ⇒ l₂R h
Transitive, false as l₁ R l₂, l₂ R l₃
⇒ l₁ || l₃ . l₁ R l₂.

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.

(i) (a) : Let x be the number of extra days after 1^st July.
\(\therefore\) Price = Rs.(300 - 3Xx) = Rs.(300 - 3x)
Quantity = 80 quintals + x(1 quintal per day)
= (80 + x) quintals
(ii) (b) : R(x) = Quantity x Price
= (80 + x) (300 - 3x) = 24000 - 240x + 300x -3x²
= 24000 + 60x - 3x²
(iii) (a) : We have, R(x) = 24000 + 60x - 3x²
\(\begin{equation} \Rightarrow R^{\prime}(x)=60-6 x \Rightarrow R^{\prime \prime}(x)=-6 \end{equation}\)
For R(x) to be maximum, R'(x) = 0 and R"(x) < 0
\(\begin{equation} \Rightarrow 60-6 x=0 \Rightarrow x=10 \end{equation}\)
(iv) (a) : Shyams father will attain maximum revenue after 10 days.
So, he should harvest the onions after 10 days of 1^st July i.e., on 11^th July.
(v) (c) : Maximum revenue is collected by Shyams father when x = 10
\(\therefore\) Maximum revenue = R(10)
= 24000 + 60(10) - 3(10)2 = 24000 + 600 - 300 = 24300

(I) (d) : In factory A, number of units of types I, II and III for boys are 80, 70, 65 respectively and for girls number of units of types I, II and III are 80, 75, 90 respectively.

(ii) (a) : In factory B, number of units of types I, II and III for boys are 85, 65, 72 respectively and for girls number of units of types I, II and III are 50, 55, 80 respectively.
 
(iii) (c) : Let X be the matrix that represent the number of units of each type produced by factory A for boys, and Y be the matrix that represent the number of units of each type produced by factory B for boys.
Then, X = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[170} & 130 & 130] \end{array}\) and Y = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[85} & 65 & 72] \end{array}\) 
Now, required matrix = X + Y = [80 70 65] + [85 65 72]
=  [165 135 137]
(iv) (a): Required matrix = [80 75 90] + [50 55 80]
= [130 130 170]
(v) (a) : Clearly,R = P+Q
\(=\left[\begin{array}{ll} 80 & 80 \\ 70 & 75 \\ 65 & 90 \end{array}\right]+\left[\begin{array}{ll} 85 & 50 \\ 65 & 55 \\ 72 & 80 \end{array}\right]=\left[\begin{array}{ll} 165 & 130 \\ 135 & 130 \\ 137 & 170 \end{array}\right]\) 
\(\therefore \quad R^{\prime}=\left[\begin{array}{lll} 165 & 135 & 137 \\ 130 & 130 & 170 \end{array}\right]\)

(d) R is correct; A is incorrect

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.