The relation R defined on the set of all positive integers by "mRn" if m divides n".
Given relation is "mRn if m divides n".
Reflexivity :  mRm since m divides m for all positive integers m.
\(\therefore\)  R is reflexive.
Symmetricity: mRn \(\Rightarrow\)  nRm.
m divides n \(\Rightarrow\)  n divides m but 'n' does not divide 'm'
\(\therefore\) R is not symmetric
Transitive : mRn and nRp \(\Rightarrow\)  mRp.
m divides n and n divides p \(\Rightarrow\) m divides p.
\(\therefore\) R is transitive.
\(\therefore\) R is reflexive, and transitive.

(B\(\cup \)C) = {3, 4, 5, 6 ,7, 9}
Now, A\(\times\)(B\(\cup \)C) = {1, 2, 3} \(\times\){3, 4, 5, 6, 7, 9}
= {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(1)
Now A\(\times\)B = {1,2,3} \(\times\) {4,5,6,7}
= {(1,4)(1,5)(1,6)(1,7)(2,4)(2,5)(2,6)(2,7)(3,4)(3,5)(3,6)(3,7)}
A\(\times\)C = {1,2,3} \(\times\) {3,4,5,9}
= {(1,3)(1,4)(1,5)(1,9)(2,3)(2,4)(2,5)(2,9)(3,3)(3,4)(3,5)(3,9)}
RHS(A\(\times\)B)\(\cup \)(A\(\times\)C) = {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(2)
From (1) & (2), LHS = RHS
Hence verified

{x \(\in \)​ ​​​​​​N : 4x + 9 < 52}
Let C = {x\(\in \)N:4x + 9 < 52}
\(\Rightarrow\) C = {x\(\in \)N:4x < 52 - 9}
\(\Rightarrow\) C = {x\(\in \)N: 4x < 43}
\(\Rightarrow\) C = \(\left\{ x\in N:x<\frac { 43 }{ 4 } \right\} \) \(\Rightarrow\) C = {x \(\in \) N : x < 10.75}
\(\Rightarrow\) C = {1,2,3,4,5,6,7,8,9,10}.

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
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

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5, 1) (4, 2) (3, 3) (2, 4) and (1, 5).
Reflexivity: R is not reflexive since (5, 5) \(\notin \) R. 

For any two distinct elements x₁, x₂ in \([0,{\pi\over 2}]\)
in x₁ = sin x₂ ⇒ x₁ = x₂
∴ f is one - one...(1)
Also cos x₁ = cos x₂ ⇒ x₁ = x₂
∴ g is also one-one...(2)
Now, (f + g) (x) = f(x) + g(x) = sin x + cos x
⇒ (f+g)(0) = sin0 + cos0 = 0+1=1
and \(f+g\left(\pi\over 2\right)=sin\frac{\pi}{2}+cos{\pi}{2}=1+0=1\)
(f+g)(0) = (f+g)\((\frac{\pi}{2})\) ⇒ 0 ≠ \(\frac{\pi}{2}\)
∴ (f + g) is not one-one...(3)
From (1), (2) and (3), we get f and g are one-one but (f +g) is not one-one.

For any [-1, 0], we have
fog(x) = f(g(x)) = f([x]) = f( -1) = -1
gof(x) = g(f(x)) = g(-1) = [-1] = -1
⇒ fog(x) = gof(x) for all x ∈ [-1,0)
Hence fog and gof coincide on [-1, 0)

Given f: R ⟶ R, given by f(x) = X²
and g: R ⟶ R given by g(x) = 2x + 1.
∴ fog (x) = f(g(x))
= f(2x + 1)
= (2x + 1)² = 4x² + 1+ 4x (1)
gof(x) = g(f(x) = g(x²) = 2x² + 1 (2)
From (1) and (2), fog(x) ≠ gof(x).

2a + 3b = 30
R = {(3,8), (6,6), (9,4). (12,2)}
Not reflexive, Not Symmetric, transitive, hence not an equivalence relation.

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly,f(x) is defined for all x, satisfying x + 2 > 0 ⇒x > -2 ⇒ x ∈[-2, ∞)
∴ Domain of f = [-2, ∞]
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
fg : [-2, 2] ⟶ R is given by \((f g)(x) =f(x) - g(x) =\sqrt{x+2}+\sqrt{4-x^2}=\sqrt{(x+2)^2+(2-x)}\)
\(=(x+2)\sqrt{2-x}\)

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly,f(x) is defined for all x, satisfying x + 2 > 0 ⇒x > -2 ⇒ x ∈[-2, ∞)
∴ Domain of f = [-2, ∞)
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
(f + g) : [-2, 2] ⟶ R is given by \((f+ g)(x) =f(x) + g(x) =\sqrt{x+2}+\sqrt{4-x^2}\)

Given R = {(a, b) : |a - b| is a multiple of 4}
Reflexivity: Where a, b ∈ A = {0, 1, 2, ... 12}. For any a ∈ A, we have |a - a| = 0 which is a multiple of 4.
⇒ (a, a) ∈ A for all a ∈ A
∴ R is reflexive
Symmetry: Let (a, b) ∈ R. Then
(a, b) ∈ R
⇒ |a - b| is a multiple of 4.
⇒ |a - b| = 4⋋- for some ⋋∈N.
⇒ Ib - al = 4⋋- for some ⋋∈N.
⇒ (b, a) ∈N
∴ R is symmetricTransitivity: Let (a, b) ∈ Rand (b, c) ∈ R
Then (a, b) ∈ R and (b, c) ∈ R
⇒ |a - b| is a multiple of 4 and Ib - c| is a multiple of 4.
⇒ |a - b| = 4⋋ and |b - c| = 4μ for some ⋋ μ∈ N
⇒ a-b = ±4 -and b-c = ±4μ for some ⋋, μ ∈ N
⇒ a-c = ±4⋋ ±4μ for some ⋋, μ ∈ N
⇒ |a - c| is a multiple of 4.
⇒ (a-c)∈R
∴ R is transitive.
Hence, R is an equivalence relation.

Given R = {(a, b): a < b²} where a, b ∈ R
reflexivity: We know that \(\left(1\over 2\right)\le\left(1\over 2\right)^2\)is not true
\(⇒\ \left({1\over 2},{1\over 2}\right)∉R\)
⇒ R is not reflexive
Symmetry: We know that -1< 3² but 3 ≰ (-1)² is not true
⇒ (-1, 3) ∈ R but (3, -1)∉R
∴ R is not symmetric
Transitive: We observe that 2< (-3)² and -3< (1)² but 2 ≰ (1)² is not true
⇒ (2, -3) ∈ R and (-3, 1) ∈ R but (2, 1) ∉ R
⇒ R is not transitive
∴ R is neither reflexive nor symmetric nor transitive.

Given (-1, 0) ∈ A x A and (0, 1) ∈ A \(\times\) A
(-1,0) ∈ A \(\times\) A ⇒ -1,0 ∈ A and (0, 1) ∈ A \(\times\) A ⇒ 0, 1 ∈ A
∴ -1,0,1 ∈ A
∴ A = {-1, 0, 1}
Also it is given that A\(\times\) A has 9 elements.
∴ A has exactly three elements.
∴ A {-1, 0, 1}
∴  A \(\times\) A = {(-1, -1) (-1, 0) (-1, 1) (0, -1) (0, 0) (0, 1) (1, -1) (1, 0) (1, 1)}

B ⋂ C = {4, 6}
A\(B ⋂ C) = {0, 1, 2, 3}
A\B = {0,2}
A\B = {0, 1, 3}
(A\B) U (A\C)= {a, 1,2, 3}
From (1) and (2), A\(B ∩ C) = (A\B) U (A\C)
Venn diagram:

From (ii) and (v), we get
A\(B ∩ C) = (A\B) U (A\C)

Given A = {a, b, c, d}, B = {a, c, e}, C = {a, e}
B∩C = {a, c}
A∩(B∩C) = {a}
A ∩ B = {a, e}
(A ∩ B) ∩ C = {a}
From (1) and (2), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

(i) y = sin (-x)

Let y = sin x.
Then sin(-x) is the reflection of the graph of sin x, about y-axis.
(ii) y = -sin(-x)

-sin(-x) is the reflection of the graph of sin(-x) about the x-axis.
(iii) \(y=sin\left( {\pi\over 2}+x\right)\)
Let y = sinx.
Then \(sin\left( {\pi\over 2}+n\right)\)causes the shift to the left for \(\pi\over 2\) unit to the sin x curve.
(iv) \(y=sin\left({\pi\over 2}-x \right)\)

Let y = sin x. Then \(sin\left( {\pi\over 2}-n\right)\) causes the shift to the left for \(\pi\over 2\) unit to the sin (-x) curve.

As m - m = 0,
m - m is divisible by 7 \(\Rightarrow\)  mRm
\(\therefore\) R is reflexive.
Let mRn. Then m - n = 7k for some integer k 
Thus n-m = 7 (-k) and hence nRm
\(\therefore\) R is symmetric.
Let mRn and nRp
\(\Rightarrow\) m-n = 7k and n - p = 7l for some
\(\Rightarrow\) m = 7k + n and - p = 7l- n integers k and l
so m-p = 7k+n+7l-n
\(\Rightarrow\) m = p = 7(k+l) \(\Rightarrow\) mRp
\(\therefore\) R is transitive.
Thus, R is an equivalence relation.

Given A\(\times\)B = {(x, 1)(y, 2) (z, 1)}
Since n(A) = 3 and n(B) = 2,
A \(\times\) B will have 6 elements.
The remaining elements of A \(\times\) B will be (x, 2) (y, 1) (z, 2)
\(\therefore\) A\(\times\) B = {(x,1)(y,2)(z,1)(x,2)(y,1)(z,2)}
\(\therefore\) A = {x, y, z} and B = {1, 2}

Given n(p(A)) = 1024 = 2¹⁰
\(\Rightarrow\) n(A) = 10      [\(\therefore\) if n(A) = n, then n(p(A)) = 2ⁿ]
n(p(B)) = 32 = 2⁵
\(\Rightarrow\) n(B) = 5.
We know that,
n(A\(\cup\) B) = n(A) + n(B) - n(A\(\cap\)B)
\(\Rightarrow\) 15 = 10+5 - (A\(\cap\)B)
\(\Rightarrow\) n(A\(\cap\)B) = 0.