f(x) = 3 - 2x
Given f (x²) = [f (x)]²
3 - 2x² = (3 - 2x)²
3 - 2x² = 9 - 12x + 4x²
6x² - 12x + 6 = 0
6(x² - 2x + 1) = 0
(x - 1)² = 0 
x = 1

f(x) = 2x + 5, x ≠ 0.
\(\frac{f(x+2)-f(2)}{x} =\frac{[2(x+2)+5]-[2(2)+5]}{x} \) 
\(=\frac{2 x+4+5-9}{x}=\frac{2 x+9-9}{x} \)
\(=\frac{2 x}{x}=2\)

Given X = {3, 4, 6, 8}
Relation R = {(x, f(x)) | x \(\in \) X, f(x) = x² + 1}
When x = 3 ⇒ f(x) = f(3)² = 9 + 1 = 10 \(\in \) N
When x = 4 ⇒ f(x) = f(4)² = 16 + 1 = 17 \(\in \) N
When x = 6 ⇒ f(x) = f(6)² = 36 + 1 = 37 \(\in \) N
When x = 8 ⇒ f(x) = f(8)² = 64 + 1 = 65 \(\in \) N
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Since, all the elements of X are having natural numbers as images, it is a function from X to N.

f = {(x, y) / x, y \(\in \) N and y = 2x}
Given that y = 2x
x = {1,2,3,..}

f = {(1,2), (2, 4), (3, 6), (4, 8)..}
Domain of f = {1, 2, 3, 4...........}
Codomain = {1,2,3, 4.......}
Range of f = {2,4,6, 8......}
Here, the first elements (x) are having unique images. So, this relation is a function.

R₁ = {(–5, a), (1, a), (3, b)}
We may represent the relation R₁ in an arrow diagram 
R₁ is not a function as 4 \(\in\) X does not have an image in y.

f(x) = x² - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)² - 2 = 2; f( -1) = ( -1)² - 2 = -1
f(0) = (0)² - 2 = - 2 ; f(3) = (3)² - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.

Pictorial representation of R . From the diagram, we see that for each x \(\in \) X, there exists only one y \(\in \) Y. Thus all elements in X have only one image in Y. Therefore R is a function Domain X = {1, 2, 3, 4}; Co-domain Y = {2, 3, 6, 8,10}; Range of f = {2, 4, 6, 8}.

Given Set =  {(x, y) / y = x + 3, x \(\in \) {0, 1, 2, 3, 4, 5}}
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
When x = 2,y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
Relation R = {(0, 3), (1,4), (2,5), (3,6), (4,7), (5,8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

A = {1, 2, 3, 4,  ..., 45}
Relation is "is square of a number" and A \(\rightarrow\) A on A
A x A = {(1, 1), (1,2), (1, 3), (1, 4).....( 45, 45)}
The square of 1 is 1 ∈ A and (1, 1) ∈ A x A 
The square of 2 is 4 ∈ A and (4, 2) ∈ A x A
The square of 3 is 9 ∈ A and (9, 3) ∈ A x A  
The square of 4 is 16 ∈ A and (16, 4) ∈ A x A 
The square of 5 is 25 ∈ A and (25, 5) ∈ A x A 
The square of 6 is 36 ∈ A and (36, 6) ∈ A x A 
The square of 7 is 49 \(\notin\) A.
R = {(1, 1), (4, 2),(9, 3), (16, 4),(25, 5), (36, 6)}
Domain of R = {1, 4, 9,16, 25, 36 }
Range of R = {1,2, 3, 4, 5, 6}

A = { 1, 2, 3, 7}, B = { 3, 0, -1, 7}
A x B = {(1, 3), (1, 0), (1, - 1), (1, 7),(2,3), (2, 0), (2, -1), (2, 7), (3,3), (3, 0), (3, - 1), (3, 7), (7, 3)., (7, 0), (7, -1), (7 ,7)}
R₁ = {(2, 1), (7, 1)}
Since (2, 1) and (7,1) are not the elements of A x B, R₁ is not a relation from A to B. Moreover \(1 \notin B .\)

(i) Set builder form of R = ((x, y) | y = x - 2, x \(\in \) P, y \(\in \) Q}
(ii) Roster form R = {(5 , 3),(6 , 4)(7 , 5)}
(iii) Domain of R = {5, 6, 7} and range of R = {3, 4, 5}

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

A = {1,2,3} B = {2,3,5,7} 
A x B = {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7)}
B x A = {(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(7,1),(7,2),(7,3)}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Given that A = {1,3,5} and B = {2,3}
(i) A x B = {1,3,5} x {2,3} = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)}     ...(1)
B x A = {2,3} x {1,3,5} = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)}         ...(2)
(ii) From (1) and (2) we conclude that A x B ≠ B x A as (1,2) ≠ (2,1) and (1,3) ≠ (3,1). etc
(iii) n(A) = 3; n (B) = 2.
From (1) and (2) we observe that, n (A x B) = n (B x A) = 6;
we see that, n(A) x n(B) = 3 x 2 = 6 and n (B) x n (A) = 2 x 3 = 6
Hence, n (A x B) = n (B x A) = n(A) x n(B) = 6.
Thus, n(A x B) = n (B x A) = n(A) x n(B).