The curves in Fig.1.18(a) and Fig.1.18(c) do not represent a function as the vertical lines meet the curves in two points P and Q.
The curves in Fig.1.18(b) and Fig.1.18(d) represent a function as the vertical lines meet the curve in at most one point.

Let the distance be 'd'
Speed = 500 km/hr
Time = 't' hours
Distance = Time x Speed
d(t) = 500 t

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.

(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.

Given A = {2, – 2, 3}, B = {1, – 4}.
A x B = {2,-2,3) x {1 ,-4}
= {(2, 1), (2, - 4), (- 2, 1), (- 2, - 4), (3, 1), (3, - 4)}
A x A = {2,-2,3} x {2,-2,3}
= {(2,2), (2, - 2), (2,3), (-2,2), (- 2, - 2), (- 2,3),(3,2),(3, - 2), (3, 3)}
B x A = {1, -4} x {2, -2, 3}
= {(1,2),(1, -2 ),(1, 3)}, (- 4, 2), (- 4, - 2), (- 4, 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).