An arrow diagram
 

A = {1,2,3},B =  {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) =  2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(ii) An arrow diagram

(iv) A Graph f = \(\{(x, f(x) / x \in A\}\)
= {(0, 1), (1, 3), (2, 5), (3, 7)}

f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2

f(x) = 2x + 1, g(x) = x² - 2
f o g(x) = f(g(x)) = f(x² - 2) = 2(x² - 2) + 1 = 2x² - 3
g o f(x) = g(f(x)) = g(2x + 1) = (2x + 1)² - 2 = 4x² + 4x - 1
Thus f o g = 2x² - 3, g o f = 4x² + 4x - 1. From the above, we see that f o g ≠ g o f.

A = {1, 2, 3}, B = {4, 5, 6, 7}; f = {(1, 4),(2, 5),(3, 6)}
Then f is a function from A to B and for different elements in A, there are different images in B. Hence f is one–one function. Note that the element 7 in the co-domain does not have any pre-image in the domain. Hence f is not onto.
Therefore f is one–one but not an onto 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}.

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 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).

The distance travelled by the particle in time t hours is given by S(t)=\(\frac { { t }^{ 2 }+t }{ 2 } \).
(i) t = 3.5 hours. Therefore, S(3.5)=\(\frac { (3.5)^{ 2 }+3.5 }{ 2 } =\frac { 15.75 }{ 2 } \)=7.875
The distance travelled in 3.5 hours is 7.875 kms.
(ii) t = 8.25 hours. Therefore, S(8.25)=\(\frac { (8.25)^{ 2 }+8.25 }{ 2 } =\frac { 76.3125 }{ 2 } \)=38.15625
The distance travelled in 8.25 hours is 38.16 kms, approximately.

(i) x = 1, we get
f(1) = 2(1) - (1)² = 2 - 1 = 1
(ii) x = x + 1, we get
f(x + 1) = 2(x + 1) - (x + 1)² = 2x + 2 - (x² + 2x + 1) = -x² + 1
(iii) f(x) + f(1) = (2x - x²) + 1= - x² + 2x + 1
[Note that f(x) + f(1) ≠ f(x + 1). In general f(a + b) is not equal to f(a) + f(b)]

Ordered Pair : The Domain of the relation is about the salaries given to person
Relation is R = {(10000, A₁), (10000, A₂), (10000, A₃),
(10000, A₄), (10000, A₅), (25000, C₁),
(25000, C₂), (25000, C₃), (25000, C₄),
(50000, M₁), (50000, M₂), (50000, M₃),
(100000, E₁), (100000, E₂)}
Relation R defined by x R y
'x' is the salary given to person y
Arrow diagram

Given A = {5,6} , B = {4,5,6} , C = {5,6,7}
L.H.S: A x A = {5,6} x {5,6}
= {(5,6),(5,6),(6,5),(6,6)}          
R.H.S: B x B = {4,5,6} x {4,5,6}
= {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}
C x C = {5,6,7} x {5,6,7}
= {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}
(B x B) ∩ (C x C) = {(5,5),(5,6),(6,5),(6,6)}
LHS = RHS
A x A = (B x B) ∩ (C x C)
Hence proved