RD Chapter 2 Relations Ex 2.1 Solutions
Question - 1 : - (i) If (a/3 + 1, b – 2/3) = (5/3, 1/3), find the values of a and b.
(ii) If (x + 1, 1) = (3y, y – 1), find the values of x and y.
Answer - 1 : -
Given:
(a/3 + 1, b – 2/3) = (5/3, 1/3)
By the definition of equality of ordered pairs,
Let us solve for a and b
a/3 + 1 = 5/3 and b – 2/3 = 1/3
a/3 = 5/3 – 1 and b = 1/3 + 2/3
a/3 = (5-3)/3 and b = (1+2)/3
a/3 = 2/3 and b = 3/3
a = 2(3)/3 and b = 1
a = 2 and b = 1
∴ Values of a and b are, a = 2 and b = 1
(ii) If (x + 1, 1) = (3y, y – 1), find the values of x and y.
Given:
(x + 1, 1) = (3y, y – 1)
By the definition of equality of ordered pairs,
Let us solve for x and y
x + 1 = 3y and 1 = y – 1
x = 3y – 1 and y = 1 + 1
x = 3y – 1 and y = 2
Since, y = 2 we can substitute in
x = 3y – 1
= 3(2) – 1
= 6 – 1
= 5
∴ Values of x and y are, x = 5 and y = 2
Question - 2 : - If the ordered pairs (x, – 1) and (5, y) belong to the set {(a, b): b = 2a – 3}, find the values of x and y.
Answer - 2 : -
Given:
The ordered pairs (x, – 1) and (5, y) belong to the set {(a, b): b = 2a – 3}
Solving for first order pair
(x, – 1) = {(a, b): b = 2a – 3}
x = a and -1 = b
By taking b = 2a – 3
If b = – 1 then 2a = – 1 + 3
= 2
a = 2/2
= 1
So, a = 1
Since x = a, x = 1
Similarly, solving for second order pair
(5, y) = {(a, b): b = 2a – 3}
5 = a and y = b
By taking b = 2a – 3
If a = 5 then b = 2×5 – 3
= 10 – 3
= 7
So, b = 7
Since y = b, y = 7
∴ Values of x and y are, x = 1 and y = 7
Question - 3 : - If a ∈ {- 1, 2, 3, 4, 5} and b ∈ {0, 3, 6}, write the set of all ordered pairs (a, b) such that a + b = 5.
Answer - 3 : -
Given: a ∈ {- 1, 2, 3, 4, 5} and b ∈ {0, 3, 6},
To find: the ordered pair (a, b) such that a + b = 5
Then the ordered pair (a, b) such that a + b = 5 are as follows
(a, b) ∈ {(- 1, 6), (2, 3), (5, 0)}
Question - 4 : - If a ∈ {2, 4, 6, 9} and b ∈ {4, 6, 18, 27}, then form the set of all ordered pairs (a, b) such that a divides b and a
Answer - 4 : -
Given:
a ∈ {2, 4, 6, 9} and b ∈{4, 6, 18, 27}
Here,
2 divides 4, 6, 18 and is also less than all of them
4 divides 4 and is also less than none of them
6 divides 6, 18 and is less than 18 only
9 divides 18, 27 and is less than all of them
∴ Ordered pairs (a, b) are (2, 4), (2, 6), (2, 18), (6, 18), (9, 18) and (9, 27)
Question - 5 : - If A = {1, 2} and B = {1, 3}, find A x B and B x A.
Answer - 5 : -
Given:
A = {1, 2} and B = {1, 3}
A × B = {1, 2} × {1, 3}
= {(1, 1), (1, 3), (2, 1), (2, 3)}
B × A = {1, 3} × {1, 2}
= {(1, 1), (1, 2), (3, 1), (3, 2)}
Question - 6 : - Let A = {1, 2, 3} and B = {3, 4}. Find A x B and show it graphically
Answer - 6 : -
Given:
A = {1, 2, 3} and B = {3, 4}
A x B = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
Steps to follow to represent A × B graphically,
Step 1: One horizontal and one vertical axis should be drawn
Step 2: Element of set A should be represented in horizontal axis and on vertical axis elements of set B should be represented
Step 3: Draw dotted lines perpendicular to horizontal and vertical axes through the elements of set A and B
Step 4: Point of intersection of these perpendicular represents A × B
Question - 7 : - If A = {1, 2, 3} and B = {2, 4}, what are A x B, B x A, A x A, B x B, and (A x B) ∩ (B x A)?
Answer - 7 : -
Given:
A = {1, 2, 3} and B = {2, 4}
Now let us find: A × B, B × A, A × A, (A × B) ∩ (B × A)
A × B = {1, 2, 3} × {2, 4}
= {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {2, 4} × {1, 2, 3}
= {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
A × A = {1, 2, 3} × {1, 2, 3}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
B × B = {2, 4} × {2, 4}
= {(2, 2), (2, 4), (4, 2), (4, 4)}
Intersection of two sets represents common elements of both the sets
So,
(A × B) ∩ (B × A) = {(2, 2)}
Question - 8 : - 
Answer - 8 : -
Question - 9 : - If A = { 1, 2, 3}, show that a ono-one function f : A → A must be onto.
Answer - 9 : -
Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.
Hence, f has to be onto.
Question - 10 : - If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Answer - 10 : -
Suppose f is not one-one.
Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.
Also, the image of 3 under f can be only one element.
Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}
i.e f is not an onto function, a contradiction.
Hence, f must be one-one.