Relations and Functions
NCERT Solution
Exercise 2
Question 1: Let A = {1, 2, 3, . . . . 14}.
Define a relation R from A to A by `R={x,y}:3x-y=0`,where x,y ∈A}.
Write down its domain, co-domain and range.
Solution: By definition of the relation,
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Hence, the corresponding arrow diagram is as follows:
The set of first element, i.e. the domain = {1, 2, 3, 4}
Similarly, the set of second elements, i.e. the range = {3, 6, 9, 12}
And the co-domain = {1, 2, 3, ….. 14}
Question 2: Define a relation R on the set N of natural numbers by
R={(x,y):y=x+5,x is a natural number less than 4,x,y ∈N}.
Depict this relationship using (i) roster form (ii) an arrow diagram.
Write down the domain and the range.
Solution: Given, `R = {(x,y):y =x + 5`, x is a natural number less than 4,x,y ∈N}.
(i) Therefore, in roster form `R = {(1, 6), (2, 7), (3, 8)}`
(ii) The arrow diagram
Domain is the set of first element, i.e. {1, 2, 3}
Range is the set of second element, i.e. {6, 7, 8}
Question 3: `A = {1, 2, 3, 5}` and `B = {4, 6, 9}`. Define a relation R from A to B by
R={(x,y):the difference between x and y is odd,x∈A,y∈B}. Write R in roster form.
Solution: Given, A = {1, 2, 3, 5} and B = {4, 6, 9}
Relation R from A to B is given by
R={(x,y):the difference between x and y is odd,x∈A,y∈B}
Therefore, R = {(5, 4)}
Question 4: The figure shows a relationship between the sets P and Q. Write the relation
(i) in set builder form,
(ii) roster form
What is the domain and range?
Solution: It is clear that the relation R is “the difference between x and y, which is 2”
In set builder form R={(x,y):x-y=2,x∈P,y∈Q}
In roster form, R = {(5, 3), (6, 4), (7, 5)}
The domain of this relation is {5, 6, 7}
The range of this relation is {3, 4, 5}
Question 5: Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by : {(a,b):a,b∈A,a divides b}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R
Solution: A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by : {(a,b):a,b∈A,a divides b}
(i) In roster form R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of R= {1, 2, 3, 4, 6)
(iii) Range of R= {1, 2, 3, 4, 6}
Question 6: Determine the domain and range of the relation R defined by R={(x,x+5):x∈{0,1,2,3,4,5}}.
Solution: Given, R={(x,x+5):x∈{0,1,2,3,4,5}}
Therefore,
On substitutin x = 0, 1, 2, 3, 4, 5 and on putting x = 0, 1, 2, 3, 4, 5 in x+5
Domain R = {0, 1, 2, 3, 4, 5}
Range R = {5, 6, 7, 8, 9, 10}
Question 7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Solution: Given, R = {(x, x3) : x is a prime number less than 10}
Therefore,
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
Question 8: Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A into B.
Solution: Given, A = {x, y, z} and B = {1, 2}
Therefore, A x B = {x, y, z} x {1, 2}
= {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Here, n(A x B) = 6
Therefore,
Number of relations in the set A x B = 26 = 64
Question 9: Let R be the relation on Z defined by R={(a,b):a,b∈Z,a-b is an integer}. Find the domain and range of R.
Solution: Given, R={(a,b):a,b∈Z,a-b is an integer}
Now (a-b) is an integer, provided a and b are both even or both odd.
Therefore,
Domain R = Z, (Because a is an integer)
Range R = Z, (Because a is an integer)