Exercise 2.3
Q1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
Solution:
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
Let be R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
There is no same component in the domain of given relation, so this relation is a function.
Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}
Solution:
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Let be R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
There are also unique domain components or order pair in the given relation, so this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
Solution:
(iii) {(1, 3), (1, 5), (2, 5)}
Since the domain has the same first element i.e., 1 corresponds to two different images i.e.,
3 and 5, this relation is not a function.
Q2. Find the domain and range of the following real functions:
(i) f(x) = – |x|
Solution2:
Q3: A function f is defined by f(x) = 2x – 5
(i) f (0) (ii) f(7) (iii) f(-3)
Solution 3:
The given function is f(x) = 2x – 5
Therefore,
(i) f(0) = 2 × 0 – 5 = 0 – 5 = - 5
(ii) f(7) = 2 × 7 – 5 = 14 – 5 = - 9
(iii) f(-3) = 2 × -3 – 5 = -6 – 5 = - 11
Solution:
Solution:
(iii) f(x) = x is real number.
It is clear that the range of f is the set of all real number.
Therefore, range of f = R