Chapter 2- Relations and Functions Exercise 2.3 (Page No-44)
1. 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)}
2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.
Hence, domain = {2, 5, 8, 11, 14, 17} and range = {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.
Hence, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
Here 1, corresponds to two different images, i.e., 3 and 5;
So, this relation is not a function.
2. Find the domain and range of the following real function:
(i) f(x) = –|x| (ii) f(x) = √(9 – x2)
Solution:
(i) Given,
f(x) = –|x|, x ∈ R
We know that,
As f(x) is defined for x ∈ R,
So, the domain of f is R.
Now,We can see that the range of f(x) = –|x| is all real numbers except positive real numbers.
Hence, the range of f(x) is (–∞, 0].
(ii) f(x) = √(9 – x2)
We can see that
√(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0.
Therefore, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].
Now we can see that
For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.
Hence, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].
3. A function f is defined by f(x) = 2x – 5. Write down the values of
(i) f(0), (ii) f(7), (iii) f(–3)
Solution:
It is given,
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
9C= 900
Therefore,
C=100
5. Find the range of each of the following functions:
(i) f(x) = 2 – 3x, x ∈ R, x > 0
(ii) f(x) = x2 + 2, x is a real number
(iii) f(x) = x, x is a real number
Solution:
(i) It is given,
f(x) = 2 – 3x, x ∈ R, x > 0
We can write the values of f(x) for various values of real numbers x > 0 in tabular form as following :
Thus, we can see that the range of f(x) is the set of all real numbers less than 2.
Hence, range of f (x) = (–∞, 2)
Method (2)
Let x > 0
⇒ 3x > 0
⇒ 2 –3x < 2
⇒ f(x) < 2
Hence, Range of f = (–∞, 2)
(ii) It is given,
f(x) = x2 + 2, x is a real number
We can write the values of f(x) for various values of real numbers x > 0 in tabular form as following :
Thus, we can see that the range of f(x) is the set of all real numbers greater than 2.
Hence, range of f (x) = [2,∞)
Method (2)
We know that,
x2 ≥ 0
Therefore ,
x2 + 2 ≥ 2 [Adding 2 on both sides]
Therefore, the value of x2 + 2 will be always greater or equal to 2, for x is a real number.
Therefore, Range = [2, ∞)
(iii) Given,
f(x) = x, x is a real number
So, the range of f is the set of all real numbers.
Thus,
Range of f = R