NCERT Solution for Class 11 Maths Chapter 2: Relations and Functions Exercise 2.1
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Question 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)} Answer
A function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
Since 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.
Here,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)}
Since 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.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, So this relation is not a function.
Question 2
Find the domain and range of the following real function:
(i) f(x) = –|x|
(ii) f(x) = √9 − x2 Answer
(i) f(x) = –|x|, x ∈ R
This can be written as
f(x) = −|x| =
{ −x, if x ≥ 0
x, if x < 0
Here we see that f(x) is defined for x ∈ R. So, the domain of f is R.
From the Function, we can see that, it returns negative real number for all positive and negative real number
Therefore, the range of f is (−∞, 0].
(ii) f(x) = √(9 – x2)
It means 9 – x2 ≥ 0
=> 3 ≤ x ≤ 3
So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].
For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between
0 and 3. ∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].
Question 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) Answer
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
Question 4
The function ‘t’ which maps temperature in degree Celsius into temperature
in degree Fahrenheit is defined by
t(C) = 9C/5+ 32.
Find
(i) t (0) (ii) t (28) (iii) t (–10)
(iv) The value of C, when t(C) = 212 Answer
The given function is
t(C) = 9C/5+ 32.
Therefore,
(i) t (0) = 9(0)/5+ 32 =32
(ii) t (28) = 9(28)/5+ 32 =412/5
(iii) t (-10) = 9(-10)/5+ 32 =14
(iv) It is given that t(C) = 212
9C/5+ 32=212
9C/5=180
C=100
Hence, the value of t, when t(C) = 212, is 100.
Question 5
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x Î R, x > 0.
(ii) f(x) = x2 + 2, x, is a real number.
(iii) f(x) = x, x is a real number Answer
(i) f(x) = 2 – 3x, x Î R, x > 0
Let x > 0
⇒ 3x > 0
⇒ 2 –3x < 2
⇒ f(x) < 2
So, Range of f = (-∞ , 2)
(ii) f(x) = x2 + 2, x, is a real number
For any real number, we know that
x2 ≥ 0
⇒ x2 + 2 ≥ 2
⇒ f(x) ≥ 2
So, Range of f = [2, ∞ )
(iii) f(x) = x, x is a real number
Range of f is the set of all real numbers. So, Range of f = R
