NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3
Ex 2.3 Class 11 Maths 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)}
Solution.
(i) Given relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1),
(17, 1)}
Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their
unique images.
∴ The given relation is a function.
Hence, domain = {2, 5, 8, 11, 14, 17) and Range = {1}.
(ii) Given relation R = {(2, 1), (4, 2),
(6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their
unique images.
∴ The given relation is a function.
Hence, domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.
(iii) Given relation R = {(1, 3), (1, 5),
(2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have the same first element,
i.e., 1 does not have a unique image under R.
∴ It is not a function.
Ex 2.3 Class 11 Maths Question 2.
Find the domain and range of the following real functions:(i) f(x) = −|x|
(ii) f(x) = √9 − x2
Solution.
Ex 2.3 Class 11 Maths 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)
Solution.
Given f(x) = 2x – 5
(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
Ex 2.3 Class 11 Maths 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.
Solution.
Ex 2.3 Class 11 Maths Question 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) Given
f(x) = 2 – 3x, x ∈ R, x > 0
∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f(x) < 2
∴ The range of f(x) is (-2).
(ii) Given f(x) = x2 +
2, x is a real number
We know x2 ≥ 0 ⇒
x2 + 2 ≥ 0 + 2
⇒ x2 + 2 > 2 ∴ f(x) ≥ 2
∴ The range of f(x) is [2, ∞).
(iii) Given f(x) = x, x is a real number.
Let y = f(x) = x ⇒ y = x
∴ Range of f(x) = Domain of f(x)
∴ Range of f(x) is R.
NCERT Solutions for Maths Class 9