TOPICS

Exercise - 2.3

Relations and Functions

**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)}.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} R = {2, 5, 8, 11, 14, 17} are the elements of the domain of the given relation having their unique images, this relation is a function. So, 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)} R = {2, 4, 6, 8, 10, 12, 14} are the elements of the domain of the given relation having their unique images, this relation is a function. So, Domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7} (iii) {(1,3), (1,5), (2,5)}. This relation is not a function because there is an element 1 which is associated to two elements 3 and 5.

**Question-2 :-** Find the domain and range of the following real functions:

(i) f(x) = –|x|

(ii) f(x) = √9 - x²

(i) f(x) = –|x|, x ∈ R Since f(x) = –|x| is defined for x ∈ R, the domain of f is R. It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers. (ii) f(x) = √9 - x² Since √9 - x² is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, Domain of f(x) = {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. Range of f(x) = {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).

Given that 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

**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.

Give that t(C) = 9C/5 + 32. (i) t(0) = (9 x 0)/5 + 32 = 0 + 32 = 32 (ii) t(28) = (9 x 28)/5 + 32 = 252/5 + 32 = (252 + 160)/5 = 412/5 (iii) t(-10) = (9 x (-10))/5 + 32 = -18 + 32 = 14 (iv) t(C) = 212 212 = 9C/5 + 32 212 - 32 = 9C/5 180 = 9C/5 C = (180 x 5)/9 C = 20 x 5 = 100

**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.

(i) f(x) = 2 – 3x, x ∈ R, x > 0. Let x > 0 ⇒ 3x > 0 ⇒ 2 –3x < 2 ⇒ f(x) < 2 Range of f = (–∞, 2) (ii) f(x) = x² + 2, x is a real number. Let x be any real number. Accordingly, x² ≥ 0 ⇒ x² + 2 ≥ 0 + 2 ⇒ x² + 2 ≥ 2 ⇒ f(x) ≥ 2 Range of f = [2,∞) (iii) f(x) = x, x is a real number. It is clear that the range of f is the set of all real numbers. Range of f = R

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