# Relations and Functions – Class XI – Exercise 2.3

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)}

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, this relation is not a function.

2: Find the domain and range of the following real function:

(i) f(x) = β|x|

(ii) π(π₯) = β(9 β π₯2)

Solution:

(i) f(x) = β|x|, x β R

We know that |π₯| = { π₯, ππ π₯ β₯ 0

{βπ₯, ππ π₯ < 0

β΄ π(π₯) = β|π₯| = { βπ₯, ππ π₯ β₯ 0

{ Β π₯, ππ π₯ < 0

Since f(x) is defined for x β R, the domain of f is R.

It can be observed that the range of f(x) = β|x| is all real numbers except positive real numbers.

β΄ The range of f is (ββ, 0]. (ii) π(π₯) = β(9 β π₯2)

Since β(9 β π₯2) is defined for all real numbers that are greater than or equal to β3 and less than or equal to 3, the domain of f(x) is {x : β3 β€ x β€ 3} or [β3, 3].

For any value of x such that β3 β€ x β€ 3, the value of f(x) will lie between 0 and 3. β΄The range of f(x) is {x: 0 β€ x β€ 3} or [0, 3].

3: A function f is defined by f(x) = 2x β 5. Write down the values of (i) f(0), (ii) f(7), (iii) f(β3)

Solution:

The given function is f(x) = 2x β 5.

Therefore,

(i) f(0) = 2 Γ 0 β 5

= 0 β 5

= β5

(ii) f(7) = 2 Γ 7 β 5

= 14 β 5

= 9

(iii) f(β3) = 2 Γ (β3) β 5

= β 6 β 5

= β11

1. The function βtβ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by

π(πΆ) = 9πΆ/5 + 32. Find

(i) t (0)

(ii) t (28)

(iii) t (β10)

(iv) The value of C, when t(C) = 212

Solution:

The given function is π(πΆ) = 9πΆ 5 + 32. Therefore,

(i) t(0) = 9×0/5 + 32 = 0 + 32 = 32

(ii) t(28) = 9×28/5 + 32 = 252+160/5 = 412/5

(iii) t(-10) = 9x(-10)/5 + 32 = 9(-2)+32 = -18+32 = 14

(iv) it is given that t(c) = 212

212 = 9c/5 + 32

9c/5 = 212 β 32

9c/5 = 180

9C = 180 x 5

C = 180×5/9 = 100

Thus the value of t, when tΒ© = 212, is 100

1. Find the range of each of the following functions

(i) f(x) = 2 β 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) f(x) = 2 β 3x, x R, x > 0

The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as

 x 0.01 0.1 0.9 1 2 2.5 4 5 β¦ f(x) 1.97 1.7 -0.7 -1 -4 -5.5 -10 -13 β¦

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

i.e., range of f = (ββ , 2)