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) = –|x|

(ii) 𝑓(π‘₯) = √(9 βˆ’ π‘₯2)

Solution:

(i) f(x) = –|x|, x ∈ R

We know that |π‘₯| = { π‘₯, 𝑖𝑓 π‘₯ β‰₯ 0

{βˆ’π‘₯, 𝑖𝑓 π‘₯ < 0

∴ 𝑓(π‘₯) = βˆ’|π‘₯| = { βˆ’π‘₯, 𝑖𝑓 π‘₯ β‰₯ 0

{ Β π‘₯, 𝑖𝑓 π‘₯ < 0

Since f(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.

∴ The range of f is (βˆ’βˆž, 0]. (ii) 𝑓(π‘₯) = √(9 βˆ’ π‘₯2)

Since √(9 βˆ’ π‘₯2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, 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].


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:

The given function is f(x) = 2x – 5.

Therefore,

(i) f(0) = 2 Γ— 0 – 5

= 0 – 5

= –5

(ii) f(7) = 2 Γ— 7 – 5

= 14 – 5

= 9

(iii) f(–3) = 2 Γ— (–3) – 5

= – 6 – 5

= –11


  1. The function β€˜t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by

𝑓(𝐢) = 9𝐢/5 + 32. Find

(i) t (0)

(ii) t (28)

(iii) t (–10)

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

Solution:

The given function is 𝑓(𝐢) = 9𝐢 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 – 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 – 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 – 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

x0.010.10.9122.545…
f(x)1.971.7-0.7-1-4-5.5-10-13…

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€“βˆž , 2)