# Class 11 Maths NCERT Solutions for Chapter 2 Relations and Functions Exercise 2.3

### Relations and Functions Exercise 2.3 Solutions** **

**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(x) = √(9 – x**

^{2})

**Solution**

(i) f(x) = - |x| , x ∈ R

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 (–∞, 0].

(ii) f(x) = √(9 – x^{2})

Since √(9 – x^{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*) = 2*x* – 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

**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**

The given function is t(C) = 9C/5 + 32 .

Therefore,

(iv) It is given that t(C) = 212

Thus, the value of t, when t(C) = 212, is 100.

**5. Find the range of each of the following functions.(i) f(x) = 2 – 3x, x ∈ R, x > 0.(ii) f(x) = x**

^{2}+ 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

x |
0.01 |
0.1 |
0.9 |
1 |
2 |
2.5 |
4 |
5 |
… |

f(x) |
1.97 |
1.7 |
–0.7 |
–1 |
–4 |
–5.5 |
–10 |
–13 |
… |

i.e., range of f = (–∞, 2)

Alter:

Let x > 0

⇒ 3x > 0

⇒ 2 –3x < 2

⇒ f(x) < 2

∴Range of f = (–∞, 2)

^{2}+ 2, x, is a real number

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

x |
0 |
±0.3 |
±0.8 |
±1 |
±2 |
±3 |
… |

f(x) |
2 |
2.09 |
2.64 |
3 |
6 |
11 |
... |

i.e., range of f = [2, ∞)

Alter:

Let x be any real number.

Accordingly,

x

^{2}≥ 0

⇒ x

^{2}+ 2 ≥ 0 + 2

⇒ x

^{2}+ 2 ≥ 2

⇒ f(x) ≥ 2

∴ Range of f = [2,)

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R