Our ncert solutions for Exercise 2.3 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.3 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

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## Exercise 2.3 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

CBSE board students who preparing for **class 11 ncert solutions maths and Mathematics** solved exercise **chapter 2. Relations and Functions** available and this helps in upcoming exams
2024-2025.

### You can Find Mathematics solution Class 11 Chapter 2. Relations and Functions

- All Chapter review quick revision notes for chapter 2. Relations and Functions Class 11
- NCERT Solutions And Textual questions Answers Class 11 Mathematics
- Extra NCERT Book questions Answers Class 11 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 2.3 class 11 Mathematics Chapter 2. Relations and Functions

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- Exercise 2.3 Class 11 Maths 2. Relations And Functions - Ncert Solutions - Toppers Study
- Class 11 Ncert Solutions
- Solution Chapter 2. Relations And Functions Class 11
- Solutions Class 11
- Chapter 2. Relations And Functions Exercise 2.3 Class 11

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## 2. Relations and Functions

### | Exercise 2.3 |

## Exercise 2.3 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

**Exercise 2.3 **

**Q1. 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)} **

Let be R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

There is no same component in the domain of given relation, so this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

**Solution:**

**(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)} **

Let be R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

There are also unique domain components or order pair in the given relation, so this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

**Solution: **

**(iii) {(1, 3), (1, 5), (2, 5)} **

Since the domain has the same first element i.e., 1 corresponds to two different images i.e.,

3 and 5, this relation is not a function.

**Q2. Find the domain and range of the following real functions:**

(i) f(x) = – |x|

**Solution2: **

**Q3: ****A function f is defined by f(x) = 2x – 5**

**(i) f (0) (ii) f(7) (iii) f(-3)**

**Solution 3: **

The given function is f(x) = 2x – 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

**Solution:**

**Solution:**

**(iii) f(x) = ****x is real number**.

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

Therefore, range of f = R

##### Other Pages of this Chapter: 2. Relations and Functions

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