# NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions Ex 2.3

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 2 |

Exercise | Class 11 Maths Chapter 2 Relations and Functions Exercise 2.3 |

Number of Questions Solved | 5 |

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

**NCERT TEXTBOOK EXERCISES**

**Ex 2.3 Class 11 Maths 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)}.

**Solution.**

**(i) **We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.

∴ The given relation is a function.

Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

**(ii)** We have a relation

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

Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.

∴ The given relation is a function.

Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

**(iii)** We have a relation R = {(1, 3), (1, 5), (2, 5)}

Since the distinct ordered pairs (1, 3) and (1, 5) have same first element i.e., 1 does not have a unique image under R.

∴ It is not a function.

**Ex 2.3 Class 11 Maths Question 2. Find the domain and range of the following real functions:**

(i) f(x) = −|x|

(ii) f(x) = $\sqrt { 9-{ x }^{ 2 } }$

**Solution.**

**Ex 2.3 Class 11 Maths 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)

**Solution.**

We are given 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.

**Ex 2.3 Class 11 Maths Question 4. The function T which maps temperature in degree Celsius into temperature in degree by $t(C)=\frac { 9C }{ 5 } +32$**

Find

(i) t (0)

(ii) t (28)

(iii) t (-10)

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

**Solution.**

**Ex 2.3 Class 11 Maths Question 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)** Given f (x) = 2 – 3x, x ∈ R, x > 0

∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2

∴ The range of f (x) is (-2).

**(ii)** Given f (x) = x^{2} + 2, x is a real number

We know x^{2}≥ 0 ⇒ x^{2} + 2 ≥ 0 + 2

⇒ x^{2} + 2 > 2 ∴ f (x) ≥ 2

∴ The range of f (x) is [2, ∞).

**(iii)** Given f (x) = x, x is a real number.

Let y =f (x) = x ⇒ y = x

∴ Range of f (x) = Domain of f (x)

∴ Range of f (x) is R.

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

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