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# NCERT solutions for class 11 Maths exercise 2.3 chapter 2 Relations and Functions

NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions are solved here for helping the students of 11 class maths students to clear their doubts on solving the NCERT unsolved questions of class 11 maths exercise 2.3 of chapter 2 Relations and functions. The proper understanding of exercise 2.3 is needed for every student for attempting the questions of higher class maths. All NCERT solutions of exercise 2.3 are prepared by an expert of maths who has huge experience in maths teaching. Each solution of the maths question is explained beautifully in easy language, therefore, every student can understand it.

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## NCERT solutions for class 11 Maths exercise 2.3 chapter 2 Relations and Functions

Exercise 2.1

Exercise 2.2

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

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## NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions

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)}

Ans. In the ordered pair (2, 1),(5,1),(8,1),(11,1),(14,1),(17,1), all the elements 2,5,8,11,14 and 17 has have a unique image,therefore given relationship is a function.

In the set R = {(2, 1),(5,1),(8,1),(11,1),(14,1),(17,1)},the first element of the ordered pair is the domain

Therefore

Domain of the function= {2,5,8,11,14, 17}

In the set R = {(2, 1),(5,1),(8,1),(11,1),(14,1),(17,1)},the second element of the ordered pair is the range

Therefore

Range of the function= {1}

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

In the given relation the elements  2,4,6,8,10,12 and 14 have a unique image,therefore given relation is a function

In the set of ordered pair {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)},first element shows the domain of the function

Therefore, domain of the function = {2,4,6,8,10,12,14}

In the set of ordered pair {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)},second element shows the range of the function

Range of the function = {1,2,3,4,5.6,7}

(iii)R={1,3),(1,5),(2,5)}

In the given relation the element  1 has two images 3 and 5, in this relation every element have not a unique image, therefore given relation is not a function

NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions

Q2.Find the domain and range of the following real function:

$(i)\: f\left ( x \right )=-\left | x \right |\: \: \left ( ii \right )\: f\left ( x \right )=\sqrt{9-x^{2}}$

Ans. (i) The given function f(x) is defined for x ∈ R

$f\left ( x \right )=-\left | x \right |$

In the value of   $\left | x \right |$  ,  x can have 0, positive and negative value,so it can be expressed as follows

$\fn_cm \left | x \right |=\left\{\begin{matrix}x, & x\geq 0\\ -x,& x<0 \end{matrix}\right.$

Since the function is defined as $f\left ( x \right )=-\left | x \right |$, therefore the function can have 0 or negative value, the function can be expressed as follows

$\fn_cm f(x)=-\left | x \right |=\left\{\begin{matrix}-x, & x\geq 0\\ x,& x<0 \end{matrix}\right.$

In both cases  x≥0 and x<0, f(x) have 0 and negative value

Therefore the range of the function is (-∞,0]

$\fn_cm \left ( ii \right )\: f\left ( x \right )=\sqrt{9-x^{2}}$

Since for x ∈ R, √x ≥ 0

⇒9 – x² ≥ 0

⇒9 ≥ x²

⇒x²≤ 9

The domain is [-3,3]

The given function is f(x) = √(9 -x²)

For any value of x within the domain [-3.3], the minimum and maximum value of f(x) decides the range of f(x)

It can be observed that for x = 0, f(x) = 3 and for x = 3, f(x) = 0

The possibel values of f(x) lies between 0 to 3

Therefore the range of f(x) is [0,3]

NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions

Q3. A function f is defined by f(x) = 2x -5. Write down the values of

(i) f(0)   (ii) f(7)    (iii) f(-3)

Ans. The given function is

f(x) = 2x -5

(i) f(0) = 2×0-5 = -5

(ii) f(7) = 2× 7- 5 = 9

(iii) f(-3) = 2×-3 -5=-11

NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions

Q4. The function ‘t’ which maps temperature in degree Celcius into temperature in degree Fahrenheit is defined by

$\fn_cm t\left ( C \right )=\frac{9C}{5}+32$

Find (i) t(0)  (ii) t(28)  (iii) t(-10)  (iv) The value of C when t(C) =212

Ans. The given function is

$\fn_cm t\left ( C \right )=\frac{9C}{5}+32$

$\fn_cm \left ( i \right )\: t\left ( 0 \right )=\frac{9\times 0}{5}+32=32$

$\fn_cm \left ( ii \right )\: t\left ( 28 \right )=\frac{9\times 28}{5}+32=\frac{252}{5}+32=\frac{252+160}{5}=\frac{412}{5}=82.4$

$\fn_cm \left ( iii \right )\: t\left ( -10 \right )=\frac{9\times -10}{5}+32=\frac{-90}{5}+32=-18+32=14$

(iv) We are given t(C) =212

$\fn_cm \therefore 212=\frac{9C}{5}+32$

$\fn_cm \frac{9C}{5} = 212-32$

$\fn_cm \frac{9C}{5} = 180$

$\fn_cm C=\frac{5\times 180}{9}$

C = 100

NCERT solutions for class 11 exercise 2.3 chapter 2 Relations and Functions

Q5.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

Ans.(i) The given function is f(x) = 2 -3x

Also given

x > 0

Multiplying by x in both sides of inequality

3x > 0

⇒2-3x < 2-0

⇒2-3x <2

⇒f(x) <2

It indicates that all possible values of f(x) are less than 2

Therefore range of f(x) is (-∞,2)

(ii) f(x) = x²+2 , x is a real number

We are given that x is a real number

Therefore

x² ≥ 0

x²+2 ≥ 0+2

⇒x²+2 ≥ 2

⇒f(x) ≥ 2

It indicates that all possible values of f(x) are greater and equal to 2

Therefore range of f(x) is [2,∞)

(iii) The given function is f(x) = x

Also given x is a real number

Since f(x) = x, It indicates that all possible values of f(x) are all real numbers

Therefore range of the function is f(x) =R

### NCERT Solutions of class 9 maths

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### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals