NCERT solutions for class 11 Maths exercise 2.3 chapter 2 Relations and Functions - Future Study Point

# 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

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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 elementsย  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 elementย  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:

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

In the value ofย  ย ย  ,ย  x can have 0, positive and negative value,so it can be expressed as follows

Since the function is defined as , therefore the function can have 0 or negative value, the function can be expressed as follows

In both casesย  xโฅ0 and x<0, f(x) have 0 and negative value

Therefore the range of the function is (-โ,0]

Since for x โ R, โx โฅ 0

โ9 – xยฒ โฅ 0

โ9 โฅ xยฒ

โxยฒโค 9

The domain is [-3,3]

The given function is f(x) = โ(9 -xยฒ)

For any value of xย 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 ofย

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

Ans. The given function is

f(x) = 2x -5

(i) f(0) = 2ร0-5 = -5

(ii) f(7) = 2ร 7- 5 = 9

(iii) f(-3) = 2ร-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

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

Ans. The given function is

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

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 โ R, x > 0

(ii) f(x) = xยฒ+2 , x is a real number

(iii)f(x) = x , x is a real number

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

Also given

x > 0

Multiplying by x in both sides of inequality

3x > 0

โ2-3x < 2-0

โ2-3x <2

โf(x) <2

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

Therefore range of f(x) is (-โ,2)

(ii) f(x) = xยฒ+2 , x is a real number

We are given that x is a real number

Therefore

xยฒ โฅ 0

xยฒ+2 โฅ 0+2

โxยฒ+2 โฅ 2

โf(x) โฅ 2

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

Therefore range of f(x) is [2,โ)

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

Also givenย 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

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