NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

Q: How to find which relation is a function in class 11 maths chapter 2 NCERT solutions?

The definition of a function in class 11 chapter 2 states that each element in the domain can be related to only one element in the range. This means that if a student draws a vertical line on a graph, it can intersect the x-axis only once. To determine if a relation is a function, one can use vertical line tests or various formulas.

Q: What are important topics of the chapter Relations And Functions?

Basic definitions of relations and functions, Cartesian products of sets, and domain and range of the functions are the important topic of this chapter. you can practice them to get maximum benifit. Practice problems form class 11 maths chapter 2 NCERT solutions pdf after downloading using link given in this article.

Q: What is the meaning of relations in Chapter 2 of NCERT Solutions for Class 11 Maths?

Relations are just a bunch of ordered pairs, with one object from each set. Functions can also be considered relations, but they have different concepts. The NCERT Solutions for Class 11 Maths Chapter 2 provide students with a clear definition and analysis of relations according to the CBSE Syllabus 2023. The solutions have many examples to help students solve relation-related problems with ease.

Q: Explain the basic steps for the Cartesian product of sets in NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions.

To learn how to solve problems from class 11 Maths NCERT solutions chapter 2 that are related to the Cartesian product of sets, it is essential for students to understand the first exercise of the chapter well. Before each set of problems in the exercise, solved examples are given to help students learn how to solve problems quickly. By practicing problems from the NCERT textbook, students can improve their understanding of the concepts, which is necessary for performing well in exams.

Q: Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. Practice problems from class 11 maths chapter 2 solutions to get command and in-depth understanding of concepts.

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

Edited By Ramraj Saini | Updated on Sep 21, 2023 08:28 PM IST

Relations And Functions Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions are discussed here. In the previous chapter you have learned about sets. This NCERT book chapter relation and function class 11 is the continuation of chapter 1 sets. In this article, you will get NCERT solutions for class 11 maths chapter 2 relations and functions. The answers to the exercises in the NCERT textbook are made to help students get ready for their exams and do really well. NCERT Solution are written by very knowledgeable teachers who explain each answer in an easy-to-understand way that follows the latest CBSE Syllabus 2023. Using these answers can help Class 11 students get really good at understanding Relations and Functions Important topics like domain, co-domain, and range of functions are covered in this chapter relation and function class 11.

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Relations And Functions Class 11 Questions And Answers
Relations And Functions Class 11 Questions And Answers PDF Free Download
Relations And Functions Class 11 Solutions - Important Formulae
Relations And Functions Class 11 NCERT Solutions (Intext Questions and Exercise)
Relations and Functions Class 11 - Topics
2.1 Introduction
Summary Of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions
NCERT Solutions for Class 11 Mathematics - Chapter Wise
Key Features of Relations and Functions Class 11 Solutions
NCERT solutions for class 11 - Subject wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

In NCERT solutions for class 11 maths chapter 2 relations and functions, questions related to these topics are covered. Also, you will learn different types of specific real-valued functions and their graphs in ch 2 maths class 11. NCERT solutions for class 11 maths chapter 2 relations and functions will build your fundamentals of functions which will be helpful in the 12th board exam also. Check all NCERT solutions from class 6 to 12 to learn science and Maths. Here you will get NCERT solutions for class 11 also.

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Relations And Functions Class 11 Questions And Answers PDF Free Download

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Relations And Functions Class 11 Solutions - Important Formulae

Relations:

R is a relation between sets A and B: R ⊆ A × B

Inverse of Relation R: R^⁻¹ = {(b, a) : (a, b) ∈ R}

Domain of R = Range of R^⁻¹

Range of R = Domain of R^⁻¹

Functions:

A function f: A → B maps every element of A to one and only one element in B.

Cartesian product A × B: A × B = {(a, b) : a ∈ A, b ∈ B}

(a, b) = (x, y) implies a = x and b = y

n(A) = x, n(B) = y, then n(A × B) = xy and A × ∅ = ∅

A × B ≠ B × A

Function f: A → B can be denoted as f(x) = y.

Algebra of Functions:

For functions f: X → R and g: X → R:

(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
(f * g)(x) = f(x) * g(x)
(kf)(x) = k * (f(x)), where k is a real number
{f/g}(x) = f(x)/g(x), g(x) ≠ 0

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Relations And Functions Class 11 NCERT Solutions (Intext Questions and Exercise)

Relations and functions class 11 questions and answers - Exercise: 2.1

Question:1 If $\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$ , find the values of x and y.

Answer:

It is given that
$\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
$\frac{x}{3}+1= \frac{5}{3} \ \ \ and \ \ \ y - \frac{2}{3}= \frac{1}{3}$
$\frac{x}{3}= \frac{5}{3}-1 \ \ \ and \ \ \ y = \frac{1}{3}+ \frac{2}{3}$
$\frac{x}{3}= \frac{5-3}{3} \ \ \ and \ \ \ y = \frac{1+2}{3}$
$\frac{x}{3}= \frac{2}{3} \ \ \ and \ \ \ y = \frac{3}{3}$
$x= 2 \ \ \ and \ \ \ y = 1$

Therefore, values of x and y are 2 and 1 respectively

Question:2 If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in $( A \times B )$ .

Answer:

It is given that set A has 3 elements and the elements in set B are 3 , 4 , and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in $(A \times B)$
= ( Number of elements in set A ) $\times$ ( Number of elements in set B)
= 3 $\times$ 3
= 9
Therefore, number of elements in $(A \times B)$ is 9

Question:3 If G = {7, 8} and H = {5, 4, 2}, find $G \times H \: \: and \: \: H \times G$

Answer:

It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
P $\times$ Q = {(p,q) , where p $\epsilon$ P , q $\epsilon$ Q }
Therefore,
G $\times$ H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
H $\times$ G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question:4 (i) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If P = {m, n} and Q = { n, m}, then $P \times Q$ = {(m, n),(n, m)}.

Answer:

FALSE
If P = {m, n} and Q = { n, m}
Then,
$P \times Q = \left \{ (m,m),(m,n),(n,m),(n,n) \right \}$

Question:4(ii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

Answer:

It is a TRUE statement

$\because$ If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

Question:4 (iii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If A = {1, 2}, B = {3, 4}, then $A \times (B \cap \phi ) = \phi$

Answer:

This statement is TRUE

$\because$ If A = {1, 2}, B = {3, 4}, then

$B\cap\phi=\phi$

There for

$A \times (B \cap \phi ) = \phi$

Question:5 If A = {–1, 1}, find $A \times A \times A$

Answer:

It is given that
A = {–1, 1}
A is an non-empty set
Therefore,
Lets first find $A \times A$
$A \times A = \left \{ -1,1 \right \} \times \left \{ -1,1 \right \} = \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$
Now,
$A\times A \times A=A\times (A \times A) = \left \{ -1,1 \right \} \times \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$ $=\left \{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right \}$

Question:6 If $A \times B$ = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Answer:

It is given that
$A \times B$ = {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty set P and Q is defined as

$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
$A= \left \{ a,b \right \} \ \ \ and \ \ \ B = \left \{ x , y \right \}$

Question:7 (i) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$ .

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$B \cap C = \left \{ 1,2,3,4 \right \} \cap \left \{ 5,6 \right \} = \Phi$
Now,
$A \times ( B \cap C ) = A \times \phi = \phi \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

$A \times B = \left \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4) \right \}$
And
$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
Now,
$(A \times B)\cap (A \times C) =\phi \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
From equation (i) and (ii) it is clear that
$L.H.S. = R.H.S.$
Hence,
$A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$

Question:7 (ii) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times C$ is a subset of $B \times D$

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,

$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
And
$B \times D = \left \{ (1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8) \right \}$
We can clearly observe that all the elements of the set $A \times C$ are the elements of the set $B \times D$
Therefore, $A \times C$ is a subset of $B \times D$

Question:8 Let A = {1, 2} and B = {3, 4}. Write $A \times B$ . How many subsets will $A \times B$ have?
List them.

Answer:

It is given that
A = {1, 2} and B = {3, 4}
Then,
$A \times B = \left \{ (1,3),(1,4),(2,3),(2,4) \right \}$
$\Rightarrow n\left ( A \times B \right ) = 4$
Now, we know that if C is a set with $n(C) = m$
Then,
$n[P(C)]= 2^m$
Therefore,
The set $A \times B$ has $2^4=16$ subsets.

Question:9 Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in $A \times B$ , find A and B, where x, y and z are distinct elements.

Answer:

It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of Cartesian product of two non-empty Set P and Q:
$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we can see that
P = set of all first elements.
And
Q = set of all second elements.
Now,
$\Rightarrow$ (x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}

Question:10 The Cartesian product $A \times A$ has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of $A \times A$

Answer:

It is given that Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
$n(A \times A) = n(A) \times n(A)$
It is given that $n(A \times A) = 9$
Therefore,
$n(A) \times n(A) = 9$
$\Rightarrow n(A) = 3$
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore,
the remaining elements of set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0)

NCERT class 11 maths chapter 2 question answer - Exercise: 2.2

Question:1 Let A = {1, 2, 3,...,14}. Define a relation R from A to A by $R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$ .Write down its domain, codomain and range.

Answer:

It is given that
$A = \left \{ 1, 2, 3, ..., 14 \right \} \ and \ R = \left \{ (x, y) : 3x - y = 0, \ where \ x, y \ \epsilon \ A \right \}$
Now, the relation R from A to A is given as
$R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$
Therefore,
the relation in roaster form is , $,R = \left \{ (1, 3), (2, 6), (3, 9), (4, 12) \right \}$
Now,
We know that Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of $R = \left \{ 1, 2, 3, 4 \right \}$
And
Codomain of R = the whole set A
i.e. Codomain of $R = \left \{ 1, 2, 3, ..., 14 \right \}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of $R = \left \{ 3, 6, 9, 12 \right \}$

Question:2 Define a relation R on the set N of natural numbers by R = { ( x,y ) : y = x +5 , x is a natural number less than $4 ; x , y \epsilon N \left. \right \}$ . Depict this relationship using roster form. Write down the domain and the range.

Answer:

It is given that
$R = \{ ( x,y ) : y = x +5 , x$ is a natural number less than $4 ; x , y \epsilon N \left. \right \}$

As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is, $R = \left \{ (1,6), (2,7), (3,8) \right \}$
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ 1, 2, 3 \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ 6, 7, 8 \right \}$

Therefore, domain and the range are $\left \{ 1,2,3 \right \} \ \ and \ \ \left \{ 6, 7, 8 \right \}$ respectively

Question:3 A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by $R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$ . Write R in roster form.

Answer:

It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the difference which are odd we get,

Therefore,
the relation in roaster form, $R = \left \{ (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6) \right \}$

Question:4 (i) The Fig2.7 shows a relationship between the sets P and Q. Write this relation in set-builder form

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Answer:

It is given in the figure that

P = {5,6,7}, Q = {3,4,5}

Therefore,
the relation in set builder form is ,
$R = \left \{ {(x, y): y = x-2; x \ \epsilon \ P} \right \}$
OR
$R = \left \{ {(x, y): y = x-2; \ for \ x = 5, 6, 7} \right \}$

Question:4 (ii) The Fig2.7 shows a relationship between the sets P and Q. Write this relation roster form. What is it domain and range?

1646202396333

Answer:

From the given figure. we observe that

P = {5,6,7}, Q = {3,4,5}

And the relation in roaster form is , $R = \left \{ {(5,3), (6,4), (7,5)} \right \}$

As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {5, 6, 7} \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ {3, 4, 5} \right \}$

Question:5 (i) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$ Write R in roster form

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)} \right \}$

Question:5 (ii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$ Find the domain of R

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

Question:5 (iii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$ Find the range of R.

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

Question:6 Determine the domain and range of the relation R defined by $R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Answer:

It is given that
$R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)} \right \}$

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {0, 1, 2, 3, 4, 5} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of $R =\left \{ {5, 6, 7, 8, 9, 10} \right \}$

Therefore, the domain and range of the relation R is $\left \{ 0,1,2,3,4,5 \right \} \ \ and \ \ \left \{ {5, 6, 7, 8, 9, 10} \right \}$ respectively

Question:7 Write the relation $R = \left \{ \right.(x, x^3) : x\: \: is\: \: a\: \: prime\: \: number \: \: less\: \: than\: \: 10\: \: \left. \right \}$ in roster form.

Answer:

It is given that
$R = \left \{ \right.(x, x^3) : x\: \: is\: \: a\: \: prime\: \: number \: \: less\: \: than\: \: 10\: \: \left. \right \}$
Now,
As we know the prime number less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is , $R =\left \{ {(2,8), (3,27), (5,125), (7,343)} \right \}$

Question:8 Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer:

It is given that
A = {x, y, z} and B = {1, 2}
Now,
$A \times B = \left \{ {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)} \right \}$
Therefore,
$n(A \times B) = 6$
Then, the number of subsets of the set $(A \times B) = 2^n = 2^6$

Therefore, the number of relations from A to B is $2^6$

Question:9 Let R be the relation on Z defined by $R = \left \{ ( a,b) : a , b \epsilon Z , a-b\: \: is \: \: an \: \: integer \right \}$
Find the domain and range of R.

Answer:

It is given that
$R = \left \{ ( a,b) : a , b \epsilon Z , a-b\: \: is \: \: an \: \: integer \right \}$
Now, as we know that the difference between any two integers is always an integer.
And
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of R = Z

Therefore, the domain and range of R is Z and Z respectively

NCERT class 11 maths chapter 2 question answer - Exercise: 2.3

Question:1 (i) Which of the following relations are functions? Give reasons. If it is a function,determine its domain and range. {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

Answer:

Since, 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {2, 5, 8, 11, 14, 17} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R =\left \{ {1} \right \}$

Therefore, domain and range of R are $\left \{ {2, 5, 8, 11, 14, 17} \right \} \ and \ \left \{ 1 \right \}$ respectively

Question:1 (ii) Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

Answer:

Since, 2, 4, 6, 8, 10,12 and 14 are the elements of domain R having their unique images. Hence, this relation R is a function.

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {2, 4, 6, 8, 10,12, 14} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R =\left \{ {1, 2, 3, 4, 5,6, 7} \right \}$

Therefore, domain and range of R are $\left \{ {2, 4, 6, 8, 10,12, 14} \right \} \ and \ \left \{ {1, 2, 3, 4, 5,6, 7} \right \}$ respectively

Question:1 (iii) Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. {(1,3), (1,5), (2,5)}.

Answer:

Since the same first element 1 corresponds to two different images 3 and 5. Hence, this relation is not a function.

Question:2 (i) Find the domain and range of the following real functions:

$f (x ) = - |x|$

Answer:

Given function is
$f (x ) = - |x|$
_{Now, we know that}

$|x|\left\{\begin{matrix} x &if \ x> 0 \\ -x& if \ x<0 \end{matrix}\right.$
$\Rightarrow f(x)=-|x|\left\{\begin{matrix} -x &if \ x> 0 \\ x& if \ x<0 \end{matrix}\right.$

Now, for a function f(x),
Domain: The values that can be put in the function to obtain real value. For example f(x) = x, now we can put any value in place of x and we will get a real value. Hence, the domain of this function will be Real Numbers.
Range: The values that we obtain of the function after putting the value from domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1. This 1 is a value of Range that we obtained.

Since f(x) is defined for $x \ \epsilon \ R$ , 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. Because will always get a negative number when we put a value from the domain.
Therefore, the range of f is $(-\infty , 0]$

Question:2 (ii) Find the domain and range of the following real functions:

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

Answer:

Given function is
$f ( x ) = \sqrt { 9- x ^2 }$
Now,
Domain: These are the values of x for which f(x) is defined.
for the given f(x) we can say that, f(x) should be real and for that,9 - x ² ≥ 0 [Since a value less than 0 will give an imaginary value]
$\Rightarrow 3^2-x^2= (3-x)(3+x)\geq 0$
$\Rightarrow -3\leq x\leq 3$
Therefore,
The domain of f(x) is $[-3,3]$
Now,
If we put the value of x from $[-3,3]$ we will observe that the value of function $f ( x ) = \sqrt { 9- x ^2 }$ varies from 0 to 3
Therefore,
Range of f(x) is $[0,3]$

Question:3 (i) A function f is defined by f(x) = 2x –5. Write down the values of f (0),

Answer:

Given function is
$f(x) = 2x-5$
Now,
$f(0) = 2(0)-5=0-5 = -5$
Therefore,
Value of f(0) is -5

Question:3 (ii) A function f is defined by f(x) = 2x –5. Write down the values of f (7)

Answer:

Given function is
$f(x) = 2x-5$
Now,
$f(7) = 2(7)-5=14-5 = 9$
Therefore,
Value of f(7) is 9

Question:3 (iii) A function f is defined by f(x) = 2x –5. Write down the values of f (-3)

Answer:

Given function is
$f(x) = 2x-5$
Now,
$f(-3) = 2(-3)-5=-6-5 = -11$
Therefore,
Value of f(-3) is -11

Question:4(i) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t ( C ) = \frac{9 C }{5} + 32$ t (0)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 0 ) = \frac{9 (0) }{5} + 32= 0+ 32 = 32$
Therefore,
Value of t(0) is 32

Question:4(ii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t ( C ) = \frac{9 C }{5} + 32$ t (28)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 28 ) = \frac{9 (28) }{5} + 32= \frac{252}{5}+ 32 = \frac{252+160}{5}= \frac{412}{5}$
Therefore,
Value of t(28) is $\frac{412}{5}$

Question:4(iii) The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t ( C ) = \frac{9 C }{5} + 32$ t (-10)

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( -10 ) = \frac{9 (-10) }{5} + 32= \frac{-90}{5}+ 32 = -18+32= 14$
Therefore,
Value of t(-10) is 14

Question:4(iv) The function ‘t’ which maps temperature in degree Celsius into temperature in
degree Fahrenheit is defined by $t ( C ) = \frac{9 C }{5} + 32$ The value of C, when t(C) = 212.

Answer:

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$212 = \frac{9 (C) }{5} + 32$
$212 \times 5= {9 (C) } + 160$
${9 (C) } =1060-160$
$C = \frac{900}{9} = 100$
Therefore,
When t(C) = 212 , value of C is 100

Question:5 (i) Find the range of each of the following functions.

$f (x) = 2 - 3x, x \epsilon R, x > 0.$

Answer:

Given function is

$f (x) = 2 - 3x, x \epsilon R, x > 0.$
It is given that $x > 0$
Now,
$\Rightarrow 3x > 0$
$\Rightarrow -3x < 0$
Add 2 on both the sides
$\Rightarrow -3x+2 < 0+2$
$\Rightarrow 2-3x < 2$
$\Rightarrow f(x) < 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = 2-3x)$
Therefore,
Range of function $f(x) = 2 -3x$ is $(-\infty,2)$

Question:5 (ii) Find the range of each of the following functions

$f ( x ) = x ^2 +2$ , x is a real number.

Answer:

Given function is

$f ( x ) = x ^2 +2$
It is given that x is a real number
Now,
$\Rightarrow x^2 \geq 0$
Add 2 on both the sides
$\Rightarrow x^2+2 \geq 0+2$
$\Rightarrow f(x) \geq 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = x^2+2)$
Therefore,
Range of function $f ( x ) = x ^2 +2$ is $[2,\infty)$

Question:5 (iii) Find the range of each of the following functions.

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

Answer:

Given function is

$f ( x ) = x$
It is given that x is a real number
Therefore,
Range of function $f ( x ) = x$ is R

Relations and functions class 11 NCERT solutions - Miscellaneous Exercise

Question:1 The relation f is defined by $f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.$ The relation g is defined by $g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.$ Show that f is a function and g is not a function.

Answer:

It is given that
$f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.$
Now,
$f(x) = x^2 \ for \ 0\leq x\leq 3$
And
$f(x) = 3x \ for \ 3\leq x\leq 10$

At x = 3, $f(x) = x^2 = 3^2 = 9$

Also, at x = 3, $f(x) = 3x = 3\times 3 = 9$

We can see that for $0\leq x\leq 10$ , f(x) has unique images.

Therefore, By definition of a function, the given relation is function.

Now,
It is given that
$g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.$
Now,
$g(x) = x^2 \ for \ 0\leq x\leq 2$
And
$g(x) = 3x \ for \ 2\leq x\leq 10$

At x = 2, $g(x) = x^2 = 2^2 = 4$
Also, at x = 2, $g(x) = 3x = 3\times2 = 6$
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus, f(x) does not have unique images
Therefore, by definition of a function, the given relation is not a function

Hence proved

Question:2 If $f (x)= x^2$ find $\frac{f ( 1.1)- f (1)}{(1.1-1)}$

Answer:

Given function is
$f(x)= x^2$
Now,
$\frac{f ( 1.1)- f (1)}{(1.1-1)} = \frac{(1.1)^2-1^2}{(1.1-1)} = \frac{1.21-1}{0.1}= \frac{0.21}{0.1}= 2.1$

Therefore, value of $\frac{f ( 1.1)- f (1)}{(1.1-1)}$ is 2.1

Question:3 Find the domain of the function $f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$

Answer:

Given function is
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
Now, we will simplify it into
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
$= \frac{x^2+2x+1}{x^2-6x-2x+12}$
$= \frac{x^2+2x+1}{x(x-6)-2(x-6)}$
$= \frac{x^2+2x+1}{(x-2)(x-6)}$
Now, we can clearly see that $x \neq 2 , 6$
Therefore, the Domain of f(x) is $(R-\left \{ 2,6 \right \})$

Question:4 Find the domain and the range of the real function f defined by $f (x) = \sqrt{(x-1)}$

Answer:

Given function is
$f (x) = \sqrt{(x-1)}$
We can clearly see that f(x) is only defined for the values of x , $x\geq 1$
Therefore,
The domain of the function $f (x) = \sqrt{(x-1)}$ is $[1,\infty)$
Now, as
$\Rightarrow x\geq 1$
$\Rightarrow x-1\geq 1-1$
$\Rightarrow x-1\geq 0$
take square root on both sides
$\Rightarrow \sqrt{x-1}\geq 0$
$\Rightarrow f(x)\geq 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x)= \sqrt{x-1})$
Therefore,
Range of function $f (x) = \sqrt{(x-1)}$ is $[0,\infty)$

Question:5 Find the domain and the range of the real function f defined by $f (x) = |x-1|$

Answer:

Given function is
$f (x) = |x-1|$
As the given function is defined of all real number
The domain of the function $f (x) = |x-1|$ is R
Now, as we know that the mod function always gives only positive values
Therefore,
Range of function $f (x) = |x-1|$ is all non-negative real numbers i.e. $[0,\infty)$

Question:6 Let $f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \epsilon R \right \}$ R be a function from R into R. Determine the range of f.

Answer:

Given function is
$f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \epsilon R \right \}$
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is Range of the function.
Let's take
$y = \frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow y+yx^2=x^2$
$\Rightarrow y=x^2(1-y)$
$\Rightarrow x^2= \frac{y}{1-y}$
$\Rightarrow x= \pm \sqrt{\frac{y}{1-y}}$
Now, 1 - y should be greater than zero and y should be greater than and equal to zero for x to exist because other than those values the x will be imaginary
Thus, $1 - y > 0 , y < 1 \ and \ y \geq 0$
Therefore,
Range of given function is $[0,1)$

Question:7 Let f, g : R $\rightarrow$ R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g

Answer:

It is given that
$f,g : R \rightarrow R$
$f(x)=x+1 \ \ and \ \ g(x) = 2x - 3$
Now,
$(f+g)x = f(x)+g(x)$
$= (x+1)+(2x-3)$
$= 3x-2$
Therefore,
$(f+g)x= 3x-2$

Now,
$(f-g)x = f(x)-g(x)$
$= (x+1)-(2x-3)$
$= x+1-2x+3$
$= -x+4$
Therefore,
$(f-g)x= -x+4$

Now,
$\left ( \frac{f}{g} \right )x = \frac{f(x)}{g(x)} , g(x)\neq 0$
$=\frac{x+1}{2x-3} \ , x \neq \frac{3}{2}$
Therefore, values of $(f+g)x,(f-g)x \ and \ \left ( \frac{f}{g} \right )x$ are $(3x-2),(-x+4) \ and \ \frac{x+1}{2x-3}$ respectively

Question:8 Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Answer:

It is given that
$f =\left \{ {(1,1), (2,3), (0,-1), (-1, -3)} \right \}$
And
$f(x) = ax+b$
Now,
At x = 1 , $f(x) = 1$
$\Rightarrow f(1)= a(1)+b$
$\Rightarrow a+b = 1 \ \ \ \ \ \ \ \ \ \ -(i)$
Similarly,
At $x = 0$ , $f(x) = -1$
$\Rightarrow f(0) = a(0)+b$
$\Rightarrow b = -1$
Now, put this value of b in equation (i)
we will get,
$a = 2$
Therefore, values of a and b are 2 and -1 respectively

Question:9 (i) Let R be a relation from N to N defined by $R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$ . Are the following true?

$( a,a ) \epsilon R ,$ for all $a \epsilon N$

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,a ) \epsilon R ,$ for all $a \epsilon N$
Now, it can be seen that $2 \ \epsilon \ N$ But, $2 \neq 2^ 2 = 4$
Therefore, this statement is FALSE

Question:9 (ii) Let R be a relation from N to N defined by $R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$ . Are the following true?

$( a,a ) \epsilon R ,$ implies (b,a) $\epsilon$ R

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R ,$ implies $(b,a) \ \epsilon \ R$
Now , it can be seen that $( 2,4 ) \ \epsilon \ R ,$ and $4 = 2^2 = 4$ , But $2 \neq 4^2 =16$
Therefore, $(2,4) \ \notin \ N$
Therefore, given statement is FALSE

Question:9 (iii) Let R be a relation from N to N defined by $R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$ . Are the following true?

(a,b) $\epsilon$ R, (b,c) $\epsilon$ R implies (a,c) $\epsilon$ R.

Answer:

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R , (b,c) \ \epsilon \ R$ implies $(a,c) \ \epsilon \ R$
Now, it can be seen that $(16,4) \ \epsilon \ R , ( 4,2 ) \ \epsilon \ R$ because $16 = 4^2 = 16$ and $4 = 2^2 = 4$ , But $16 \neq 2^2 =4$
Therefore, $(16,2) \ \notin \ N$
Therefore, the given statement is FALSE

Question:10 (i) Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? f is a relation from A to B Justify your answer

Answer:

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$ Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of $A \times B$
Hence f is a relation from A to B
Therefore, given statement is TRUE

Question:10 (ii) Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? f is a function from A to B justify your answer

Answer:

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$
As we can observe that same first element i.e. 2 corresponds to two different images that is 9 and 11.
Hence f is not a function from A to B
Therefore, given statement is FALSE

Question:11 Let f be the subset of $Z \times Z$ defined by $f = {(ab, a + b) : a, b \epsilon Z}$ . Is f a function from Z to Z? Justify your answer.

Answer:

It is given that

Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now, for value 2, 6, -2, -6 $\epsilon \ Z$
$f = { (2 \times 6, 2 + 6), (-2 \times -6, -2 - 6), (2 \times -6, 2 - 6), (-2 \times 6, -2 + 6)}$ $\Rightarrow f = \left \{ (12, 8), (12, -8), (-12, -4), (-12, 4) \right \}$
Now, we can observe that same first element i.e. 12 corresponds to two different images that are 8 and -8.
Thus, f is not a function

Question:12 Let A = {9,10,11,12,13} and let f : A $\rightarrow$ N be defined by f (n) = the highest prime factor of n. Find the range of f.

Answer:

It is given that
A = {9,10,11,12,13}
And
f : A → N be defined by f(n) = the highest prime factor of n.
Now,
Prime factor of 9 = 3

Prime factor of 10 = 2,5

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

f(n) = the highest prime factor of n.

Hence,

f(9) = the highest prime factor of 9 = 3

f(10) = the highest prime factor of 10 = 5

f(11) = the highest prime factor of 11 = 11

f(12) = the highest prime factor of 12 = 3

f(13) = the highest prime factor of 13 = 13

As the range of f is the set of all f(n), where $n \ \epsilon \ A$

Therefore, the range of f is: {3, 5, 11, 13}.

Relations and Functions Class 11 - Topics

2.1 Introduction

2.2 Cartesian Products of Sets

2.3 Relations

2.4 Functions

Summary Of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions

When two things are placed in a particular order, it is called an ordered pair.
The cartesian product of two sets, A and B, is a new set that contains all possible ordered pairs (a,b) where a is from set A and b is from set B.
A relation R between two sets, A and B, is a way to explain how some of the ordered pairs in A × B are related. For example, we can say that x is related to y if (x,y) is in R.
The image of an element x under a relation R is the y value in (x,y) that is in R.
The domain of R is the set of all first elements in the ordered pairs of R, and the range of R is the set of all second elements in the ordered pairs of R.
A function f is a special type of relation where every element in the set A has exactly one image in set B. We write f: A→B, where f(x) = y means that x is related to y through the function f.
The range of a function is the set of all images of the elements in the domain of the function.
A real function has a set of real numbers or a subset of real numbers as both its domain and range.

If you are interested in Relation And Function Class 11 exercises solutions then these are listed below.

relation and function class 11 Exercise 2.1 10 Questions

relation and function class 11 Exercise 2.2 9 Questions

relation and function class 11 Exercise 2.3 5 Questions

relation and function class 11 Miscellaneous Exercise 12 Questions

NCERT Solutions for Class 11 Mathematics - Chapter Wise

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Principle of Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Introduction to Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

Key Features of Relations and Functions Class 11 Solutions

Comprehensive Coverage: The class 11 relations and functions NCERT solutions provide a comprehensive explanation of all the topics covered in the chapter, ensuring a strong understanding of relations and functions.

Step-by-Step Solutions: Each problem and example in the NCERT textbook is solved step by step, making it easier for students to follow the logical progression of concepts.

Clear and Concise Language: The maths chapter 2 class 11 solutions are presented in clear and concise language, making complex mathematical concepts more accessible to students.

NCERT solutions for class 11 - Subject wise

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for Class 11 physics

Benefits of NCERT solutions

NCERT class 11 maths ch 2 question answer will build your basics of functions which will be helpful in 12th board exams also.
All these questions are prepared and explained in a detailed manner so it will be very easy for you to understand the concepts
NCERT relations and functions class 11 solutions are develops yours basic concept which will be helpful in further studies relational algebra, relational calculus, statistics, machine learning, etc.
Tip- Only reading the solutions won't help, you should try to solve on your own. If you are not able to solve, you can take the help of relations and functions class 11 NCERT solutions.

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. How to find which relation is a function in class 11 maths chapter 2 NCERT solutions?

The definition of a function in class 11 chapter 2 states that each element in the domain can be related to only one element in the range. This means that if a student draws a vertical line on a graph, it can intersect the x-axis only once. To determine if a relation is a function, one can use vertical line tests or various formulas.

2. What are important topics of the chapter Relations And Functions?

Basic definitions of relations and functions, Cartesian products of sets, and domain and range of the functions are the important topic of this chapter. you can practice them to get maximum benifit. Practice problems form class 11 maths chapter 2 NCERT solutions pdf after downloading using link given in this article.

3. What is the meaning of relations in Chapter 2 of NCERT Solutions for Class 11 Maths?

Relations are just a bunch of ordered pairs, with one object from each set. Functions can also be considered relations, but they have different concepts. The NCERT Solutions for Class 11 Maths Chapter 2 provide students with a clear definition and analysis of relations according to the CBSE Syllabus 2023. The solutions have many examples to help students solve relation-related problems with ease.

4. Explain the basic steps for the Cartesian product of sets in NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions.

To learn how to solve problems from class 11 Maths NCERT solutions chapter 2 that are related to the Cartesian product of sets, it is essential for students to understand the first exercise of the chapter well. Before each set of problems in the exercise, solved examples are given to help students learn how to solve problems quickly. By practicing problems from the NCERT textbook, students can improve their understanding of the concepts, which is necessary for performing well in exams.

5. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. Practice problems from class 11 maths chapter 2 solutions to get command and in-depth understanding of concepts.

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NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

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NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
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Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

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