NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions
In every school, each student has a unique ID number, and in the school’s database, that ID number represents the relation to that specific student. Similarly, in Mathematics, the Relations and Functions chapter discusses the connection between elements of two sets and their mapping. Also, this chapter includes domain, co-domain, range and graphing of a function. This is the continuation of the last chapter, which is Chapter 1: Sets. Relations and functions class 11 NCERT solutions are useful for calculus chapters (Differentiation and Integration) as well as algebra and coordinate geometry. In day-to-day life, Relations and Functions can be helpful in Science, Engineering, Business, Medicine, and many other fields.
This Story also Contains
- NCERT Solution for Class 11 Maths Chapter 2 Solutions: Download PDF
- NCERT Solutions for Class 11 Maths Chapter 2: Exercise Questions
- Class 11 Maths NCERT Chapter 2: Extra Question
- Relations and Functions Class 11 Chapter 2: Topics
- Relations And Functions Class 11 Solutions: Important Formulae
- Approach to Solve Questions of Relations And Functions Class 11
- What Extra Should Students Study Beyond NCERT for JEE?
- NCERT Solutions for Class 11 Maths: Chapter Wise
NCERT Solutions for Class 11 Maths are well-detailed and ensure a comprehensive grasp of each concept mentioned in the chapters. Relations and functions class 11 solutions of NCERT are prepared by experienced Subject matter experts from Careers360 following the latest NCERT 2025-26 guidelines. Before analysing the solutions, students should also read the latest NCERT Syllabus to make their study plan accordingly. NCERT relations and functions class 11 questions and answers will build the fundamentals of functions for all types of students, which will be helpful in the 12th board exam as well as higher-level competitive exams.
NCERT Solution for Class 11 Maths Chapter 2 Solutions: Download PDF
Careers360 brings you NCERT Solutions for Class 11 Maths Chapter 2, carefully prepared by subject experts to simplify your studies and help in exams. A downloadable PDF is also available.
NCERT Solutions for Class 11 Maths Chapter 2: Exercise Questions
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Relations and functions class 11 questions and answers: Exercise: 2.1 |
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,
$\frac{x}{3}+1= \frac{5}{3} \ \ \ \text{and} \ \ \ y - \frac{2}{3}= \frac{1}{3}$
$⇒\frac{x}{3}= \frac{5}{3}-1 \ \ \ \text{and} \ \ \ y = \frac{1}{3}+ \frac{2}{3}$
$⇒\frac{x}{3}= \frac{5-3}{3} \ \ \ \text{and} \ \ \ y = \frac{1+2}{3}$
$⇒\frac{x}{3}= \frac{2}{3} \ \ \ \text{and} \ \ \ y = \frac{3}{3}$
$⇒x= 2 \ \ \ \text{and} \ \ \ y = 1$
Therefore, the 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, the 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 $\in$ P , q $\in$ 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)}
Answer:
This statement is FALSE.
If P = {m, n} and Q = {n, m}
Then,
$P \times Q = \left \{ (m,m),(m,n),(n,m),(n,n) \right \}$
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 $x \in A$ and $y \in B$
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 a non-empty set.
Therefore,
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 \}$ $=\{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),$
$(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \}$
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 sets P and Q is defined as:
$P \times Q = \left \{ (p,q) : p \ \in \ P , q \ \in \ Q \right \}$
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 equations (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
$B \times D$
$ =\{ (1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),$
$(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8) \}$
We can 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 the Cartesian product of two non-empty sets P and Q, we get:
$P \times Q = \left \{ (p,q) : p \ \in \ P , q \ \in \ Q \right \}$
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}
Answer:
It is given that the Cartesian product (A × A) has 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 the 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 Page number: 29-30 Total Questions: 9 |
Answer:
It is given that
$A = \left \{ 1, 2, 3, ..., 14 \right \} \ and \ R = \left \{ (x, y) : 3x - y = 0, \ where \ x, y \ \in \ A \right \}$
Now, the relation R from A to A is given as
$R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \in A \right \}$
Therefore,
the relation in roaster form is, $ R = \left \{ (1, 3), (2, 6), (3, 9), (4, 12) \right \}$
Now,
We know that the Domain of R = the set of all first elements of the ordered pairs in the relation
Therefore,
Domain of $R = \left \{ 1, 2, 3, 4 \right \}$
And
Co-domain of R = the whole set A
i.e., Co-domain of $R = \left \{ 1, 2, 3, ..., 14 \right \}$
Now,
Range of R = set of all second elements of the ordered pairs in the relation.
Therefore,
range of $R = \left \{ 3, 6, 9, 12 \right \}$
Answer:
It is given that
$R=(x, y): y=x+5, x$ is a natural number less than $4 ; x , y \in N \left. \right \}$
As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is, $R = \left \{ (1,6), (2,7), (3,8) \right \}$
Domain of R = set of all first elements of the ordered pairs in the relation.
Therefore,
Domain of $R = \left \{ 1, 2, 3 \right \}$
Now,
Range of R = set of all second elements of the ordered 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 $R = \left \{ ( x,y ): \text{the difference between x and y is odd}; x \in A, y \in B \right \}$. Write R in roster form.
Answer:
It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R=\{(x, y):$ the difference between $x$ and $y$ is odd $; x \in A, y \in B\}$
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 odd differences, 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.
Answer:
It is given in the figure that
P = {5,6,7}, Q = {3,4,5}
Therefore,
the relation in set builder form is,
$R = \left \{ {(x, y): y = x-2; x \ \in \ P} \right \}$
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 its domain and range?
Answer:
From the given figure, we observe that
P = {5,6,7}, Q = {3,4,5}
And the relation in roaster form is , $R = \left \{ {(5,3), (6,4), (7,5)} \right \}$
Domain of R = set of all first elements of the ordered pairs in the relation.
Therefore,
Domain of $R = \left \{ {5, 6, 7} \right \}$
Now,
Range of R = set of all second elements of the ordered 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 $\left \{ ( a,b): a,b \in A,\text{b is exactly divisible by a} \right \}$. Write R in roster form.
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And$R=\{(a, b): a, b \in A, b$ is exactly divisible by $a\}$
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 $\left \{ ( a,b): a,b \in A, \text{b is exactly divisible by a} \right \}$. Find the domain of R.
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \in A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
Domain of R = set of all first elements of the ordered 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 $\{(a,b): a, b\in A,b$ is exactly divisible by $a\}$. Find the range of R.
Answer:
It is given that
A = {1, 2, 3, 4, 6}
And
$R=\{(a, b): a, b \in A, b$ is exactly divisible by $a\}$
Now,
As the range of R = set of all second elements of the ordered pairs in the relation.
Therefore,
Range of $R = \left \{ {1, 2, 3, 4, 6} \right \}$
Answer:
It is given that
$R = \left \{ ( x , x +5 ): x \in \left \{ 0,1,2,3,,4,5 \right \} \right \}$
Therefore,
the relation in roaster form is, $R = \left \{ {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)} \right \}$
Now,
Domain of R = set of all first elements of the ordered 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 ordered pairs in the relation.
Range of $R =\left \{ {5, 6, 7, 8, 9, 10} \right \}$
Therefore, the domain and range of the relation R are $\left \{ 0,1,2,3,4,5 \right \} \ \ and \ \ \left \{ {5, 6, 7, 8, 9, 10} \right \}$ respectively
Answer:
It is given that
$R=\left\{\left(x, x^3\right): x\right.$ is a prime number less than 10$\}$
Now,
As we know, the prime numbers 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 $2^6$.
Answer:
It is given that
$R = \left \{ ( a,b) : a , b \in Z , a-b\: \: is \: \: an \: \: integer \right \}$
Now, as we know, the difference between any two integers is always an integer.
And
Domain of R = set of all first elements of the ordered pairs in the relation.
Therefore,
The domain of R = Z
Now,
Range of R = set of all second elements of the ordered pairs in the relation.
Therefore, the range of R = Z
Therefore, the domain and range of R are Z and Z, respectively
| NCERT class 11 maths chapter 2 question answer - Exercise: 2.3 Page number: 38 Total Questions: 5 |
Answer:
Since 2, 5, 8, 11, 14 and 17 are the elements of the domain R having their unique images. Hence, this relation R is a function.
Now,
Domain of R = set of all first elements of the ordered 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 ordered pairs in the relation.
Therefore,
Range of $R =\left \{ {1} \right \}$
Therefore, the domain and range of R are $\left \{ {2, 5, 8, 11, 14, 17} \right \} \ and \ \left \{ 1 \right \}$, respectively.
Answer:
Since 2, 4, 6, 8, 10,12 and 14 are the elements of the 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 ordered 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, the 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 a 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 the Real Numbers.
Range: The values that we obtain from the function after putting the value from the domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1.
This 1 is a value of the Range that we obtained.
Since f(x) is defined for $x \ \in \ R$, 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 we 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, the 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, the 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, the value of f(-3) is -11.
Answer:
Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 0 ) = \frac{9 (0) }{5} + 32= 0+ 32 = 32$
Therefore, the Value of t(0) is 32.
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, the value of t(28) is $\frac{412}{5}$.
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, the value of t(-10) is 14.
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, the value of C is 100.
Question 5 (i): Find the range of each of the following functions.
$f (x) = 2 - 3x, x \in R, x > 0.$
Answer:
Given function is:
$f (x) = 2 - 3x, x \in R, x > 0.$
It is given that $x > 0$
Now,
$\Rightarrow 3x > 0$
$\Rightarrow -3x < 0$
Add 2 on both 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 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, the Range of the function $f ( x ) = x$ is R.
| Relations and functions class 11 NCERT solutions - Miscellaneous Exercise Page number: 40-41 Total Questions: 12 |
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 the definition of a function, the given relation is a 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 the 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, the 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
$\Rightarrow x\geq 1$
$\Rightarrow x-1\geq 1-1$
$\Rightarrow x-1\geq 0$
Taking the square root on both sides, we get,
$\Rightarrow \sqrt{x-1}\geq 0$
$\Rightarrow f(x)\geq 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x)= \sqrt{x-1})$
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 for all real numbers.
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. $[0,\infty)$
Answer:
Given function is
$f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \in R \right \}$
The 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 the 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 the given function is $[0,1)$.
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.
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 get,
$a = 2$
Therefore, the values of a and b are 2 and -1, respectively.
Answer:
It is given that
$R = \left \{ ( a,b): a,b \in N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,a ) \in R ,$ for all $a \in N$
Now, it can be seen that $2 \ \in \ N$ But, $2 \neq 2^ 2 = 4$
Therefore, this statement is FALSE.
Answer:
It is given that
$R = \left \{ ( a,b): a,b \in N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \in \ R ,$ implies $(b,a) \ \in \ R$
Now , it can be seen that $( 2,4 ) \ \in \ R ,$ and $4 = 2^2 = 4$ , But $2 \neq 4^2 =16$
Therefore, $(2,4) \ \notin \ N$
Therefore, the given statement is FALSE.
Answer:
It is given that
$R = \left \{ ( a,b): a,b \in N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \in \ R , (b,c) \ \in \ R$ implies $(a,c) \ \in \ R$
Now, it can be seen that $(16,4) \ \in \ R , ( 4,2 ) \ \in \ R$ 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.
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,
$A \times B$
$= \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9),$
$ (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) \}$
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, the 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,
$A \times B$
$ = \{ (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) \}$
As we can observe, the 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, the given statement is FALSE.
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 $\in \ 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 the same first element, i.e. 12, corresponds to two different images that are 8 and -8.
Thus, f is not a function.
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 $n \ \in \ A$
Therefore, the range of f is: {3, 5, 11, 13}.
Also, read,
- NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Exercise 2.1
- NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Exercise 2.2
- NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Exercise 2.3
- NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Miscellaneous Exercise
Class 11 Maths NCERT Chapter 2: Extra Question
Question:
The number of relations on the set $\mathrm{A}=\{1,2,3\}$ containing at most 6 elements including $(1,2)$, which are reflexive and transitive but not symmetric, is ______
Solution:
$\begin{aligned}
& A=\{1,2,3\} \\
& (1,1),(2,2),(3,3),(1,2) \in R
\end{aligned}$
The remaining elements are
$(2,1),(2,3),(1,3),(3,1),(3,2)$
(1) If the relation contains exactly 4 elements, $=1$ way
(2) If the relation contains exactly 5 elements
It can be $(1,3),(3,2) \Rightarrow 2$ ways
(3) If a relation contains exactly 6 elements
It can be:
$\begin{aligned} & ((2,3),(1,3)),((1,3),(3,2)),((3,1),(3,2)) \\ & \Rightarrow 3 \text { ways. } \\ & \therefore \text { Total = }1+2+3= 6 \text { ways }\end{aligned}$
Hence, the correct answer is 6 ways.
Relations and Functions Class 11 Chapter 2: Topics
The topics discussed in the NCERT Solutions for class 11 chapter 2 Relations and Functions are:
- Introduction
- Cartesian Products of Sets
- Relations
- Functions
- Some functions and their graphs
- Algebra of real functions
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
-
$\{\frac fg \}x=\frac{f(x)}{g(x)},g(x)\neq0$
Approach to Solve Questions of Relations And Functions Class 11
Here are some approaches that students can use to approach the questions related to relations and functions.
- Master concepts of ordered pairs, as it is fundamental in defining relations correctly.
(a, b) ≠ (b, a) unless a = b - For a better visual and structured understanding, represent relations as a subset of Cartesian products A × B.
- Cardinality-based questions are very common in this chapter. Use the following trick for this type of question.
If A has m elements and B has n elements, then the number of relations from A to B is $2^{(m×n)}$. - Master the definitions of One-One, Onto, and Bijective functions to differentiate between them. You can make a chart of the properties of these functions for a better understanding.
- Use graphs and diagrams where necessary and visualise properties, especially in MCQ-based questions.
What Extra Should Students Study Beyond NCERT for JEE?
Here are a few key topics outside the NCERT syllabus that students should study for JEE. Learning these will give you an extra edge during the exam.
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NCERT Solutions for Class 11 Maths: Chapter Wise
We at Careers360 compiled all the NCERT class 11 Maths solutions in one place for easy student reference. The following links will allow you to access them.
Also, read,
NCERT solutions for class 11: Subject-wise
Students can check the following links for more in-depth learning.
| NCERT Solutions for Class 11 Biology |
| NCERT Solutions for Class 11 Chemistry |
| NCERT Solutions for Class 11 Physics |
NCERT Books and NCERT Syllabus
Here is the latest NCERT syllabus, which is very useful for students before strategising their study plan.
Also, links to some reference books which are important for further studies.
Frequently Asked Questions (FAQs)
Q: How do you solve Relation and Function questions in Class 11 Maths?
A:
Problems related to Relations:
- First, understand the problem and identify the sets A and B from which the relation has to be defined.
- List down all possible pairs of values (a,b) where a∈A and b∈B
- "R" defines relation and it is the subset of A × B. Check which pairs of values satisfy the given conditions.
- Also, all the relation properties like Reflexive, Symmetric, and Transitive should be checked.
Problems related to Functions:
- First, verify if the given relation is a function or not. After confirming it, find the domain and range. Domain is the set of input values and range is the set of output values.
- Find out which type of function it is. Injective, surjective, or bijective.
- Then, students can use algebraic manipulation to find composition, inverse and function properties.
Q: How Do You Determine if a Relation is a Function?
A:
We know that every function is a relation, but not every relation is a function.
To check if a relation is a function, for every element x in one domain, there must be exactly one element y in the co-domain.
If any input has more than one output, then it is not considered a function.
Also, if a vertical line cuts a graph at more than one point, then it is also not a function.
Example: f=(1,2),(2,3),(3,4). Here, each input has a unique output, and one input is not related to multiple outputs.
So it is a function.
f≠(1,2),(2,3),(2,4). Here, input 2 has multiple outputs, i.e., 3 and 4. So it is not a function.
Q: What are the Domain and Range of a Function in Class 11 Maths?
A:
For f(x)=x,
Domain = [0,inf) as negative numbers can't be under the square root.
So, Domain is the possible set of values for which a function is defined.
Range = [0,inf)
So, Range is all possible output values that a function can produce.
For f(x)=x2,
Domain = All real numbers
Range = [0,inf) as the square of a number can't be negative.
Q: What is the Difference Between Relations and Functions in Class 11 Maths?
A:
| Relation | Function |
|---|---|
| 1. A relationship between two or more sets of values or a subset of a Cartesian product A × B is called a relation. | 1. A function is defined as for every input of A, there should be a unique output B. |
| 2. One element of set A can be related to more than one element of set B. | 2. One element of set A can map with only one element of set B. |
| 3. It is denoted by "R". | 3. It is denoted by "F". |
| 4. Mapping is not necessarily unique. | 4. It will have unique mapping for each input. |
| 5. Every relation is not a function. | 5. Every function is considered a relation. |
| Example: R = (1, 2), (1, 3), (4, 6) | Example: F = (1, x), (2, y), (3, z) |
Q: What are the Types of Functions Covered in Class 11 Chapter 2?
A:
Various types of functions are covered in Class 11 Chapter 2. These are:
- Injective function(One-One)
- Surjective function (Onto)
- Bijective function (One-One and Onto)
- Constant function
- Identity function
- Modulus function
- Polynomial function
- Signum function
- Greatest integer function
- Rational function
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