Have you ever tried to make a chart linking your classmates to their favorite subjects or your friends to their birthdays? That’s the core idea behind relations in mathematics! The relation is defined as the link between a pair of objects from two sets and the function is a special type of relation where each input is related to exactly one output. These ideas serve as the cornerstone for understanding the relationships between variables in both mathematical and practical contexts. In this exercise of NCERT, you will get questions related to basic definitions of relations and Cartesian products of sets.
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The NCERT Solutions for Chapter 2 Exercise 2.1 will help you grasp these concepts by offering detailed explanations and step-by-step calculations. The terms like domain, codomain and range are introduced in this exercise in order to identify different types of relations. Students can also download the PDF available for the NCERT Solutions for feasible learning.
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)}
Answer:
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 $x \epsilon A$ and $y \epsilon B$
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 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,
$A \times A \times A=A \times(A \times A)=\{-1,1\} \times\{(-1,-1),(-1,1)$,$(1,-1),(1,1)\}$
$=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1)$,$(1,-1,1),(1,1,-1),(1,1,1)\}$
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 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
$B \times D$ = {(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 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.
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}
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)
Also read,
1) Cartesian product of sets
If $A$ and $B$ are two non-empty sets, the Cartesian product is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
$$
A \times B=\{(a, b) \mid a \in A, b \in B\}
$$
If set $A$ has $m$ elements and set $B$ has $n$ elements, then $A \times B$ will have $m \times n$ elements.
2) Definition of a relation
A relation from set $A$ to set $B$ is a subset of the Cartesian product $A \times B$.
Each element in the relation is an ordered pair $(a, b)$ where $a \in A, b \in B$, and $a$ is related to $b$.
So, relation $R \subseteq A \times B$.
Domain, codomain, and range of a relation
Domain- Set of all first elements (inputs) in the relation.
Codomain- The entire set $B$ (i.e., all possible outputs).
Range- Actual set of second elements (outputs) that are related to elements of $A$.
$$
\text { Range } \subseteq \text { Codomain }
$$
3) Types of relations (empty, universal, identity, etc.)
1. Empty (Void) Relation
No element of set $A$ is related to any element of $A$.
$$
R=\emptyset
$$
2. Universal Relation
Every element of $A$ is related to every element of $A$.
$$
R=A \times A
$$
3. Identity Relation
Each element is related only to itself.
$$
R=\{(a, a) \mid a \in A\}
$$
4. Reflexive Relation
Every element is related to itself, i.e., $(a, a) \in R$ for all $a \in A$.
5. Symmetric Relation
If $(a, b) \in R$, then $(b, a) \in R$ must also be true.
6. Transitive Relation
If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ should also be in $R$.
Students can also check the NCERT textbook and exemplar solutions from the links below and can excel in their exam preparations.
NCERT Solutions for Class 11 Maths |
NCERT Solutions for Class 11 Physics |
NCERT Solutions for Class 11 Chemistry |
NCERT Solutions for Class 11 Biology |
NCERT Exemplar Solutions for Class 11 Maths |
NCERT Exemplar Solutions for Class 11 Physics |
NCERT Exemplar Solutions for Class 11 Chemistry |
NCERT Exemplar Solutions for Class 11 Biology |
Frequently Asked Questions (FAQs)
No, two ordered pairs are equal, if and only if the corresponding elements of both the pairs are equal.
x +1 = 3
x = 3 -1 =2
No, cartesian product P x Q and Q x P are not same. The number of elements in each set are the same.
P = { 1,2}
P x P = { (1,1), (1,2), (2,1), (2,2) }
P = { 1,3}
P x P x P= {(1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3)}.
n(A) = 2 and n(B) = 3
n(A × B) = 2x3 = 6
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