NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1 - Relations and Functions

Question1: If $(\frac{x}{3}+1,y-\frac{2}{3})=(\frac{5}{3},\frac{1}{3})$ , find the values of x and y.

Answer:

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

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 $G\times HandH\times G$

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 $ϵ$ P , q $ϵ$ 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,
$P\times Q=\{(m,m),(m,n),(n,m),(n,n)\}$

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 $xϵA$ and $yϵB$

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 $xϵA$ and $yϵB$

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

Answer:

This statement is TRUE

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

$B\cap \varphi =\varphi$

There for

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

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 $A\times A$
$A\times A=\{-1,1\}\times \{-1,1\}=\{(-1,-1),(-1,1),(1,-1),(1,1)\}$
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

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

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,
$B\cap C=\{1,2,3,4\}\cap \{5,6\}=\mathrm{\Phi }$
Now,
$A\times (B\cap C)=A\times \varphi =\varphi -(i)$

$A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$
And
$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$
Now,
$(A\times B)\cap (A\times C)=\varphi -(ii)$
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,

$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$
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,
$A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$
$\Rightarrow n(A\times B)=4$
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:
$P\times Q=\{(p,q):pϵP,qϵQ\}$
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)