# Class 11 Maths Ncert Solutions Ex 2.1

## Class 11 Maths Ncert Solutions Chapter 2 Ex 2.1

Q.1: If (a3+1,b23)$(\frac{ a }{ 3 } \; + \; 1, \; b \; – \; \frac{ 2 }{ 3 })$ = (53,13)$(\frac{ 5 }{ 3 }, \; \frac{ 1 }{ 3 })$, what is the value of a and b?

Sol:

(a3+1,b23)$(\frac{ a }{ 3 } \; + \; 1, \; b \; – \; \frac{ 2 }{ 3 })$ = (53,13)$(\frac{ 5 }{ 3 }, \; \frac{ 1 }{ 3 })$

As the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, a3+1=53$\frac{ a }{ 3 } \; + \; 1 \; = \; \frac{ 5 }{ 3 }\\$

a3=531$\frac{ a }{ 3 } \; = \; \frac{ 5 }{ 3 } \; – \; 1$ a3=23$\frac{ a }{ 3 } \; = \; \frac{ 2 }{ 3 }$

Therefore,   a = 2

Now, b23=13$b \; – \; \frac{ 2 }{ 3 } \; = \; \frac{ 1 }{ 3 }$

b=13+23$b \; = \; \frac{ 1 }{ 3 } \; + \; \frac{ 2 }{ 3 }$

Therefore,   b = 1

Hence, a = 2 and b = 1

Q:2. If the set X has 4 elements and the set Y = {2, 3, 4, 5}, then find the number of elements in X × Y

Sol:

There are 4 elements in set X and the elements of set X are 2, 3, 4, and 5.

No. of elements in X × Y = (No. of elements in X) × (No. of elements in Y)

= 4 × 4

= 16

Therefore, the no. of elements in (X×Y)$(X \; \times \; Y)$ is 16.

Q.3: If A = {8, 9} and B = {4, 5, 2}, what is the value of A × B and B × A?

Sol:

A = {8, 9}

B = {4, 5, 2}

As we know that Cartesian product ‘P × Q’ of two non-empty sets ‘P’ and ‘Q’ is defined as P × Q = {(p, q): p $\in$ P, q $\in$ Q}

Therefore,

A × B = {(8, 4), (8, 5), (8, 2), (9, 4), (9, 5), (9, 2)}

B × A = {(4, 8), (4, 9), (5, 8), (5, 9), (2, 8), (2, 9)}

Q.4: State whether the given statements are True or False. If the statement is false, write that statement correctly.

(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}

(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x $\in$ P and b $\in$ Q.

(iii). If M = {2, 3}, N = {4, 5}, then M × (N $\cap$Ø ) = Ø.

Sol:

(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}

The given statement is False.

X = {a, b}

Y = {b, a}

Therefore, X × Y = {(a, b), (a, a), (b, b), (b, a)}

(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x $\in$ P and b $\in$ Q.

The given statement is True.

(iii). If M = {2, 3}, N = {4, 5}, then M × (N $\cap$Ø ) = Ø.

The given statement is True.

Q.5: If M = {-2, 2}, then find M × M × M.

Sol:

For any non – empty set ‘M’, M × M × M is defined as:

M × M × M = {(x, y, z): x, y, z $\in$ M}

Since, M = {-2, 2}             [ Given]

Therefore, M × M × M = {(–2, –2, –2), (–2, –2, 2), (–2, 2, –2), (–2, 2, 2), (2, –2, –2), (2, –2, 2), (2, 2, –2), (2, 2, 2)}

Q.6: If X × Y = {(a, m), (a, n), (b, m), (b, n)}. Find X and Y.

Sol:

X × Y = {(a, m), (a, n), (b, m), (b, n)}

As we know, that Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p $\in$ P, q $\in$ Q}

Therefore, ‘X’ is the set of all the first elements and ‘Y’ is the set of all the second elements.

Therefore, X = {a, b} and Y = {m, n}

Q.7: Let P = {2, 3}, Q = {2, 3, 4, 5}, R = {6, 7} and S = {6, 7, 8, 9}. Verify the following:

(i). P×(QR)$P \; \times \; (Q \; \cap \; R)$ = (P×Q)(P×R)$(P \; \times \; Q) \; \cap \; (P \; \times \; R)$

(ii). P × R is a subset of Q × S

(i). P×(QR)$P \; \times \; (Q \; \cap \; R)$ = (P×Q)(P×R)$(P \; \times \; Q) \; \cap \; (P \; \times \; R)$

Taking LHS:

(QR)$(Q \; \cap \; R)$ = {2, 3, 4, 5} $\cap$ {6, 7}

= Ø

P×(QR)$P \; \times \; (Q \; \cap \; R)$ = P × Ø = Ø

Now Taking RHS:

P×Q$P \; \times \; Q$ = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (3, 5)}

P×R$P \; \times \; R$ = {(2, 6), (2, 7), (3, 6), (3, 7)}

(P×Q)(P×R)$(P \; \times \; Q) \; \cap \; (P \; \times \; R)$ = Ø

Therefore, LHS = RHS

P×(QR)$P \; \times \; (Q \; \cap \; R)$ = (P×Q)(P×R)$(P \; \times \; Q) \; \cap \; (P \; \times \; R)$

(ii).   P × R is a subset of Q × S

P × R = {(2, 6), (2, 7), (3, 6), (3, 7)}

Q × S = {(2, 6), (2, 7), (2, 8), (2, 9), (3, 6), (3, 7), (3, 8), (3, 9), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9)}

We can see that all the elements of set P × R are the elements of the set Q × S.

Therefore, P × R is a subset of Q × S.

Q.8: Let P = {2, 3} and Q = {4, 5}. Find P × Q and then find how many subsets will P × Q have? List them.

Sol:

P = {2, 3}

Q = {4, 5}

P × Q = {(2, 4), (2, 5), (3, 4), (3, 5)}

n (P × Q) = 4

As we know, that If ‘A’ is a set with n(A) = m,

Then, n[P(A)] = 2m$2^{ m }$

Therefore,

For the set P × Q = 24$2^{ 4 }$

= 16 subsets

The subsets are as following:

Ø, {(2, 4)}, {(2, 5)}, {(3, 4)}, {(3, 5)}, {(2, 4), (2, 5)}, {(2, 4), (3, 4)}, {(2, 4), (3, 5)}, {(2, 5), (3, 4)}, {(2, 5), (3, 5)}, {(3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4)}, {(2, 4), (2, 5), (3, 5)}, {(2, 4), (3, 4), (3, 5)}, {(2, 5), (3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4), (3, 5)}

Q.9: Let M and N be two sets where n (M) = 3 and n (N) = 2. If (a, 1), (b, 2), (c, 1) are in M × N, find M and N, where a, b and c are different elements.

Sol:

n(M) = 3

n(N) = 2

Since, (a, 1), (b, 2), (c, 1) are in M × N

[M = Set of first elements of the ordered pair elements of M × N]

[N = Set of second elements of the ordered pair elements of M × N]

Therefore, a, b and c are the elements of M

And, 1 and 2 are the elements of N

As n(M) = 3 and n(N) = 2, hence M = {a, b, c} and N = {1, 2}

Q.10: The Cartesian product Z × Z has 9 elements among which are found (-2, 0) and (0, 2). Find the set Z and also the remaining elements of Z × Z.

Sol:

As we know that, If n(M) = p and n(N) = q, then n(M × N) = pq.

Now,

n (Z × Z) = n(Z) × n(Z)

But, it is given that, n(Z × Z) = 9

Therefore, n(Z) × n(Z) = 9

n(Z) = 3

The pairs (-2, 0) and (0, 2) are two of the nine elements of Z × Z

As we know, Z × Z = {(x, x): x $\in$ Z}

Therefore, –2, 0, and 2 are elements of Z

Since, n(Z) = 3, we can see that Z = {–2, 0, 2}

Therefore, the remaining elements of the set Z × Z are (–2, –2), (–2, 2), (0, –2), (0, 0), (2, –2), (2, 0), and (2, 2).