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#### Page No 2.12:

#### Question 1:

Given A = {1, 2, 3}, B = {3, 4}, C ={4, 5, 6}, find (A × B) ∩ (B × C).

#### Answer:

Given:

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

Now,

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(B × C) = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

∴ (A × B) ∩ (B × C) = {(3, 4)}

#### Page No 2.12:

#### Question 2:

If *A* = {2, 3}, *B* = {4, 5}, *C* ={5, 6}, find *A* × (*B* ∪ *C*), A × (*B* ∩ *C*), (*A* × *B*) ∪ (*A* × *C*).

#### Answer:

Given:

*A* = {2, 3}, *B* = {4, 5} and *C* ={5, 6}

Also,

(*B* ∪ *C*) = {4, 5, 6}

Thus, we have:

*A* × (*B* ∪ *C*) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,6)}

And,

(*B* ∩ *C*) = {5}

Thus, we have:

A × (*B* ∩ *C*) = {(2, 5), (3, 5)}

Now,

(*A* × *B*) = {(2, 4), (2, 5), (3, 4), (3, 5)}

(*A* × *C*) = {(2, 5), (2, 6), (3, 5), (3, 6)}

∴ (*A* × *B*) ∪ (*A* × *C*) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

#### Page No 2.12:

#### Question 3:

If *A* = {1, 2, 3}, *B* = {4}, *C* = {5}, then verify that:

(i) *A* × (*B* ∪ *C*) = (*A* × *B*) ∪ (*A* × *C*)

(ii) *A* × (*B* ∩ *C*) = (*A* × *B*) ∩ (*A* × *C*)

(iii) *A* × (*B* − *C*) = (*A* × *B*) − (*A* × *C*)

#### Answer:

Given:

*A* = {1, 2, 3}, *B* = {4} and *C* = {5}

(i) *A* × (*B* ∪ *C*) = (*A* × *B*) ∪ (*A* × *C*)

We have:

(*B* ∪ *C*) = {4, 5}

LHS: *A* × (*B* ∪ *C*) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Now,

(*A* × *B*) = {(1, 4), (2, 4), (3, 4)}

And,

(*A* × *C*) = {(1, 5), (2, 5), (3, 5)}

RHS: (*A* × *B*) ∪ (*A* × *C*) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

∴ LHS = RHS

(ii) *A* × (*B* ∩ *C*) = (*A* × *B*) ∩ (*A* × *C*)

We have:

(*B* ∩ *C*) = $\varphi $

LHS: *A* × (*B* ∩ *C*) = $\varphi $

And,

(*A* × *B*) = {(1, 4), (2, 4), (3, 4)}

(*A* × *C*) = {(1, 5), (2, 5), (3, 5)}

RHS: (*A* × *B*) ∩ (*A* × *C*) = $\varphi $

∴ LHS = RHS

(iii) *A* × (*B* − *C*) = (*A* × *B*) − (*A* × *C*)

We have:

(*B* − *C*) = $\varphi $

LHS: *A* × (*B* − *C*) = $\varphi $

Now,

(*A* × *B*) = {(1, 4), (2, 4), (3, 4)}

And,

(*A* × *C*) = {(1, 5), (2, 5), (3, 5)}

RHS: (*A* × *B*) − (*A* × *C*) = $\varphi $

∴ LHS = RHS

#### Page No 2.12:

#### Question 4:

Let *A* = {1, 2}, *B* = {1, 2, 3, 4}, *C* = {5, 6} and *D* = {5, 6, 7, 8}. Verify that:

(i) *A* × *C* ⊂ *B* × *D*

(ii) *A* × (*B* ∩ *C*) = (*A* × *B*) ∩ (*A* × *C*)

#### Answer:

Given:

*A* = {1, 2}, *B* = {1, 2, 3, 4}, *C* = {5, 6} and *D* = {5, 6, 7, 8}

(i) *A* × *C* ⊂ *B* × *D*

LHS: *A* × *C = *{(1, 5), (1, 6), (2, 5), (2, 6)}

RHS:* B* × *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)}

∴ *A* × *C* ⊂ *B* × *D*

(ii) *A ×* (*B* ∩ *C*) = (*A* × *B*) ∩ (*A* × *C*)

We have:

(*B* ∩ *C*) = $\varphi $

LHS: *A ×* (*B* ∩ *C*) = $\varphi $

Now,

(*A* × *B*) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

(*A* × *C*)* = *{(1, 5), (1, 6), (2, 5), (2, 6)}

RHS: * * (*A* × *B*) ∩ (*A* × *C*) = $\varphi $

∴ LHS = RHS

#### Page No 2.12:

#### Question 5:

If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find

(i) A × (B ∩ C)

(ii) (A × B) ∩ (A × C)

(iii) A × (B ∪ C)

(iv) (A × B) ∪ (A × C)

#### Answer:

Given:

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

(i) A × (B ∩ C)

Now,

(B ∩ C) = {4}

∴ A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}

(ii) (A × B) ∩ (A × C)

Now,

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

And,

(A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

∴ (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}

(iii) A × (B ∪ C)

Now,

(B ∪ C) = {3, 4, 5, 6}

∴ A × (B ∪ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv) (A × B) ∪ (A × C)

Now,

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

And,

(A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

∴ (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

#### Page No 2.12:

#### Question 6:

Prove that: (i) (*A* ∪ *B*) × *C* = (*A* × *C*) ∪ (*B* × *C*) (ii) (*A* ∩ B) × C = (*A* × *C*) ∩ (B×*C*)

#### Answer:

(i) (*A* ∪ *B*) × *C* = (*A* × *C*) ∪ (*B* × *C*)

Let* *(*a, b*) be an arbitrary element of (*A* ∪ *B*) × *C.*

Thus, we have:

$(a,b)\in (A\cup B)\times C\phantom{\rule{0ex}{0ex}}\Rightarrow a\in (A\cup B)andb\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (a\in Aora\in B)andb\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (a\in Aandb\in C)or(a\in Bandb\in C)\phantom{\rule{0ex}{0ex}}\Rightarrow (a,b)\in (A\times C)or(a,b)\in (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (a,b)\in (A\times C)\cup (B\times C)\phantom{\rule{0ex}{0ex}}\therefore (A\cup B)\times C\subseteq (A\times C)\cup (B\times C)...\left(\mathrm{i}\right)$

Again, let (*x*,* y*)* *be an arbitrary element of (*A* × *C*) ∪ (*B* × *C*).

Thus, we have:

$(x,y)\in (A\times C)\cup (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x,y)\in (A\times C)or(x,y)\in (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x\in Ay\in C)or(x\in By\in C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x\in Aorx\in B)ory\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (x\in A\cup B)y\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (x,y)\in (A\cup B)\times C\phantom{\rule{0ex}{0ex}}\therefore (A\times C)\cup (B\times C)\subseteq (A\cup B)\times C...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}$

From (i) and (ii), we get:

(*A* ∪ *B*) × *C* = (*A* × *C*) ∪ (*B* × *C*)

(ii) (*A* ∩ B) × C = (*A* × *C*) ∩ (B×*C*)

Let* *(*a*,* b*)* *be an arbitrary element of (*A* ∩ B) × C.

Thus, we have:

$(a,b)\in (A\cap B)\times C\phantom{\rule{0ex}{0ex}}\Rightarrow a\in (A\cap B)b\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (a\in Aa\in B)b\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (a\in Ab\in C)(a\in Bb\in C)\phantom{\rule{0ex}{0ex}}\Rightarrow (a,b)\in (A\times C)(a,b)\in (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (a,b)\in (A\times C)\cap (B\times C)\phantom{\rule{0ex}{0ex}}\therefore (A\cap B)\times C\subseteq (A\times C)\cap (B\times C)...\left(\mathrm{iii}\right)$

Again, let (*x*,* y*) be an arbitrary element of (*A* × *C*) ∩ (*B* × *C*).

Thus, we have:

$(x,y)\in (A\times C)\cap (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x,y)\in (A\times C)(x,y)\in (B\times C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x\in Ay\in C)(x\in By\in C)\phantom{\rule{0ex}{0ex}}\Rightarrow (x\in Ax\in B)y\in C\phantom{\rule{0ex}{0ex}}\Rightarrow x\in (A\cap B)y\in C\phantom{\rule{0ex}{0ex}}\Rightarrow (x,y)\in (A\cap B)\times C\phantom{\rule{0ex}{0ex}}\therefore (A\times C)\cap (B\times C)\subseteq (A\cap B)\times C...\left(\mathrm{iv}\right)$

From (iii) and (iv), we get:

(*A* ∩ *B*) × *C* = (*A* × *C*) ∩ (*B* × *C*)

#### Page No 2.12:

#### Question 7:

If *A* × *B* ⊆ *C* × *D* and *A* × *B* ≠ ϕ, prove that *A* ⊆ *C* and *B* ⊆ *D*.

#### Answer:

$\mathrm{Let}:\phantom{\rule{0ex}{0ex}}(x,y)\in (A\times B)\phantom{\rule{0ex}{0ex}}\therefore x\in A,y\in B\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\because (A\times B)\subseteq (C\times D)\phantom{\rule{0ex}{0ex}}\therefore (x,y)\in (C\times D)\phantom{\rule{0ex}{0ex}}\mathrm{Or},\phantom{\rule{0ex}{0ex}}x\in Candy\in D\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}A\subseteq CB\subseteq D\phantom{\rule{0ex}{0ex}}$

#### Page No 2.20:

#### Question 1:

If A = [1, 2, 3], B = [4, 5, 6], which of the following are relations from A to B? Give reasons in support of your answer.

(i) [(1, 6), (3, 4), (5, 2)]

(ii) [(1, 5), (2, 6), (3, 4), (3, 6)]

(iii) [(4, 2), (4, 3), (5, 1)]

(iv) A × B.

#### Answer:

Given:

A = {1, 2, 3} and B = {4, 5, 6}

Thus, we have:

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(i) {(1, 6), (3, 4), (5, 2)}

Since it is not a subset of A × B, it is not a relation from A to B.

(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}

Since it is a subset of A × B, it is a relation from A to B.

(iii) {(4, 2), (4, 3), (5, 1)}

Since it is not a subset of A × B, it is not a relation from A to B.

(iv) A × B

Since it is a subset (equal to) of A × B, it is a relation from A to B.

#### Page No 2.20:

#### Question 2:

A relation R is defined from a set A = [2, 3, 4, 5] to a set B = [3, 6, 7, 10] as follows:

(*x*, *y*) ∈ R ⇔ *x* is relatively prime to *y*

Express R as a set of ordered pairs and determine its domain and range.

#### Answer:

Given:

(*x*, *y*) ∈ R ⇔ *x* is relatively prime to y.

Here,

2 is co-prime to 3 and 7.

3 is co-prime to 7 and 10.

4 is co-prime to 3 and 7.

5 is co-prime to 3, 6 and 7.

Thus, we get:

R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Domain of R = {2, 3, 4, 5}

Range of R = {3, 7, 6, 10}

#### Page No 2.20:

#### Question 3:

Let A be the set of first five natural numbers and let R be a relation on A defined as follows:

(*x*, *y*) ∈ R ⇔ *x* ≤ *y*

Express R and R^{−1} as sets of ordered pairs. Determine also (i) the domain of R^{−1} (ii) the range of R.

#### Answer:

Given:

A is the set of the first five natural numbers.

∴ A = {1, 2, 3, 4, 5}

The relation is defined as:

(*x*, *y*) ∈ R ⇔ *x* ≤ *y*

Now,

R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}

R^{-1} = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4), (5, 5)}

(i) Domain of R^{-1} = {1, 2, 3, 4, 5}

(ii) Range of R = {1, 2, 3, 4, 5}

#### Page No 2.20:

#### Question 4:

Find the inverse relation R^{−1} in each of the following cases:

(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

(ii) R = {(*x*, *y*), : *x*, *y* ∈ N, *x* + 2*y* = 8}

(iii) R is a relation from {11, 12, 13} to (8, 10, 12] defined by *y* = *x* − 3.

#### Answer:

(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

R^{−1}^{ }= {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}

(ii) R = {(*x*, *y*) : *x*, *y* ∈ N, *x* + 2*y* = 8}

On solving *x* + 2*y* = 8, we get:

*x* = 8 $-$ 2*y*

On putting *y* = 1, we get *x* = 6.

On putting *y* = 2, we get *x* = 4.

On putting *y* = 3, we get *x* = 2.

∴ R = {(6, 1), (4, 2), (2, 3)}

Or,

R^{−1}^{ }= {(1, 6), (2, 4), (3, 2)}

(iii) R is a relation from {11, 12, 13} to {8, 10, 12} defined by *y* = *x* − 3.

*x* belongs to {11, 12, 13} and y belongs to {8, 10, 12}.

Also, 11 − 3 = 8 and 13 − 3 = 10

∴ R = {(11, 8), (13,10)}

Or,

R^{−1}^{ }= {(8, 11), (10,13)}

#### Page No 2.20:

#### Question 5:

Write the following relation as the sets of ordered pairs:

(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] defined by *x* = 2*y*.

(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (*x*, *y*) ∈ R ⇔ *x* is relatively prime to *y*.

(iii) A relation R on the set [0, 1, 2, ....., 10] defined by 2*x* + 3*y* = 12.

(iv) A relation R from a set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16,18] defined by (*x*, *y*) ∈ R ⇔ *x* divides *y*.

#### Answer:

(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] is defined by *x* = 2*y*.

Putting *y* = 1, 2, 3 in *x* = 2*y*, we get:

*x* = 2, 4, 6

∴ R = {(2, 1), (4, 2), (6, 3)}

(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (*x*, *y*) ∈ R ⇔ *x* is relatively prime to *y.*

Here,

2 is relatively prime to 3, 5 and 7.

3 is relatively prime to 2, 4, 5 and 7.

4 is relatively prime to 3, 5 and 7.

5 is relatively prime to 2, 3, 4, 6 and 7.

6 is relatively prime to 5 and 7.

7 is relatively prime to 2, 3, 4, 5 and 6.

∴ R = {(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), (6, 7), (7, 2), (7, 3), (7,4), (7, 5), (7, 6)}

(iii) A relation R on the set [0, 1, 2,..., 10] is defined by 2*x* + 3*y* = 12.

$x=\frac{12-3y}{2}$

Putting *y* = 0, 2, 4, we get:

*x* = 6, 3, 0

∴ R = {(0, 4), (3, 2), (6, 0)}

(iv) A relation R from the set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16, 18] defined by (*x*, *y*) ∈ R ⇔ *x* divides *y*.

Here,

5 divides 10 and 15.

6 divides 12 and 18.

8 divides 16.

∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8,16)}

#### Page No 2.20:

#### Question 6:

Let *R* be a relation in N defined by (*x*, *y*) ∈ R ⇔ *x* + 2*y* =8. Express R and R^{−1} as sets of ordered pairs.

#### Answer:

Let *R* be a relation in N defined by (*x*, *y*) ∈ R ⇔ *x* + 2*y* = 8.

We have:

*x* = 8$-$2*y*

For *y* = 3, 2, 1, we have:

*x* = 2, 4, 6

∴ *R* = {(2, 3), (4, 2), (6, 1)}

And,

*R*^{−1}^{ }= {(3, 2), (2, 4), (1, 6)}

#### Page No 2.21:

#### Question 7:

Let A = (3, 5) and B = (7, 11). Let R = {(*a*, *b*) : *a* ∈ A, *b* ∈ B, *a* − *b* is odd}. Show that R is an empty relation from A into B.

#### Answer:

Given:

A = (3, 5) and B = (7, 11)

Also,

R = {(*a*, *b*) : *a* ∈ A, *b* ∈ B, *a* − *b* is odd}

*a* are the elements of A and *b* are the elements of B.

$\therefore a-b=3-7,3-11,5-7,5-11\phantom{\rule{0ex}{0ex}}\Rightarrow a-b=-4,-8,-2,-6\phantom{\rule{0ex}{0ex}}\mathrm{Here},a-b\mathrm{is}\mathrm{always}\mathrm{an}\mathrm{even}\mathrm{number}.$

So, R is an empty relation from A to B.

Hence proved.

#### Page No 2.21:

#### Question 8:

Let A = [1, 2] and B = [3, 4]. Find the total number of relation from A into B.

#### Answer:

We have:

A = {1, 2} and B = {3, 4}

Now,

$n(A\times B)=n\left(A\right)\times n\left(B\right)=2\times 2=4$

There are 2* ^{n}* relations from A to B, where

*n*is the number of elements in their Cartesian product.

∴ Number of relations from A to B is 2

^{4}= 16.

#### Page No 2.21:

#### Question 9:

Determine the domain and range of the relation R defined by

(i) R = [(*x*, *x* + 5): *x* ∈ (0, 1, 2, 3, 4, 5)]

(ii) R = {(*x*, *x*^{3}) :* x* is a prime number less than 10}

#### Answer:

(i) R = {(*x*, *x* + 5): *x* ∈ (0, 1, 2, 3, 4, 5)}

We have:

R = {(0, 0 + 5), (1, 1 + 5), (2, 2 + 5), (3, 3 + 5), (4, 4 + 5), (5, 5 + 5)}

Or, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴ Domain (R) = {0, 1, 2, 3, 4, 5}

Range (R) = {5, 6, 7, 8, 9, 10}

(ii) R = {(*x*, *x*^{3}) :* x* is a prime number less than 10}

We have:

*x* = 2, 3, 5, 7

*x*^{3}^{ }= 8, 27, 125, 343

Thus, we get:

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Domain (R) = {2, 3, 5, 7}

Range (R) = {8, 27, 125, 343}

#### Page No 2.21:

#### Question 10:

Determine the domain and range of the following relations:

(i) R = {(*a*, *b*) : *a* ∈ N, *a* < 5, *b* = 4}

(ii) $S=\left\{\left(a,b\right):b=\left|a-1\right|,a\in Z\mathrm{and}\left|a\right|\le 3\right\}$

#### Answer:

(i) *R *= {(*a*, *b*) : *a* ∈ N, *a* < 5, *b* = 4}

We have:

*a *= 1, 2, 3, 4

*b *= 4

*R* = {(1, 4), (2, 4), (3, 4), (4, 4)}

Domain (*R*) = {1, 2, 3, 4}

Range (*R*) = {4}

(ii) $S=\left\{\left(a,b\right):b=\left|a-1\right|,a\in Z\mathrm{and}\left|a\right|\le 3\right\}$

Now,

*a *= $-$3, $-$2, $-$1, 0, 1, 2, 3

$b=\left|-3-1\right|=4\phantom{\rule{0ex}{0ex}}b=\left|-2-1\right|=3\phantom{\rule{0ex}{0ex}}b=\left|-1-1\right|=2\phantom{\rule{0ex}{0ex}}b=\left|0-1\right|=1\phantom{\rule{0ex}{0ex}}b=\left|1-1\right|=0\phantom{\rule{0ex}{0ex}}b=\left|2-1\right|=1\phantom{\rule{0ex}{0ex}}b=\left|3-1\right|=2$

Thus, we have:

*b* = 4, 3, 2, 1, 0, 1, 2

Or,

S = {($-$3, 4), ($-$2, 3), ($-$1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}

Domain (*S*) = {$-$3, $-$2, $-$1, 0, 1, 2, 3}

Range (*S*) = {0, 1, 2, 3, 4}

#### Page No 2.21:

#### Question 11:

Let A = {*a*, *b*}. List all relations on A and find their number.

#### Answer:

Any relation in *A* can be written as a set of ordered pairs.

The only ordered pairs that can be included are (*a*, *a*), (*a, b*), (*b, a*) and (*b, b*).

There are four ordered pairs in the set, and each subset is a unique combination of them.

Each unique combination makes different relations in A.

{ } [the empty set]

{(*a, a*)}

{(*a, b*)}

{(*a, a*), (*a, b*)}

{(*b, a*)}

{(*a, a*), (*b, a*)}

{(*a, b*), (*b, a*)}

{(*a, a*), (*a, b*), (*b, a*)}

{(*b, b*)}

{(*a, a*), (*b, b*)}

{(*a, b*), (*b, b*)}

{(*a, a*), (*a, b*), (*b, b*)}

{(*b, a*), (*b, b*)}

{(*a, a*), (*b, a*), (*b, b*)}

{(*a, b*), (*b, a*), (*b, b*)}

{(*a ,a*), (*a, b*), (*b, a*), (*b, b*)}

Number of elements in the Cartesian product of *A* and *A* = $2\times 2=4\phantom{\rule{0ex}{0ex}}$

∴ Number of relations = ${2}^{4}=16$

#### Page No 2.21:

#### Question 12:

Let A = (*x*, *y*, *z*) and B = (*a*, *b*). Find the total number of relations from A into B.

#### Answer:

Given:

*A* = (*x*, *y*, *z*) and *B* = (*a*, *b*)

Now,

Number of elements in the Cartesian product of $A\mathrm{and}B=3\times 2=6\phantom{\rule{0ex}{0ex}}$

Number of relations from *A* to *B* = ${2}^{6}=64$

#### Page No 2.21:

#### Question 13:

Let R be a relation from N to N defined by R = [(*a*, *b*) : *a*, *b* ∈ N and *a* = *b*^{2}].

Are the following statements true?

(i) (*a*, *a*) ∈ R for all *a* ∈ N

(ii) (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R

(iii) (*a*, *b*) ∈ R and (*b*, *c*) ∈ R ⇒ (*a*, *c*) ∈ R

#### Answer:

Given: R = [(*a*, *b*) : *a*, *b* ∈ N and *a* = *b*^{2}]

(i) (*a*, *a*) ∈ R for all *a* ∈ N.

Here, 2 ∈N, but $2\ne {2}^{2}$.

∴ (2,2)$\notin $R

False

(ii) (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R

∵ 4 = 2^{2}

(4, 2) ∈ R, but (2,4)$\notin $R.

False

(iii) (*a*, *b*) ∈ R and (*b*, *c*) ∈ R ⇒ (*a*, *c*) ∈ R

∵ 16 = 4^{2} and 4 = 2^{2}

∴ (16, 4) ∈ R and (4, 2) ∈ R

Here,

(16,2)$\notin $R

False

#### Page No 2.21:

#### Question 14:

Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by

R = {(*x*, *y*) : 3*x* − *y* = 0, where *x*, *y* ∈ A}.

Depict this relationship using an arrow diagram. Write down its domain, co-domain and range.

#### Answer:

A = [1, 2, 3,..., 14]

R = {(*x*, *y*) : 3*x* − *y* = 0, where *x*, *y* ∈ A}

Or,

R = {(*x*, *y*) : 3*x* = *y*, where *x*, *y* ∈ A}

As

$3\times 1=3\phantom{\rule{0ex}{0ex}}3\times 2=6\phantom{\rule{0ex}{0ex}}3\times 3=9\phantom{\rule{0ex}{0ex}}3\times 4=12$

Or,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Domain (R) = {1, 2, 3, 4}

Range (R) = {3, 6, 9, 12}

Co-domain (R) = A

#### Page No 2.21:

#### Question 15:

Define a relation R on the set N of natural number by R = {(*x*, *y*) : *y* = *x* + 5, *x* is a natural number less than 4, *x*, *y* ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R.

#### Answer:

R = {(*x*, *y*) : *y* = *x* + 5, *x* is a natural number less than 4, *x*, *y* ∈ N}

(i) ∵ *x = *1, 2, 3

*∴ y *= 1 + 5, 2 + 5, 3 + 5

y = 6, 7, 8

Thus, we have:

R = {(1, 6), (2, 7), (3, 8)}

(ii)

Now,

Domain (R) = {1, 2, 3}

Range (R) = {6, 7, 8}

#### Page No 2.21:

#### Question 16:

A = [1, 2, 3, 5] and B = [4, 6, 9]. Define a relation R from A to B by R = {(*x*, *y*) : the difference between *x* and *y* is odd, *x* ∈ A, *y* ∈ B}. Write R in Roster form.

#### Answer:

A = [1, 2, 3, 5] and B = [4, 6, 9]

R = {(*x*, *y*) : the difference between *x* and *y* is odd, *x* ∈ A, *y* ∈ B}

For* x* = 1,

4$-$1 = 3 and 6$-$1 = 5

*y* = 4, 6

For *x* = 2,

9$-$2 = 7

*y* = 9

For *x* = 3,

4$-$3 = 1 and 6$-$3 = 3

*y* = 4, 6

For *x* = 5,

5$-$4 =1 and 6$-$5 =1

*y* = 4, 6

Thus, we have:

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

#### Page No 2.21:

#### Question 17:

Write the relation R = {(*x*, *x*^{3}) : *x* is a prime number less than 10} in roster form.

#### Answer:

R = {(*x*, *x*^{3}) : *x* is a prime number less than 10}

*x *= 2, 3, 5, 7

*x*^{3} = 8, 27, 125, 343

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

#### Page No 2.21:

#### Question 18:

Let A = [1, 2, 3, 4, 5, 6]. Let R be a relation on A defined by

{(*a*, *b*) : *a*, *b* ∈ A, *b* is exactly divisible by *a*}

(i) Writer R in roster form

(ii) Find the domain of R

(ii) Find the range of R.

#### Answer:

A = [1, 2, 3, 4, 5, 6]

R = {(*a*, *b*) : *a*, *b* ∈ A, *b* is exactly divisible by *a*}

(i) Here,

2 is divisible by 1 and 2.

3 is divisible by 1 and 3.

4 is divisible by 1 and 4.

5 is divisible by 1 and 5.

6 is divisible by 1, 2, 3 and 6.

∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}

(ii) Domain (R) = {1, 2, 3, 4, 5, 6}

(iii) Range (R) = {1, 2, 3, 4, 5, 6}

#### Page No 2.21:

#### Question 19:

The adjacent figure shows a relationship between the sets P and Q. Write this relation in (i) set builder form (ii) roster form. What is its domain and range?

Figure

#### Answer:

(i) We have:

5$-$2 = 3

6$-$2 = 4

7$-$2 = 5

∴ R = $\left\{\right(x,y):y=x-2,x\in P,y\in Q\}$

(ii) R = {(5, 3), (6, 4), (7, 5)}

(iii) Domain (R) = {5, 6, 7}

Range (R) = {3, 4, 5}

#### Page No 2.21:

#### Question 20:

Let R be the relation on Z defined by

R = {(*a*, *b*) : *a*, *b* ∈ Z, *a* − *b* is an integer}

Find the domain and range of R.

#### Answer:

R = {(*a*, *b*) : *a*, *b* ∈ Z, *a − b* is an integer}

We know:

Difference of any two integers is always an integer.

Thus, for all *a*, *b* ∈ Z, we get *a* − *b *as an integer.

∴ Domain (R) = Z

And,

Range (R) = Z

#### Page No 2.21:

#### Question 21:

For the relation R_{1} defined on R by the rule (*a*, *b*) ∈ R_{1} ⇔ 1 + *ab* > 0.

Prove that: (*a*, *b*) ∈ R_{1} and (*b* , *c*) ∈ R_{1} ⇒ (*a*, *c*) ∈ R_{1} is not true for all *a*, *b*, *c* ∈ R.

#### Answer:

We have:

(*a*, *b*) ∈ R_{1} ⇔ 1 + *ab* > 0

Let:

a = 1, b = $-\frac{1}{2}$ and c = $-$4

Now,

$\left(1,-\frac{1}{2}\right)\in {R}_{1}and\left(-\frac{1}{2},-4\right)\in {R}_{1}\phantom{\rule{0ex}{0ex}}$, as $1+\left(-\frac{1}{2}\right)0and1+\left(-\frac{1}{2}\right)\left(-4\right)0$.

But $1+1\times \left(-4\right)0$.

∴ (1,$-$4) $\notin {R}_{1}$

And,

(*a*, *b*) ∈ R_{1} and (*b* , *c*) ∈ R_{1}

Thus, (*a*, *c*) ∈ R_{1} is not true for all *a*, *b*, *c* ∈ R.

#### Page No 2.21:

#### Question 22:

Let R be a relation on N × N defined by

(*a*, *b*) R (*c*, *d*) ⇔ *a* + *d* = *b* + *c* for all (*a*, *b*), (*c*, *d*) ∈ N × N

Show that:

(i) (*a*, *b*) R (*a*, *b*) for all (*a*, *b*) ∈ N × N

(ii) (*a*, *b*) R (*c*, *d*) ⇒ (*c*, *d*) R (*a*, *b*) for all (*a*, *b*), (*c*, *d*) ∈ N × N

(iii) (*a*, *b*) R (*c*, *d*) and (*c*, *d*) R (*e*, *f*) ⇒ (*a*, *b*) R (*e*, *f*) for all (*a*, *b*), (*c*, *d*), (*e*, *f*) ∈ N × N

#### Answer:

We are given ,

(*a*, *b*) R (*c*, *d*) ⇔ *a* + *d* = *b* + *c* for all (*a*, *b*), (*c*, *d*) ∈ N × N

(i) (*a*, *b*) R (*a*, *b*) for all (*a*, *b*) ∈ N × N

$\because a+b=b+a\mathrm{for}\mathrm{all}a,b\in N\phantom{\rule{0ex}{0ex}}\therefore (a,b)R(a,b)\mathrm{for}\mathrm{all}a,b\in N\phantom{\rule{0ex}{0ex}}$

(ii) (*a*, *b*) R (*c*, *d*) ⇒ (*c*, *d*) R (*a*, *b*) for all (*a*, *b*), (*c*, *d*) ∈ N × N

$(a,b)R(c,d)\Rightarrow a+d=b+c\phantom{\rule{0ex}{0ex}}\Rightarrow c+b=d+a\phantom{\rule{0ex}{0ex}}\Rightarrow (c,d)R(a,b)$

(iii) (*a*, *b*) R (*c*, *d*) and (*c*, *d*) R (*e*, *f*) ⇒ (*a*, *b*) R (*e*, *f*) for all (*a*, *b*), (*c*, *d*), (*e*, *f*) ∈ N × N

$(a,b)R(c,d)\mathrm{and}(c,d)R(e,f)\phantom{\rule{0ex}{0ex}}\Rightarrow a+d=b+c\mathrm{and}c+f=d+e\phantom{\rule{0ex}{0ex}}\Rightarrow a+d+c+f=b+c+d+e\phantom{\rule{0ex}{0ex}}\Rightarrow a+f=b+e\phantom{\rule{0ex}{0ex}}\Rightarrow (a,b)R(e,f)$

#### Page No 2.24:

#### Question 1:

If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A − C) × (B − C).

#### Answer:

Given:

A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}

Now,

(A − C) = {1, 4}

(B − C) = {4}

Thus, we have:

(A − C) × (B − C) = {(1, 4), (4, 4)}

#### Page No 2.24:

#### Question 2:

If *n*(A) = 3, *n*(B) = 4, then write *n*(A × A × B).

#### Answer:

Given:

*n*(*A*) = 3 and *n*(*B*) = 4

Now, we have:

*n*(*A* × *A* × *B*) = $n\left(A\times A\right)\times n\left(B\right)=3\times 3\times 4=36$

#### Page No 2.24:

#### Question 3:

If R is a relation defined on the set Z of integers by the rule (*x*, *y*) ∈ R ⇔ *x*^{2} + *y*^{2} = 9, then write domain of R.

#### Answer:

We need to find (*x*, *y*) ∈ R such that *x*^{2} + *y*^{2} = 9.

$\mathrm{Now},\phantom{\rule{0ex}{0ex}}{\left(3\right)}^{2}+{0}^{2}=9\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(-3\right)}^{2}+{0}^{2}=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

*x* can take values $-$3, 0 and 3.

∴ Domain (R) = {$-$3, 0, 3}

#### Page No 2.25:

#### Question 4:

If R = {(*x*, *y*) : *x*, *y* ∈ Z, *x*^{2} + *y*^{2} ≤ 4} is a relation defined on the set Z of integers, then write domain of R.

#### Answer:

Given:

R = {(*x*, *y*) : *x*, *y* ∈ Z, *x*^{2} + *y*^{2} ≤ 4}

We know:

${\left(-2\right)}^{2}+{0}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(2\right)}^{2}+{0}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(-1\right)}^{2}+{0}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(1\right)}^{2}+{0}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(-1\right)}^{2}+{\left(1\right)}^{2}\le 4\phantom{\rule{0ex}{0ex}}{0}^{2}+{0}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(1\right)}^{2}+{\left(1\right)}^{2}\le 4\phantom{\rule{0ex}{0ex}}{\left(-1\right)}^{2}+{\left(-1\right)}^{2}\le 4$

∴ Domain (R) = {$-$2, $-$1, 0, 1, 2}

#### Page No 2.25:

#### Question 5:

If R is a relation from set A = (11, 12, 13) to set B = (8, 10, 12) defined by *y* = *x* − 3, then write R^{−1}.

#### Answer:

Given:

A = (11, 12, 13) and B = (8, 10, 12)

R is defined by (*y* = *x* − 3) from A to B.

We know:

8 = 11$-$3

10 = 13$-$3

∴ R = {(11, 8), (13, 10)}

Or,

R^{-1} = {(8, 11), (10, 13)}

#### Page No 2.25:

#### Question 6:

Let A = {1, 2, 3} and $R=\left\{\left(a,b\right):\left|{a}^{2}-{b}^{2}\right|\le 5,a,b\in A\right\}$. Then write R as set of ordered pairs.

#### Answer:

Given:

A = {1, 2, 3}

$R=\left\{\left(a,b\right):\left|{a}^{2}-{b}^{2}\right|\le 5,a,b\in A\right\}$

We know that

$\left|{1}^{2}-{1}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{2}^{2}-{2}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{3}^{2}-{3}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{1}^{2}-{2}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{2}^{2}-{1}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{2}^{2}-{3}^{2}\right|\le 5,\phantom{\rule{0ex}{0ex}}\left|{3}^{2}-{2}^{2}\right|\le 5$

Thus, R ={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}

#### Page No 2.25:

#### Question 7:

Let R = [(*x*, *y*) : *x*, *y* ∈ Z, *y* = 2*x* − 4]. If (*a*, -2) and (4, *b*^{2}) ∈ R, then write the values of *a* and *b*.

#### Answer:

R = [(*x*, *y*) : *x*, *y* ∈ Z, *y* = 2*x* − 4]

(*a*, $-$2) and (4, *b*^{2}) ∈ R

$\mathrm{So,}-2\; =\; 2(a)\; -4\phantom{\rule{0ex}{0ex}}\Rightarrow 2=2a\phantom{\rule{0ex}{0ex}}\Rightarrow a=1$

$\mathrm{Also},{b}^{2}=2\left(4\right)-4\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow b=\pm 2\phantom{\rule{0ex}{0ex}}$

Thus, a =1 and b = $\pm 2$

#### Page No 2.25:

#### Question 8:

If R = {(2, 1), (4, 7), (1, −2), ...}, then write the linear relation between the components of the ordered pairs of the relation R.

#### Answer:

Given:

R = {(2, 1), (4, 7), (1, −2), ...}

We can observe that

$1=3\times 2-5\phantom{\rule{0ex}{0ex}}7=3\times 4-5\phantom{\rule{0ex}{0ex}}-2=3\times 1-5\phantom{\rule{0ex}{0ex}}$

Thus, the linear relation between the components of the ordered pairs of the relation R is *y* = 3*x** *$-$ 5.

#### Page No 2.25:

#### Question 9:

If A = [1, 3, 5] and B = [2, 4], list of elements of R, if

R = {(*x*, *y*) : *x*, *y* ∈ A × B and *x* > *y*}

#### Answer:

Given:

A = {1, 3, 5} and B = {2, 4}

R = {(*x*, *y*) : *x*, *y* ∈ A × B and *x* > *y*}

A × B = {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)}

As 3 > 2, 5 > 2 and 5 > 4,

we have R = {(3,2),(5,2),(5,4)}

#### Page No 2.25:

#### Question 10:

If R = [(*x*, *y*) : *x*, *y* ∈ W, 2*x* + *y* = 8], then write the domain and range of R.

#### Answer:

R = {(*x*, *y*) : *x*, *y* ∈ W, 2*x* + *y* = 8}

$\mathrm{As}\mathrm{y}=8-2\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=0,\mathrm{y}=8\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=1,\mathrm{y}=6\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=2,\mathrm{y}=4\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=3,\mathrm{y}=2\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=4,\mathrm{y}=0\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{x}=5,\mathrm{y}0\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{y}0\mathrm{for}\mathrm{all}\mathrm{x}5$

∴ Domain (R) = {0,1,2,3,4} and Range (R) = {0,2,4,6,8}

#### Page No 2.25:

#### Question 11:

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 × B, write A and B.

#### Answer:

Given:

(*x*, 1), (*y*, 2), (*z*, 1) are in A × B

*n*(A) = 3 and *n*(B) = 2

$(x,1)\in A\times B\Rightarrow x\in A,1\in B\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},y\in A,2\in B\phantom{\rule{0ex}{0ex}}\mathrm{and}z\in A,1\in B\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

So, A = {*x,y,z*} and B = {1,2}

#### Page No 2.25:

#### Question 12:

Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(*x*, *y*) : *x* − *y* is odd}. Write R in roster form.

#### Answer:

Given:

A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(*x*, *y*) : *x* − *y* is odd}

Since 1$-$4 = $-$3 is odd, we have:

1$-$6 = $-$5 is odd

2$-$9 = $-$7 is odd

3$-$4 =$-$1 is odd

3$-$6 = $-$3 is odd

5$-$4 = 1 is odd

5$-$6 = $-$1 is odd

∴ R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}

#### Page No 2.25:

#### Question 1:

If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A − B) × (B − C) is

(a) {(1, 2), (1, 5), (2, 5)}

(b) [(1, 4)]

(c) (1, 4)

(d) none of these

#### Answer:

(b) {(1, 4)}

A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}

(A − B) = {1}

(B − C) = {4}

So, (A − B) × (B − C) = {(1,4)}

#### Page No 2.25:

#### Question 2:

If R is a relation on the set A = [1, 2, 3, 4, 5, 6, 7, 8, 9] given by *x* R *y* ⇔ *y* = 3*x*, then R =

(a) [(3, 1), (6, 2), (8, 2), (9, 3)]

(b) [(3, 1), (6, 2), (9, 3)]

(c) [(3, 1), (2, 6), (3, 9)]

(d) none of these

#### Answer:

(d) none of these

A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

*x* R *y* ⇔ *y* = 3*x*

For *x *= 1,* y* = 3

For *x* = 2, *y* = 6

For *x* = 3, *y* = 9

Thus, R = {(1,3),(2,6),(3,9)}

#### Page No 2.25:

#### Question 3:

Let A = [1, 2, 3], B = [1, 3, 5]. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then R^{−1} is

(a) {(3, 3), (3, 1), (5, 2)}

(b) {(1, 3), (2, 5), (3, 3)}

(c) {(1, 3), (5, 2)}

(d) None of these

#### Answer:

(a) {(3, 3), (3, 1), (5, 2)}

A = {1, 2, 3}, B ={1, 3, 5}

R = {(1, 3), (2, 5), (3, 3)}

∴ R^{−1} = {(3,1),(5,2),(3,3)}

#### Page No 2.25:

#### Question 4:

If A = [1, 2, 3], B = [1, 4, 6, 9] and R is a relation from A to B defined by '*x*' is greater than *y*. The range of R is

(a) {1, 4, 6, 9}

(b) (4, 6, 9)

(c) [1]

(d) none of these.

#### Answer:

(c) {1}

A = {1, 2, 3} and B = {1, 4, 6, 9}

R is a relation from A to B defined by: *x* is greater than *y*.

Then R = {(2,1),(3,1)}

∴ Range (R) = {1}

#### Page No 2.25:

#### Question 5:

If R = {(*x*, *y*) : *x*, *y* ∈ Z, *x*^{2} +* **y*^{2} ≤ 4} is a relation on Z, then the domain of R is

(a) [0, 1, 2]

(b) [0, −1, −2]

(c) {−2, −1, 0, 1, 2]

(d) None of these

#### Answer:

(c) {−2, −1, 0, 1, 2}

R = {(*x*, *y*) : *x*, *y* ∈ Z, *x*^{2} + *y*^{2} ≤ 4}

We know that,

Hence, domain (R) = {$-$2,$-$1,0,1,2,}

#### Page No 2.25:

#### Question 6:

A relation R is defined from [2, 3, 4, 5] to [3, 6, 7, 10] by : *x* R *y* ⇔ *x* is relatively prime to *y*. Then, domain of R is

(a) [2, 3, 5]

(b) [3, 5]

(c) [2, 3, 4]

(d) [2, 3, 4, 5]

#### Answer:

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

Given:

From {2, 3, 4, 5} to {3, 6, 7, 10}, *x* R *y* ⇔ *x* is relatively prime to y

2 is relatively prime to 3,7

3 is relatively prime to 7,10

4 is relatively prime to 3,7

5 is relatively prime to 3,6,7

So, domain of R is {2,3,4,5}

#### Page No 2.26:

#### Question 7:

A relation ϕ from C to R is defined by *x* ϕ *y* ⇔ $\left|x\right|$ = *y*. Which one is correct?

(a) (2 + 3*i*) ϕ 13

(b) 3ϕ (−3)

(c) (1 + *i*) ϕ 2

(d) *i* ϕ 1

#### Answer:

(d) *i* ϕ 1

We have $\left|i\right|=\sqrt{{1}^{2}+{0}^{2}}=1$

Thus, *i* ϕ 1 satisfies *x* ϕ *y* ⇔ $\left|x\right|$ = *y*.

#### Page No 2.26:

#### Question 8:

Let R be a relation on N defined by *x* + 2*y* = 8. The domain of R is

(a) [2, 4, 8]

(b) [2, 4, 6, 8]

(c) [2, 4, 6]

(d) [1, 2, 3, 4]

#### Answer:

(c) {2, 4, 6}

*x* + 2*y* = 8

⇒* x* = 8 $-$ 2y

For *y* = 1, *x* = 6

*y* = 2, *x* = 4

*y* = 3, *x* = 2

Then R = {(2,3),(4,2),(6,1)}

∴ Domain of R = {2,4,6}

#### Page No 2.26:

#### Question 9:

R is a relation from [11, 12, 13] to [8, 10, 12] defined by* y* = *x* − 3. Then, R^{−1} is

(a) [(8, 11), (10, 13)]

(b) [(11, 8), (13, 10)]

(c) [(10, 13), (8, 11), (12, 10)]

(d) none of these

#### Answer:

(a) [(8, 11), (10, 13)]

R is a relation from [11, 12, 13] to [8, 10, 12], defined by* y* = *x* − 3

Now, we have:

11$-$ 3 = 8

13 $-$ 3 = 10

So, R = {(13,10),(11,8)}

∴ R^{−1} = {(10,13),(8,11)}

#### Page No 2.26:

#### Question 10:

If the set A has *p* elements, B has *q* elements, then the number of elements in A × B is

(a) *p* + *q*

(b) *p* + *q* + 1

(c) *pq*

(d) *p*^{2}

#### Answer:

(c) *pq
n*(

*A × B*)

*= n*(

*A*)

*× n*(

*B*)

*= p*×

*q*=

*pq*

#### Page No 2.26:

#### Question 11:

Let R be a relation from a set A to a set B, then

(a) R = A ∪ B

(b) R = A ∩ B

(c) R ⊆ A × B

(d) R ⊆ B × A

#### Answer:

(c) R ⊆ A × B

If R is a relation from set A to set B, then R is always a subset of A × B.

#### Page No 2.26:

#### Question 12:

If R is a relation from a finite set A having *m* elements of a finite set B having *n* elements, then the number of relations from A to B is

(a) 2^{mn}

(b) 2^{mn} − 1

(c) 2*mn*

(d) *m ^{n}*

#### Answer:

(a) 2^{mn}

Given: n(A) = m

n(B) = n

∴ $n\left(A\times B\right)=mn$

Then, the number of relations from A to B is 2^{mn}.

#### Page No 2.26:

#### Question 13:

If R is a relation on a finite set having *n* elements, then the number of relations on A is

(a) 2^{n}

(b) ${2}^{{n}^{2}}$

(c) *n*^{2}

(d) *n ^{n}*

#### Answer:

(b) ${2}^{{n}^{2}}$

Given : A finite set with *n* elements

Its Cartesian product with itself will have n^{2}^{ }elements.

∴ Number of relations on A = ${2}^{{n}^{2}}$

#### Page No 2.8:

#### Question 1:

(i) If $\left(\frac{a}{3}+1,b-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)$, find the values of *a* and *b*.

(ii) If (*x* + 1, 1) = (3, *y* − 2), find the values of *x* and *y*.

#### Answer:

(i) $\left(\frac{a}{3}+1,b-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)$

By the definition of equality of ordered pairs, we have:

$\left(\frac{a}{3}+1,b-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)$

$\Rightarrow \left(\frac{a}{3}+1\right)=\frac{5}{3}\mathrm{and}\left(b-\frac{2}{3}\right)=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{3}=\frac{5}{3}-1\mathrm{and}b=\frac{1}{3}+\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{3}=\frac{2}{3}\mathrm{and}b=1\phantom{\rule{0ex}{0ex}}\Rightarrow a=2\mathrm{and}b=1$

(ii) (*x* + 1, 1) = (3, *y* − 2)

By the definition of equality of ordered pairs, we have:

$(x+1)=3\mathrm{and}1=(y-2)\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\mathrm{and}y=3$

#### Page No 2.8:

#### Question 2:

If the ordered pairs (*x*, −1) and (5, *y*) belong to the set {(*a*, *b*) : *b* = 2*a* − 3}, find the values of *x* and *y*.

#### Answer:

The ordered pairs (*x*, −1) and (5, *y*) belong to the set {(*a*, *b*) : *b* = 2*a* − 3}.

Thus, we have:

*x = a *and −1 *= b *such that *b* = 2*a* − 3.

∴ −1 = 2*x** *− 3

or, 2*x* = 3 − 1 = 2

or,* x* = 1

Also,

5 = *a *and *y = b* such that *b* = 2*a* − 3.

∴ *y* = 2(5) − 3

or, *y *= 10 − 3 = 7

Thus, we get:

*x *= 1 and *y = *7

#### Page No 2.8:

#### Question 3:

If *a* ∈ [−1, 2, 3, 4, 5] and *b* ∈ [0, 3, 6], write the set of all ordered pairs (*a*, *b*) such that *a* + *b* = 5.

#### Answer:

Given:

*a* ∈ [−1, 2, 3, 4, 5] and *b* ∈ [0, 3, 6]

We know:

−1 + 6 = 5, 2 + 3 = 5 and 5 + 0 = 5

Thus, possible ordered pairs (*a*, *b*) are {(−1, 6), (2, 3), (5, 0)} such that *a* + *b* = 5.

#### Page No 2.8:

#### Question 4:

If *a* ∈ [2, 4, 6, 9] and *b* ∈ [4, 6, 18, 27], then form the set of all ordered pairs (*a*, *b*) such that a divides *b* and *a* < *b*.

#### Answer:

Given:

*a* ∈ [2, 4, 6, 9] and *b* ∈ [4, 6, 18, 27]

Here,

2 divides 4, 6 and 18 and 2 is less than all of them.

6 divides 18 and 6 and 6 is less than 18.

9 divides 18 and 27 and 9 is less than 18 and 27.

Now,

Set of all ordered pairs (*a*, *b*) such that a divides *b* and *a* < *b* = {(2, 4), (2, 6), (2, 18), (6, 18), (9, 18), (9, 27)}

#### Page No 2.8:

#### Question 5:

If *A* = {1, 2} and *B* = {1, 3}, find *A* × *B* and *B* × *A*.

#### Answer:

Given:

*A* = {1, 2} and *B* = {1, 3}

Now,

*A* × *B =* {(1, 1), (1, 3), (2, 1), (2, 3)}

*B* × *A = *{(1, 1), (1, 2), (3, 1), (3, 2)}

#### Page No 2.8:

#### Question 6:

Let *A* = {1, 2, 3} and *B* = {3, 4}. Find *A* × *B* and show it graphically.

#### Answer:

Given:

*A* = {1, 2, 3} and *B* = {3, 4}

Now,

*A* × *B = *{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

To represent *A* × *B *graphically, follow the given steps:

(a) Draw two mutually perpendicular lines—one horizontal and one vertical.

(b) On the horizontal line, represent the elements of set A; and on the vertical line, represent the elements of set B.

(c) Draw vertical dotted lines through points representing elements of set A on the horizontal line and horizontal lines through points representing elements of set B on the vertical line.

The points of intersection of these lines will represent *A* × *B* graphically.

#### Page No 2.8:

#### Question 7:

If *A* = {1, 2, 3} and *B* = {2, 4}, what are *A* × *B*, *B* × *A*, *A* × *A*, *B* × *B* and (*A* × *B*) ∩ (*B* × *A*)?

#### Answer:

Given :

*A* = {1, 2, 3} and *B* = {2, 4}

Now,

*A* × *B =* {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}

*B* × *A =* {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

*A* × *A = *{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

*B* × *B = *{(2, 2), (2, 4), (4, 2), (4, 4)}

We observe:

(*A* × *B*) ∩ (*B* × *A*) = {(2, 2)}

#### Page No 2.8:

#### Question 8:

If *A* and *B* are two set having 3 elements in common. If *n*(*A*) = 5, *n*(*B*) = 4, find *n*(*A* × *B*) and *n*[(*A* × *B*) ∩ (*B* × *A*)].

#### Answer:

Given:

*n(A)* = 5 and *n(B)* = 4

Thus, we have:

*n(A* × *B)* = 5*(*4*)* = 20

*A* and *B* are two sets having 3 elements in common.

Now,

Let:

*A *=* **(a, a, a, b, c)* and *B *= *(a, a, a, d)*

Thus, we have:

*(A × B)* = {(*a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (b, a), (b, a), (b, a), (b, d), (c, a), (c, a), (c, a), (c, d)*}

*(B × A)* = {*(a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (d, a), (d, a), (d, a), (d, b), (d, c)*}

[*(A × B) ∩ (B × A)*] = {*(a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a)}*

∴ *n*[*(A × B) ∩ (B × A)*] = 9

#### Page No 2.8:

#### Question 9:

Let *A* and *B* be two sets. Show that the sets *A* × *B* and *B* × *A* have elements in common *iff* the sets *A* and *B* have an elements in common.

#### Answer:

Case (i): Let:

*A =* (a, b, c)

*B* = (e, f)

Now, we have:

*A* × *B = *{(*a, e*}), (*a, f*),* *(*b*,*e*), (*b, f*), (*c, e*), (*c, f*)}

*B* × *A* = {(*e, a*), (*e, b*), (*e, c*), (*f, a*), (*f, b*), (*f, c*)}

Thus, they have no elements in common.

Case (ii): Let:

*A =* (a, b, c)

*B = *(*a, f*)

Thus, we have:

*A* × *B = *{(*a, a*}), (*a,f*), (*b, a*), (*b, f*), (*c,a*), (*c, f*)}

*B* × *A* = {(*a, a*), (*a, b*), (*a, c*), (*f, a*), (*f, b*), (*f, c*)}

Here, *A* × *B *and* B* × *A* have two elements in common.

Thus, *A* × *B* and *B* × *A* will have elements in common *iff* sets *A* and *B* have elements in common.

#### Page No 2.8:

#### Question 10:

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* × *B*, find *A* and *B*, where *x*, *y*, *z* are distinct elements.

#### Answer:

A is the set of all first entries in ordered pairs in *A × B* and *B* is the set of all second entries in ordered pairs in *A* × *B*.

Also*,
n*(

*A*) = 3 and

*n*(

*B*) = 2

∴

*A =*{

*x*,

*y*,

*z*} and

*B =*{1, 2}

#### Page No 2.8:

#### Question 11:

Let *A* = {1, 2, 3, 4} and R = {(*a*, *b*) : *a* ∈ *A*, *b* ∈ *A*, *a* divides *b*}. Write *R* explicitly.

#### Answer:

Given:

*A* = {1, 2, 3, 4}

*R* = {(*a*, *b*) : *a* ∈ *A*, *b* ∈ *A*, *a* divides *b*}

We know:

1 divides 1, 2, 3 and 4.

2 divides 2 and 4.

3 divides 3.

4 divides 4.

∴* R *= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}

#### Page No 2.8:

#### Question 12:

If *A* = {−1, 1}, find *A* × *A* × *A*.

#### Answer:

Given:

*A* = {−1, 1}

Thus, we have:

*A* × *A = *{(−1, −1), (−1, 1), (1, −1), (1, 1)}

And,

*A* × *A* × *A = *{(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}

#### Page No 2.8:

#### Question 13:

State whether each of the following statements are true or false. If the statements is false, re-write the given statements correctly:

(i) If P = {*m*, *n*} and Q = {*n*, *m*}, then P × Q = {(*m*, *n*), (*n*, *m*)}

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (*x*, *y*) such that *x* ∈ B and *y* ∈ A.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ ϕ) = ϕ.

#### Answer:

(i) False

Correct statement:

If P = {*m*, *n*} and Q = {*n*, *m*}, then P × Q = {(*m*, *n*), (*m, m*)*, *(*n, n*), (*n, m*)}.

(ii) False

Correct statement:

If A and B are non-empty sets, then A × B is a non-empty set of an ordered pair (*x*, *y*) such that *x* ∈ A and *y* ∈ B.

(iii) True

A = {1, 2} and B = {3, 4}

Now,

(B ∩ ϕ) = ϕ

The Cartesian product of any set and an empty set is an empty set.

∴ A × (B ∩ ϕ) = ϕ

#### Page No 2.8:

#### Question 14:

If *A* = {1, 2}, from the set *A* × *A* × *A*.

#### Answer:

Given:

*A* = {1, 2}

Now,

*A* × *A =* {(1, 1), (1, 2), (2, 1), (2, 2)}

∴ *A* × *A* × *A = *{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

#### Page No 2.8:

#### Question 15:

If A = {1, 2, 4} and B = {1, 2, 3}, represent following sets graphically:

(i) A × B

(ii) B × A

(iii) A × A

(iv) B × B

#### Answer:

Given:

A = {1, 2, 4} and B = {1, 2, 3}

(i) A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

(ii) B × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 4)}

(iii) A × A = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}

(iv) B × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

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