ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.2

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.2

Question 1.
If A and B are two sets such that n(A) = 2 and n(B) = 3, find the number of relations from
(i) A to B
(ii) B to A
(iii) A to A.
Solution:
(i) Given n (A) = 2 ;
n (B) = 3
some relation R be a subset of A × B.
∴ n (A × B) = n (A) × n (B)
= 2 × 36
Thus total number of relations from A to B = 2^{n (A × B)}
= 2⁶
= 64

(ii) Thus, total number of relations from B to A = 2^{n (B × A)}
= 2^{n (B) × n (A)}
= 2^{3 × 2}
= 2⁶ = 64

(iii) Thus, total no. of relation from A to A = 2^{n (A × A)}
= 2^{n (A) × n (A)}
= 2^{2 × 2}
= 2⁴ = 16

Question 2.
Let A = {1, 2} and B = {3, 4}. Find
(i) A × B
(ii) number of relations from A to B.
Solution:
Given A = {1, 2)
and B = {3, 4}
(i) ∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

(ii) Total no. of relatiøn from A to B = 2^{n (A) × n (B)}
= 2^{2 × 2}
= 2⁴ = 16

Question 3.
If a relation R = {(- 2, 1), (0, 2), (3, 1), (0, – 1), (4, 2), (5, 1)}, then write its domain and range.
Solu
Given R = {(- 2, 1), (0, 2), (3, 1), (0, – 1), (4, 2), (5, 1)}
Then domain of R is the first element of each ordered pair and range of R is the second element of each ordered pair.
Thus, domain of R = {- 2, 0, 3, 4, 5}
Range (R) = {1, 2, – 1}

Question 4.
If A = {2, 3, 5}, B = {2, 4, 6} and R is the relation from A to B defined by R = {(x, y) : x ∈ A, y ∈ B and x < y},then write R in the roster form.
Solution:
Given A = {2, 3, 5}
and B = {2, 4, 6}
Given R = {(x, y} : x ∈ A, y ∈ B and x < y}
since 2 < 4
∴ (2, 4) ∈ R ;
5 < 6 ⇒ (5, 6) ∈ R
since 2 < 6
∴ (2, 6) ∈ R
3 < 4
∴ (3, 4) ∈ R
3 < 6
∴ (3, 6) ∈ R
Thus R = {(2,4),(2, 6) (3, 4), (3, 6), (5, 6)}.

Question 5.
If A = {1, 3, 5, 7, 8} and B = {2,3, 4, 6, 8, 10} and R be the relation is one less than’ from A to B, then write R in the roster form.
Solution:
Given A = {1, 3, 5, 7, 8}
and B = {2, 3, 4, 6, 8, 10}
given R be the relation is one less than from A to B
since 1 < 2 = 1 + 1 ;
3 < 3 + 1 = 4 ;
5 < 5 + 1 = 6 ;
7 < 7 + 1 = 8
∴ R = {(1, 2), (3, 4), (5, 6), (7, 8)}.

Question 6.
If A = {2, 3, 4}, B = {4, 6, 9, 10} and R = {(x, y) : (x, y) ∈ A × B such that x is a factor of y}, then write R in roster form.
Solution:
Given A = {2, 3, 4}
and B = {4, 6, 9, 10}
and R = {(x, y) : (x, y) ∈ A × B s.t. x is a factor of y}
since 2/4, 2/6, 2/10, 3/6, 3/9, 4/4
Thus R = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.

Short Answer questions (7 to 11) :

Question 7.
If A = (1, 2, 3, ………………, 17} and R is a relation on A defined by R ={(x, y) : 3x – y = 0, x, y ∈ A), then write R ¡n the roster form.
Solution:
Given A = {1, 2, 3, …………….., 17}
and R be a relation defined by
R = {(x, y) : 3x – y = 0, x, y ∈ A}
since 3x – y = 0
⇒ y = 3x
When x = 1
⇒ y = 3 × 1 = 3 ;
When x = 2
⇒ y = 3 × 2 = 6 ;
When x = 3
⇒ y = 3 × 3 = 9 ;
When x = 4
⇒ y = 3 × 4 = 12 ;
When x = 5
⇒ y = 3 × 5 = 15 ;
When x = 6
⇒ y = 3 × 6 = 18 ∉ A
and for all other values of x ∈ A gives no value of y that in A.
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12), (5, 15)}.

Question 8.
Write the following relations in the roster form :
(i) R = {(x, x³) : x is a prime number less than 10}
(ii) R={(x – 2, x²) : x is a prime number less than 10}.
Solution:
(i) Given R = {(x, x³) : x is a prime number less than 10}
Now prime number less than 10 are 2, 3, 5, 7
∴ (2, 8), (3, 27), (5, 125), (7, 343) ∈ R
Thus R = {(2, 8), (3, 27), (5, 125), (7, 343)}.

(ii) Given R = {(x – 2, x²) : x is a prime number less than 10}
Now prime numbers less than 10 are 2, 3, 5 and 7.
R = {(0, 2²), (3 – 2, 3²), (5 – 2, 5²), (7 – 2, 7²)}
⇒ R = {(0, 4), (1, 9), (3, 25), (5, 49)}

Question 9.
Let a relation R = {(1, – 1), (2, 0), (3, 1), (4, 2), (5, 3)}, then
(i) write the domain and the range of R.
(ii) write R in the builder form.
Solution:
Given relation R = {(1, – 1), (2, 0), (3, 1), (4, 2), (5, 3)}
(i) domain of R = {1, 2, 3, 4, 5}
and Range of R = {- 1, 0, 1, 2, 3}

(ii) Clearly x ∈ N, 1 ≤ x ≤ 5
When x = 1, y = – 1
⇒ y = x – 2
When x = 2, y = 0
⇒ y = x – 2
similarly the result y = x – 2 holds for other pairs of (x, y).
Thus R = {(x, y) : x ∈ N, 1 ≤ x ≤ 5, y = x – 2}
be the relation ¡n set builder form.

Question 10.
Let A = {2, 4, 6}, B = {4, 6, 18} and R be the relation ‘is a factor of’ from A to B. Find R as a set of ordered pairs and represent it by an arrow diagram.
Solution:
Given A = {2, 4, 6}
and B = {4, 6, 18}
and R be the relation from A to B defined by ‘is a factor of’.
since, 2 be a factor of 4
i.e. 2/4 ⇒ (2, 4) ∈ R
2/6 ⇒ (2, 6) ∈ R
2/18 ⇒ (2, 18) ∈ R
4/4 ⇒ (4, 4) ∈ R
6/6 ⇒ (6, 6) ∈ R
and 6/18 ⇒ (6, 18) ∈ R
Thus R = {(2, 4), (2, 6), (2, 18), (4, 4), (6, 6), (6, 18)}.

Question 11.
If R₁ = {(x, | x |) : x is a real number}, then find the domain and the range of the relation R₁. (Exemplar)
Solution:
Since x can be any real number
∴ D_R1 = R
Range of R₁ = {|x|, x ∈ R}
Since x ≥ 0 ∀ x ∈ R
∴ Range of R₁ = [0, ∞).

Long answer questions (12 to 21) :

Question 12.
Determine the domain and the range of the relation R defined by R = {(x, x + 5) : x ∈ (0, 1, 2, 3, 4, 5)}.
Solution:
Given R = {(x, x + 5) : x ∈ (0, 1, 2, 3, 4, 5)}
given y = x + 5
When x = 0
⇒ y = 0 + 5 = 5
⇒ (0, 5) ∈ R
When x = 1
⇒ y = 1 + 5 = 6
⇒(1, 6) ∈ R
When x = 2
⇒ y = 2 + 5 = 7
⇒ (2, 7) ∈ R
When x = 3
⇒ y = 3 + 5 = 8
⇒ (3 8) ∈ R
When x = 4
⇒ y = 4 + 5 = 9
⇒ (4, 9) ∈ R
When x = 5
⇒ y = 5 + 5 = 10
⇒ (5, 10) ∈ R
Thus, R = {(0, 5),(1,6),(2,7), (3, 8), (4, 9), (5, 10)}.
Hence domain of R = (0, 1, 2, 3, 4, 5)
and Range of R = {5, 6, 7, 8, 9, 10}.

Question 13.
The adjoining diagram shows a relationship between the sets P and Q. Write this relation in
(i) roster form
(ii) set builder form.
What is its domain and range?

Solution:
(i) ∴ R = {(5, 3), (6, 4), (7, 5)}

(ii) In set builder form,
R = {(x, y) ; 5 ≤ x ≤ 7 ; y = x – 2}
given y = x – 2
When x = 5
⇒ y = 5 – 2 = 3
⇒ (5, 3) ∈ R
When x = 6
⇒ y = 6 – 2 = 4
⇒ (6, 4) ∈ R
When x = 7
⇒ y = 7 – 2 = 5
⇒ (7, 5) ∈ R
Domain of R = {5, 6, 7}
and Range of R= {3, 4, 5}

Question 14.
Let R be the relation on N defined by R = {(a, b) : a ∈ N, b ∈ N and a +3b = 12}, then
(i) List the elements of R
(ii) find the domain of R
(iii) find the range of R.
Solution:
(i) Given R be a relation on N defined by
R = {(a, b) : a ∈ N, b ∈ N and a + 3b = 12}
Given a + 3b = 12
⇒ b = \(\frac{12-a}{3}\)
since a ∈ N
When a = 1
⇒ b = \(\frac{12-1}{3}=\frac{11}{3}\) ∉ N
When a = 2
⇒ b = \(\frac{10}{3}\) ∉ N
When a = 3
⇒ b = \(\frac{9}{3}\) = 3 ∈ N
⇒ (3, 3) ∈ N
When a = 4
⇒ b = \(\frac{12-4}{3}=\frac{8}{3}\) ∉ N
When a = 5
⇒ b = \(\frac{12-5}{3}=\frac{7}{3}\) ∉ N
When a = 6
⇒ b = \(\frac{12-6}{3}=\frac{6}{3}\) = 2 ∈ N
⇒ (6, 2) ∈ N
When a = 7
⇒ b = \(\frac{12-7}{3}=\frac{5}{3}\) ∉ N
When a = 8
⇒ b = \(\frac{12-8}{3}=\frac{4}{3}\) ∉ N
Wben a = 9
⇒ b = \(\frac{12-9}{3}=\frac{3}{3}\) = 1 ∉ N
⇒ (9, 1) ∈ N
For all other values of a, we do not get b ∈ N
Thus, R = {(3, 3), (6, 2), (9, 1)}
(ii) ∴ domain of R = {3, 6, 9}
(iii) Range of R = {3, 2, 1}

Question 15.
If R = {(x, y) : x, y ∈ W, x² + y² = 100}, then find the domain and the range of R. Also write R in roster form.
Solution:
Given R = {(x, y) : x, y ∈ W, x² + y² = 100}
since x, y ∈ W
and x² + y² = 100
⇒ y² = 100 – x²
When x = 0
⇒ y² = 100
⇒ y = ± 10,
since y ∈ W
∴ y = 10
∴ (0, 10) ∈ R
When x = 1
⇒ y² = 100 – 1 = 99
⇒ y = ± \(\sqrt{99}\) ∉ W
When x = 2
⇒ y² = 100 – 4 = 96
⇒ y = ± \(\sqrt{96}\) ∉ W
When x = 3
⇒ y² = 100 – 9 = 91
⇒ y = ± \(\sqrt{91}\) ∉ W
When x = 4
⇒ y² = 100 – 16 = 84
⇒ y = ± \(\sqrt{84}\) ∉ W
When x = 5
⇒ y² = 100 – 25 = 75
⇒ y = ± \(\sqrt{75}\) ∉ W
When x = 6
⇒ y² = 100 – 36 = 64
⇒ y = ± 8
Since y ∈ W
∴ y = 8 ∈ W
⇒ (6, 8) ∈ R
When x = 7
⇒ y² = 100 – 49 = 51
⇒ y = ± \(\sqrt{51}\) ∉ W
When x = 8
⇒ y² = 100 – 64 = 36
⇒ y = ± 6,
since y ∈ W
∴ y = 6
⇒ (8, 6) ∈ R
When x = 9
⇒ y² = 100 – 81 = 19
y = ± \(\sqrt{19}\) ∉ W
When x = 10
⇒ y² = 100 – 100 = 0
⇒ y = 0 ∈ W
⇒ (10, 0) ∈ R
For all other values of x, we donot get y ∈ W
Thus, R = {(0, 10), (6, 8), (8, 6), (10, 0)}
∴ domain of R = {0, 6, 8, 10}
and Range of R = {10 8, 6, 0}.

Question 16.
If R is the relation on N defined by R = {(x, y) : y = x + \(\frac{12}{x}\), x, y ∈ N), then find
(i) R in roster form
(ii) domain of R
(iii) range of R
Solu:
(i) Given relation R is defined by
R = {(x, y) : y = x + \(\frac{12}{x}\), x, y ∈ N}
When x = 1
⇒ y = 1 + \(\frac{12}{1}\)
= 13 ∈ N
⇒ (1, 13) ∈ R
When x = 2
⇒ y = 2 + \(\frac{12}{2}\) = 8
⇒ 8 ∈ N
⇒ (2, 8) ∈ R
When x = 3
⇒ y = 3 + 4 = 7 ∈ N
⇒ (3, 7) ∈ R
When x = 4
⇒ y = 4 + 3 = 7 ∈ N
⇒ (4, 7) ∈ R
When x = 6
⇒ y = 6 + 2 = 8 ∈ N
⇒ (6, 8) ∈ R
When x = 12
⇒ y = 12 + 1 = 13 ∈ N
⇒ (12, 13) ∈ R
While at other values of x, we donot get y ∈ N.
Hence R = {(1, 13), (2, 8), (3, 7), (4, 7), (6, 8), (12, 13)}
(ii) domain of R = {1, 2, 3, 4, 6, 12}
range of R = {13, 8, 7}

Question 17.
Write down the domain and the range of the relation (x, y) : x = 3y and x and y are natural numbers less than lo.
Solution:
Given relation R = {(x, y) : x = 3y ; 1 ≤ x, y < 10, x, y ∈ N}
Here 1 ≤ x, y < 10, x, y ∈ N} since x = 3y
⇒ y = \(\frac{x}{3}\) When x = 3
⇒ y = 1 ∈ N ⇒ (3, 1) ∈ R When x = 6
⇒ y = 2 ∈ N ⇒ (6, 2) ∈ R When x = 9
⇒ y = 3 ∈ N ⇒ (9, 3) ∈ R
While at all other values of x, we donot get y ∈ N
Thus, R = {(3, 1), (6, 2), (9, 3)}
∴ domain of R = {3, 6, 9}
and Range of R = {1, 2, 3}.

Question 18.
Let A = {- 2, – 1, 0, 1, 2}, list the ordered pairs satisfying each of the following relations on A :
(i) ‘is greater than’
(ii) ‘is the square of’
(iii) ‘is the negative of’.
Solution:
Given A = {- 2. – 1, 0, 1, 2}
and given relation be ‘is greater than’ (1)
since – 1 > – 2
⇒ (- 1, – 2) ∈ R ;
1 > – 2
⇒ (1 – 2) ∈ R
o > – 2
⇒ (0, – 2) ∈ R ;
1 > – 1
⇒ (1, – 1) ∈ R
0 > – 1
⇒ (0, – 1) ∈R;
1 > 0
⇒ (1, 0) ∈ R
2 > – 2
⇒ (2, – 2) ∈ R ;
2 > – 1
⇒ (2, – 1) ∈ R
2 > 0
⇒ (2, 0) ∈ R ;
2 > 1
⇒ (2, 1) ∈ R
∴ R = {(- 1, – 2), (1, – 2), (0, – 2), (1, – 1), (0, – 1), (1. 0), (2, -2), (2, – 1), (2, 0), (2, 1)}.

(ii) Here relation be ‘is the square of’
Since 0 = 0²
⇒ (0, 0) ∈ R
1 = (- 1)²
⇒ (1, – 1) ∈ R
Also, 1 = 1²
⇒ (1, 1) ∈ R
∴ R = {(0, 0), (1, – 1), (1, 1)}

(iii) Here, relation be ‘is the negative of’
Here, – 1 be the negative of 1
∴ (- 1, 1) ∈ R
1 be the negative of – 1
∴ (1, – 1) ∈ R
0 be the negative of 0
∴ (0, 0) ∈ R
2 be the negative of – 2
∴ (2, – 2) ∈ R
and – 2 be the negative of 2
∴ (- 2, 2) ∈ R
Hence, R = {(- 1, 1), (1, – 1), (0, 0), (2, – 2), (- 2, 2)}.

Question 19.
If A = {1, 3, 5, 6) and B = {3, 4, 5}, write the relation R as a set of ordered pairs if
(i) R = {(x, y) : (x, y) ∈ A × B : x + y is even)
(ii) R = {(x, y) : (x, y) ∈ A × B : x y is odd}.
Solution:
Given A = { 1, 3, 5, 6}
and B = {3, 4, 5}
(i) and Relation R = {(x, y) : (x, y) ∈ A × B : x + y is even}
Here A × B = {(1, 3), (1, 4),(1, 5), (3, 3), (3, 4), (3, 5), (5, 3), (5, 4), (5, 5), (6, 3), (6, 4), (6, 5)}
since 1 + 3 = 4 (even)
⇒ (1, 3) ∈ R
since 1 + 5 = 6 (even)
⇒ (1, 5) ∈ R
since 3 + 3 = 6 (even)
⇒ (3, 3) ∈ R
since 3 + 5 = 8 (even)
⇒ (3, 5) ∈ R
since 5 + 3 = 8 (even)
⇒ (5, 3) ∈ R
since 5 + 5 = 10 (even)
⇒ (5, 5) ∈ R
since 6 + 4 = 10 (even)
⇒ (6, 4) ∈ R
R = {(1, 3), (1, 5), (3, 3), (5, 3), (3, 5), (5, 5), (6, 4)}

(ii) Given R = {(x, y) : (x, y) ∈ A × B : xy is odd}
since 1 . 3 = 3 (odd)
⇒ (1, 3) ∈ R
1 . 5 = 5 (odd)
⇒ (1, 5) ∈ R
3 . 3 = 9 (odd)
⇒ (3, 3) ∈ R
3 . 5 = 15 (odd)
⇒ (3, 5) ∈ R
5 . 3 = 15 (odd)
⇒ (5, 3) ∈ R
5 . 5 = 25 (odd)
⇒ (5, 5) ∈ R
Thus, R = {(1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5)}

Question 20.
Let R = {(x, y) : x, y ∈ Z, y = 2x – 4}. If (a, – 2) and (4, b²) belong to R, find the values of a and b.
Solution:
Given R = {(x, y) : x, y ∈ Z, y = 2x – 4}
since (a, – 2) ∈ R
⇒ – 2 = 2 × a – 4
⇒ – 2 = 2a – 4
⇒ 2a = 2
⇒ a = 1
and (4, b²) ∈ R
⇒ b² = 2 × 4 – 4 = 4
⇒ b = ± 2.

Question 21.
Find the linear relation between the components of the ordered pairs of the relation R where
(i) R = {(- 1,- 1), (0, 2), (1, 5), ………………}.
(ii) R = {(0, 2), (- 1, 5), (2, – 4), ……………..}.
Solution:
(i) Given R = ((- 1,- 1), (0, 2), (1, 5), ……………..}
Let the linear relation be
y = ax + b ………….(1)
since (- 1, – 1) ∈ R
∴ When x = – 1, y = – 1
∴ from (1) ;
– 1 = – a + b ………………..(2)
Also (0, 2) ∈ R
i.e. When x = 0, y = 2
∴ from (1) ;
2 = a × 0 + b
⇒ b = 2
∴ from (2) ;
– 1 = – a + 2
⇒ – a = – 3
⇒ a = 3
Hence the required relation be, y = 3x + 2
Clearly (1, 5) ∈ R.
Since 5 = 3 ∈ 1 + 2

(ii) Given R = {(0, 2), (- 1, 5), (2, – 4), …………….}
Let the required linear relation be
y = ax + b ………………….(1)
since (0, 2) ∈ R
i.e. when x = 0, y = 2
∴ from (1) ;
2 = 0 + b
⇒ b = 2
Now (- 1, 5) ∈ R
i.e. When x = – 1, y = 5
∴ from (1) ;
5 = – a + b
⇒ a = b – 5
= 2 – 5 = – 3
Hence the required relation be
y= – 3x + 2 …………………(2)
Also, (2, – 4) satisfies eqn. (2).