Mahesh

Referring to the AP Inter 1st Year Maths Study Material Chapter 2 Relations and Functions Exercise 2b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Relations and Functions Solutions Exercise 2b

I.

Question 1.
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 e A, y ∈ B}. Write R in roster form.
Solution:
Given A = {1, 2, 3, 5}, B= {4, 6, 9}
R = {(x, y): the difference between x & y is odd; x ∈ A, y ∈ B}.
Roster form of R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Question 2.
Determine the domain and 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}}
The roster form of R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}; Range of R = {5, 6, 7, 8, 9, 10}

Question 3.
Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.
Solution:
Given R = {(x, x³) ; x is a’prime number less than 10}
The prime numbers less than 10 are 2, 3, 5 & 7
∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 4.
Let A = (x, y, z) and B = (1, 2). Find the number of relations from A to B.
Solution:
Given A = {x, y, z} & B = {1, 2}
A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
n (A × B) = 6, n(B) = 2, the number of subsets of A × B is 26.
The number of relations from A to B is 26.

Question 5.
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.
Solution:
Given R = {(a, b): a, b ∈ Z, a – b is an integer}
∴ Domain of R = Z, Range of R = Z

II.

Question 1.
Let A = {1, 2, 3,…, 14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A). Write down its domain, codomain and range,
Solution:
A relation R from A to A is given by R = {(x, y) ; 3x – y = 0, where x, y ∈ A}
The roster form is given by R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R = {1, 2, 3, 4}
The whole set A is the co-domain of the relation R.
∴ Co-domain of R = {1, 2, 3, 14}
The range of R = {3, 6, 9, 12}.

Question 2.
Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution:
Given R = {(x, y): y = x + 5, x < 4;x, y ∈ N)
⇒ x = 1, 2, 3 R = {(1, 6), (2, 7), (3, 8)}
The domain of R = {1, 2, 3},
The range of R = {6, 7, 8}

Question 3.
The Fig. shows a relationship between the sets P and Q. Write this relation,
i) in set-builder form ii) roster form. What is its domain and range ?
Solution:
Given P = {5, 6, 7} Q = {3, 4, 5}
i) The set builder form
R = {(x, y): y = x – 2 for x = 5, 6, 7}

ii) The roster form R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}

Question 4.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A is defined by {(a, b): a , b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Solution:
Given A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b ∈ A, b is exactly divisible by a}
i) The roster form R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2) (2, 4) (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
ii) Domain of R = {1, 2, 3, 4, 6}
iii) Range of R = {1, 2, 3, 4, 6}

Referring to the AP Inter 1st Year Maths Study Material Chapter 2 Relations and Functions Exercise 2a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Relations and Functions Solutions Exercise 2a

I.

Question 1.
If (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = \(\left(\frac{5}{3}, \frac{1}{3}\right)\), find the values of x and y.
Solution:
Given (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = \(\left(\frac{5}{3}, \frac{1}{3}\right)\)

∴ x = 2, y = 1.

Question 2.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B),
Solution:
Given number of elements in A = 3 ⇒ n(A) = 3 ; Given B = {3, 4, 5} ⇒ n(B) = 3
Number of elements in A × B = (Number of elements in A) × (Number of elements in B)
n(A × B) = n(A) × n(B) = 3 × 3 = 9
Number of elements in (A × B) =9

Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution:
Given G = {7, 8}; H = {5, 4, 2}
G × H = {7, 8} × {5, 4, 2} = {(7, 5), (7, 4), (7, 2), (8, 5),(8, 4), (8, 2)}
H × G = {5, 4, 2} × {7,8} = {(5, 7), (5, 8), (4, 7), (4, 8), (2,7), (2, 8)}

Question 4.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
i) If P = (m, n} and 9 = In, ni}, then P × g = j(m, n),(n, m)|.
ii) If A and B are two non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.
Solution:
i) False; P × Q = {m, n} ×{n,m} = {(m, n), (m, m), (n, n), (n, m)}
ii) True; Because A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
iii) True; Because B ∩ Φ = Φ ⇒ A × (B ∩ Φ ) = A × Φ = Φ.

Question 5.
If A = (-1, 1), find A × A × A.
Solution:
Given A = {-1, 1}
A × A = {-1, 1} × {-1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
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, -1), (1,-1,1) (1,1-1), (1, 1, 1)}

Question 6.
If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
Solution:
Given A x B = {(a, x), (a, y), (b, x), (b, y)}
A = set of first elements of A × B = {a, b};
B = set of second elements of A × B = {x, y}.

Question 7.
Let A = {1, 2) and B = {3, 4). Write A × B. How many subsets will A × B have? List them.
Solution:
Given A = {(1, 2), B = {3, 4} ; A × B = {(1, 3), (1, 4), (2, 3), (2, 4)} ⇒ n (A × B) = 4
The number of subsets of A × B = 24 = 16
Subsets of A × B = { }, {(1, 3)}, {(1,4)}, {(2, 3)}, {(2, 4)}
{(1,3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}
{(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}
{(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2,3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}.

Question 8.
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 and z are distinct elements.
Solution:
Given A & B are 2 sets such that n(A) = 3 & n(B) = 2
Given (x, 1), (y, 2), (z, 1) lies in A × B
⇒ A = {x, y, z}, B = {1, 2}

Question 9.
The Cartesian product A × A has 9 elements among which are found (-1, 0) and (0,1). Find the set A and the remaining elements of A × A.
Solution:
Given (-1, 0) and (0, 1) are in the cartesian product of A × A having 9 elements.
∴ n(A × A) = 9 ⇒ n(A) × n(A) = 9 ⇒ n(A) = 3.
∴ A = {-1, 0, 1}
A × A = {-1, 0, 1} × {-1, 0, 1} = {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}

II.

Question 1.
Let A = (1, 2), B = (1, 2, 3, 4), C = (5, 6| and D = (5, 6, 7, 8). Verify that
(i) A × (B ∩ C) – (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.
Solution:
Given A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} 8s D = {5, 6, 7, 8}
i) B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = { }
A × (B ∩ C)= A × Φ = Φ
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)}
(A × B) ∩ (A × C) = Φ
∴ A × (B ∩ C) = Φ
(A × B) ∩ (A ∩ C) = Φ
∴ A × (B ∩ C) = (A × B) ∩ (A × C)

ii) A × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5) (2, 6)}
B × D = {1, 2, 3, 4} × {5, 6, 7, 8} = {(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 a is subset of B × D

Practice AP Inter 1st Year Maths Study Material Chapter 1 Sets MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Sets MCQ

Question 1.
List of elements of the set {x : x is an integer, x³ ≤ 50}
1) {2, 3, 4}
2) {1, 3, 4}
3) {1, 2, 3}
4) {1, 2, 4}
Solution:
We look integer x such that x³ ≤ 50
Cube of integers : 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64 (exceeds 50).
So the integers are : x = 1, 2, 3.

Question 2.
Set-builder form of {1, 2, 3, 4} is
1) {x : x is a positive integer, x² < 30}.
2) {x : x is a positive integer, x² + 1 < 20}.
3) {x : x is a positive integer, x² – 1 < 15}.
4) {x : x is a positive integer, 2x < 20}.
Try each :
1) x² < 30 : includes more than {1, 2, 3, 4}.
2) x² + 1 < 20 : x < 4 works, lets check : 1 → 2, 2 → 5, 3 → 10, 4 → 17.
3) x² – 1 < 15: 1 → 0, 2 3, 3 → 8, 4 → 15 but it includes more value like 5.
4) 2x < 20 : includes more values.

Question 3.
Which of the following statement is true ?
1) {a, b} ⊂ {x : x is a vowel in the English alphabet}
2) {1, 2, 3, 4} ⊂ {1, 2, 3}
3) {5} ∈ {4, 5, 6}
4) {2, 3) ∈ {1, {2, 3}, 4, 5}
Solution:
1) X “b” is not a vowel
2) X false
3) X because {5} ∈ {4, 5, 6} but {5} is not an element
4) ✓ true.

Question 4.
The set builder form of interval [-4, 5) is
1) {x : x is an integer -4 < x < 5}
2) {x : x is an integer -4 < x < 5}
3) {x : x is an integer -4 < x < 5}
4) {x : x is an integer -4 < x < 5}
Solution:
Set builder form of interval [- 4, 5)
We need : integers x such that – 4 < x < 5.

Question 5.
A = (1, 2, 3, 4, 5), B = (4, 5, 6, 7, 8), then (A – B) ∪ (B – A)
1) {1, 2, 3, 4, 5, 6, 7, 8}
2) {1, 2, 3, 6, 7, 8}
3) {1, 2, 4, 5, 6}
4) {2, 4, 6, 8}
Solution:
A – B = {1, 2, 3}; B – A = {6, 7, 8}; (A – B) ∪ (B – A) = {1, 2, 3, 6, 7, 8}.

Question 6.
If A = (1, 2, 3, 4}, B = {2, 3, 4}, C = {3, 4, 5, 6} then A ∩ (B ∪ C)
1) {2, 3, 4}
2) {1, 2, 3, 4}
3) {3, 4, 5, 6}
4) {1, 6}
Solution:
B ∪ C = {2, 3, 4, 5, 6}; A ∩ (B ∪ C) = {2, 3, 4}.

Question 7.
Which of the following is false ?
1) (A’)’ = A
2) A ∪ A’ = Φ
3) (A ∪ B)’ = A’ ∩ B’
4) Φ = U (U is universal set)
Solution:
1) (A’)’ = A – True (law of double complement)
2) A ∪ A’ = Φ – False, should be universal set.
3) (A ∪ B)’ = A’ ∩ B’- True (De Morgan’s law)
4) f = U – True

Question 8.
If U = (1, 2, 3, 4, 5) and A = (2, 3), then Φ’ ∩ A =
1) Φ
2) U
3) A
4) {1, 4, 5}
Solution:
Φ = U, so Φ ‘ ∩ A = U ∩ A = A.

Question 9.
U = {x : x is a natural number 1 < x < 15} and A = 1 < x < 15) then A’ = ____________
1) {2, 3, 5, 7, 11, 13}
2) {4, 6, 8, 9, 10, 12, 14}
3) {2, 3, 4, 5, 7, 11, 13}
4) {2, 4, 6, 8, 10, 12, 14}
Solution:
Given : U = {x : x is a natural number and 1 < x < 15}
U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
A = {x : x is a prime number and 1 < x < 15}
A = {2, 3, 5, 7, 11, 13}
So, A’ = U – A = {4, 6, 8, 9, 10, 12, 14}.

Question 10.
Which of the following is not a subset of (a, b, c, d} ?
1) Φ
2) {a, b}
3) {d, e}
4) {b, d}
Solution:
Let’s analyse :
1) Φ – ✓ Always a subset of any set.
2) {a, b} – ✓ All elements are present in the set.
3) {d, e} – X “e” is not in {a, b, c, d}.
4) {b, d} – ✓ Both elements are present in the set.
∴ {d, e} is not a subset.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1g Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1g

Question 1.
If X and Y are two sets such that n(X) & 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y),
Solution:
Given that, n(X) = 17; n(Y) = 23; n(X ∪ Y) = 38
We know that, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 40 – 38 = 2
∴ n(X ∩ Y) = 2.

Question 2.
If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements, how many elements does X ∩ Y have ?
Solution:
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 18 = 8 + 15 – n(X ∩ Y)
⇒ 18 = 23 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5
∴ n(X ∩ Y) = 5 elements.

Question 3.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English ?
Solution:
Let No. of people in a group be = 400 = n(H ∪ E)
No. of people who can speak Hindi = 250 = n(H)
No. of people who can speak English = 200 = n(E)
No. of people who can speak both = x = n(H ∩ E)
According to n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – x
400 = 450 – x
x = 450 – 400 = 50
∴ n(H ∩ E) = 50
∴ 50 people can speak both Hindi and English.

Question 4.
If S and T are two sets such that S has 21 elements. T has 32 elements and S ∩ T has 11 elements how many elements does S ∪ T have ?
Solution:
Given that n(S) = 21, n(T) = 32 and n(S ∩ T) = 11
We know that principle of inclusion & exclusion
n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
⇒ n(S ∪ T) = 21 + 32 – 11 = 42
∴ 42 elements are in S ∪ T.

Question 5.
IfX and Yare two sets such that Xhas 40 elements, XuYhas 60elements and X ∩ Y has 10 elements, how many elements does Y have ?
Solution:
Given that, X has 40 elements = n(X) = 40
(X ∪ Y) has 60 elements = n(X ∪ Y) = 60
(X ∩ Y) has 10 elements = n(X ∩ Y)= 10
We know that, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 60 – 30 = 30
∴ Y has 30 elements.

Question 6.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea ?
Solution:
Let No. of people in group n(C ∪ T) = 70
No. of people who likes coffee n(C) = 37
No. of people who likes tea n(T) = 52
Let No. of people who likes both n(C ∩ T) = x
Since everyone likes at least one, we know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – x
⇒ x = 89 – 70
⇒ x = 19
∴ ‘19’ people like both coffee and tea.

Question 7.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket ? How many like tennis ?
Solution:
Total no. of people in a group n(C ∪ T) = 65
No. of people who likes only cricket n(C) = 40
No. of people who likes both cricket and tennis
n(C ∩ T) = 10
We know that,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n (T) – 10
⇒ n(T) = 65 – 30
⇒ n(T) = 35
So, 35 people like tennis (include both cricket)
Now, n(T) = n(T – C) + n(C ∩ T)
35 = n(T – C) + 10
⇒ n(T – C) = 35 – 10 = 25
∴ So, 25 people like only tennis.

Question 8.
In a committee, 50 people speak French. 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Given that,
No. of people who speaks French n(F) = 50,
No. of people who speaks Spanish n (S) = 20
No. of people who speaks both Spanish and French n(S ∩ F) = 10
We know that
n(S ∪ F) = n(F) + n(S) – n(S ∩ F)
n(S ∪ F) = 50 + 20 – 10 = 60
∴ 60 people speak at least one of these two languages.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1f Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1f

I.

Question 1.
Decide, among the following sets, which sets are subsets of one and another.
A = {x : x ∈ R and X satisfy x² -8X+12 = 0}
B = {2, 4, 6}, C = {2, 4, 6. 8, ….. }, D = {6}.
Solution:
A = {x : x ∈ R and x satisfies x² – 8x + 12}
2 and 6 are the only solutions of x² – 8x + 12 = 0. ∴ A = {2, 6)
B = {2, 4, 6}, C = {2, 4, 6, 8, ……..}, D = {6}. ∴ D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C.

Question 2.
In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
i) If x ∈ A and A ∈ B, then x ∈ B.
ii) If A ⊂ B and B ∈ C, then A ∈ C.
iii) If A ⊂ B and B ⊂ C, then A ⊂ C.
iv) If A ⊄ B and B ⊄ C, then A ⊂ C.
v) If x ∈ A and A ⊄ B, then x ∈ B.
vi) If A ⊂ B and x ∉ B, then x ∈ A.
Solution:
i) False
Let A = {1, 2} and B = {1, {1, 2), {3}}.
Now 2 ∈ {1, 2} and {1, 2} ∈ {{3}, 1, {1, 2}}.
∴ A ∈ B. However, 2 ∉ {{3}, 1, {1, 2}}

ii) False
Let A = {2}, B = (0, 2} and C = {1, {0, 2}, 3}. As A ⊂ B.
∴ B ∈ C. However, A ∉ C.

iii) True
Let A ⊂ B and B ⊂ C.
Let x ∈ A ⇒ x ∈ B [∵ A ⊂ B]
⇒ x ∈ C [∵ B ⊂ C]
∴ A ⊂ C.

iv) False
Let A = {1, 2, 9} and B = {0, 6, 8} and C = {0, 1, 2, 6, 9}
Accordingly, A ⊄ B and B ⊄ C. However, A ⊂ C

v) False
Let A = {1, 2} and B = {1, 3}. Here x = 2 ⊄ A and A ct B but x ∉ B.

vi) True
A ⊂ B ⇒ all elements of A are in B.
∴ If x ∈ B; it means x is not an element of B.
∵ all elements of A are also in B.
If x is not in B, it cannot be in A.

Question 3.
Show that if A ⊂ B, then (C – B) ⊂ (C – A).
Solution:
Let A c B. To show that C – B ⊂ C – A.
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
∴ C – B ⊂ C – A.

Question 4.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
Let A = {0, 1}, B = {0, 2, 3} and C = {0, 4, 5}
Accordingly, A ∩ B = {0} and A ∩ C = {0}
Here, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ C].

II.

Question 1.
Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
Given that A ∪ B = A ∪ C and A ∩ B = A ∩ C
Let x ∈ B
⇒ x ∈ A ∪ B [∵ B ⊂ A ∪ B]
⇒ x ∈ A ∪ C [∵ A ∪ B = A ∪ C]
⇒ x ∈ A or x ∈ C
⇒ If x ∈ A then x ∈ A ∩ B [∵ x ∈ B]
⇒ x ∈ A ∩ C [∵ A ∩ B = A]
⇒ x ∈ C
∴ B ⊂ C
Similarly, we can show that C ⊂ B
Hence B = C.

Question 2.
Show that the following four conditions are equivalent.
i) A ⊂ B
ii) A – B = Φ
iii) A ∪ B = B
iv) A ∩ B = A
Solution:
To prove (i) is equivalent to (ii)
From (i) A ⊂ B.
It means that all elements of A are in B. i.e., there is no element of A that is not in B. i.e., A – B = Φ (which is (ii)).
Thus (i) ⇔ (ii).
To prove (i) is equivalent to (iii).
From (i) A ⊂ B.
It means that all elements of A are in B.
Then A ∪ B = set of all elements of A and B = B (which is (iii)).
Thus, (i) ⇔ (iii).
To prove (i) is equivalent to (iv) :
From (i) A ⊂ B.
It means that all elements of A are in B.
Then A ∩ B = Set of all elements common to A and B.
= A (which is (iv)).
Thus, (i) ⇔ (iv)
We have proved that (i) ⇔ (ii), (i) ⇔ (iii) and (i) ⇔ (iv).
Thus, (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).

Question 3.
Show that for any sets A and B.
Solution:
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B).
Solution:
i) To show that A = (A ∩ B) ∪(A-B).
Consider R.H.S. = (A ∩ B) ∪ (A – B).
= (A ∩ B) ∪ (A ∩ B’) (∵ By Def.of difference of sets A – B = A ∩ B’).
= A ∩ (B ∪ B’) [∵ By distributive law],
= A ∩ U [∵ A ∪ A’ – U].
= A = L.H.S
∴ A = (A ∩ B) ∪ (A – B).

ii) To show that A ∪ (B – A) = (A ∪ B)
Consider A ∪ (B – A)
= A ∪ (B ∩ A’) (∵ By def. of difference of sets, A – B = A ∩ B’).
= (A ∪ B) ∩ (A ∪ A’) [∵ By the distributive law],
= (A ∪ B) ∩ U (∵ A ∪ A’ = U).
= A ∪ B
∴ A ∪ (B – A) = A ∪ B.

Question 4.
Using properties of sets, show that
i) A ∪ (A ∩ B) = A
ii) A ∩ (A ∪ B) = A.
Solution:
i) To show that, A ∪ (A ∩ B) = A
We know that,
A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A …………(1)
Also, A ⊂ A ∪ (A ∩ B) ………….(2)
From (1) and (2), A ∪ (A ∩ B) = A

ii) To show : A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∩ (A ∪ B) = A ∪ (A ∩ B) = A {From (i)}.

Question 5.
Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints : A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use Distributive law)
Solution:
Given A and B be two sets such that A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X.
To show that A = B
It can be seen that
A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [∵ A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Distributive law]
= (A ∩ B) ∪ Φ [ A ∩ X = Φ]
= (A ∩ B) ……….(1)

Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X)
= (B ∩ A) ∪ (B ∩ X)
= (B ∩ A) ∪ Φ
= (B ∩ A) …………(2)
Hence, from (1) 8s (2), we obtain A = B.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and
A ∩ B ∩ C = Φ.
Let A = {0, 1}, B = {1, 2} and C = {2, 0}
Accordingly, A ∩ B = {1}, B ∩ C = {2} & A ∩ C = {0}
A ∩ B, B ∩ C and A ∩ C are non-empty.
However, A ∩ B ∩ C = Φ

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1e

I.

Question 1.
If U = {1. 2, 3, 4. 5. 6, 7. 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}.
Verify that (i) (A ∪ B)’ = A’ ∩ B’ (ii) (A ∩ B)’ = A’ ∪ B’
Solution:
Given U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} & B = {2, 3, 5, 7}
i) A ∪ B = {2, 3, 4, 5, 6, 7, 8}
(A ∪ B}’ = U – (A ∪ B) = {1, 9}
A = {1, 3, 5, 7, 9}, B’ = {1, 4, 6, 8, 9}
A’ ∩ B’ = {1, 9}
(A ∪ B)’ = A’ ∩ B’.

ii) A ∩ B = {2}
(A ∩ B)’ = {1, 3, 4, 5, 6, 7, 8, 9}
A’ ∪ B’ = {1, 3, 5, 7,9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
(A ∩ B)’ = A’ ∪ B’.

Question 2.
Let U be the set of all triangles in a plane. If A is the set of all triangles with atleast one angle different from 60°, what is A’?
Solution:
Given U be the set of all triangles in a plane.
A is the set of all triangles with atleast one angle different from 60°.
A’ is the set of all equilateral triangles.

Question 3.
Fill in the blanks to make each of the following a true statement.
(i) A ∪ A’ = …………….
(ii) Φ’ ∩ A= …………….
(iii) A ∩ A’ = ………………..
(iv) U’ ∩ A = …………………
Solution:
i) A ∪A’ = U.
ii) Φ’ ∩ A = U ∩ A = A.
iii) A ∩ A’ = Φ.
iv) U’ ∩ A = Φ ∩ A = Φ.

II.

Question 1.
If U = {a, b, c, d, e,f, g, h}, find the complements of the following sets.
(i) A = {a, b, c)
(ii) B = {d, e, f, g)
(iii) C = [a, c, e, g]
(iv) D = {f, g, h, a)
Solution:
Given U = {a, b, c, d, e, f, g, h}; A = {a, b, c}; B = {d, e, f, g}; C = {a, c, e, g}; D = {f, g, h, a}
i) A’ = {d, e, f, g, h}.
ii) B’ = {a, b, c, h}.
iii) C’ = {b, d, f, h}.
iv) D’ = {b, c, d, e}.

Question 2.
Draw appropriate Venn diagram for each of the following.
(i) (A ∪B)’
(ii) A’ ∩ B’
(iii) (A ∩ B)’
(iv) A’ ∪B’
Solution:
i) (A ∪ B)’

ii) A’ ∩ B’

iii) (A ∩ B)’

iv) A’ ∪ B’

III.

Question 1.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}. A = {1, 2. 3, 4}, B = {2. 4, 6, 8} and C = {3, 4, 5, 6).
Find (i) A’ (ii) B’ (iii) (A ∪ C)’ (iv) (A ∪ B)’
Solution:
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6}
i) A’ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4} = {5, 6, 7, 8, 9}.
ii) B’ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}.
iii) A ∪ C = {1, 2, 3, 4, 5, 6}
(A ∪ C)’ = U – (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6} = {7, 8, 9).
iv) A ∪ B = {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = {5, 7, 9}.
v) (A’)’ = A = {1, 2, 3, 4}.
vi) B – C = {2, 8}
(B – C)’ = {1,3, 4, 5, 6, 7, 9}.

Question 2.
Taking the set of natural numbers as the universal set, write down the • complements of the following sets.
i) {x : x is an even natural number}.
ii) {x : x is an odd natural number}
iii) {x : x is a positive multiple of 3}.
iv) (x : x is a prime number}
v) (x : x is a natural number divisible by 3 and 5}.
vi) {x : x is a perfect square}.
vii) {x : x is a perfect cube},
viii) {x : x + 5 = 8}.
ix) {x : 2x + 5 = 9}.
x) {x : x > 7}.
xi) {x : x ∈ N and 2x +1 > 10}.
Solution:
Given U = N = {1, 2, 3, }
i) (x : x is an even natural number}’ = {x : x is an odd natural number}.
ii) {x : x is an odd natural number}’ = (x : x is an even natural number}.
iii) {x : x is a positive multiple of 3}’ = {x: x ∈ N and x is not a multiple of 3}.
iv) {x : x is a prime number}’ = {x : x is a positive composite number and x = 1}.
v) {x : x is a natural number divisible by 3 & 5} = {x : x is a natural number that is not divisible by 3 & 5 }.
vi) {x : x is a perfect square} = {x : x ∈ N and x is not a perfect square}.
vii) {x : x is a perfect cube} = (x : x ∈ N and x is not a perfect cube}.
viii) {x : x + 5 = 8} = {x : x ∈ N and x ≠ 3}.
ix) { x : 2x + 5 = 9} = {x : x ∈ N and x ≠ 2}.
x) {x : x ≥ 7} = {x : x ∈ N and x < 7}.
xi) {x : x ∈ N and 2x + 1 > 10} = {x : x ∈ N and x ≤ 9/2}.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1d

I.

Question 1.
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Solution:
Given A {a, b) and B {a, b, C). Yes, A ⊂ B
∴ A ∪ B = {a, b, c) = B.

Question 2.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution:
Given A and B are two sets such that A ⊂ B then A ∪ B = B.

Question 3.
Which of the following pairs of sets are disjoint.
i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6}.
Solution:
Let A = {1, 2, 3, 4), B = {4, 5, 6).
A ∩ B = {4} ∴ Given pair of sets are not disjoint.

ii) {a, e, i, o, u} and (c, d, e, f}.
Solution:
Let A = {a, e, i, o, u), B = {c, d, e, f)
A ∩ B = {e} ∴ Given pair of sets are not disjoint.

iii) {x: x is an even Integer} and {x: x is an odd Integer}.
Solution:
Let A = {x: x is an even integer), B = {x : x is an odd integer)
A ∩ B = Φ ∴ Given pair of sets are disjoint.

Question 4.
If x = {a, b, c, d} and Y = {f, b, d, g} find
i) X – Y
ii) Y – X
iii) X ∩ Y.
Solution:
Given X = {a,b, c, d}; Y = {f, b, d, g)
i) X – Y = {a, c}.
ii) Y – X = {f, g).
iii) X ∩ Y = (b, d}.

Question 5.
If R is the set of real numbers and 9 is the set of rational numbers, then what is R – Q ?
Solution:
Given, R : set of real numbers.; Q : set of rational numbers.
R – Q = set of irrational numbers.

Question 6.
State whether each of the following statement is true or false. Justify your answer.
i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
ii) {a, e, i, o, u) and {a, b, c, d} are disjoint sets.
iii) (2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
iv) }2, 6, 10} and {3, 7, 11} are disjoint sets.
Solution:
i) False, because 3e { 2, 3, 4, 5} & 3 ∈ {3, 6}.
ii) False, because a € {a, e, i, o, u} & a ∈ (a, b, c, d}.
iii) True, because {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ.
iv) True, because {2, 6, 10} ∩ {3, 7, 11} = Φ.

II.

Question 1.
Find the union of each of the following pairs of sets.
i) X = {1, 3, 5}, Y = {1, 2, 3}.
ii) A = [a, e, i, o, u}, B = {a, b, c}.
iii) A = {x : x is a natural number and multiple of 3},
B = {x : x is a natural number less than 6}.
iv) A = {x : x is a natural number and 1 < x < 6},
B = {x : x is a natural number and 6 < x < 10}.
v) A= {1, 2, 3}, B = f
Solution:
i) X = (1, 3, 5}, Y = {1, 2, 3}
X ∪ Y = {1, 3, 5} ∪ {1, 2, 3} = {1, 2, 3, 5}.

ii) A = {a, e, i, o, u}, B = {a, b, c}
∴ A ∪ B = {a, e, i, o, u} ∪ {a, b, c} = {a, b, c, e, i, o, u}.

iii) A = {x : x is a natural number and multiple of 3} = {3, 6, 9, }
B = {x : x is a natural number less than 6} = {1, 2, 3, 4, 5}
A ∪ B = {x : x = 1, 2, 3, 4, 5 and a multiple of 3}.

iv) A = {x : x is a natural number and 1 < x < 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∪ B = {2,3, 4, 5, 6, 7, 8, 9}.

v) A = {1, 2, 3}, B = Φ; A ∪ B = {1, 2, 3}.

Question 2.
Find the intersection of each pair of sets of question 1 above.
Solution:
i) X = {1, 3, 5}, Y = {1, 2, 3} ⇒ X ∩ Y = {1, 3, 5} ∩ {1, 2, 3} = {1, 3}.
ii) A = {a, e, i, o, u}, B = {a, b, c} A ∩ B = {a}.
iii) A = {3, 6, 9, }, B = {1, 2, 3, 4, 5} ⇒ A ∩ B = {3}.
iv) A = {2, 3, 4, 5, 6}, B = {7, 8, 9} ⇒ A ∩ B = {} = Φ.
v) A = {1, 2, 3}, B = Φ ⇒ A ∩ B = Φ.

Question 3.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10); find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Solution:
Given A = {.1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
i) A ∪ B = {1, 2, 3, 4, 5, 6}.
ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.
iii) B ∪ C = {3,4, 5, 6, 7, 8}.
iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.
v) A ∪ B ∪ C = {l,2,3, 4, 5, 6} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}.
vi) A ∪ B ∪ D = {1,2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} ={1,2, 3, 4, 5, 6, 7, 8, 9, 10}
vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}.

Question 4.
If A = {3. 5, 7. 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17};
find (i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution:
Given A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
i) A ∩ B = {7, 9, 11}.
ii) B ∩ C = {11, 13}.
iii) A ∩ C ∩ D = (A ∩ C) ∩ D = {11} ∩ {15, 17} = Φ
iv) A ∩ C = {11}.
v) B ∩ D = Φ (or) { }
vi) A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
vii) A ∩ D = { } = Φ.
viii) A ∩ (B ∪ D) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17} = {7, 9, 11}.
ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17} ∩ {7,9, 11, 13, 15} = {7, 9, 11, 15}.

III.

Question 1.
If A = {x : x is a natural number}, B = {x : x is an even natural number)
C = {x : x is an odd natural number} and D = [x : x is a prime number},
Find (i) A ∩ B (ii) A ∩ C (iii) A ∩ D (iv) B ∩ C (v) B ∩ D (vi) C ∩ D
Solution:
A = {x : x is a natural number};
B = {x : x is an even natural number}
C = {x ; x is an odd natural number};
D = {x ; x is a prime number}

i) A ∩ B = {x : x is an even natural number} = B.
ii) A ∩ C = {x : x is an odd natural number) = C.
iii) A ∩ D = {x : x is a prime number} = D.
iv) B ∩ C = Φ.
v) B ∩ D = { 2 }.
vi) C ∩ D = {x : x is an odd prime number}.

Question 2.
If A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20};
Find (i) A – B (ii) A – C (iii) A – D (iv) B – A (v) C – A (vii) B – C (viii) B – D (ix) C – B (x) D – B (xi) C – D
Solution:
Given A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20}.
i) A – B = {3, 6, 9, 15, 18, 21}.
ii) A – C = {3, 9, 15, 18, 21}.
iii) A – D = {3, 6, 9, 12, 18, 21}.
iv) B – A = {4, 8, 16, 20}.
v) C – A = {2, 4, 8, 10, 14, 16}.
vi) D – A = {5, 10, 20}.
vii) B – C = {20}.
viii) B – D = {4, 8, 12, 16}.
ix) C – B = {2, 6, 10, 14}.
x) D – B = {5, 10, 15}.
xi) C – D = {2, 4, 6, 8, 12, 14, 16}.
xii) D – C = {5, 15, 20}.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1c

Question 1.
Make correct statements by filling the symbol a or <z in the blank spaces,
i) {2, 3, 4 }…{1, 2, 3, 4, 5}.
Solution:
⊂

ii) {a, b, c }…{b, c, d}.
Solution:
⊄

iii) {x : x is a student of Class XI of your school}…{x : x is a student of your school}.
Solution:
⊂

iv) {r : x is a circle in the plane} … {x : r is a circle in the same plane with radius 1 unit}.
Solution:
⊄

v) {x : x is a triangle in a plane} … {x : x is a rectangle in the plane}.
Solution:
⊄

vi) (x : x is an equilateral triangle in a plane}…{x : x is a triangle in the same plane}
Solution:
⊂

vii) {x : x is an even natural number} . . . {x : x is an integer}.
Solution:
⊂

Question 2.
Examine whether the following statements are true or false.
i) ( a, b } ⊄ { b, c, a }.
ii) { a, e } ⊂ { x : x is a vowel in the english alphabet}.
iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }.
iv) { a } ⊂ { a, b, c }.
v) { a } ∈ { a, b, c }.
vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}.
Solution:
i) False, each element of {a, b} is also an element of {b, c, a}.
ii) {a, e} ⊂ {x : x is a vowel in the english alphabet} is true.
iii) {1, 2, 3} ⊂ {1, 3, 5} is false because 2 ∉ {1, 3, 5}.
iv) True because each element of {a} is also an element of {a, b, c}.
v) False the elements of {a, b, c} are a, b, c therefore {a} ⊂ {a, b, c}.
vi) True {2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}.

Question 3.
Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?
i) {3, 4} ⊂ A
Solution:
The statement {3, 4} ⊂ A is incorrect because 3e {3, 4} but 3 ∈ A.

ii) {3, 4} ∈ A
Solution:
The statement {3, 4} ∈ A is correct because {3, 4} is an element of A.

iii) {{3, 4}} ⊂ A
Solution:
The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} ∈ A.

iv) 1 ∈ A
Solution:
The statement 1 ∈ A is correct because 1 is an element of A.

v) 1 ⊂ A
Solution:
The statement 1 ⊂ A is incorrect because an element of a set can never be a subset of itself.

vi) {1, 2, 5} ⊂ A
Solution:
The statement {1,2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.

vii) {1, 2, 5} ∈ A
Solution:
The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.

viii) {1, 2, 3} ⊂ A
Solution:
The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3} but 3 ∉ A.

ix) Φ ∈ A
Solution:
The statement Φ ∈ A is incorrect because Φ is not an element of A.

x) Φ ⊂ A
Solution:
The statement Φ ⊂ A is correct because Φ is a subset of every set.

xi) {Φ} ⊂ A
Solution:
The statement {Φ} ⊂ A is incorrect because Φ ∈ {Φ} but Φ ∉ A.

Question 4.
Write down all the subsets of the following sets.
i) {a}
Solution:
The subsets of { a} are Φ and {a}.

ii) {a, b}
Solution:
The subsets of {a, b} are Φ, { a}, { b } & {a, b}.

iii) {1, 2, 3}
Solution:
The subsets of {1, 2, 3} are (j) , {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3}.

iv) Φ
Solution:
The subset of Φ is Φ .

Question 5.
Write the following as intervals.
(i) (x : x ∈ R, – 4 < x < 6}.
(ii) {x : x ∈ R, – 12 < x < -10}.
(iii) {x : x ∈ R, 0 < x < 7}.
(iv) {x : x ∈ R, 3 < x < 4}.
Solution:
i) {x : x ∈ R, -4 < x < 6} = (-4, 6]
ii) {x : x ∈ R, -12 < x < – 10} = (-12, -10)
iii)(x : x ∈ R, 0 < x < 7} = [0, 7).
iv) {x : x ∈ R, 3 < x < 4} = [3, 4],

Question 6.
Write the following intervals in set-builder form.
(i) (- 3, 0)
(ii) [6,12]
(iii) (6, 12]
(iv) [-23, 5)
Solution:
i) (-3, 0) = {x : x ∈ R and -3 < x < 0 }.
ii) [6, 12] = {x : x ∈ R and 6 < x < 12 }.
iii) (6, 12] = {x : x ∈ R and 6 < x < 12}.
iv) [-23, 5) = {x : x ∈ R and -23 < x < 5}.

Question 7.
What universal set(s) would you propose for each of the following?
i) The set of right triangles.
ii) The set of isosceles triangles.
Solution:
i) The universal set for the set of right triangles is the set of all triangles.
ii) The universal set for the set of isosceles triangles is the set of all polygons.

Question 8.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C.
i) (0, 1, 2, 3, 4, 5, 6).
ii) Φ
iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
iv) {1, 2, 3, 4, 5, 6, 7, 8}.
Solution:
i) A ⊂ {0, 1, 2, 3, 4, 5, 6}, B ⊂ {0, 1, 2, 3, 4, 5, 6}, C ⊄ {0, 1, 2, 3, 4, 5, 6}
∴ {0, 1, 2, 3, 4, 5, 6} cannot be the universal set of the sets A, B, & C.

ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
∴ Φ cannot be the universal set of the sets A, B & C.

iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
∴ The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B & C.

iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8};
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
The set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B & C.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1b

Question 1.
Which of the following are examples of the null set.
i) Set of odd natural numbers divisible by 2.
ii) Set of even prime numbers.
iii) {x : x is a natural numbers, x < 5 and x > 7}.
iv) {y : y is a point common to any two parallel lines}.
Solution:
i) Set of odd natural numbers divisible by 2 is a null set because no odd number is divisible by 3.
ii) Set of even prime numbers is not a null set because 2 is an even prime number.
iii) {x : x is a natural numbers, x < 5 & x > 7}isa null set.
iv) {y : y is a point common to any two parallel lines} is a null set.

Question 2.
Which of the following sets are finite or infinite ?
i) The set of months of a year.
ii) {1, 2, 3, …….}
iii) {1, 2, 3, ……. 99, 100).
iv) The set of positive integers greater than 100.
v) The set of prime numbers less than 99.
Solution:
i) The set of months of a year is a finite set because it has 12 elements.
ii) {1, 2, 3, ……..} is an infinite set because it has infinite number of natural numbers.
iii) Given set is a finite set because it has 100 elements.
iv) Given set is an infinite set because it has infinite elements.
v) Given set is a finite set because prime numbers < 99 are finite.

Question 3.
State whether each of the following set is finite or infinite.
i) The set of lines which are parallel to the X-axis.
ii) The set of letters in the English alphabet.
iii) The set of numbers which are multiple of 5.
iv) The set of animals living on the earth.
v) The set of circles passing through the origin (0, 0).
Solution:
i) The set of lines which are parallel to the X-axis is an infinite set because lines parallel to the X-axis are infinite.
ii) The set of letters in the English alphabet is a finite set because it has 26 elements.
iii) The set of numbers which are multiples of 5 is an infinite set because multiples of 5 are infinite in numbers.
iv) The set of animals living on the earth is a finite set because animals living on the earth is finite.
v) The set of circles passing through the origin (0, 0) is an infinite set because the number of circles passing through the origin is infinite.

Question 4.
In the following state whether A = B or not.
i) A = {a, b, c, d} B = (d, c, b, a)
ii) A = {4, 8, 12, 16} B = {8, 4, 16, 18}
iii) A = {2, 4, 6, 8, 10} B = {x : x is a positive even integer and x < 10}
iv) A = {x : x is a multiple of 10} B = {10, 15, 20, 25, 30, ….}
Solution:
i) A = B because every element in A is element in B and every element in B is element in A.
ii) A ≠ B because 12 ∈ A but 12 ∉ B.
iii) A = {2, 4, 6, 8, 10}, B = {2, 4, 6, 8, 10} ∴ A = B
iv) A = {10, 20, 30, …………}, B = {10, 15, 20, 25, 30, …….}
A ≠ B because 15 ∈ B, but 15 ∉ A.

Question 5.
Are the following pair of sets equal ? Give reasons,
i) A = {2, 3}, B = {x : x is a solution of x² + 5x + 6 = 0}.
ii) A = {x : x is a letter in the word FOLLOW},
B = {y: y is a letter in the word WOLF}.
Solution:
i) A = {2, 3}, B = {x : x is a solution of x² + 5x + 6 = 0} = {-2, -3}
B = {-2, -3} ∴ A ≠ B
ii) A = {F, O, L, W}, B = [W, O, L, F} ∴ A = B.

Question 6.
From the sets given below, select equal sets.
A = {2, 4, 8, 12},
B = {1, 2, 3, 4},
C = {4, 8, 12, 14},
D = {3, 1, 4, 2}
E = (-1, 1}, F = {0, a},
G = {1, -1}, H = (0, 1}
Solution:
Two sets are equal if they contain exactly the same elements, regardless of order.
B = D, E = G.

Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sets Solutions Exercise 1a

Question 1.
Which of the following are sets? Justify your answer.
i) The collection of all the months of a year beginning with the letter J.
Solution:
The collection of all the months of a year beginning with the letter J is a set because it is well defined.

ii) The collection of ten most talented writers of India.
Solution:
The collection of ten most talented writers of India is not a set because it is not well defined.

iii) A team of eleven best cricket batsmen of the world.
Solution:
A team of eleven best cricket batsmen of the world is not a set because it is not well defined.

iv) The collection of all boys in your class.
Solution:
The collection of all boys in my class is a set because it is well-defined.

v) The collection of all natural numbers less than 100.
Solution:
The collection of all natural numbers less than 100 is a set because it is well-defined.

vi) A collection of novels written by the writer Munshi Prem Chand.
Solution:
A collection of novels written by the writer Munshi Prem Chand is a set because it is well defined.

vii) The collection of all even integers.
Solution:
The collection of all even integers is a set because it is well defined.

viii) The collection of questions in this chapter.
Solution:
The collection of questions in this chapter is a set because it is well defined,

ix) A collection of most dangerous animals of the world.
Solution:
A collection of most dangerous animals of the world is not a set because it is not well defined.

Question 2.
Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces,
i) 5 … A
ii) 8 … A
iii) 0 … A
iv) 4 … A
v) 2 … A
vi) 10 … A
Solution:
i) ∈
ii) ∉
iii) ∉
iv) ∈
v) ∈
vi) ∉

Question 3.
Write the following sets in roster form.
i) A = (x : x is an integer and – 3 < x < 7}.
ii) B = {x: x is a natural number less than 6}.
iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}.
iv) D = {x: x is a prime number which is divisor of 60).
v) E = {The set of all letters in the word TRIGONOMETRY}
vi) F = {The set of all letters in the word BETTER}.
Solution:
i) A = {-3, -2,-1, 0, 1, 2, 3, 4, 5, 6}.
ii) B = {1, 2, 3, 4, 5}.
iii) C = {17, 26, 35, 44, 53, 62, 71, 80}.
iv) D = {2, 3, 5}.
v) E = {T, R, I, G, O, N, M, E, Y}.
vi) F = {B, E, T, R}.

Question 4.
Write the following sets in the set-builder form.
i) {3, 6, 9, 12}
ii) {2, 4, 8, 16, 32}
iii) {5, 25, 125, 625}
iv) {2, 4, 6 , ………..}
v) {1, 4, 9, …………. 100}
Solution:
i) {x : x is a natural number multiple of 3 and x ≤ 15}.
ii) {x : x = 2n where n ∈ N & 1 ≤ n ≤ 5}.
iii) {x : x = 5n where ne N & 1 ≤ n ≤ 4}.
iv) { x : x is an even natural number}.
v) { x : x = n², 1 ≤ n ≤ 10}.

Question 5.
List all the elements of the following sets.
i) A = {x : x is an odd natural number).
ii) B = {x : x is an integer, –\(\frac{1}{2}\) < x < \(\frac{9}{2}\)}.
iii) C = {x : x is an integer, x² ≤ 4).
iv) D = {x : x is a letter in the word “LOYAL”}.
v) E = {x : x is a month of a year not having 31 days},
vi) F = {x : x is a consonant in the English alphabet which precedes k}.
Solution:
i) A = {1,3, 5, 7, ……….}.
ii) B = {0, 1,2, 3, 4}.
iii) C = {-2, -1, 0, 1, 2}.
iv) D = {L, O, Y, A},
v) E = {February, April, June, September, November},
vi) F = { b, c, d, f, g, h, j}

Question 6.
Match each of the set on the left in the roster form with the same set on the right described in set-builder form.

Answer: