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1.

If A = {2, 3, 5, 7,11} and B = {5, 7, 9, 11, 13}, find A ∆ B.

Answer»

A ∆ B = {2, 3, 9, 13}
Discussion
2.

(i) If n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B’) = 25, find n(A ∩ B) and n(U).(ii) If n(A) = 300, n(A ∪ B) = 500, n(A ∩ B) = 50 and n(B’) = 350, find n(B) and n(U).

Answer»

(i) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

n(A ∩ B) = 25 + 40 – 50 = 65 – 50 = 15

n(U) = n(B) + n(B’) = 40 + 25 = 65

(ii) n(U) = n(B) + n(B’)

n(A ∩ B) = n(A) + n(B) – n(A B)

n(B) = n(A ∪ B) + n(A ∩ B) – n(A) = 500 + 50 – 300 = 250

n(U) = 250 + 350 = 600
Discussion
3.

If U = {x : x ∈ N and x < 10}, A = {1, 2, 3, 5, 8} and B = {2, 5, 6, 7, 9}, then n[(A ∪ B)’] is (1) 1 (2) 2 (3) 4 (4) 8

Answer»

(1) 1 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

A = {1, 2, 3, 5, 8} 

B = {2, 5, 6, 7, 9} 

A ∪ B = {1, 2, 3, 5, 6, 7, 8, 9} 

(A ∪ B)’ = {4}, 

n(A ∪ B)’ = 1
Discussion
4.

Which of the following is true? (1) A – B = A ∩ B (2) A – B = B – A (3) (A ∪ B)’ = A’ ∪ B’ (4) (A ∩ B)’ = A’ ∪ B’

Answer»

(4) (A ∩ B)’ = A’ ∪ B’

(1) (A – B) = A ∩ B ✘ 

(2) A – B = B – A ✘ 

(3) (A ∪ B) = A’ ∪ B’ ✘ 

(4) (A ∩ B)’ = A’ ∪ B’ ✓
Discussion
5.

If A ∪ B = A ∩ B, then (1) A ≠ B (2) A = B (3) A ⊂ B (4) B ⊂ A

Answer»

Answer is (2) A = B
Discussion
6.

In a class of 50 boys, 35 boys play carom and 20 boys play chess then the number of boys play both games is (1) 5 (2) 30 (3) 15(4) 10

Answer»

(1) 5 

n(A ∪ B) = n(A) + n(B) – n(A n B) ⇒ 50 = 35 + 20 – n(A ∩ B) ⇒ n(A ∩ B) = 5
Discussion
7.

From the adjacent diagram n[P(A ∆ B)] is (1) 8 (2) 16 (3) 32 (4) 64

Answer»

(3) 32 

A ∆ B = {60, 85, 75, 90, 70}

⇒ n(A ∆ B) = 5 

⇒ n(P(A ∆ B)) = 25 = 32
Discussion
8.

If n(A) = 10 and n(B) = 15 then the minimum and maximum number of elements in A ∩ B is (1) (10, 15) (2) (15, 10) (3) (10, 0) (4) (0, 10)

Answer»

Answer is (4) (0, 10)
Discussion
9.

For any three sets P, Q and R, P – (Q ∩ R) is (1) P – (Q ∪ R) (2) (P ∩ Q) – R (3) (P – Q) ∪ (P – R) (4) (P – Q) ∩ (P – R)

Answer»

(3) (P – Q) ∪ (P – R) 

P – (Q ∩ R) = (P – Q) ∪ (P – R)
Discussion
10.

Verify n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) for the following sets. (i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f} (ii) A = {1, 3, 5}, B = {2, 3, 5, 6} and C = {1, 5, 6, 7}.

Answer»

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

(i) A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f} 

n (A) = 5, n (B) = 4, n (C) = 4

n( A ∩ B) = 3 

n(B ∩ C) = 2 

n( A ∩ C) = 3 

n( A ∩ B ∩ C) = 2 

A ∩ B = {c, e, f} 

B ∩ C = {c, f} 

A ∩ C = {a, c, f} 

A ∩ B ∩ C = {c, f} 

A ∪ B ∪ C = {a, c, d, e, f, b, h} 

∴ n(A ∪ B ∪ C) = 7 … (1) 

n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) 

= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8 = 7 … (2) 

∴ (1) = (2) 

⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) 

Hence it is verified.

(ii) A = {1, 3, 5}, B = {2, 3, 5, 6 }, C = {1, 5, 6,7} = 3

n (B) = 4, 

n (C) = 4 

n(A ∩ B) = 2 

n(B ∩ C) = 2 

n(C ∩ A) = 2 

n(A ∩ B ∩ C) = 1 

n(A ∪ B ∪ C) = 6 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) 

6 = 3 + 4 + 4 – 2 – 2 – 2 + 1 = 12 – 6 = 6 

Hence it is verified.
Discussion
11.

If B – A is B, then A ∩ B is (1) A (2) B (3) U (4) Ø

Answer»

(4) Ø 

B – A = B ⇒ A and B are disjoint sets.
Discussion
12.

If K = {a, b, d, e,f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:(i) K ∪ (L ∩ M)(ii) K ∩ (L ∪ M)(iii) (K ∪ L) ∩ (K ∪ M)(iv) (K ∩ L) ∪ (K ∩ M) and verify distributive laws.

Answer»

K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}

(i) K ∪ (L ∩ M)

L ∩ M = {b, c, d, g} ∩ {a, b, c, d, h} = {b, c, d}

K ∪ (L ∩ M) = {a, b, d, e, f } ∪ {b, c, d) = {a, b, c, d, e, f}

(ii) K ∩ (L ∪ M)

L ∪ M = {a, b, c, d, g, h}

K ∩ (L ∪ M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h} = {a, b, d}

(iii) (K ∪ L) ∩ (K ∪ M)

K ∪ L = {a, b, c, d, e, f, g}

K ∪ M = {a, b, c, d, e, f, h}

(K ∪ L) ∩ (K ∪ M) = {a, b, c, d, e,f}

(iv) (K ∩ L) ∪ (K ∩ M)

(K ∩ L) = {b, d}

(K ∩ M) = {a,b,d}

(K ∩ L) ∪ (K ∩ M) = {b, d} ∪ {a, b, d} = {a, b, d}

Distributive laws

K ∪ (L ∩ M) = (K ∪ L) ∩ (K ∪ M)

{a, b, c, d, e, f) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}

= {a, b, c, d, e, f}

Thus Verified.

K ∩ (L ∪ M) = (K ∩ L) ∪ (K ∩ M)

{a, b, d} = {a, b, c, d, e, f, g} ∪ {a, b, c, d, e, f, h}

= {a, b, d}

Thus Verified.
Discussion
13.

Let A = {Ø} and B = P(A) then A ∩ B is (1) {Ø, {Ø}}(2) {Ø} (3) Ø(4) {0}

Answer»

(2) {Ø} 

P(A) = {Ø {Ø}}
Discussion
14.

If A = {x : x = 2n, n ∈ W and n < 4}, B = {x : x = 2 n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.

Answer»

A = {x : x = 2n, n ∈ W, n < 4}

⇒ x = 20 = 1

x = 21 = 2

x = 22 = 4

x = 23 = 8

∴ A = {1, 2, 4, 8}

B = {x : x = 2n, n ∈ N and n ≤ 4}

⇒ x = 2 x 1 = 2

x = 2 x 2 = 4

x = 2 x 3 = 6

x = 2 x 4 = 8

∴ B = {2, 4, 6, 8}

C = {0, 1, 2, 5, 6}

Associative property of intersection of set

A ∩ (B ∩ C) = (A ∩ B) ∩ C

B ∩ C = {2, 6}

A ∩ (B ∩ C) = {1, 2, 4, 8} ∩ {2, 6} = {2} … (1)

A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8} = {2, 4, 8}

(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6} = {2} … (2)

From (1) and (2),

It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C
Discussion
15.

Find A ∪ B, A ∩ B, A – B and B – A for the following sets.(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x &lt; 6}(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”

Answer»

(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}

A ∪ B = {2, 6, 10, 14} ∪ {2, 5, 14, 16} = {2, 5, 6, 10, 14, 16}

A ∩ B = {2, 6, 10, 14} ∩ {2, 5, 14, 16} = {2, 14}

A – B = {2, 6, 10, 14} – {2, 5, 14, 16} = {6, 10}

B – A = {2, 5, 14, 16} – {2, 6, 10, 14} = {5, 16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}

A ∪ B = {a, b, c, e, u} ∪ {a, e, i, o, u} = {a, b, c, e, i, o, u}

A ∩ B = {a, b, c, e, u} ∩ {a, e, i, o, u} {a, e, u}

A – B = {a, b, c, e, u} – {a, e, i, o, u} = {b, c}

B – A = {a, e, i, o, u} – {a, b, c, e, u} = {i, o}

(iii) x ∈ {1, 2, 3, ...}; x ∈ {0, 1, 2, 3, 4, 5, ...}

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B = {0, 1, 2, 3, 4, 5}

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5}

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5}

= {1, 2, 3, 4, 5}

A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5}

= {6, 7, 8, 9, 10}

B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

= {0}

(iv) A= {m, a, t, h, e, i, c, s}, B = {g, e, o, m, t, r, y}

A ∪ B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {m, a, t, h, e, i, c, s, g, o, r, y)

A ∩ B = {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t,r,y} = {m, t, e}

A – B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {a, h, i, c, s}

B – A = {m, a, t, h, e, i, c, 5} ∩ {g, e, o, m, t,r,y} = {g,o, r, y}
Discussion
16.

If U = {4, 7, 8, 10, 11, 12, 15, 16}, A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.

Answer»

U = {4, 7, 8, 10, 11, 12, 15, 16}

A = {7, 8, 11, 12}, B = {4, 8, 12, 15}

De Morgan’s Laws for complementation.

(A ∪ B)’ = A’ ∩ B’

A ∪ B = {4, 7, 8, 11, 12, 15}

(A ∪ B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15} 

= {10, 16} … (1)

A’ = {4, 10, 15, 16}

B’ = {7, 10, 11, 16}

A’ ∩ B’ = {10, 16} … (2)

From (1) and (2) it is verified that (A ∪ B)’ = A’ ∩ B’
Discussion
17.

If A = {x : x ∈ Z, -2 &lt; x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}, then verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Answer»

A = {x : x ∈ Z, -2 < x ≤ 4} = {-1, 0, 1, 2, 3, 4}

B = {x : x ∈ W, x ≤ 5} = {0, 1, 2, 3, 4, 5}

C = {-4, -1, 0, 2, 3, 4}

A ∪ (B ∩ C)

B ∩ C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4} = {0, 2, 3, 4}

A ∪ (B ∩ C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4} = {-1, 0, 1, 2, 3, 4} … (1)

(A ∪ B) ∩ (A ∪ C)

A ∩ B = {0, 1, 2, 3, 4}

A ∩ C = {-1, 0, 2, 3, 4}

(A ∩ B) ∪ (A ∩ C) = {0, 1, 2, 3, 4} ∪ {-1, 0, 2, 3, 4} = {-1, 0, 1, 2, 3, 4} … (2)

From (1) and (2), it is verified that

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Discussion
18.

If A = {b, c, e, g, h}, B = {a, c, d, g, i} and C = {a, d, e, g, h}, then show that A – (B ∩ C) = (A – B) ∪ (A – C).

Answer»

A = {b, c, e, g, h}

B = {a, c, d, g, i}

C = {a, d, e, g, h}

B ∩ C = {a, d, g}

A – (B ∩ C) = {b, c, e, g, h} – {a, d, g} = {b, c, e, h} … (1)

A - B = {b, c, e, g, h} – {a, c, d, g, i} = {b, e, h}

A – C = {b, c, e, g, h} – {a, d, e, g, h} = {b, c}

(A – B) ∪ (A – C) = {b, c, e, h} … (2)

From (1) and (2) it is verified that

A – (B ∩ C) = (A – B) ∪ (A – C)
Discussion
19.

If A = {2, 5, 6, 7} and B = {3, 5, 7, 8}, then verify the commutative property of (i) union of sets (ii) intersection of sets

Answer»

Given, A = {2, 5, 6, 7} and B = {3, 5, 7, 8} 

(i) A ∪ B = {2, 3, 5, 6, 7, 8} … (1) 

B ∪ A = {2, 3, 5, 6, 7, 8} … (2) 

From (1) and (2) we have A ∪ B = B ∪ A 

It is verified that union of sets is commutative.

(ii) A n B = {5, 7} … (3)

B n A = {5, 7} … (4) 

From (3) and (4) we get, A ∩ B = B ∩ A 

It is verified that intersection of sets is commutative.
Discussion
20.

Find the union of the following sets. (i) A = {1, 2, 3, 5, 6} and B = {4, 5, 6, 7, 8} (ii) X = {3, 4, 5} and Y = Ø

Answer»

(i) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} 

(ii) X ∪ Y = {3, 4, 5}
Discussion
21.

If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.

Answer»

Associative Property of union of sets 

A ∪ (B ∪ C) = (A ∪ B) ∪ C) 

B ∪ C = {m, n, q, s, t} ∪ {m, n, p, q, s} = {m, n, p, q, s, t} 

A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n, p, q, s, t} = {m, n, p, q, r, s, t} … (1)

(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t} = {p, q, r, s, m, n, t}

(A ∪ B) ∪ C = {p, q, r, s, m, n, t} ∪ {m, n, p, q, s} = {p, q, r, s, m, n, t} ... (2) 

From (1) & (2) 

It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C
Discussion
22.

Using the given venn diagram, write the elements of(i) A(ii) B(iii) A ∪ B(iv) A ∩ B(v) A – B(vi) B – A(vii) A’(viii) B’(ix) U

Answer»

(i) A = {2, 4, 7, 8, 10}

(ii) B = {3, 4, 6, 7, 9, 11}

(iii) A ∪ B = {2, 3, 4, 6, 7, 8, 9, 10, 11}

(iv) A ∩ B = {4, 7}

(v) A – B = {2, 8, 10}

(vi) B – A = {3, 6, 9, 11}

(vii) A’ = {1, 3, 6, 9, 11, 12}

(viii) B’ = {1, 2, 8, 10, 12}

(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}.
Discussion
23.

If P = {x : x ∈ N and 1 &lt; x &lt; 11}, Q = {x : x = 2n, n ∈ N and it &lt; 6} and R = {4, 6, 8, 9, 10, 12}, then verify P – (Q ∩ R) = (P – Q) ∪ (P – R).

Answer»

The roster form of sets P, Q and R are P = {2, 3, 4, 5, 6, 7, 8, 9, 10}, Q = {2, 4, 6, 8, 10} and R = {4, 6, 8, 9, 10, 12} 

First, we find Q ∩ R = {4, 6, 8, 10} 

Then, P – (Q ∩ R) = {2, 3, 5, 7, 9} … (1) 

Next, P – Q = {3, 5, 7, 9} and 

P – R = {2, 3, 5, 7} 

And so, (P – Q) ∪ (P – Q) = {2, 3, 5, 7, 9} ... (2) 

Hence from (1) and (2), it verified that P – (Q ∩ R) = (P – Q) ∪ (P – R) 

Finding the elements of set Q 

Given, x = 2n 

n = 1 → x = 2(1) = 2 

n = 2 → x = 2(2) = 4

n = 3 → x = 2(3) = 6 

n = 4 → x = 2(4) = 8 

n = 5 → x = 2(5) = 10 

Therefore, x takes values such as 2, 4, 6, 8, 10
Discussion
24.

Test for the commutative property of union and intersection of the sets P = {x : x is a real number between 2 and 7} and Q = {x : x is an irrational number between 2 and 7}

Answer»

Commutative Property of union of sets 

(A ∪ B)’ = (B ∪ A) 

Here P = {3, 4, 5, 6}, Q = {√3, √5, √6} 

P ∪ Q = {3, 4, 5, 6} ∪ {√3, √5, √6} 

= {3, 4, 5, 6, √3, √5, √6} ... (1) 

Q ∪ P = {√3, √5, √6} ∪ {3, 4, 5, 6} = {√3, √5, √6, 3, 4, 5, 6} … (2) 

(1) = (2) 

∴ P ∪ Q = Q ∪ P

∴ It is verified that union of sets is commutative. 

Commutative Property of intersection of sets 

(P ∩ Q) = (Q ∩ P) P ∩ Q = {3, 4, 5, 6} ∩ {√3, √5, √6} = { } … (1) 

Q ∩ P = {√3, √5, √6} ∩ {3, 4, 5, 6} = { } … (2)

From (1) and (2) 

P ∩ Q = Q ∩ P 

∴ It is verified that intersection of sets is commutative.
Discussion
25.

If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}, then find (i) (P ∪ Q) ∪ R (ii) (P ∩ Q) ∩ S (iii) (Q ∩ S) ∩ R

Answer»

(i) (P ∪ Q) ∪ R 

(P ∪ Q) = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11} = {1, 2, 3, 5, 7, 9, 11} 

(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9} 

= {1, 2, 3, 4, 5, 7, 9, 11}

(ii) (P ∩ Q) ∩ S 

(P ∩ Q) = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11} = {2, 5, 9} 

(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8} 

= {2, 5}

(iii) (Q ∩ S) ∩ R 

(Q ∩ S) = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8} = {2, 3, 5} 

(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9} 

= {3, 5}
Discussion
26.

If A = {a, {a, b}}, write all the subsets of A.

Answer»

A = {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.
Discussion
27.

If A = {x, y, z} then the number of non-empty subsets of A is (1) 8 (2) 5 (3) 6 (4) 7

Answer»

(4) 7 

Number of non-empty subsets = 2 – 1 = 8

= 8 – 1 = 7
Discussion
28.

Let U = {0, 1, 2 , 3, 4, 5, 6, 7}, A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.(i) A’(ii) B’(iii) A‘ ∪ B’(iv) A’ ∩ B’(v) (A ∪ B)’(vi) (A ∩ B)’(vii) (A’)’(viii) (B’)’

Answer»

(i) A’ = U – A = {0, 1 ,2, y, 4, 5, 6, 7} – {1, 3, 5, 7} = {0, 2, 4, 6}

(ii) B’ = U – B = {0, 1, 2, 3, 4, 5, 6 ,7} – {0, 2, 3, 5, 7} = {1, 4, 6}

(iii) A’ ∪ B’ = {0, 2, 4, 6} ∪ {1, 4, 6} 

= {0, 1, 2, 4, 6}

(iv) A’ ∩ B’ = {0, 2, 4, 6} ∩ {1, 4, 6} = {4, 6}

(v) (A ∪ B)’ = U – (A ∪ B) = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7} 

= {4, 6}

(vi) (A ∩ B)’ = U – (A ∩ B) = {0, 1, 2, 3, 4, 5, 6, 7} – {3,5,7} = {0, 1, 2, 4, 6}

(vii) (A’)’ = U – A’ = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6} = {1, 3, 5, 7}

(viii) (B’)’ = U – B’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6} 

= {0, 2, 3, 5, 7}.
Discussion
29.

Represent the following sets in descriptive form.(i) P = {January, June, July}(ii) Q = {7, 11, 13, 17, 19, 23, 29}(iii) R = {x : x ∈ N, x &lt; 5}(iv) S = {x : x is a consonant in English alphabets}

Answer»

(i) P is the set of English Months beginning with J.

(ii) Q is the set of all prime numbers between 5 and 31.

(iii) R is the set of all natural numbers less than 5.

(iv) S is the set of all English consonants.
Discussion
30.

Identify the following sets as finite or infinite.(i) X = The set of all districts in Tamilnadu.(ii) Y = The set of all straight lines passing through a point.(iii) A = {x : x ∈ Z and x &lt; 5}(iv) B = {x : x2 – 5x + 6 = 0, x ∈ N}

Answer»

(i) Finite set

(ii) Infinite set

(iii) A = { ... , -2, -1, 0, 1, 2, 3, 4}

∴ Infinite set

(iv) x2 – 5x + 6 = 0

(x – 3) (x – 2) = 0

B = {3, 2}

∴ Finite set.
Discussion
31.

Find the number of subsets and the number of proper subsets of the following sets.(i) W = {red,blue, yellow}(ii) X = {x2 : x ∈ N, x2 ≤ 100}.

Answer»

(i) Given W = {red, blue, yellow}

Then n(W) = 3

The number of subsets = n[P(W)] = 23 = 8

The number of proper subsets = n[P(W)] – 1 = 23 – 1 = 8 – 1 

= 7

(ii) Given X = {1,2,3, }

X2 = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

n(X) = 10

The Number of subsets = n[P(X)] = 210 = 1024

The Number of proper subsets = n[P(X)] – 1 = 210 – 1 = 1024 – 1 

= 1023.
Discussion
32.

Fill in the blanks with appropriate cardinal numbers.

Answer»
S. Non(A)n(B)n(A ∪ B)n(A ∩ B)n(A - B)n(B - A)
1304565102035
2204555101035
3506287252537
430437032740
Discussion
33.

State whether the following sets are finite or infinite. (i) A = {x : x is a multiple of 5, x ∈ N}. (ii) B = {0, 1, 2, 3, 4, 75}. (iii) The set of all positive integers greater than 50.

Answer»

(i) A = {5, 10, 15, 20, …… } 

∴ A is an infinite set 

(ii) Finite

(iii) Let X be the set of all positive integers greater than 50 

Then X = {51, 52, 53, ….. } 

∴ X is an infinite set.
Discussion
34.

State which of the following sets are disjoint. (i) A = {2, 4, 6, 8}, B = {x : x is an even number &lt; 10, x ∈ N} (ii) X = {1, 3, 5, 7, 9}, Y = {0, 2, 4, 6, 8, 10} (iii) R = {a, b, c, d, e}, S = {d, e, b, c, a}

Answer»

(i) A = {2, 4, 6, 8}, B = {2, 4, 6, 8}

A ∩ B = {2, 4, 6, 8} ≠ Ø 

∴ A and B are not disjoint sets.

(ii) X ∩ Y = { } = Φ, X and Y are disjoint sets. 

(iii) R ∩ S = {a, b, c, d, e} ≠ Ø 

∴ R and S are not disjoint sets.
Discussion
35.

Find the cardinal number of the following sets. (i) A = {x : x is a prime factor of 12}. (ii) B = {x : x ∈ W, x ≤ 5}.(iii) X = {x : x is an even prime number}

Answer»

(i) Factors of 12 are 1, 2, 3, 4, 6, 12. So, the prime factors of 12 are 2, 3.

We write the set A in roster form as A = {2, 3} and hence n(A) = 2. 

(ii) In Tabular form B = {0, 1, 2, 3, 4, 5} 

The set B has six elements and hence n(B) = 6 

(iii) X = {2} [2 is the only even prime number] 

∴ n (X) = 1
Discussion
36.

The set P = {x | x ∈ Z, -1 &lt; X &lt; 1} is a (1) Singleton set (2) Power set (3) Null set (4) Subset

Answer»

(1) Singleton set 

P = {0}
Discussion
37.

Identify the following sets as null set or singleton set.(i) A = (x : x ∈ N, 1 &lt; x &lt; 2}(ii) B = The set of all even natural numbers which are not divisible by 2.(iii) C = {0}(iv) D = The set of all triangles having four sides.

Answer»

(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.

∴ Null set

(ii) B = { } ∵ All even natural numbers are divisible by 2.

∴ B is Null set

(iii) C = {0}

∴ Singleton set

(iv) D = { }

∵ No triangle has four sides.

∴ D is a Null set.
Discussion