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Correct option is (A) A – B

Answer is (B)

Because all the elements of set B = {1, 5, 9} is present in set C = (1, 5, 7, 9} 

Hence, B ⊂ C

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

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

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

(A ∪B)’ = U – (A ∪ B) 

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

and A’ ∩ B’ = {7, 8, 9, 10} ∩ {1, 5, 6, 7, 8, 9, 10} = {7, 8, 9, 10} 

(A ∪B)’ = A’ ∩ B’ 

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

(A ∩ B)’ = U – (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10} 

A’ ∪ B’ = {7, 8, 9, 10} ∪ {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10} 

(A ∩ B)’ = A’ ∪ B’

(i) (A ∪ B) ∩ B 

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

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

(ii) (A ∩ B) ∪ C 

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

(A ∩ B) ∪ C = {2, 3, 4} ∪ {4, 6, 8, 10} = {2, 3, 4, 6, 8, 10} 

(iii) A’ ∪ B’ 

A = {1, 2, 3, 4, 5, 6} 

U = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} 

A’ = U – A = {1,2,…..10} – {1, 2,… 6} = {7, 8, 9, 10}

Similarly, B’ = {1, 5, 6, 7, 8, 9, 10} 

A’ ∪ B’ = {7, 8, 9, 10} ∪ {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10}

(iv) (A ∪ B)’ 

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

(A ∪ B)’ = U – (A ∪ B) = {1, 2,… 10} – {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}

(i) {x : x ∈ R, a < x < b} = (a, b) 

(ii) {x : x ∈ R, 3 < x ≤ 5} = [3, 5] 

(iii) {x : x ∈ R, 0 ≤ x < 8}= [0, 8] 

(iv) {x : x ∈ R, -1 ≤ x ≤ 5} = [-1, 5]

False

{1, 2, 3, 4} and {3, 4, 5, 6}

Since, every element of both the sets are not same

They are not equal

So, the given statement is false

Answer is True

Answer is False

True

We know, every integer is a rational number

Hence, Z ⊂ Q and Q ∪ Z = Q

So, the given statement is true.

In generally X ∪ Y = {a: a ∈ X or a ∈ Y}

X ∩ Y = {a: a ∈ X and a ∈ Y}.

(i) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

A ∪ B = {x: x ∈ A or x ∈ B}

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

(ii) A = {1, 2, 3, 4, 5}

C = {7, 8, 9, 10, 11}

A ∪ C = {x: x ∈ A or x ∈ C}

= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

(iii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(iv) B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

B ∪ D = {x: x ∈ B or x ∈ D}

= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(v) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

A ∪ B = {x: x ∈ A or x ∈ B}

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

A ∪ B ∪ C = {x: x ∈ A ∪ B or x ∈ C}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(vi) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

A ∪ B = {x: x ∈ A or x ∈ B}

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

A ∪ B ∪ D = {x: x ∈ A ∪ B or x ∈ D}

= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(vii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

B ∪ C ∪ D = {x: x ∈ B ∪ C or x ∈ D}

= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

(viii) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

A ∩ B ∪ C = {x: x ∈ A and x ∈ B ∪ C}

= {4, 5}

(ix) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {4, 5}

(B ∩ C) = {x: x ∈ B and x ∈ C}

= {7, 8}

(A ∩ B) ∩ (B ∩ C) = {x: x ∈ (A ∩ B) and x ∈ (B ∩ C)}

= ϕ

(x) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

A ∪ D = {x: x ∈ A or x ∈ D}

= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(A ∪ D) ∩ (B ∪ C) = {x: x ∈ (A ∪ D) and x ∈ (B ∪ C)}

= {4, 5, 10, 11}

(i) According to the question,

Y = {1, 2, 3,…, 10} where a represents any element of Y

Y = {1, 2, 3,…, 10}

1² = 1, 2² = 4, 3² = 9

1, 4, 9 ∈ Y ⇒ 1, 2, 3 do not satisfy given condition

Hence,

{a: a ∈ Y and a²∉ Y} = {4, 5, 6, 7, 8, 9, 10}

(ii) According to the question,

Y = {1, 2, 3,…, 10} where a represents any element of Y

Y = {1, 2, 3,…, 10}

a + 1 = 6 ⇒ a = 5

⇒ 5 satisfies the given condition

Hence,

{a: a + 1 = 6, a ∈ Y } = {5}

(iii) According to the question,

Y = {1, 2, 3,…, 10} where a represents any element of Y

Y = {1, 2, 3,…, 10}

a is less than 6 ⇒ 1, 2, 3, 4, 5

1, 2, 3, 4, 5 satisfy the given condition

Hence,

{a: a is less than 6, a ∈ Y } = {1, 2, 3, 4, 5}

Given; U = {a, b, c} and B = {a, c, d, e} 

(i) (A ∪ B)’ = (A’ ∩ B’) 

(ii) (A ∩ B)’ = (A’ ∪ B’)

(A ∩ B) = {x:x ϵ A and x ϵ B}. 

= {2} 

(A∩B)’ means Complement of (A∩B) with respect to universal set U. 

So, 

(A∩B)’ = U – (A∩B) 

U – (A∩B)’ is defined as {x ϵ U : x ∉ (A∩B)’} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

(A∩B)’ = {2} 

U – (A∩B)’ = {1, 3, 4, 5, 6, 7, 8, 9} 

A’ means Complement of A with respect to universal set U. 

So, 

A’ = U – A 

U – A is defined as {x ϵ U : x ∉ A} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

A = {2, 4, 6, 8} 

A’ = {1, 3, 5, 7, 9} 

B’ means Complement of B with respect to universal set U. 

So, 

B’ = U – B 

U – B is defined as {x ϵ U : x ∉ B} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

B = {2, 3, 5, 7}. 

B’ = {1, 4, 6, 8, 9} 

A’ ∪ B’ = {x: x ϵ A or x ϵ B} 

= {1, 3, 4, 5, 6, 7, 8, 9} 

Hence verified

A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. 

Now, 

B ∪ C = {5, 6} 

A ∩ (B ∪ C) = {5} 

Similarly finding out R.H.S we get, 

A ∩ B = {2, 5} 

A ∩ C = {4, 5} 

(A ∩ B) ∩ (A ∩ C) = {5} 

L.H.S = R.H.S 

Hence, Verified.

A = All the natural numbers i.e. {1, 2, 3…..}

B = All the even natural numbers i.e. {2, 4, 6, 8…}

C = All the odd natural numbers i.e. {1, 3, 5, 7……}

D = All the prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …}

(i) A ∩ B

Here, A contains all elements of B.

∴ B ⊂ A = {2, 4, 6, 8…}

∴ A ∩ B = B

(ii) A ∩ C

Here, A contains all elements of C.

∴ C ⊂ A = {1, 3, 5…}

∴ A ∩ C = C

(iii) A ∩ D

Here, A contains all elements of D.

∴ D ⊂ A = {2, 3, 5, 7..}

∴ A ∩ D = D

(iv) B ∩ C

B ∩ C = ϕ

Since, there is no natural number which is both even and odd at same time.

(v) B ∩ D

B ∩ D = 2

Since, {2} is the only natural number which is even and a prime number.

(vi) C ∩ D

C ∩ D = {1, 3, 5, 7…}

= D – {2}

Hence, every prime number is odd except {2}.

A ∪ {1, 2} = {1, 2, 3, 5, 9}

The elements of A and {1, 2} together give us the result

Therefore smallest set of A can be

A = {1, 2, 3, 5, 9} – {1, 2}

A = {3, 5, 9}

Given; U is the universal set and \(A\subset U\)

(i) A ∪ A’ = U 

(ii) A ∩ A’ = Φ or {} 

(iii) ϕ ‘∩ A = Φ 

(iv) U’ ∩ A = Φ or {}

B – C is defined as {x ϵ B : x ∉ C} 

B = {2, 4, 6, 8} 

C = {3, 4, 5, 6} 

B – C = {2, 8} 

Now, (B – C)’ = U – (B – C) 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

B – C = {2, 8} 

(B – C)’ = {1, 3, 4, 5, 6, 7, 9}.

A’ = U – A 

((A’)’ = (U – A)’ 

= U – (U – A) 

= A 

A = {1, 2, 3, 4}

B’ means Complement of B with respect to universal set U. 

So, 

B’ = U – B 

U – B is defined as {x ϵ U : x ∉ B} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

B = {2, 4, 6, 8} 

B’ = {1, 3, 5, 7, 9}

(A ∩ C) = {x:x ϵ A and x ϵ C}. 

= {3, 4} 

(A∩C)’ means Complement of (A∩C) with respect to universal set U. 

So, 

(A∩C)’ = U – (A∩C) 

U – (A∩C)’ is defined as {x ϵ U : x ∉ (A∩C)’} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

(A∩C)’ = {3, 4} 

U – (A∩C)’ = {1, 2, 5, 6, 7, 8, 9}

Correct option is C) The set of integers < 10

The set of integers < 10 

{….., -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

i) x² = 4 ⇒ x = ± 2 

3x = 9 ⇒ x = 3 

The value of ‘x’ is not same in both cases, so this is a null set. 

ii) This is a null set, because the sum of the three angles of a triangle is equal to 180°.

Here, we have,

Suppose, the total number of people be n (P) = 950

The people who can speak English n (E) = 460

The people who can speak Hindi n (H) = 750

(i) How many people can speak both Hindi and English.

The people who can speak both Hindi and English = n (H ∩ E)

As we know,

n (P) = n (E) + n (H) – n (H ∩ E)

By substituting the values we get

950 = 460 + 750 – n (H ∩ E)

950 = 1210 – n (H ∩ E)

n (H ∩ E) = 260

∴ Number of people who can speak both English and Hindi are 260.

(ii) How many people can speak Hindi only.

As we can see that H is disjoint union of n (H–E) and n (H ∩ E).

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ H = n (H–E) ∪ n (H ∩ E)

n (H) = n (H–E) + n (H ∩ E)

750 = n (H – E) + 260

 n (H–E) = 490

∴ 490 people can speak only Hindi.

(iii) How many people can speak English only.

As we can see that E is disjoint union of n (E–H) and n (H ∩ E)

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ E = n (E–H) ∪ n (H ∩ E).

n (E) = n (E–H) + n (H ∩ E).

460 = n (H – E) + 260

n (H–E) = 460 – 260 = 200

Hence, 200 people can speak only English.

The well defined collections are sets.

For examples: The collection of good cricket players of India is not a set. However, the collection of all good player is a set.

Collection of vowels in English alphabets is a set.

Hence, we can say that every set is a collection, but every collection is not necessarily a set. Only well defined collections are sets.

i. A = {a| a is a natural number smaller than zero} 

Natural numbers begin from 1.

∴ A = { } 

∴ A is an empty set.

ii. B = {x | x² = 0}

Here, x = 0 

∴ x = 0 … [Taking square root on both sides] 

∴ B = {0} 

∴B is not an empty set.

iii. C = {x | 5x – 2 = 0, x ∈ N} 

Here, 5x – 2 = 0 

∴ 5x = 2

∴ x = 2/5

Given, x ∈ N

But, x = is not a natural number. 

∴ C = { } 

∴ C is an empty set.

Let 

Total number of People n(P) = 1000 

People who speak Hindi n(H) = 750 

People who speak Bengali n(B) = 400 

We have P = n(H∪B) 

⇒ n(P) = n(H)+n(B)– n (H ∩B) 

= 1000 = 750+400 – n (H ∩B) 

⇒n (H ∩B) = 150. 

∴ 150 People can speak both languages. 

H = (H–B) ∪ (H ∩B) (union is disjoint) 

∴ n(H) = n(H–B) +n(H ∩B) 

750 = n(H–B)+150 

n(H–B) = 600 

∴ 600 people speak Hindi. 

B = (B–H) ∪ (H ∩B) (union is disjoint) 

∴ n(B) = n(B–H) +n(H ∩B) 

400 = n(B–H)+150 

n(B–H) = 250 

∴ 250 people speak Bengali.

B = {x: x + 5 = 5} is not an empty set 

let x ∈ Z or x ∈ W 

then for x = 0 ⇒ x + 5 = 0 + 5 = 5 

So if x ∈ W, or x ∈ Z then for x = 0, 

x + 5 = 5 is true. 

Then the set B = {0} which is not an empty set. 

Note: But if x ∈ N 

We will have no ‘x’ such that x + 5 = 5 

then ‘B’ will be an empty set. 

But in the textbook it is not given whether x ∈ N (or) x ∈ W (or) x ∈ Z. 

Hence we consider first one.