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False 

Since 6 is divisible by both 3 and 2. 

Thus R ∩ S ≠ φ

True 

Since Q ⊂ R 

So Q ∩ R = Q

True

Since Q ⊂ R

So, Q ∩ R = Q

∴ x is an integer between -1/2 and 9/2 

So all integers between given values -0.5<x<4.5

0,1,2,3,4,

∴ C = {0,1,2,3,4}

All months of a year do not have 31 days : 

February, April, June, September, November. 

E: { February, April, June,September,November}

All vowels in equation are E,U,A,I,O 

∴ D = {A,E,I,O,U}

Substituting the values of n we will get the solutions 

At n = 1, x = \(\frac{1}{2(1)-1}\) = \(\frac{1}1\)

At n = 1,   x = \(\frac{1}{2(2)-1}\) = \(\frac{1}3\) 

At n = 1,   x = \(\frac{1}{2(3)-1}\) = \(\frac{1}5\) 

At n = 1,   x = \(\frac{1}{2(4)-1}\) = \(\frac{1}7\) 

At n = 1,  x = \(\frac{1}{2(5)-1}\) = \(\frac{1}9\)

∴  x = 1,\(\frac{1}3\),\(\frac{1}5\),\(\frac{1}7\),\(\frac{1}9\)

Correct option is A) Infinite set

Correct option is B) {2, 4, 6, 8}

A ⊄ B 

Explanation: we have, A = {2,4,6} and B = {1,2,4,8,16,32}. 

Thus , A ⊄ B , since 6 ∉ B.

A ⊄ B

We have, A = set of triangles and B = set of rectangles. 

Now, we can see elements of A does not belong to B since the set of rectangles does not include a set of triangles.

A ⊂ B 

Explanation: Set of squares is a subset of set of rectangles since all squares are rectangles.

A ⊄ B 

Explanation: since all isosceles triangles are not equilateral triangles. Therefore set of isosceles triangle is not contained in the set of equilateral triangle.

(i) A = {x: x is an integer and –3 < x < 7}

The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.

Therefore, the given set can be written in roster form as

A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x: x is a natural number less than 6}

The elements of this set are 1, 2, 3, 4, and 5 only.

Therefore, the given set can be written in roster form as

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

(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8} The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.

Therefore, this set can be written in roster form as

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = {x: x is a prime number which is a divisor of 60}

60 = 2 × 2 × 3 × 5

The elements of this set are 2, 3, and 5 only.

Therefore, this set can be written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY

There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.

Therefore, this set can be written in roster form as

E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER

There are 6 letters in the word BETTER, out of which letters E and T are repeated.

Therefore, this set can be written in roster form as

F = {B, E, T, R}

A = {x | 3x – 1 = 2}

Here, 3x – 1 = 2 

∴ 3x = 3 

∴ x = 1 

∴ A = {1} …(i)

B = {x | x is a natural number but x is neither prime nor composite} 

1 is the only number which is neither prime nor composite, 

∴ x = 1 

∴ B = {1} …(ii)

C = {x | x G N, x < 2}

1 is the only natural number less than 2.

∴ x = 1 

∴ C = {1} …(iii)

∴ The element in sets A, B and C is identical. …

[From (i), (ii) and (iii)] 

∴ A, B and C are equal sets

(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identify a month that belongs to this collection.

  Hence, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection because the criteria for determining a writer’s talent may vary from person to person.

  Hence, this collection is not a set.

(iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person. 

Hence, this collection is not a set.

(iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection.

Hence, this collection is a set.

(v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection.

Hence, this collection is a set.

(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection.

Hence, this collection is a set.

(vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection.

Hence, this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter.

Hence, this collection is a set.

(ix) The collection of most dangerous animals of the world is not a welldefined collection because the criteria for determining the dangerousness of an animal can vary from person to person.

Hence, this collection is not a set.

The correct answer is (C).

Since, let A and B be such sets, i.e., n (A) = m, n (B) = n

So,  n (P(A)) = 2^m, n (P(B)) = 2ⁿ

Thus, n (P(A)) – n (P(B)) = 56, i.e., 2^m – 2ⁿ = 56

⇒ 2ⁿ (2^{m – n} – 1) = 2³ 7

⇒ n = 3 , 2^{m – n} – 1 = 7

⇒ m = 6

Let F = set of people speaking French 

S = set of people speaking Spanish. 

Then n(S) = 20, n(F) = 50 n(F∩S) = 10 

∴ n(F∪S) = n(S) + n(F)-n(F∩S) 

= 20 + 50-10 = 60 

∴ 60 people speak French or Spanish or both.

Let C = set of students who like cricket F = set of students who like football. 

Then n(C uf) = 35, n(C) = 24, n(F) 

= 16 n(C n F) = n(C) + n(F) – n(C u F) 

= 24 + 16-35 = 5 

∴ 5 Students like to play both games.

Given: 

n(P) = 40 

n (P ∪ Q) = 60 

n (P ∩ Q) =10 

n(Q) = ? 

We know, 

n (P ∪ Q) = n(P) + n(Q) – n (P ∩ Q) 

Substituting the values we get 

60 = 40+n(Q)–10 

60 = 30+ n(Q) 

n(Q) =30 

∴ Q = has 30 elements.

Natural numbers = 1, 2, 3, 4, 5,…

 Odd Natural numbers = 1, 3, 5, 7, 9, 11, … 

No odd natural number is divisible by 2. 

∴ no elements in this set 

∴ It is a null set.

Prime numbers = Those numbers which are divisible by 1 and number itself. 

Prime numbers = 2, 3, 5, 7, 11, 13,…, 83, 89, 97, … 

Prime number greater than 90 = 97 

Prime number less than 96 = 89 

Prime number less than 96 but greater than 90 = ɸ 

∴ The set is empty 

∴ It is a null set

Prime numbers = Those numbers which are divisible by 1 and number itself. 

Prime numbers = 2, 3, 5, 7, 11, 13,… 

Even Prime number = 2 

∴ set is not empty. 

∴ It is not a null set

Natural numbers = 1, 2, 3, 4, 5, 6,… 

If x = 1, then 2x + 3 = 2(1) + 3 = 2 + 3 = 5 ≠ 4 

∴ no elements in the set B because the equation 2x + 3 = 4 is not satisfied by any natural number of x. 

∴ It is a null set.

Natural numbers = 1, 2, 3, 4, 5, 6, 7,…

Natural number greater than 1 (1 < x) = 2, 3, 4, 5, .. 

Natural number less than or equal to 2 (x ≥ 2) = 1 

A number cannot be simultaneously greater than 1 and less than equal to 2 

∴ no elements in this set 

∴ It is a null set.

Natural numbers = 1, 2, 3, 4, 5, 6,… 

If x = 1, then x² + 1 = (1)² + 1 = 1 + 1 = 2 ≠ 0 

∴ no elements in the set B because the equation 2x + 3 = 4 is not satisfied by any natural number of x.

∴ It is a null set.

Given: x = 2n + 1 and n ≤ 5 

⇒ n = 0, 1, 2, 3, 4 and 5 [∵ n ∈ W] 

Given x = 2n + 1 

n = 0, x = 2 × 0 + 1 = 1 

n = 1, x = 2 × 1 + 1 = 3 

n = 2, x = 2 × 2 + 1 = 5 

n = 3, x = 2 × 3 + 1 = 7 

n = 4, x = 2 × 4 + 1 = 9 

n = 5, x = 2 × 5 + 1 = 11

 So, the elements of B are 1, 3, 5, 7, 9 and 11 

∴, B = {1, 3, 5, 7, 9, 11}

Given: x = 2n and n ≤ 5 

⇒ n = 1, 2, 3, 4 and 5 [∵ n ∈ N] 

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 

So, the elements of A are 2, 4, 6, 8 and 10 

∴, A = {2, 4, 6, 8, 10}

Whole numbers = 0, 1, 2, 3, … 

If we take x = 0 then x + 3 = 0 + 3 = 3 

If we take x = 1 then x + 3 = 1 + 3 = 4 > 3 

So, 0 is the element of set E because it satisfies the given equation. 

∴ It is not a null set.

Given; A = {2x : x ϵ N}, 1 ≤ x < 4}, B = {x + 2) : x ϵ N and 2 ≤ x < 5} and C = {x : x ϵ N and 4 < x < 8} 

According to the given conditions; A = {2, 4, 6}, B = {4, 5, 6} and C = {5, 6, 7} 

(i) A ∩ B = {4, 6}

(ii) A ∪ B = {2, 4, 5, 6} 

(iii) (A ∪ B) ∩ C = {5, 6}

: Natural numbers start from 1 

A = {1, 2, 3, 4, 5, 6, 7} 

B = {2, 3, 5, 7} 

C = {1, 3, 5, 7, 9} 

(i) B∩C = {3, 5, 7} 

AU(B∩C) = {1, 2, 3, 4, 5, 6, 7} 

A∪B = {1, 2, 3, 4, 5, 6, 7} 

A∪C = {1, 2, 3, 4, 5, 6, 7, 9}

(A∪B)∩(A∪C) = {1, 23, 4, 5, 6, 7} 

A∪(B∩C) = (A∪B)∩(A∪C) 

Hence proved 

(ii) B∪C = {1, 2, 3, 5, 7, 9} 

A∩(B∪C) = {1, 2, 3, 5, 7} 

A∩B = {2, 3, 5, 7} 

A∩C = {1, 3, 5, 7} 

(A∩B)∪(A∩C) = {1, 2, 3, 5, 7} 

A∩(B∪C) = (A∩B)∪(A∩C)

(i) A = (-4,0) 

Explanation: All the points between -4 and 0 belong to the open interval (-4,0) but -4 ,0 themselves do not belong to this interval. 

(ii) B = [0,3) 

Explanation: B = {x : x ϵ R, 0 ≤ x < 3} is an open interval from 0 to 3, including 0 but excluding 3. 

(iii) C = (2,6] 

Explanation: C = {x : x ϵ R, 2 < x ≤ 6} is an open interval from 2 to 6, including 6 but excluding 2.

(iv) D = [-5,2] 

Explanation: D = {x : x ϵ R, –5 ≤ x ≤ 2} is a closed interval from -5 to 2 and contains the end points. 

(v) E = [-3,2) 

Explanation: E = {x : x ϵ R, –3 ≤ x < 2} is an open interval from -3 to 2, including -3 but excluding 2. 

(vi) F = [-2,0) 

Explanation: F = {x : x ϵ R, –2 ≤ x < 0} is an open interval from -2 to 0, including -2 but excluding 0.

Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements & we write A = B

We have, 

E = {x : x ϵ Z, x² ≤ 4}

Here, x ∈ Z and x² ≤ 4 

If x = -2, then x² = (-2)² = 4 = 4 

If x = -1, then x² = (-1)² = 1 < 4 

If x = 0, then x² = (0)² = 0 < 4 

If x = 1, then x² = (1)² = 1 < 4 

If x = 2, then x² = (2)² = 4 = 4 

So, 

E = {-2, -1, 0, 1, 2} 

and F = {x : x ϵ Z, x² = 4} 

Here, x ∈ Z and x² = 4 

If x = -2, then x² = (-2)² = 4 = 4 

If x = 2, then x² = (2)² = 4 = 4 

So, F = {-2, 2} 

∴ E ≠ F because the elements in the both the sets are not equal.