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(i) {x:x ϵ N,x<7} 

This is read as x is such that x belongs to natural number and x is less than 7. It satisfies all condition of roster form. 

(ii) {x ϵ z: x=1/n+1,n ϵ W} 

This is read as x is such that x is an integer greater than or equal to 0. And it’s value is 1/x+1. 

(iii) {x N: x = 3n, n W} 

(iv) {x:x ϵ N,9<x<16)

(v) {x:x=0} 

(vi) {x ϵ N: x = n²,n≤10, n ϵ N} 

(vii) {x N: x = 2n, n N} 

(viii) {x N: x = 5ⁿ,0 < n<6 N)

(i) A = {x : x² ≤ 10, x ∈ Z}

First of all, x is an integer thus it can be positive and negative also.

x² ≤ 10

(-3)² = 9 < 10

(-2)² = 4 < 10

(-1)² = 1 < 10

0² = 0 < 10

1² = 1 < 10

2² = 4 < 10

3² = 9 < 10

The square root of next integers are greater than 10.

x ≤ √10

x = 0, ±1, ±2, ±3

A = {0, ±1, ±2, ±3}

(ii) B = {x : x = 1/(2n-1), 1 ≤ n ≤ 5}

Let us substituting the value of n to find the values of x.

At n = 1, x = 1/(2(1) - 1) = 1/1 

At n = 2, x = 1/(2(2) - 1) = 1/3 

At n = 3, x = 1/(2(3)-1) = 1/5 

At n=4, x = 1/(2(4)-1) = 1/7 

At n=5, x = 1/(2(5)-1) = 1/9 

x = 1, 1/3, 1/5, 1/7, 1/9 

∴ B = {1, 1/3, 1/5, 1/7, 1/9}

(iii) C = {x : x is an integer, -1/2 < x < 9/2}

Given as, x is an integer between -1/2 and 9/2

Therefore,  all integers between -0.5<x<4.5 are = 0, 1, 2, 3, 4

∴ C = {0, 1, 2, 3, 4}

(iv) D={x : x is a vowel in the word “EQUATION”}

Here, all vowels in the word ‘EQUATION’ are E, U, A, I, O

∴ D = {A, E, I, O, U}

(v) E = {x : x is a month of a year not having 31 days}

Since, a month has either 28, 29, 30, 31 days.

Out of 12 months in a year which are not having 31 days are:

February, April, June, September, November.

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

(vi) F = {x : x is a letter of the word “MISSISSIPPI”}

Since, letters in word ‘MISSISSIPPI’ are M, I, S, P.

∴ F = {M, I, S, P}.

(i) A = {1, 2, 3, 4, 5, 6}

Since, {x : x ∈ N, x < 7}

This is read as x is such that x belongs to natural number and x is less than 7. It satisfied the all condition of roster form.

(ii) B = {1, 1/2, 1/3, 1/4, 1/5, …}

Here, {x : x = 1/n, n ∈ N}

This is read as x is such that x = 1/n, here n ∈ N.

(iii) C = {0, 3, 6, 9, 12, ….}

Since, {x : x = 3n, n ∈ Z⁺, is the set of positive integers}

This is read as x is such that C is the set of multiples of 3 including 0.

(iv) D = {10, 11, 12, 13, 14, 15}

Here, {x : x ∈ N, 9 < x < 16}

This is read as x is such that D is the set of natural numbers which are more than 9 but less than 16.

(v) E = {0}

Since, {x : x = 0}

This is read as x is such that E is an integer equal to 0.

(vi) {1, 4, 9, 16,…, 100}

Here,

1² = 1

2² = 4

3² = 9

4² = 16

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10² = 100

Therefore, above set can be expressed in set-builder form as {x²: x ∈ N, 1≤ x ≤10}

(vii) {2, 4, 6, 8,….}

Here, {x: x = 2n, n ∈ N}

This is read as x is such that the given set are multiples of 2.

(viii) {5, 25, 125, 625}

Here,

5¹ = 5

5² = 25

5³ = 125

5⁴ = 625

Therefore, above set can be expressed in set-builder form as {5ⁿ: n ∈ N, 1≤ n ≤ 4}

(i) According to the question,

A = {x : x ∈ R, 2x + 11 = 15}

Roster form,

2x + 11 = 15

⇒ 2x = 15 – 11

⇒ 2x = 4

⇒ x = 2

Hence, A = {2}

(ii) According to the question,

B = {x | x² = x, x ∈ R}

Roster form,

x² = x

⇒ x² – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 or 1

Hence, B = {0, 1}

(iii) According to the question,

C = {x | x is a positive factor of a prime number p}

Roster form,

Only possible positive factors of a prime number p = 1 and p itself.

Therefore,

x = 1 or p

Hence, C = {1, p}

(i) {x : x is a letter before e in the English alphabet}

Therefore, when we read whole sentence it becomes x is such that x is a letter before ‘e’ in the English alphabet. Then letters before ‘e’ are a,b,c,d.

∴ Roster form will be {a,b,c,d}.

(ii) {x ∈ N: x² < 25}

x ∈ N that implies x is a natural number.

x² < 25

x < ±5

Here, as x belongs to the natural number that means x < 5.

Here, also all numbers less than 5 are 1,2,3,4.

∴ Roster form will be {1,2,3,4}.

(iii) {x ∈ N: x is a prime number, 10 < x < 20}

Here, X is a natural number and is between 10 and 20.

Since, X is such that X is a prime number between 10 and 20.

Prime numbers between 10 and 20 are 11,13,17,19.

∴ Roster form will be {11,13,17,19}.

(iv) {x ∈ N: x = 2n, n ∈ N}

X is a natural number also x = 2n

Hence, roster form will be {2,4,6,8…..}.

This an infinite set.

(v) {x ∈ R: x > x}

Any real number is equal to its value it is neither less nor greater.

Therefore, Roster form of such real numbers which has value less than itself has no such numbers.

∴ Roster form will be ϕ. This is called a null set.

(vi) {x : x is a prime number which is a divisor of 60}

All numbers which are divisor of 60 are = 1,2,3,4,5,6,10,12,15,20,30,60.

Then, prime numbers are = 2, 3, 5.

Hence, roster form will be {2, 3, 5}.

(vii) {x : x is a two digit number such that the sum of its digits is 8}

Here, numbers which have sum of its digits as 8 are = 17, 26, 35, 44, 53, 62, 71, 80

Hence, roster form will be {17, 26, 35, 44, 53, 62, 71, 80}.

(viii) The set of all letters in the word ‘Trigonometry’

Since, as repetition is not allowed in a set, now the distinct letters are

Trigonometry = t, r, i, g, o, n, m, e, y

Hence, roster form will be {t, r, i, g, o, n, m, e, y}

(ix) The set of all letters in the word ‘Better.’

Since, as repetition is not allowed in a set, now the distinct letters are

Better = b, e, t, r

Hence, roster form will be {b, e, t, r}

Here x:x it is read as x is such that x 

So now when we read whole sentence it becomes x is such that x is a letter before e in the English alphabet. Now letters before e are a,b,c,d. 

∴ Roster form will be {a,b,c,d}.

First thing analyze the given data. x ϵ N that implies x is a natural number.

x² < 25 

⇒ x < ±5 

As x belongs to the natural number that means x < 5. 

All numbers less than 5 are 1,2,3,4. 

∴ Roster form will be {1,2,3,4}.

Let M,P and C denote the set of students who opting mathematics ,physics and chemistry respectively. 

Number of students who opted. 

mathematics only = n(M) - n(M⋂P) - n(M⋂C) + n(M⋂P⋂C) 

= 100 - 30 - 28 +18 

= 60

A. A

Given: A and B are two given sets

To find: A ∩ (A ∪ B)

Let x ∈ A ∩ (A ∪ B)

⇒ x ∈ A and x ∈ (A ∪ B)

⇒ x ∈ A and (x ∈ A or x ∈ B)

⇒ (x ∈ A and x ∈ A) or (x ∈ A and x ∈ B)

⇒ x ∈ A or x ∈ A ∩ B

⇒ x ∈ A

Hence, A ∩ (A ∪ B) = A

(i) A ⊂ B

To prove that the following four statements are equivalent, we need to prove (i)=(ii), (ii)=(iii), (iii)=(iv), (iv)=(v)

Now, firstly let us prove (i)=(ii)

As we know, A–B = {x ∈ A: x ∉ B} as A ⊂ B,

Therefore, each element of A is an element of B,

∴ A–B = ϕ

Thus, (i)=(ii)

(ii) A – B = ϕ

As we need to show that (ii)=(iii)

On assuming A–B = ϕ

To prove: A∪B = B

∴ Every element of A is an element of B

Therefore, A ⊂ B and so A∪B = B

Thus, (ii)=(iii)

(iii) A ∪ B = B

As we need to show that (iii)=(iv)

On assuming A ∪ B = B

To prove: A ∩ B = A.

∴ A⊂ B and so A ∩ B = A

Thus, (iii)=(iv)

(iv) A ∩ B = A

Finally, we need to show (iv)=(i)

On assuming A ∩ B = A

To prove: A ⊂ B

Thus, A ∩ B = A, therefore A ⊂ B

Thus, (iv) = (i)

= A∩ (A ∪ B’) 

= A∩( A’∩ B’) [By De–morgan’s law] 

= (A ∩ A’) ∩ B’ [∴ A ∩ A’ = ϕ ] 

= ϕ ∩ B’ 

= ϕ 

= RHS

To find sets A, B and C such that A ∩ B = ϕ , A ∩ C = ϕ and A ∩ B ∩ C = ϕ

Take A = {1, 2, 3} 

B = {2, 4, 6} 

C = {3, 4, 7} 

Then, 

A ∩ B = {2} 

∴ A ∩ B ≠ ϕ 

A ∩ C = {3} 

∴ A ∩ C ≠ ϕ 

C ∩ B = {4} 

∴ C ∩ B ≠ ϕ

 But A, B and C do not have no elements in common, 

∴ A ∩ B ∩ C = ϕ.

A = {3, {2}} 

We have to find P(A) which is power set of A 

The power set of set A is collection of all possible subsets of A 

The possible subsets of A are {ϕ}, {3}, {{2}}, {3, {2}} 

Hence the power set P(A) will be 

P(A) = {{ϕ}, {3}, {{2}}, {3, {2}}}

A = {5, 6, 7} 

We have to find P(A) which is power set of A 

The power set of set A is collection of all possible subsets of A 

The possible subsets of A are {ϕ}, {5}, {6}, {7}, {5,6}, {5,7}, {6, 7}, {5, 6, 7} 

Hence the power set P(A) will be 

P(A) = {{ϕ}, {5}, {6}, {7}, {5,6}, {5,7}, {6, 7}, {5, 6, 7}}

Here, 

n(A) = 3 and n(B) = 6 

Now, 

n(A⋃B) = n(A) + n(B) - n(A⋂B) 

= 3 + 6 - n(A⋂B) 

= 9 - n(A⋂B) 

So,

n(A⋃B) is minimum whenever n(A⋂B) is maximum and it is possible only when A⊂B 

Now,A⊂B then max(n(A⋂B)) = n(A) = 3. 

∴ min(n(A⋃B) ) = 9-3 = 6

A = {x : x = 6n ∀ n ∈ N) 

As x = 6n hence for n = 1, 2, 3, 4, 5, 6… x = 6, 12, 18, 24, 30, 36… 

Hence A = {6, 12, 18, 24, 30, 36…} 

B = {x : x = 9n ∀ n ∈ N) 

As x = 9n hence for n = 1, 2, 3, 4… x = 9, 18, 27, 36… 

Hence B = {9, 18, 27, 36…} 

A ∩ B means common elements to both sets 

The common elements are 18, 36, 54, … 

Hence A ∩ B = {18, 36, 54, …} 

All the elements are multiple of 18 

Hence A ∩ B = {x: x = 18n ∀ n ∈ N}

Here, 

A = {X:XϵN,X is a multiple of 3} and 

B = {X:XϵN,X is a multiple of 5}. 

A⋂B = {X:XϵN,X is a multiple of 3} and{X:XϵN,X is a multiple of 5} 

= {X:XϵN,X is a multiple of 3 and 5} 

= {X:XϵN,X is a multiple of 3*5 = 15} 

= {X:XϵN,X is a multiple of 15}

Here, we have,

Since, teachers for teaching physics or math = 20

Teachers for teaching physics and math = 4

Teachers for teaching maths = 12

Suppose teachers who teach physics be ‘n (P)’ and for Maths be ‘n (M)’

Then,

Since, teachers who teach for physics or math = n (P ∪ M) = 20

4 teachers who teach for physics and math = n (P ∩ M) = 4

12 teachers who teach for maths = n (M) = 12

As we know,

n (P ∪ M) = n (M) + n (P) – n (P ∩ M)

On substituting the values we get,

20 = 12 + n (P) – 4

20 = 8 + n (P)

n (P) =12

Hence, there are 12 physics teachers.

Let M = set of teachers who teach Mathematics 

P = set of teachers who teach Physics 

Then, n(M∪P) = 20, n(M) = 12, n(M∩P) = 4 

n(P) = n(M∪P) + n(M∩P)- n(M) 

= 20 + 4-12 = 12.