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# Solutions of Miscellaneous Exercise

**1. Decide among the following sets, which sets are subset of one and another:**

**A { x : x ? R and x satisfy x ^{2} – 8x + 12 = 0}**

x^{2} – 8x + 12 = 0

x^{2} – 6x – 2x + 12 = 0

x(x – 6) - 2(x – 6) = 0

(x – 6) (x – 2) = 0

X = 6, 2

∴ A = 6, 2

**B = {2, 4, 6}, C = {2, 4, 6, 8…..}, D = {6}**

A ⊂ B

A ⊂ C

B ⊂ C

D ⊂ A

D ⊂ B

D ⊂ C

**2. (****i) if x ****? A and A ? B, then, x ? B**

Let A = {1, 2}

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

A ? B

{1, 2 } ? B

1 ∉ B, 2 ∉ B

x ∉ B

∴ False

**(ii) if A ⊂ B and B ? C then, A ? C**

x ? A, Let A = {1}

B = {1, 2}

C = {{1, 2}, 3}

A ∉ C

∴ False

**(iii) If A ⊂ B and B ⊂ C then, A ⊂ C**

x ? A x ? C

A ⊂ B A ⊂ C

? x ? B

B ⊂ C

∴ True

**(iv) If A ⊄ B and B ⊄ C, then A ⊄ C**

Let A = {1, 2}

B = {1, 3}

C = {1, 2, 4}

∴ False

**(v) If x ? A and A ⊄ B, then x ? C**

Let A = {1}

B = {1, 3},

∴ False

**(vi) If A ⊂ B and x ∉ B, then x ∉ A**

x ? A x ∉ B

A ? B A ⊂ B

x ? B x ∉ A

∴ True

**3. Let A, B and C be the sets such that A ****∪**** B = A ****∪**** C and A ∩ B = A ∩ C Show that B = C.**

A ∪ B = A ∪ B ? (1)

A ∩ B) = A ∩ B ? (2)

Show that:- B = C.

Sol:-

Taking L.H.S

B = B ∩ (A ∪ B)

= B ∩ (A ∪ C) from (1)

= (B ∩ A) ∪ (B ∩ C)

= (A ∩ C) ∪ (B ∩ C) from (2)

= (C ∩ A) ∪ (C ∩ B)

= C ∩ (A ∪ B)

= C ∩ (A ∪ C) from (1)

∴ B = C

Hence proved

**4. Show that the following four conditions are equivalent.**

**(i) A ⊂ B**

(i) ? (ii) All elements of A is in B then, A – B = ????

**(ii) A – B = ????**

(ii) ? (iii) All elements of A is in B then, A ∪ B = B

**(iii) A ****∪ B = B**

(iii) ? (iv) All elements of A is in B then, A ∩ B = A

**(iv) A ∩ B = A**

(iv) ? (i) All elements of A is in B then, A ⊂ B

**5. Show that if A ⊂ B, then C – B ⊂ C – A**

Let x ? C - A

x ? C and x ∉ B

x ? C and x ∉ A [? A ⊂ B]

x ? C – A

then, C – B ⊂ C – A

Hence proved

**6. Assume that P(A) = P(B). Prove that A = B.**

P(A) = P(B) ?(1)

Show that:- A = B

Sol:-

x ? A |
X ? B |

{x} ⊂ A |
{x) ⊂ B |

{x} ? P(A) |
{x} ? P(B) |

{x} ? P(B) from (1) |
{x} ? P(A) from (1) |

{x} ⊂ B |
{x} ⊂ A |

x ? B |
x ? A |

A ⊂ B |
B ⊂ A |

∴ A = B

Hence proved

**7. Is it true, that for any sets A and B, P(A) ****∪**** P(B) = P(A ****∪ B)? ****Justify your answer**

P(A) ∪ P(B) = P(A ∪ B) (given)

Let A = {1}

P(A) = { ????, {1} }

Let B = {2}

P(B) = { ????, {2}}

P(A) ∪ P(B) = {????, {1}, {2}} -(1)

Let A ∪ B = {1, 2}

P(A ∪ B) = {????, {1}, {2}, {1, 2}}

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

**8. ****Show that you any sets A and B, A = (A ****∩ B) ****∪ (A – B) ****and A ∪ (B – A) = A ∪ B.**

A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = A ∪ B

Taking R.H.S Taking L.H.S

(A ∩ B) ∪ (A ∩ B’) A ∪ (B ∩ A’)

A ∩ (B ∪ B’) (A ∪ B) ∩ (A ∪ A’)

A ∩ ? (A ∪ B) ∩ ?

A A ∪ B

Hence proved

**9. using properties of sets, show that**

**(i) A ∪ (A ****∩ B) = A**

Taking L.H.S

(A ∪ A) ∩ A ∪ B)

A ∩ (A ∪ B)

= A Hence proved.

**(ii) A ∩ (A ****∪ B) = A**

Taking L.H.S

(A ∩ A) ∪ (A ∩ B)

A ∪ (A ∩ B)

= A

Hence proved

**10. Show that** **A**** ∩ B = ****A**** ∩ C need not imply B = C**

Let A = {1, 2, 3}

B = {3, 4, 5}

C = {3, 5, 6}

A ∩ B = A ∩ C need not imply B = C (given)

{3} = {3} ? B ≠ C

Hence proved

**11. Let A and B be sets. If A ∩ X = B ∩ X = ???? and A** **∪ X = B ∪**** X for some set X, prove that A = B**

given:- A ∩ X = B ∩ X = ???? -(1)

A ∪ X = B ∪ X -(2)

to prove:- A = B

Proof:- Now,

Let A = A ∪ ????

= A ∪ (B ∩ X) from (1)

= (A ∪ B) ∩ (A ∪ X)

= (B ∪ A) ∩ (B ∪ X) from (2)

= B ∪ (A ∩ X)

= B ∪ ???? from (1)

A = B

∴ A = B

Hence proved

**12. find sets A, B and C such that A ∩ B, A ∩ C and B ∩ C are non-empty sets and A ∩ B ∩ C = ????**

Let A = {1, 2} A ∩ B = {2} ≠ ????

B = {2, 3} B ∩ C = {3} ≠ ????

C = {1, 3} A ∩ C = {1} ≠ ????

∴ A ∩ B ∩ C = ????

**13. In a survey of 600 students in a school, 150 students were found to be drinking tea and 225 drinking coffee, 100 were drinking both tea and coffee. Find how many students were drinking neither tea nor coffee**

? = 600, n(t) = 150, n(C) = 225, n(T ∩ C)

By using formula,

n(T ∪ C) = n(T) + n(C) – n(T ∩ C)

n(T ∪ C) = 150 + 225 - 100

n(T ∪ C) = 175

No. of students who take neither tea nor coffee = 600 – 275 = 325

**14. In a group of students, 100 students know Hindi, 50 know English and 25 know both each of the students knows either Hindi or English. How many students are in the group.**

n(H) = 100, n(E) = 150, n(H ∩ E) = 25

By using formula,

**owH** n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

n(H ∪ E) = 100 + 50 - 25

n(H ∪ E) = 125

∴ There are 125 students who knows either Hindi or English.

**15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 newspaper I, 9 read both H & I, 11 read both H & T, 8 read both T and I, 3 read all three newspapers. Find**

**(i) The no. of people who read at least one of the newspaper.**

**(ii) The no. of people who read exactly one newspaper**

N(H) = 25, n(T) = 26, n(I) = 26, n(H ∩ I) = 9, n(H ∩ I) = 11, n(T ∩ I) = 8, n(H ∩ T ∩ I) = 3

At least 1 of the newspaper = 8 + 8 + 10 + 6 + 3 + 5 + 12 = 52

Exactly 1 newspaper = 8 + 10 + 12 = 30

**16. In a survey it was found that 21 people liked product A, 26 liked product B, and 29 liked product C. If 14 people liked products B and A, 12 liked product C and A, 14 people liked products B & C and 8 liked all the three products. Find how many liked product C only.**

Liked product A = 21

Liked product B = 26

Liked product C = 29

Liked product A + B = 14

Liked product A + C = 12

Liked product B + C = 14

Liked product A + B + C = 8

∴ 11 people liked product C.