# Class 11 NCERT Solutions- Chapter 1 Sets – Miscellaneous Exercise on Chapter 1 | Set 1

**Question 1: Decide, among the following sets, which sets are subsets of one and another:**

**A = {x : x âˆˆ R and x satisfy x ^{2}â€“ 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8, . . . }, D = {6} **

**Solution:**

At first, simplifying for set A

x^{2}â€“ 8x + 12 = 0

(x-6)(x-2) = 0

x= 6 or 2

Now, A = {2,6}

A set X is said to be a subset of a set Y if every element of X is also an element of Y.

Therefore, we can write : A âŠ‚ B, A âŠ‚ C, B âŠ‚ C, D âŠ‚ A, D âŠ‚ B, D âŠ‚ C

**Question 2: In each of the following, determine whether the statement is true or false.**

**If it is true, prove it. If it is false, give an example.**

**(i) If x âˆˆ A and A âˆˆ B, then x âˆˆ B**

**(ii) If A âŠ‚ B and B âˆˆ C, then A âˆˆ C**

**(iii) If A âŠ‚ B and B âŠ‚ C, then A âŠ‚ C**

**(iv) If A âŠ„ B and B âŠ„ C, then A âŠ„ C**

**(v) If x âˆˆ A and A âŠ„ B, then x âˆˆ B**

**(vi) If A âŠ‚ B and x âˆ‰ B, then x âˆ‰ A**

**Solution:**

(i) False

Example : Let x=1, A={1,2,3} and B = {{1,2,3},{4,5,6}}

Here x âˆˆ A and A âˆˆ B but x âˆ‰ B(ii) False

Example : Let A = {1}, B = {1,3} and C = {{1,3},{5,7}}

Here A âŠ‚ B and B âˆˆ C but A âˆ‰ C(iii) True

Proof : A âŠ‚ B : Set B contains all the elements in set A

B âŠ‚ C : Set C contains all the elements in set B

Hence, by transitivity property, set C contains all the elements in set A i.e. A âŠ‚ C(iv) False

Example :Let A = {1,2}, B = {3,4,5} and C = {1,2,6,7}

Here A âŠ„ B and B âŠ„ C but A âŠ‚ C(v) False

Example : Let x=1, A = {1,2} and B = {3,4,5}

Here x âˆˆ A and A âŠ„ B but x âˆ‰ B(vi) True

Proof : A âŠ‚ B : Set B contains all the elements in set A

So if x âˆ‰ B then also x âˆ‰ A

**Question 3: Let A, B, and C be the sets such that A âˆª B = A âˆª C and A âˆ© B = A âˆ© C. Show that B = C.**

**Solution:**

Given : A âˆª B = A âˆª C …(1)

A âˆ© B = A âˆ© C …(2)

A âˆª B = A + B – (A âˆ© B) // by principle of inclusion-exclusion

A âˆª C = A + C – (A âˆ© C) // by principle of inclusion-exclusion

A + B – (A âˆ© B) = A + C – (A âˆ© C) // from (1)

A + B – (A âˆ© B) = A + C – (A âˆ© B) // from (2)B = C

**Question 4: Show that the following four conditions are equivalent :**

**(i) A âŠ‚ B (ii) A â€“ B = Ï† (iii) A âˆª B = B (iv) A âˆ© B = A**

**Solution:**

//Showing (i)=(ii)

A âŠ‚ B means Set B contains all the elements in set A i.e. set A does not have any different element from B

It means A – B = Ï†

//Showing (i)=(iii)

All elements of set A are there in Set B so A âˆª B = B

//Showing (i)=(iv)

All elements of set A are there in Set B so A âˆ© B = A

From above explanation we can say that all aboveconditions are equivalent.

**Question 5: Show that if A âŠ‚ B, then C â€“ B âŠ‚ C â€“ A**

**Solution:**

//By taking an example

Let A = {1,2} and B = {1,2,3,4,5}

and C = {2,5,6,7,8}

Set B contains all the elements in set A so A âŠ‚ B

Now C-B = {6,7,8} //elements present in C but not in B

C-A = {5,6,7,8} //elements present in C but not in A

It is clearly seen that, Set C-A contains all the elements in Set C-B henceC â€“ B âŠ‚ C â€“ Ais proved.

**Question 6: Assume that P (A) = P (B). Show that A = B**

**Solution:**

P(X) represents power set of set X

To prove A = B we have to prove that A âŠ‚ B and B âŠ‚ A

(eg : if A = {1,2} then P(A) = {Ï†, {1}, {2}, {1,2}})

Power set of any set contains all the possible subsets of it.

A âˆˆ P(A)

as P(A)=P(B) so A âˆˆ P(B)

If A is P(B) then clearly A is subset of B.

A âŠ‚ B …(1)

Repeating above process for B âˆˆ P(B) we get

B âŠ‚ A …(2)

From above equations,A = B

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

**Solution:**

It is False

Let A = {1,2} and B = {2,3}

A âˆª B = {1,2,3}

P(A) = {Ï†, {1}, {2}, {1,2}}

P(B) = {Ï†, {2}, {3}, {2,3}}

P(A) âˆª P(B) = {Ï†, {1}, {2}, {3}, {1,2}, {2,3}} …(1)

P(A âˆª B) = {Ï†, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}} …(2)P(A) âˆª P(B) â‰ P(A âˆª B)//from (1) and (2)

**Question 8: Show that for any sets A and B, A = (A âˆ© B) âˆª (A â€“ B) and A âˆª (B â€“ A) = (A âˆª B)**

**Solution:**

(A âˆ© B) âˆª (A â€“ B)

(A âˆ© B) âˆª (A âˆ© Bâ€™) //(A â€“ B) = (A âˆ© Bâ€™)

A âˆ© (B âˆª Bâ€™)

A âˆ© U //B âˆª Bâ€™ = U where U represents universal set

A //by identity property

Hence, it is proved thatA = (A âˆ© B) âˆª (A â€“ B)A âˆª (B – A)

A âˆª (B âˆ© Aâ€™)

//B â€“ A = B âˆ© Aâ€™

(A âˆª B) âˆ© (A âˆª Aâ€™)

//distributive law

(A âˆª B) âˆ© U

//A âˆª Aâ€™ = U where U represents universal set

(A âˆª B)

Hence, it is proved thatA âˆª (B â€“ A) = (A âˆª B)

**Question 9: Using properties of sets, show that**

**(i) A âˆª (A âˆ© B) = A (ii) A âˆ© (A âˆª B) = A**

**Solution:**

(i)A âˆª (A âˆ© B)

(A âˆª A) âˆ© (A âˆª B) //distributive law

A âˆ© (A âˆª B) //A âˆª A = A

A //by absorption law

(ii)A âˆ© (A âˆª B)

(A âˆ© A) âˆª (A âˆ© B) //distributive law

A âˆª (A âˆ© B) //A âˆ© A = A

A //by absorption law

**Question 10: Show that A âˆ© B = A âˆ© C need not imply B = C**

**Solution:**

Let us assume that B â‰ C

Take A = {1,2}, B ={2,3} and C = {3,4} // here B â‰ C

A âˆ© B = {2}

A âˆ© C = Ï†

Here we can see that A âˆ© B â‰ A âˆ© C

Therefore, our assumption was wrong

Hence,B = Cis must for A âˆ© B = A âˆ© C

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