# 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 x2â€“ 8x + 12 = 0}, B = {2, 4, 6},  C = {2, 4, 6, 8, . . . }, D = {6}

Solution:

At first, simplifying for set A
x2â€“ 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 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 above conditions 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 hence C â€“ B âŠ‚ C â€“ A is 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© U //B âˆª Bâ€™ = U where U represents universal set
A //by identity property
Hence, it is proved that A = (A âˆ© B) âˆª (A â€“ B)

A âˆª (B – A)
//B â€“ A = B âˆ© Aâ€™
(A âˆª B) âˆ© (A âˆª Aâ€™)
//distributive law
//A âˆª Aâ€™ = U where U represents universal set
(A âˆª B)
Hence, it is proved that A âˆª (B â€“ A) = (A âˆª B)

### (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 //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