NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | MATHS IMPORTANT QUESTIONS | CHAPTER – 1 | SETS | EDUGROWN |
In This Post we are providing Chapter-1 SETS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.
NCERT MOST IMPORTANT QUESTIONS ON SETS
Q1.If A = { 1,2,3,4,5,6}, B = {2,4,6, 8} then find A – B
Ans. We are given the sets A ={ 1,2,3,4,5,6}, B = {2,4,6, 8}
A – B = { 1,2,3,4,5,6}- {2,4,6, 8} = {1, 3, 5}
Q2. Let A and B be two sets containing 3 and 6 elements respectively. Find the maximum and the minimum number of elements in A ∪ B.
Ans.There may be the case when atleast 3 elements are common between both sets
Let a set A = {a, b, c} and B = {a, b, c, d, e, f}
∴ A ∪ B = {a, b, c, d, e, f} implies that the minimum number of elements in A ∪ B are = 6
There may be the case when there are no any elements are common between both sets
lf A = {a, b, c}, B = { d, e, f, g, h, i}
A ∪ B = {a, b, c, d, e, f,g,h,i} implies that the maximum number of elements in A ∪ B are = 9
Q3.If A = {(x,y) : x² + y²= 25 where x, y ∈ W } write a set of all possible ordered pair .
Ans. We are given the set A = {(x,y) : x² + y²= 25 where x, y ∈ W }
All possible ordered pair of set A are following
For x = 0,y =5, x=3,y=4,for x =4, y =3,for x=5,y =0
A = {(0,5),(3,4),(4,3),(5,0)}
Q4.If A = {1,2,3}, B = {4, 5, 6} and C ={5} verify that A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C.
We are given the sets A = {1,2,3}, B = {4, 5, 6} ,C = {5}
B ∩ C = {5}
LHS
A ∪ ( B ∩ C) = {1,2,3,5}
(A ∪ B) and A ∪ C
A ∪ B = {1,2,3,4,5,6} and A ∪ C = {1,2,3,5}
RHS
(A ∪ B) ∩ A ∪ C = {1,2,3,5}
Therefore
A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C, Hence proved
Q5.From the adjoining Venn diagram, write the value of the following.
(a) A ‘
(b) B’
(c) (A ∩ B)’
Ans. From the venn diagram ,we have
U = {1,2,3,4………15}
A = {7,9.11}. B ={11,12,13,14}
A’ = U – A = {1,2,3,4………15} – {7,9.11} = {1,2,3,4,5,6,8,10,12,13,14,15}
B’ = U – B = {1,2,3,4………15} – {11,12,13,14}= {1,2,3,4,5,6,7,8,9,10,,15}
We have,(A ∩ B) = 11
(A ∩ B)’ = U – (A ∩ B) = {1,2,3,4………15} – {11} = (1,2,3,4,5,6,7,8,9,10,12,13,14,15}
Q6. If P(A) = P(B) show that A = B.
Ans. P(A} and P(B) implies that both are power sets of A and B respectively
Every set is an element of its power set , so A ∈ P(A)
Since, we are given that
P(A) = P(B)
Therefore, A ∈ P(B)
Indicates that every element of A belongs to the set B
So, A ⊂ B……(i)
Similarly B ∈ P(B)
Since, we are given that
P(A) = P(B)
Indicates that every element of B belongs to the set A
So, B ⊂ A……(ii)
From (i) and (ii), we get
A = B, Hence proved
Q7. Let A and B be sets ; if A∩X = B∩X = ∅ and A∪X = B∪X for some set X. Show that A=B.
Ans. We are given that A∩X = B∩X = ∅ and A∪X = B∪X
To prove A=B
Proof. A∪X = B∪X (given)
Multiplying both sides by A∩
A∩ (A∪X) =A∩ (B∪X)
Using distributive property
(A∩ A) ∪ (A ∩ X) = (A∩ B )∪ (A∩ X)
A∩Φ = (A∩ B) ∪ Φ
A = (A∩ B) …………(i)
A∪X = B∪X
Multiplying both sides by B∩
B∩(A∪X) = B∩(B∪X)
(B∩ A) ∪ (B ∩ X) = (B∩ B )∪ (B∩ X)
(B∩ A) ∪φ = B ∪ φ
B = (B∩ A)
B = (A∩ B) …………(ii)
From (i) and (ii)
A = B, Hence proved
Q8.If A ={1,2,3,4,5},then write the proper subsets of A.
Ans. The number of elements in the given sets A ={1,2,3,4,5} are =5
The number of proper subsets of any set are = 2n – 1
Where n = number of elements = 5
The number of proper subsets of any set are = 25 – 1 =32 – 1 = 31
Q9.Write the following sets in the Roster form
(i) A={x : x ∈ R, 2x+11 =15}
(ii)B={x |x² =x, x ∈ R}
(iii)C={x = x is a positive factor of the prime number p}
Ans.(i)We have, A={x : x ∈ R, 2x+11 =15}
2x+1= 15 ⇒x= 2
∴ A = {2}
(ii)We have,B={x |x² =x, x ∈ R}
∴ x² = x ⇒ x²-x= 0 ⇒x(x-1)= 0 ⇒x=0,1
∴ B ={0,1}
(iii)We have, C={x = x is a positive factor of the prime number p}
Sice positive factors of a prime u∪mer are 1 ad the number itself,we have
C={1,p}
Q10.For all sets A,B and C show that (A – B) ∩(A – C) = A – (B ∪ C).
Ans. Considering that
x ∈ (A – B)∩(A – C)
⇒ x ∈ (A – B) and x ∈ (A – C)
⇒ (x ∈ A and x∉ B) and (x ∈ A and x∉ C)
⇒ (x ∈ A ) and (x ∉B and x∉ C )
⇒(x ∈ A ) x∉ (B ∪C)∈
⇒x ∈ A – (B ∪C)
⇒(A – B) ∩(A – C)⊂A – (B ∪ C)…….(i)
Now,Considering that
y ∈A – (B ∪ C)
⇒ y ∈A and y ∉ (B ∪ C)
⇒y ∈A and (y ∉B and y ∉ C)
⇒(y ∈A and y ∉B) and (y ∈A and y ∉ C)
⇒y ∈ (A – B) and y ∈ (A – C)
⇒y ∈ (A – B) ∩ y ∈ (A – C)
⇒A – (B ∪ C) ⊂(A – B) ∩(A – C)………(ii)
From (i) and (ii)
(A – B) ∩(A – C)= A – (B ∪ C), Hence proved
Q11.Let A,B and C be the sets such that A∪B = A∪C and A ∩B = A ∩C,show that B = C.
Ans. According to question, A ∪ B = A ∪ C and A ∩ B = A ∩ C
To show, B = C
Let us assume, x ∈ B So, x ∈ A ∪ B
x ∈ A ∪ C
Hence, x ∈ A or x ∈ C
when x ∈ A, then x ∈ B
∴ x ∈ A ∩ B
As, A ∩ B = A ∩ C
So, x ∈ A ∩ C
∴ x ∈ A or x ∈ C
x ∈ C
∴ B ⊂ C
similarly, it can be shown that C ⊂ B
Hence, B = C
Q12.Show that for any sets A and B
A = (A∩B)∪(A-B)
Ans.We have to prove
A = (A∩B)∪(A-B)
Taking RHS and solving it
(A∩B)∪(A-B)
Using the property
A -B = A -(A∩B)
A-B = A∩B’
= (A∩B) ∪ (A∩B’)
Applying distributive property
A∩(B∪C) = (A∩B) ∪(A∩C)
Replacing C = B’ in LHS
A-B = A∩(B∪B’)
= A∩ (U) [since B∪B’ = U]
= A [since A∩ U = A]
= RHS, Hence proved
Q13. Write the following sets in the roster form:
(i) A = {x : x ∈ R, 2x + 15 = 15}
(ii) B = {x : x² = x, x ∈ R}
Ans(i) It is given to us that set
A = {x : x ∈ R, 2x + 11 = 15}
2x + 11 = 15
2x = 15 – 11
2x = 4
x = 2
Therefore in roster form it is written as A = {2}
(ii) It is given to us that
B = {x : x² =x, x ∈ R}
x² = x
x² – x = 0
x(x – 1) = 0
x = 0, x = 1
Therefore in roster form it is written as
B = {0, 1}
Q14. If A and B are subsets of the universal set U, then show that
(i) A ⊂ A ∪ B
Ans. Let’s prove that
A ⊂ A ∪ B
Let x ∈ A or x ∈ B
If x ∈ A then x ∈ A ∪ B
Hence A ⊂ A ∪ B
Q15. A,B and C are subsets of universal set if A = {2,4,6,8,12,20}, B ={3,6.9.12.15},C ={5,10,15,20} and U is the set of all whole numbers, draw a venn diagram showing the relation of U,A,B and C.
Ans.
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