Math Assignment Class VIII | Square & Square Root

Basic concepts, definitions and formulas of mathematics, mathematics assignments for 9th standard to 10+2 standard, maths study material for 8th, 9th, 10th, 11th, 12th classes, Mathematics lesson plan for 10th and 12th standard, Interesting maths riddles and maths magic, Class-wise mathematics study material for students from 9th to 12
NCERT solution of chapter 1, class 11, miscellaneous exercise, Union of sets, intersection of sets, complement of sets and Venn diagram of sets, word problems on sets.
Q1) A = {x : x ∈ R and x satisfy x2-8x+12 = 0}
B = {2, 4, 6}, C = {2, 4, 6, 8, ....}, D = {6}
Solution: x2 - 8x + 12 = 0
⇒ x2 - 6x - 2x + 12 = 0
⇒ x(x - 6) -2(x - 6) = 0
⇒ (x - 2)(x - 6) = 0
⇒ x = 2, x = 6
⇒ A = {2, 6}
A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
Q2) (i) If x ∈ A and A∈B, then x ∈ B
Ans: False Statement
Reason: If A = {1,2}, B={{1,2}, 3}
Here 1∈A and A ∈ B but 1 ∉ B
So given statement is false
(ii) If A ⊂ B and B ∈ C, then A ∈ C
Let A = {1}, B={1, 2}, C ={{1, 2}, 3}
Here A ⊂ B and B ∈ C, But 1∉ C
So this is a false statement
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
Ans: True Statement
Explanation: Let A={1}, B={1,2}, C = {1,2,3}
Here 1 ∈ A and 1 ∈ B ⇒ A ⊂ B .... (1)
1, 2 ∈ B and 1,2 ∈ C ⇒ B ⊂ C ....... (2)
From (1) and (2) we get A ⊂ C
So this statement is true
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
Ans: False Statement
Let A = {1,2}, B = {2,3}, {1,2,3}
Here A⊄B and B ⊄C but A⊂C
So this statement is False
(v) If x ∈ A and A⊄B, then x∈ B
Ans: False Statement
Explanation: Let A = {1,2}, B = {2,3}
Here 1 ∈A and 1∉B ⇒ A⊄B
⇒ x ∈ A ⇒ x ∉ B
So given statement is false
(vi) If A ⊂ B and x ∉B, then x ∉ A
Ans: True Statement
Explanation: Let A = {1, 2}, B = {1, 2, 3}
Here 1, 2 ∈ A and 1, 2 ∈ B ⇒ A ⊂ B
Here 4 ∉ B ⇒ 4 ∉ A
So this statement is true
Question 3: If A ⋃ B = A ⋃ C and A ⋂ B = A ⋂ C then prove that B = C
Solution: A ⋃ B = A ⋃ C ...... (Given)
Taking B ⋂ on both side
B ⋂(A ሀ B) = B ⋂(A ⋃ C)
B = (B ⋂ A) ሀ (B ⋂ C)
= (A ⋂ B) ሀ (B ⋂ C)
= (A ⋂ C) ⋃ (B ⋂ C) (∵ A ⋂ B = A ⋂ C) ...... (1)
A ሀ B = A ⋃ C ...... (Given)
Taking C ⋂ on both side
C⋂(A ⋃ B) = C⋂(A ⋃ C)
(C⋂A) ⋃ (C⋂B) = C
C = (A⋂C) ⋃ (B⋂C) ....... (2)
From (1) and (2) we get
B = C
8. Show that for any sets A and B,
(i) A = ( A ⋂ B ) ⋃ ( A – B ) (ii) A ⋃ ( B – A ) = ( A ⋃ B )
Solution (i)
A = ( A ⋂ B ) ⋃ ( A – B )
RHS = ( A ⋂ B ) ⋃ ( A – B )
=( A ⋂ B ) ⋃ ( A ⋂ B' )
= A⋂(B⋃B')
= A ⋂ U = A = LHS
Solution (ii) A ⋃ ( B – A ) = ( A ⋃ B )
LHS = A ⋃ ( B – A )
= A ⋃ ( B – A )
= A ⋃ ( B⋂ A' )
= (A ⋃B) ⋂ (A ⋃ A')
= (A ⋃B) ⋂ U
= (A ⋃B) = RHS
9. Using properties of sets, show that
(i) A ⋃ ( A ⋂ B ) = A
(ii) A ⋂ ( A ⋃ B ) =
A.
Solution: (i) A ⋃ ( A ⋂ B ) = A
LHS = A ⋃ ( A ⋂ B )
= (A⋃A) ⋂ (A ⋃ B)
= A ⋂ (A ⋃ B) = A = RHS
Solution: (ii) A ⋂ ( A ⋃ B ) = A.
LHS = A ⋂ ( A ⋃ B )
= (A⋂A) ⋃ A⋂B
= A⋃ A⋂B = A = RHS
10. Show that A ⋂ B = A ⋂ C need not imply B = C.
Solution: Let A = {1,2}, B = {1,3}, C = {1, 4}
Here A ⋂ B = {1} and A ⋂ C = {1}
⇒ A ⋂ B = A ⋂ C but B ≠ C
11. Let A and B be sets. If A ⋂ X = B
⋂ X = Ñ„ and A ⋃ X = B
⋃ X for
some set X, show that A = B.
(Hints A = A ⋂ ( A ⋃ X ) , B = B ⋂ ( B ⋃ X )
and use Distributive law )
Solution:
Given: A ⋂ X = B ⋂ X = Ñ„ and A ⋃ X = B ⋃ X
To Prove : A = B
Proof: A ⋃ X = B ⋃ X ........ (Given)
Taking A⋂ on both side we get
A⋂(A ⋃ X) = A⋂(B ⋃ X)
A = (A⋂B)⋃(A⋂X)
= (A⋂B)⋃ Ñ„ .... (∵ A ⋂ X = Ñ„)
= (A⋂B)
⇒ A⊂B ......... (1)
A ⋃ X = B ⋃ X ........ (Given)
Taking B⋂ on both side we get
B⋂(A ⋃ X) = B⋂(B ⋃ X)
(B⋂A) ⋃ B⋂X = B
(B⋂A) ⋃Ñ„ = B
(A⋂B) = B
⇒ B⊂A ......... (2)
From (1) and (2) we get A = B
12. Find sets A, B and C such that A ⋂ B, B ⋂ C and
A ⋂ C are
non-empty sets and A ⋂ B ⋂ C =
Solution: Let A = {1,2}, B = {2,3}, C = {1,3}
A ⋂ B = {1,2} ⋂{2,3}= {2}
B ⋂ C = {2,3} ⋂{1,3}={3}
A ⋂ C = {1,2}⋂{1,3} = {1}
A ⋂ B ⋂ C = {1,2}⋂ {2,3}⋂ {1,3} = Ñ„
13. In a survey of 600 students in a school, 150 students
were found to be taking tea and 225 taking coffee, 100 were taking both tea and
coffee. Find how many students were taking neither tea nor coffee?
Solution: Total Students = 600 ⇒ n(U) = 600
No. of students taking Tea = 150 ⇒ n(T) = 150
No. of students taking Coffee = 225 ⇒ n(C) = 225
No. of students taking both Tea & Coffee = 100 ⇒ n(T⋂C) = 100
n(T⋃C) = n((T) + n(C) + n(T⋂C)
= 150 + 225 - 100 = 375 - 100 = 275
No. of students taking neither tea nor coffee
n(T' ⋂ C') = n(T⋃C)'
= n(U) - (T⋃C)
= 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 there in the group?
Solution
n(H) = 100, n(E) = 50, n(H⋂E) = 25
n(H⋃E)= n(H) + n(E) - n(H⋂E)
= 100 + 50 - 25
= 125
15. In a survey of 60 people, it was found that 25 people
read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and
I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers.
Find:
(i) the number of people who read at least one of the
newspapers.
(ii) the number of people who read exactly one newspaper.
Solution: Total People = 60
n(H) = 25, n(T) = 26, n(I) = 26
n(H⋂I) = 9, n(H⋂T) = 11, n(T⋂I)=8, n(H⋂T⋂I) = 3
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