### Common Errors in Secondary Mathematics

Common Errors Committed  by the  Students  in Secondary Mathematics   Errors  that students often make in doing secondary mathematics  during their practice and during the examinations  and their remedial measures are well explained here stp by step.  Some Common Errors in Mathematics

# Class 11  Chapter -1 Set TheoryMiscellaneous Exercise

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: x- 8x + 12 = 0

⇒ x- 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:   B = A  C  ...... (Given)

Taking B ⋂  on both side

B ⋂(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)

ሀ B = A ⋃ C  ...... (Given)

Taking C ⋂  on both side

C⋂(⋃ B) = C⋂(A ⋃ C)

(C⋂A) ⋃ (C⋂B) = C

C = (A⋂C) ⋃ (B⋂C) ....... (2)

From (1) and (2) we get

B = C

Question 4: Show that the following four conditions are equivalent
(i)  A⊂B  (ii) A - B = ф   (iii) A⋃B  = B (iv) A⋂B = A
Solution: Let A = {1, 2}, B = {1, 2, 3}
(i) 1, 2 ∈ A and 1, 2 ∈ B ⇒ A ⊂ B

(ii) A - B = {1, 2} - {1, 2, 3} = ф

(iii) 1,2 ∈ A and 1, 2 ∈ B ⇒ A⊂B
ሀ B = {1, 2} ሀ {1, 2, 3} = {1, 2, 3} = B

(iv)  1,2 ∈ A and 1, 2 ∈ B ⇒ A ⊂ B
A⋂B = {1, 2} ⋂ {1, 2, 3} = {1, 2} = A

Question 5: If A ⊂ B, then prove that C - B ⊂ C - A
Solution
Let  x ∈ C - B ⇒ x ∈ C⋂B'
⇒ x ∈ C and x ∈ B'
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A
⇒ x ∈(C-A)
Now we have x ∈ C - B ⇒ x ∈(C - A)
⇒ C - B ⊂ C - A
Question 6: If P(A) = P(B) then prove that A =B
Solution: Let X⊂A
Let x ∈ X⇒ x ∈ A
Now X ⊂ A ⇒ X ∈ P(A)
⇒ X ∈ P(B)   .... (∵ P(A) = P(B) )
⇒ X ⊂ B
⇒ x ∈ X⇒ x ∈ B
⇒x ∈ A ⇒ x ∈ B
⇒ A ⊂ B  ........  (1)
Let Y⊂B
Let y ∈ Y⇒ y ∈ B
Now Y ⊂ B ⇒ Y ∈ P(B)
⇒ Y ∈ P(A)   .... (∵ P(A) = P(B) )
⇒ Y ⊂ A
⇒ y ∈ Y⇒ y ∈ A
⇒y ∈ B ⇒ y ∈ A
⇒ B ⊂ A  ........  (2)
From equation (1) and (2) we have   A =B
Hence prove the required result

Question 7: Is it true that for any sets A and B, P(A)  P(B) = P(AB)? Justify your answer.
Solution: No it is not true to say that for any two sets A and B
P(A) ⋃ P(B) = P(A⋃B)
Reason: Let A={1,2}, B = {2, 3}
P(A) = {ф, {1}, {2}, {1, 2}}
P(B) = {ф, {2}, {3}, {2, 3}}
P(A) ⋃ P(B) = {ф, {1}, {2}, {3}, {1,2}, {2, 3}}
A⋃B = {1,2}⋃ {2, 3} = {1,2,3}
P(A⋃B) = {ф, {1}, {2}, {3}, {1,2}, {2, 3}, {1,3}, {1,2,3}}
Here we see that elements of P(A) ⋃ P(B) and P(A⋃B) are not same
∴  P(A) ⋃ P(B) ≠ P(A⋃B)

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)   ( B – A ) = ( A  B )

LHS =  ( B – A )

( B – 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  B )

= (AA⋂ (⋃ B)

= A ⋂ (⋃ B) = A = RHS

Solution: (ii)  ( A  B ) = A.

LHS =  ( 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  B = {1} and  C = {1}

⇒  B = A  C but  B ≠ C

11. Let A and B be sets. If A X = B X = ф 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:  X = B  X = ф  and   X = B  X

To Prove : A = B

Proof:     X = B  X ........ (Given)

Taking A⋂ on both side we get

A⋂( X) = A⋂( X)

A = (A⋂B)⋃(A⋂X)

= (A⋂B)⋃ ф  ....  (∵  X =  ф)

= (A⋂B)

⇒ A⊂B ......... (1)

X = B  X ........ (Given)

Taking B⋂ on both side we get

B⋂( X) = B⋂( X)

(BA) ⋃ B⋂X = B

(BA) ф = 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 = ф.

Solution: Let A = {1,2}, B = {2,3}, C  = {1,3}

B = {1,2} {2,3}= {2}

C = {2,3} {1,3}={3}

C = {1,2}{1,3} = {1}

B ⋂ C = {1,2} {2,3} {1,3} = ф

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(TC) = 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(HE)= 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

(i)  No of people who read at least one newspaper = n(H⋃T⋃I)
= 8+10+12+8+6+3+5 = 52
(ii) Number of people who read exactly one 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 A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution : n(A) = 21, n(B) = 26, n(C) = 29,
From the Venn diagram shown above it is clear that
No. of people who liked people C only = 11 Ans