RBSE Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 1 Sets Miscellaneous Exercise

Question 1.
Decide, among the following sets, which sets are subsets of one and another :
A = {x:x ∈ R and x satisfy x² - 8x +12 = 0}
B = {2, 4, 6}, C = {2, 4, 6, 8...............}, D = {6}
Answer:
Here A = {x:x ∈ R and satisfy x² - 8x + 12 =0}
Solving x² - 8x + 12 = 0
or x² - 6x -2x + 12 = 0
or x(x - 6) - 2(x - 6) = 0
or (x - 6) (x - 2) = 0
∴ x = 6, 2
Thus, A = {2, 6}
and given B = {2, 4, 6},
C = {2, 4, 6, 8, ...............} ,
D = {6}
Thus, A is subset of B and C
A ⊂ B, A ⊂ C,B ⊂ C
and D ⊂ A, D ⊂ B, D ⊂ 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 ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iiii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊂ Band B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Answer:
(i) Statement is false, since if A = {1, 2, 3} and B = {1, 2, 3}, 4}. Here A e B, since {1,2,3} is element of B but 1 ∉ B, 2 ∉ B and 3 ∉ B.

(ii) Statement is false, since if A = {1, 2}, B = {1, 2, 3} and C = {{1, 2, 3}, 4} then 4 ⊂ B,B ∈ C,then A ⊂ C since {1, 2} is not element in set C.

(iii) Statement is true, since A ⊂ B ⇒ x ∈ A and x ∈ B or B ⊂ C ⇒ x ∈ B and x ∈ C.
Also A = {a, b}, B = {a, b, c},C = {a, b, c, d}
Here x ∈ A ⇒ x ∈ B(A ⊂ B)
x ∈ B ⇒ x ∈ C(B ⊂ C)
Thus x ∈ A ⇒ x ∈ C
Thus A ⊂ C

(iv) Statement is false, if A = {a, b), B = {c, d, e} and C = {a, b,f}.
Here A ⊄ B and B ⊄ C, but A ⊂ C since both element of A occurs in C. Thus, A ⊄ B and B ⊄ C does not mean and A ⊄ C.

(v) Statement is false if
A = {3, 4}, B = {5, 6, 7}
Clearly 3 ∈ A and A ⊄ B then 3 ∈ B. Thus, x ∈ A and A ⊄ B does not means x ∈ B.

(vi) Statement is true, since ii A ⊂ B then all element of A will be element of B, x ∈ A ⇒ x ∈ B
Now, if any element not occurs in B then if will not be occurs in A, i. e.,
x ∉ B ⇒ x ∉ A
Thus, A ⊂ B then A ∉ B then x ∉ A.

Question 3.
Let A, B and C be the sets such that A ∪ B = A ∪ C and A nB = A nC, show that B = C.
Answer:
Given: A ∪ B = A ∪ C and A nB = A nC, then to prove B = C
∵ A ∪ B = A ∪ C
(A ∪ B) ∩ B = (A ∪ C) ∩ B
or (A ∩ B) ∪ (B ∩ B) = (A ∩ B) ∪ (C nB) (By distribution law)
or B = (A ∩ B) ∪ (C ∩ B) [∴(A ∩ B) ∪(B ∩ B) = B]
or B = {A ∩ B) ∪ (B ∩ C)
(∵ C ∩ B = B ∩ C) ...(1)

Again (A ∪ B) = (A ∪ C) (given)
or (A ∪ B) ∩ C = (A ∪ C) ∩ C
or (A ∩ C) (B ∩ C) = (A ∩ C) ∪ (C ∩ C) (By distribution law)
= C

or C = (A ∩ C) ∪ (B ∩ C) ...(2)
but A ∩ B = A ∩ C given,
In equation (2), replacing (A ∩ C) by (A ∩ B)
C = (A ∩ B) ∪ (B ∩ C) ...(3)
From equation (1) and (3),
B = C
Hence Proved

Question 4.
Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A - B = Φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Answer:
(i) Let A ⊂ B, then all elements of A are in set B. But A - B = {x:x ∈ A but x ∉ B}
(ii) Now A - B means elements of A which are not in B, but A ⊂ B then A - B = Φ.
Thus A ⊂ B and A - B = Φ are equivalent.
A - B = Φ ⇒ A ⊂ B
Hence Proved

(iii) Let A ∪ B = B
∵ A ⊂ B ∪ B
Thus A ⊂ B (∵ A ∪ B = B)
i.e., A ∪ B = B ⇒ A ⊂ B Hence Proved

(iv) Let A ∩ B = A
A ∩ B ⊂ B
A ⊂ B (∵ A ∩ B = A)
i.e., A ∪ B = A ⇒ A ⊂ B Hence Proved

Question 5.
Show that if AcB, then C -B ⊂ C -A
Answer:
Let x ∈ C - B ... (i)
⇒ x ∈ C and x £ B
⇒ x ∈ C and x ∈ A {∵ A ⊂ B}
⇒ x ∈ (C - A) ............(ii)
From (i) and (ii)
Hence Proved

Question 6.
Assume that P (A) = P (B), show that A = B.
Answer:
Let x ∈ A
{x} ⊂ A as well as {x} ⊂ P (A)
or {x} ⊂ P (B) {∵ P (A) = P (B)}
x ∈ B
A ⊂ B ...(1)
Again y ∈ B
{y} ⊂ B as well as {y} ⊂ P (B) or {y} ⊂ P(A) {∵ P (B) = P (A)}
y ∈ A ...(2)
From (1) and (2) A = B Hence Proved

Question 7.
Is it true that for any sets A and B .
P(A) ∪ P(B) = P(A ∪ B) Justify your answer.
Answer:
Let A and B are two sets and A = {1} and B = {a}
Now subsets of A and B are respectively Φ, {1} and Φ and {a}.
Thus, P(A) = {Φ,{1}} and
P(B) = {Φ, {a}}
Then, P(A) ∪ P(B) = {{Φ}{1} {a}} .............(1)
Now, A ∪ B = {1, a}
Subsets A ∪ B are Φ, {1}, {a} and {1, a}
Thus, P(A ∪ B)= {Φ, {1}, {a}, {1, a}} ...(2)
From Equation (1) and (2) P(A) ∪ P(B) ≠ P(A ∪ B)
Thus, given statement P(A) ∪ P(B) = P{A ∪ B) is not true.

Question 8.
Show that for any sets A and B
A = (A ∩ B) ∪ (A - B) and (A ∪ (B- A) = (A ∪ B)
Answer:
To prove for set A and B
A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = A ∪ B
(A ∩ B) ∪ (A - B) = (A ∩ B) ∪ (A ∩ B')
= A ∩ (B ∪ B') [∵ A ∩(B ∪ B') = (A ∩ B) ∪ (A ∩ B’) By distributive law]
= A ∩ U = A (∵ B ∪ B' = U) {A ∩ U = A} (∵ A ∩ U = A,v AcU)
Again thus, A = (A ∩ B) (A - B) (By distributive law)
L.H.S. = A ∪ (B - A)
Now,
A ∪ (B - A) = A ∪ (B ∩ A') (∵ B - A = B ∩ A')
= (A ∪ B) ∩ (A ∪ A') (By distributive law)
=(A ∪ B) ∩ U (∵ A ∪ A = U Universal set )
= A ∪ B
Thus, A ∪ (B - A) = A ∪ B
Hence Proved

Question 9.
Using properties of sets, show that:
(i) A ∪(A ∩ B) = A
(ii) A ∩ (A ∪ B) = A
Answer:
(i) We have to prove A ∪ (A ∩ B) = A
A ∪ (A ∩B) = (A ∪ A) ∩ (A ∪ B)
(By distubutive law)
= A ∩ (A ∪ B)
= A (∵ A ⊂ A ∪ B)
Thus, A ∪ (A ∩ B) = A
Hence Proved

(ii) We have to prove A ∪ (A ∪ B) = A
⇒ A ∩ (A ∪ B) = (A ∩ A) ∪ (A nB)
= A ∪ (A ∩ B) [∵ A ∩ BcA, A ∩A = A]
= A
Thus, A ∩ (A ∪ B) = A
Hence proved

Question 10.
Show that A ∩ B = A ∩ C need not imply B = C.
Answer:
It is not necessary for any sets A, B and C, A ∩ B = A ∩ C then B = C.
Let A = {1, 2}, B = {2, 3}, C = {2, 4}
Then, A nB = {2} and A ∩ C = {2}
Here ∴ A ∩ B = A ∩ C = {2}
Thus, it is not necessarily true that if A ∩ B = A ∩ C then B = C

Question 11.
Let A and B are sets; If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X, for some set X, show that A = B. (Hints : A = A ∩ (A ∪ A),B = B ∩ (B ∪ X) and use distributive law.)
Solution:
Given A ∩ X = B ∩ X =§ and A ∪ X = B ∪ X, then show that A=B
Y A ∪ X = B ∪ X (Given)
⇒ A ∩ (A ∪ X) = A ∩ (B ∪ X)
= {A ∩ B) ∪ (A ∩ X) (By distributive law)
⇒ A ∩ (A ∪ X) = (A ∩ B) ∪ (B ∩ X) {∵ A ∩ X = B ∩ X, given)
⇒ A ∩ (A ∪ X) = (A ∩ B) ∪ (Φ) (∵ A ∩ X -B ∩ X) = Φ, given)
A = A ∩ B ...(1) [∵ A ∩ (A ∪ X) = A]
Since in set A and A ∪ X element of set A are common.
Again AKJX = B ∪ X given
⇒ B ∩ (A ∪ X) = B ∩ (B ∪ X)
⇒ (B ∩ A) ∪ (B ∩ X) = B ∩ (B ∪ X) (By distributive law)
⇒ (B ∩ A) ∪ Φ = B (∵ B ∩ X (B ∪ X) = B.A ∩ X = B ∩ X = Φ)
⇒ (B ∩ A) ∪ Φ = B [∵ (B ∩ A) ∪ Φ =B ∩ A]
⇒ B ∩ A = B ...(2)
By commutative law,
A ∩ B = B ∩ A From equation (1) and (2),
A = B Hence Proved

Question 12.
Find sets A, B and C such that A ∩ B,B ∩ C and A ∩C are non-empty sets and A ∩ B ∩ C = Φ.
Answer:
We have to find three sets A, B and C so that A ∩ B,B ∩ C and A ∩ C be non-empty sets and A ∩ B ∩ C = Φ
Three sets A, B, C are such that
A ∩ B ≠ Φ, B ∩ C ≠ Φ and A ∩ C ≠ Φ
∴ At least one element is common in A and B, similarly there will be at least one common element in B and C, as well as A and C.
We see that no element is common in A, B and C. Thus, A ∩ B ∩ C = Φ
Since, conditions are correct. So,
∴ Three sets are :A = {1, 2}, B = {2, 3} and C = {1, 3}

Question 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?
Answer:
Let T in set of students who drink tea and C who drink coffee. .
Questionwise, n(T) = 150, n(C) =225
n(T ∩ C) = 100

By formula n(T ∪ C) = n(T) + n(C) - n(T ∩ C)
= 150 + 225 - 100
n(T ∪ C) = 275
Now, 275 students drink tea or coffee. Thus, number of students who do not drink tea nor coffee.
= (T ∪ C)' = U - (T ∪ C)
= 600 - 275 = 325
∴ 325 students neither taken tea nor coffee.

Question 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 a group?
Answer:
Let H is set of students who speaks Hndi and E is set of students who speaks English.
Given, n(H) = 100, n(E) = 50 n(H ∩ E) =25, n(H ∪ E) = ?

Using formula,
n(H ∪ E) = n(H) + n(E) - n(H ∩ E)
n(H ∪ E) = 100 + 50 - 25
∴ n(H ∪ E) = 125
Thus, 125 students are there in the group.

Question 15.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper 1,9 read both H and 1,11 read both H and T, 8 read both T and 1,3 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.
Answer:
Given n(U) = 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

Venn diagram:

Using formula, we have to find number of people who read at first one of the two newspapers.
n(H ∪ T ∪ I) = n(H) + n(T) + n(T) - n(H ∩ T) - n(T ∩ I) - n(H ∩ I) + n(H ∩ T ∩ I)
= 25 + 26 + 26 - 9 - 11 - 8 + 3 = 52
Thus, number of people who read at least one newspaper is 52.

(ii) To find number of people who read exactly one newspaper.
Number of people who read H and T newspaper = n(H ∩ T) = 11
Number of people who read only H and T newpaper = n(H ∩ T) - n(H ∩ T ∩ C)
= 11 - 3 = 8
Similarly, number of people who read only T and I newspaper
= n(T ∩ I) - n(H ∩ T ∩ I)
= 8 - 3 = 5
Number of people who read only H and I newspaper = n(H ∩ I) - n(H ∩T ∩ I)
= 9 - 3 = 6
Now, number of people who read exactly two newspapers
= n(H ∩ T) - n(H ∩ T ∩ I) + n(H ∩ I) - n(H ∩ T ∩I) + n(T ∩ I) - n(H ∩ T ∩ I)
= (11 - 3) + (9 -3) + (8 - 3)
= 8 + 6 + 5 = 19
Here, number of people who read all three newspapers is not included.
Thus, number of people who read at least 2 Newspapers =19
Thus, number of people who read' at least two newspaper or all three papers.
= 19 + 3 = 22 .

Number of people who read at the least one newspaper = n(H ∪ T ∪ I) = 52
Number of people who read only one newspaper = n(H ∪ T ∪ I) - [n(H ∩ T) + n(H ∩ I) - n(H ∩ l ∩T) + n(T ∩ I) - n{H ∩ T ∩ I)] = 52 - 22
= 30 reads only only one newspapers
OR
In Venn diagram, number of reader of 1 newspaper = 8 + 10 + 12 = 30

Question 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.
Answer:
Consider the Venn diagram:

n(A)= 21 ⇒ a + d + f + g= 21 .........(i)
n(B)= 26 ⇒ b + d + e + g =26 .......(ii)
n(C) = 29 ⇒ c + e + f + g= 29 ............(iii)
n(A ∩ B) = 14 ⇒ d + g = 14 ...........(iv)
n(B ∩ C) = 14 ⇒ e + g = 14 ..............(v)
n(C ∩ A) = 12 ⇒ f + g = 12 ...........(vi)
n (A ∩ B ∩ C) = 8 ⇒ g = 8 .........(vii)
From (vi) and (vii) f = 4
(v) and (vii) e = 6
(iv) and (vii) d=6

Put e, f g in eq. (iii)
c + 6 + 4 + 8 = 29
c = 11
Therefore, 11 people liked product C only.