Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths 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.
A={x : x ∈ R and x satisfies x² – 8x + 12 = 0}
2 and 6 are the only solutions of x² – 8x + 12 = 0.
∴ A = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
∴ D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, 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
(iii) If A ⊂B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and 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) False
Let A = {1, 2} and B={1, {1, 2}, {3}}
2 ∈ {1, 2} and {1, 2} ∈ {{3}, 1, {1, 2}}
Now, A ∈ B
∴ However, 2 ∉ {{3}, 1, {1, 2}}

(ii) False
Let A = {2}, B = {0, 2}, and C = {1, {0, 2}, 3}
As A ⊂ B
B ∈ C
However, A ∉ C

(iii) True
Let A ⊂ B and B ⊂ C.
Let x ∈ A
x ∈ B [∵ A ⊂ B]
x ∈ C [∵ B ⊂ C]
∴ A ⊂ C

(iv) False
A = {1, 2}, B = {0, 6, 8}, and C = {0, 1, 2, 6, 9}
Accordingly, A ⊄ B and B ⊄ C.
However, A ⊂ C

(v) False
A = {3, 5, 7} and B = {3, 4, 6}
5 ∈ A and A ⊄ B
However, 5 ∉ B

(vi) True
Let A ⊂ B and x ∉ B.
To show x ∉ A
If possile, suppose x ∉ A
Then, x ∉ B, which is a conadicon as x ∉ B
∴ x ∉ A.

Question 3.
Let A, B and C be the sets such that A ∪ B = A ∪ C andA ∪ B = A ∪ C. Show that B = C.
Answer:
A ∪ B = A ∪ C
⇒ (A ∪ B) ∩ C = (A ∪ C) ∩ C
⇒ (A ∩ C) ∪ (B ∩ C) = C
⇒ (A ∩ B) ∪ (B ∩ C) = C …………….(i)
Again, A ∪ B = A ∪ C
⇒ (A ∪ B) ∩ B = (A ∪ C) ∩ B
⇒ B = (A ∩ B) ∪ (C ∩ B)
⇒ B = (A ∩ B) ∪ (B ∩ C)
From eqs. (i) and (ii), we get
B = C.

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) ⟺
(ii) A ⊂ B ⟺ All elements of A are in B ⟺ A – B = Φ
(ii) ⟺
(iii) A – B = Φ ⟺ All elements of A are in B ⟺ A ∪ B = B.
(iii) ⟺ (iv)
A ∪ B = B ⟺ All the elements of A are in B.
⟺ All the elements of A are common in A and B ⟺ A ∩ B = A.
Thus, all the four given conditions are equivalent.

Question 5.
Show that if A ⊂ B, then C – B ⊂ C – A.
Answer.
Let A ⊂ B
To show: C – B ⊂ C – A
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
C – B ⊂ C – A

Question 6.
Assume that P(A) = P(B). Show that A = B.
Answer.
Let P(A) =P (B)
To show: A = B
Let x ∈ A
A ∈ P(A) = P(B)
∴ x ∈ C,
for some C ∈ P(B)
C ⊂ B
∴ x ∈B
∴ A ⊂ B
Similarly, B ⊂ A
∴ A = B.

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 = {a}, B = {b} and A ∪ B = {a, b}
∴ P(A) = {Φ, {a}}, P(B) = {Φ, {b}}
P(A ∪ B) = {Φ, {a}, {b}, {a, b}} …………….(i)
and P(A) ∪ P(B) = {Φ, {a}, {b}} …………..(ii)
From eqs. (i) and (ii), we have
P(A ∪ B) ≠ P(A) ∪ P(B).

Question 8.
Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = A ∪ B.
Answer.
(A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B) [∵ A – B = A ∩ B]
= A ∩ (B ∪ B’) [Distributive law]
= A ∩ X, where X is universal set.
= A
A ∪ (B – A) = A ∪ (B ∩ A) [∵ B – A = B ∩ A’]
= (A ∪ B) ∩ (A ∪ A) [Distributive law]
= (A ∪ B) ∩ X
where X = A ∩ A’ is universal set
= A ∪ B [∵ A∪ B ⊂ X].

Question 9.
Using properties of sets, show that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Answer.
(i) A ∪ (A ∩B) = (A ∪ A) ∩ (A ∪ B) [Distributive law]
= A ∩ (A ∪ B) [∵ A ∪ A = A]
= A [∵ A ⊆ A ∪ B]

(ii) A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
= A [∵ A ∩ B ⊂ A]

Question 10.
Show that A ∩ B=A ∩ C need not imply B = C.
Answer.
With the help of an example, we may try to establish it.
Let A = {1, 2}, B = {1, 3} and C = {1, 4}
Now A ∩ B = {1, 2} ∩{1, 3} = {1}
and A ∩ C = {1, 2} ∩{1, 4} = {1}
A ∩ B = A ∩ C
still B ≠ C.

Question 11.
Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B.
Answer. We have, A ∩ X = B ∩ X for some set X
⇒ A ∩ (A ∪ X) = A ∩(B ∪ X)
⇒ A = (A ∩ B) ∪ (A ∩ X) [∵ A ∩ (A ∪ X) = A]
⇒ A = (A ∩ B) ∪ Φ) [∵ A ∩ X = Φ (given)]
⇒ A = A ∩ B
⇒ A ⊂ B ………..(i)
Again, A ∪ X = B ∪ X
⇒ B ∩ (A ∪ X) = B ∩ (B ∪ X)
⇒ (B ∩ A) ∪ (B ∩ X) = B
⇒ (B ∩ A) ∪ Φ = B
⇒ B ∩ A = B
⇒ A ∩ B = B
⇒ B ⊂ A ……………(ii)
From eqs. (i) and (ii), we get A = B.

Question 12.
Find sets A, B and C such that A n B, B n C and A n C are non-empty sets and AnBnC=<(>.
Answer.
Let A = {1, 2}, B = {2, 3} and C = {1, 3}.
Clearly A ∩ B = {2}, B ∩ C = {3}and A ∩ C = {1}
i.e., A ∩ B, B ∩ C and A ∩ C are non-empty sets.
∴ (A ∩ B) ∩ C = {2} ∩ {1, 3}
⇒ A ∩ B ∩ C = Φ.

Question 13.
In a survey of 600 students in a school, 150 students were found to he taking tea and 225 taking coffee, 1Q0 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Answer.

Here, n(T) = 150, n(C) = 225,
n(T ∩ C) = 100
Also we know that
n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
= 150 + 225 – 100 = 275
Total mimber of students = 600
∴ Number of students who neither take tea nor coffee.
= 600 – n(T ∪ C)
= 600 – 275 = 325.

Question 14.
In a group pf students, 100 students know Hindi, 50 know English and 25 know bote. Each of the students knows either Hindi or English. How many students are there in tee group?
Answer.

Here, n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25 = 125.
Hence there are 125 students 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 bote 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.
Let A be the set of people who read newspaper H.
Let B be the set of people who read newspaper T.
Let C be the set of people who read newspaper I.
Accordingly, n (A) = 25, n(B) = 26, and n(C) = 26
n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8 n(A ∩ B ∩ C) = 3
Let U be the set of people who took part in the survey.
(i) Accordingly,
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A)
+n(A ∩ B ∩ C)
= 25 + 26 + 26 – 11 – 8 – 9 + 3 = 52
Hence, 52 people read at least one of the newspapers.

(ii)

Let a be the number of people who read newspapers H and T only.
Let b denote the number of people who read newspapers I and H only.
Let c denote the number of people who read newspapers T and I only.
Let d denote the number of people who read all three newspapers.
Accordingly, d = n(A ∩ B ∩ C) = 3
Now, n(A ∩ B) = a + d,
n(B ∩ C) = c + d,
n(C ∩ A) = b + d
∴ a + d + c + d + b + d = 11 + 8 + 9 = 28
a + b + c + d = 28 – 2d
= 28 – 6 = 22
Hence, (52 – 22) = 30
people read exactly one newspaper.

 

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.
Let A, B, and C be the set of people who like product A, product B, and product C respectively.
Accordingly, n(A) = 21, n(B) = 26,
n(C) = 29, n(A ∩ B) = 14,
n(C ∩ A) = 12, n(B ∩ C) = 14,
n(A ∩ B ∩ C) = 8
The Venn diagram for the given problem can be drawn as

It can be seen that number of people who like product C only is
{29 – (4 + 8 + 6)} = 11.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

Question 1.
The relation f is defined by f (x) =

The relation g isdefined by g(x) =
Show that f is a function and g is not a function.
Answer.
The relation f is defined as f (x) =
It is observed that for
0 ≤ x < 3, f(x) = x²
3 ≤ x ≤ 10, f(x) = 3x
Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Thus, the given relation is a function.
The relation g is defined as g (x) =
It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of the domain of the relation g corresponds to two different images te., 4 and 6. Hence, this relation is not a function.

Question 2.
If f(x) = x², find \(\frac{f(1.1)-f(1)}{(1.1-1)}\).
Ans.
f(x) = x²
∴ \(\frac{f(1.1)-f(1)}{(1.1-1)}\) = \(\frac{(1.1)^{2}-(1)^{2}}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}\)= 2.1

Question 3.
Find the domain of the function f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\).
Answer.
The given function is f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\)

f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}=\frac{x^{2}+2 x+1}{(x-6)(x-2)}\)

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.
Hence, the domain off is R – {2, 6}.

Question 4.
Find the domain and the renie of the real function f defined by f(x) = \(\sqrt{(x-1)}\).
Answer.
The given real function is f (x) = \(\sqrt{(x-1)}\).
It can be seen that \(\sqrt{(x-1)}\) is defined for (x – 1) ≥ 0.
i.e., f(x) = \(\sqrt{(x-1)}\) is defined for x ≥ 1.
Therefore, the domain off is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞).
As x ≥ 1
⇒ (x – 1) ≥ 0
⇒ \(\sqrt{(x-1)}\) ≥ 0
Therefore, the range off is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0, ∞).

Question 5.
Find the domnin and the range of the real function f defined by f(x) = |x – 1|.
Answer.
The given real function isf (x) = |x – 1|.
It is clear that |x – 1| is defined for all real numbers.
∴ Domain of f = R
Also, for x ∈ R, |x – 1| assumes all real numbers.
Hence, the range of f is the set of all non-negative real numbers.

Question 6.
Let f = {(x, \(\frac{x^{2}}{1+x^{2}}\))} : x ∈ R be a function from R into R. Determine the range of f.
Answer.
f = {(x, \(\frac{x^{2}}{1+x^{2}}\))} : x ∈ R

= {(0, 0), (±0.5, \(\frac{1}{5}\)), (±1, \(\frac{1}{2}\)), (±1.5, \(\frac{9}{13}\)), (±2, \(\frac{4}{5}\)), (3, \(\frac{9}{10}\)), (4, \(\frac{16}{17}\))}

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1, [∵ Denominator is greater than numerator]
Thus, range of f = [0,1).

Question 7.
Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and \(\frac{f}{g}\).
Answer.
f, g : R → R is defined as f(x) = x + 1, g(x) = 2x – 3
(i) (f + g) (x) = f(x) + g(x)
= (x + 1) + (2x – 3) = 3x – 2
∴ (f + g) (x) = 3x – 2

(ii) (f – g) (x) = f(x) – g(x)
= (x + 1) – (2x – 3)
= x + 1 – 2x + 3 = – x + 4
∴ (f – g) (x) = – x + 4

(iii) \(\frac{f}{g}\) (x) = \(\frac{f(x)}{g(x)}\), g(x) ≠ 0, x ∈ R.

∴ \(\left(\frac{f}{g}\right)(x)=\frac{x+1}{2 x-3}\), 2x – 3 ≠ 0 or 2x ≠ 3

∴ \(\left(\frac{f}{g}\right)(x)=\frac{x+1}{2 x-3}\), x ≠ \(\frac{3}{2}\).

Question 8.
Let f ={(1, 1), (2, 3), (0, – 1), (- 1, – 3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution.
f = {(1, 1), (2, 3), (0, – 1), (- 1, – 3)}
f(x) = ax + b
(1, 1) ∈ f
⇒ f (1) = 1
⇒ a × 1 + b = 1
⇒ a + b = 1
(0, – 1) ∈ f
⇒ f (0) = – 1
⇒ a × 0 + b = – 1
⇒ b = – 1
On substituting b= – 1 in a + b = 1, we obtain a + (- 1) = 1
⇒ a = 1 + 1 = 2.
Thus, the respective values of a and b are 2 and – 1.

Question 9.
Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
Justify your answer in each case.
Answer.
R = {(a, b) : a, b ∈ N and a = b²}
(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4.
Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².
Now, 3 ≠ 9² = 81; therefore, (3, 9) ∈ N
Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R is not true.

(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 e∈ N and 16 = 4² and 4 = 2².
Now, 16 ≠ 2² = 4; therefore, (16, 2) ∈ N
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question 10.
Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
(i) f is a relation from A to B
(ii) fis a function from A to B.
Justify your answer in each case.
Answer.
A = {1, 2, 3, 4} and B = {1, 5, 9,11, 15, 16}
∴ A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2,16), (3, 1), (3, 5), (3, 9),(3, 11),
(3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
It is given that f = {( 1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.
It is observed that f is a subset of A × B.
Thus, f is a relation from A to B.
(ii) Since the same first element i.e., 2 corresponds to two different images
i.e., 9 and 11, relation f is not a function.

Question 11.
Let f be the subset of Z × Z defined by f = {(ab, a + b) : a,b ∈ Z}. Is f a function from Z to Z ? Justify your answer.
Answer.
The relation f is defined as f = {(ab, a + b) : a, b ∈ Z}
We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.
Since 2, 6, – 2, – 6 ∈ Z, (2 × 6, 2 + 6), (- 2 × – 6, – 2 + (- 6)) ∈ f.
i.e., (12, 8), (12, – 8)
It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and – 8.
Thus, relation f is not a function.

Question 12.
Let A = {9, 10, 11, 12, 13} and let f: A→ N be defined by f(n) = the highest prime factor of n. Find the range of f.
Answer.
A = {9, 10,11, 12, 13}
f : A → N is defined as f(n) = The highest prime factor of n
Prime factor of 9 = 3;
Prime factors of 10 = 2, 5;
Prime factor of 11 = 11;
Prime factors of 12 = 2, 3;
Prime factor of 13 = 13
f(9) = The highest prime factor of 9 = 3
f(10) = The highest prime factor of 10 = 5 .
f(11) = The highest prime factor of 11 = 11
f(12) = The highest prime factor of 12 = 3
f(13) = The highest prime factor of 13 = 13
The range of f is the set of all f(n), where n ∈ A.
∴ Range of f = {3, 5, 11, 13}.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3

Question 1.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) (1, 3), (1, 5), (2, 5)}
Answer.
(i) {(2, 1), (5,1), (8, 1), (11, 1), (14, 1), (17,1)}
Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain={2, 5, 8, 11, 14, 17} and range={l}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}
Since the same first element i.e., 1 corresponds to two different images i.e.,3 and 5, this relation is not a function.

Question 2.
Find the domain and range of the following real functions:
(i) f(x) = – |x|
(ii) f(x) = \(\sqrt{9-x^{2}}\)
Answer.
(i) f(x) = – |x|, f(x) ≤ 0, ∀ x ∈ R
Domain of f = R
Range of f = {y ∈ R, y < 0}

(ii) f(x) = \(\sqrt{9-x^{2}}\)
All, f is not defined for 9 – x² ≤ 0 or x² ≤ 9 or when x ≥ 3 or x ≤ – 3.
Also, for each real number x lying between – 3 and 3 or for x = – 3, 3 f(x) is unique.
∴ Domain (f) = {x : x ∈ R and – 3 < x < 3}
Further, y = \(\sqrt{9-x^{2}}\) or y² = 9 – x²
or x = \(\sqrt{9-x^{2}}\)
Again, x is not defined for
9 – y² < 0 or y² > 9 or y > 3 or y < – 3.
But y cannot be – ve
∴ Range (f) = {y : y ∈ R and 0 < y < 3}

Question 3.
A function fis defined by f(x) = 2x – 5. Write down the values of
(i) f(0)
(ii) f(7)
(iii) f(- 3)
Answer.
The given function is f(x) = 2x – 5. Therefore,
(i) f(0) = 2 × 0 – 5 = 0 – 5 = – 5
(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9
(iii) f(- 3) = 2 × (- 3) – 5 = – 6 – 5 = -11.

Question 4.
The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = \(\frac{9 C}{5}\) + 32. Find
(i) f(0)
(ii) t(28)
(iii) t(- 10)
(iv) The value of C, when t(C) = 212.
Answer.
The given function is t (C) = \(\frac{9 C}{5}\) + 32. Therefore,
(i) t(0) = \(\frac{9 \times 0}{5}\) + 32 = 0 + 32 = 32

(ii) t(28) = \(\frac{9 \times 28}{5}\) + 32 = \(\frac{252+160}{5}=\frac{412}{5}\)

(iii) t (- 10) = \(\frac{9 \times(-10)}{5}\) + 32 = 9 × ( – 2) + 32

(iv) It is given that t(C) = 212
∴ 212 = \(\frac{9 C}{5}\) + 32 \(\frac{9 C}{5}\)
= 212 – 32 \(\frac{9 C}{5}\) = 180
9C = 180 × 5 C = \(\frac{180 \times 5}{9}\) = 100.
Thus, the value of t, when t(C) = 212, is 100.

Question 5.
Find the range of each of the following functions :
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
(ii) f(x) = x² + 2, x is a real number.
(iii) f(x) = x, x is a real number.
Answer.
(i) Let f(x) = y = 2 – 3x is x = \(\frac{2-y}{3}\)
Now x > 0 is 2 – y > 0 or y < 2
Range (f) = {y : y ∈ R and y < 2}

(ii) Let f(x) = y = x² + 2 is x² = y – 2 or x = \(\sqrt{y-2}\) ⇒ y > 2
Range (f) = {y : y ∈ R and y > 2}

(iii) f(x) = x, x is a real number
x = f(x) = y = a real number
Range (f) = {y : y ∈ R}

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2

Question 1.
Let A = {1, 2, 3, …………. , 14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.
Answer.
The relation R from A to A is given as
R = (x, y) : 3x – y = 0, where x, y ∈ A} i.e., R = { y} : 3x = y, where x, y ∈ A}
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
Codomain of R = A = {1, 2, 3, …………., 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
Range of R = {3, 6, 9, 12}.

Question 2.
Define a relation R on the set N of natural numbers by R= {(x, y): y = x + 5, x is a natural number less than 4 : x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Answer.
R = { (x, y) : y = x+ 5, x is a natural number less than 4 ,x} y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
∴ R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {6, 7, 8}

Question 3.
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Answer.
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = { (x, y) the difference between x and y is odd; x ∈ A, y ∈ B}
R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Question 4.
The given figure shows a relationship between the sets P and Question Write this relation
(i) in set-builder form
(ii) in roster form. What is its domain and range?

Answer.
According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}
(i) R = {(x, y) : y = x – 2 x ∈ P} or R = {(x, y) : y = x – 2 for x = 5, 6, 7}
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}

Question 5.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, 6) : a, b ∈ A, 6 is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Answer.
A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}.

Question 6.
Determine the domain and range of the relation R defined by R = {x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Answer.
R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 7.
Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.
Answer.
R = {(x, x³) : x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 8.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Answer.
It is given that A = {x, y, z} and B = {1, 2}.
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since h(A × B) = 6, the number of subsets of A × B is 26.
Therefore, the number of relations from A to B is 26.

Question 9.
Let R be the relation on Z defined by R = {(a, b) : a, b ∈ Z, a b is an integer}. Find the domain and range of R.
Answer.
R = {(a, b) : a, b ∈ Z, a – b is an integer}
It is known that the difference between any two integers is always an integer.
∴ Domain of R = Z
Range of R = Z.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1

Question 1.
If (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = (\(\frac{5}{3}\), \(\frac{1}{3}\))
find the values of x and y.
Answer.
It is given that (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = (\(\frac{5}{3}\), \(\frac{1}{3}\))
Since the ordered pairs are equal, the corresponding elements will also be equal.
Therefore, \(\frac{x}{3}\) + 1 = \(\frac{5}{3}\) and y – \(\frac{2}{3}\)) = (\(\frac{1}{3}\)

⇒ \(\frac{x}{3}\) = \(\frac{5}{3}\) – 1 and y = \(\frac{1}{3}\) + \(\frac{2}{3}\)

⇒ \(\frac{x}{3}\) = \(\frac{5}{3}\) and y = \(\frac{3}{3}\)
⇒ x = 2 and y= 1
∴ x = 2 and y = 1.

Question 2.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).
Answer.
It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Thus, the number of elements in (A × B) is 9.

Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Answer.
G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as
P × Q = {(p, q) : p ∈ P, q ∈ Q}
∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

Question 4.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly,
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.
Answer.
(i) False
If P = {m, n} and Q = {n, m}, then P × Q = {(m, m), (m, n), {n, m), (n, n)}
(ii) True
(iii) True.

Question 5.
If A = {- 1, 1}, find A × A × A.
Answer.
It is known that for any non-empty set A, A × A × A is defined as
A × A × A = {(a, b, c) : a, b, c ∈ A}
It is given that A = {- 1, 1}
∴ A × A × A = {(- 1, – 1, – 1), (- 1, – 1, 1), (- 1, 1, – 1), (- 1, 1, 1), (1, – 1, – 1), (1, – 1, 1), (1, 1, – 1), (1, 1, 1)}

Question 6.
If A × B = {(o, x), (a, y), (b, x), (b, y)}. Find A and B.
Answer.
It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
A is the set of all first elements and B is the set of all second elements.
Thus, A = {a, b} and B = {x, y}.

Question 7.
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D
Answer.
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
L.H.S. = A × (B ∩ C) = A × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1,6), (2, 5), (2,6)}
R.H.S. = (A × B) ∩ (A × C) = Φ
L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To verify: A × C is a subset of B × D
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
We can observe that all the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.

Question 8.
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A x B have? List them.
Answer.
A = {1, 2} and B = {3, 4}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2m.
Therefore, the set A × B has 2⁴ = 16 subsets.
These are {Φ, (1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2,4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

Question 9.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y andz are distinct elements.
Answer.
It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
We know that A=Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.

Question 10.
The Cartesian product A × A has 9 elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
Answer.
We know that if n(A) =p and n(B) =q, then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
n(A) × n(A) = 9
=> n (A) = 3
The ordered pairs (- 1, 0) and (0, 1) are two of the nine elements of A × A.
We know that A x A={(a, a) : a ∈ A}.
Therefore, – 1, 0, and 1 are elements of A.
Since n(A} = 3, it is clear that A = {- 1, 0, 1}.
The remaining elements of set A × A are (- 1, – 1), (- 1, 1), (0, – 1), (0, 0), (1, – 1), (1, 0), and (1, 1).

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Ex 1.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

Question 1.
If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Answer.
It is given that: n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38, n(X ∩ Y) = ?
We know that: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
– 38 = 17 + 23 – n(X ∩ Y)
⇒ n(X ∩ Y) = 40 – 38 = 2
n(X ∩ Y) = 2

Question 2.
If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Yhas 15 elements; how many elements doesXnY have?
Answer.
It is given that:
n(X ∪ Y) = 18, n(X) = 8, n (Y) = 15, n(X ∩Y)
We know that:
n(X ∪ Y) = n(X) + n (Y) – n(X ∩ Y)
∴ 18 = 8 + 15 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5
n(X ∩ Y) = 5.

Question 3.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Answer.
Let H be the set of people who speak Hindi, and E be the set of people who speak English
n(H ∪ E) = 400, n(H) = 250, n(E) = 200, n(H ∩ E) = ?
We know that: n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – n(H ∩ E)
⇒ 400 = 450 – n(H ∩ E)
n(H ∩ E) = 450 – 400
n(H ∩ E) = 50
Thus, 50 people can speak both Hindi and English.

Question 4.
If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?
Answer.
It is given that:-
n(S) = 21, n(T) = 32, n(S ∩ T) = 11
We know that:
n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
∴ n(S ∪ T) = 21 + 32 – 11 = 42.
Thus, the set (S ∪ T) has 42 elements.

Question 5.
If X and Y are two sets such that Xhas 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?
Answer.
It is given that:
n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10
We know that:
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
60 = 40 + n(Y) – 10
n(Y) = 60 – (40 – 10) = 30
Thus, the set Y has 30 elements.

Question 6.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?
Answer.
Let C denote the set of people who like coffee and T denote the set of people who like tea,
n(C ∪ T) = 70, n(C) = 37, n(T) = 52.
We know that:
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70 = 19
Thus, 19 people like both coffee and tea.

Question 7.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many of like tennis?
i Answer.
Let C denote the set of people who like cricket, and T denote the set of people who like tennis
n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that:
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
65 = 40 + n(T) – 10
⇒ 65 = 30 + n(T)
⇒ n(T) = 65 – 30 = 35
Therefore, 35 people like tennis.
Now, (T – C) ∪ (T ∩ C) = T
Also, (T – C) ∩ (T ∩ C) = Φ
∴ n(T) = n(T – C) + n(T ∩ C)
Thus, 25 people like only tennis.

Question 8.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Answer.
Let F be the set of people in the committee who speak French, and S be the I set of people in the committee who speak Spanish
n(F) = 50, n(S) = 20, n(S ∩ F) = 10.
We know that:
n(S ∪ F) = n(S) + n(F) – n(S ∩ F)
= 20 + 50 – 10
= 70 – 10 = 60
Thus, 60 people in the committee speak at least one of the two languages.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Ex 1.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5

Question 1.
Let U = {1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B= {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find
(i) A’
(ii) B’
(iii) (A ∪ C)’
(iv) (A ∪ B)’
(v) (A’)’
(vi) (B – C)’
Answer.
U = {1, 2, 3, 4, 5, 6, 7, 8, 9};
A = {1, 2, 3, 4};
B = {2, 4, 6, 8};
C = {3, 4, 5, 6}
(i) A’ = U – A
= {1, 2, 3, 4, S, 6, 7, 8, 9} – {1, 2, 3, 4} = {5, 6, 7, 8, 9}

(ii) B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}

(iii) (A ∪ C) = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
(A ∪ C)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6} = {7, 8, 9}

(iv) (A ∪ B) = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8} = {5, 7, 9}

(v) A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4} = {5, 6, 7, 8, 9}
(A0′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9} = {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(B – C)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8} = {1, 3, 4, 5, 6, 7, 9}.

Question 2.
If U = {a, b, c, d, e, f, g, h}, find the complements of the following
(i) A={«, b, c}
(ii) B={d, e,f, g}
(iii) C={o, c, e, g}
(iv) D={f, g, h, a}
Answer.
(i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c} = {d, e, f, g, h}
(ii) B’={a, b, c, d, e, f, g, h} – {d, e, f, g} = {a, b, c, h}
(iii) C’ = {a, b, c, d, e, f, g, h} – {a, c, e, g} = {b, d, f, h}
(iv) U = {a, b, c, d, e, f, g, h} – {f, g, h, a} = {b, c, d, e}.

Question 3.
Taking the set of natural numbers as the universal set, writedown the complements of the following sets:
(i) {x : x is an even natural number}
(ii) {x : x is an odd natural number}
(iii) {x : x is a positive multiple of 3}
(iv) {x : x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x : x is a perfect square}
(vii) {x : x is perfect cube}
(viii) (x : x + 5 = 8}
(ix) {x : 2x + 5 = 9}
(x) {x : x ≥ 7}
(xi) {x : x ∈ N and 2x + 1 > 10}
Answer.
U = N : Set of natural numbers
(i) {x : x is an even natural number}’ = {x : x is an odd natural number}
(ii) {x : x is an odd natural number}’ = {x : x is an even natural number}
(iii) {x : x is a positive multiple of 3}’ = {x : x ∈ N and x is not a multiple of 3}
(iv) {x : x is a prime number}’ = {x : x is a positive composite number and x = 1}
(v) {x : x is a natural number divisible by 3 and 5}’ = {x : x is a natural number that is not divisible by 3 or 5}
(vi) {x : x is a perfect square}’ = {x : x ∈ N and x is not a perfect square}
(vii) {x : x is a perfect cube}’ = {x : x ∈ N and x is not a perfect cube}
(viii) {x : x + 5 = 8}’ = {x : x ∈ N and x ≠ 3}
(ix) {x : 2x + 5 = 9}’ = {x: x ∈ N and x ≠ 2}
(x) {x : x ≥ 7}’ = {x : x ∈ N and x < 7} (xi) {x : x ∈ N and 2x + 1 > 10}’ = {x : x ∈ N and x < 9/2}.

Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}.Verify that
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’= A’ ∪ B’ .
Answer.
U = {1, 2, 3, 4, 5, 6,7, 8, 9}
A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
(i) (A ∪ B)’ = {2, 3, 4, 5, 6, 7, 8}’ = {1, 9}
A’ = {1, 3, 5, 7, 9}
B’ = {1, 4, 6, 8, 9}
A’ ∩ B’ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}
(A ∪ B)’ = A’ ∩ B’

(ii) (A ∩ B)’ = {2}’ = {1, 3, 4, 5, 6, 7, 8, 9}
A’ ∪ B = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9}
(A ∩ B)’ = A’ ∪ B’

Question 5.
Draw appropriate Venn diagram for each of the foUowing:
(i) (A ∪ B)’
(ii) A’ ∩ B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Answer.
(i) Shaded Area (A ∪ B)’

(ii) A’ ∩ B’ = Common shaded area

(iii) (A ∩ B)’ shaded Area

(iv) A’ ∪ B’ All shaded area formed by all horizontal and vertical lines.

Question 6.
Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A?
Answer.
A’ is the set of all equilateral triangles.

Question 7.
Fill in the blanks to make each of the following a true statement:
(i) A ∪ A’ = ……….
(ii) A’ ∩ A = ………
(iii) A ∩ A’= ………
(iv) A’ ∩ A = ………
Answer.
(i) A ∪ A’= U
(ii) Φ’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = Φ
(iv) U’ ∩ A = Φ ∩ A = Φ

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4

Question 1.
Find the union of each of the following pairs of sets:
(i) X = {1, 3, 5}
Y = {1, 2, 3}
Answer.
X = {1, 3, 5}
Y = {1, 2, 3}
X ∪ Y = {1, 2, 3, 5}

(ii) A = {a, e, i, o, u}
B = {a, b, c}
Answer.
A = {a, e, i, o, u}
B = {a, b, c}
A ∪ B = {a, b, c, e, i, o, u}

(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
Answer.
A = {x : x is a natural number and multiple of 3} = {3, 6, 9, ……..}
B = {x : x is a natural number less than 6} = {1, 2, 3, 4, 5}
A ∪ B = {1, 2, 3, 4, 5, 6, 9, 12, …………..}

(iv) A = {x : x is a natural number and 1 < x ≤ 6}
B = {x : x is a natural number and 6 < x < 10}
Answer.
A = {x : x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
A ∪ B = {x : x ∈ N and 1 < x < 10}

(v) A = {1, 2, 3}, B = Φ
Answer.
A = {1, 2, 3}, B = Φ
A ∪ B = {1, 2, 3}.

Question 2.
Let A = {a, 6}, B = {a, b, c}. Is Ac B? What is A ∪ B?
Answer.
Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B and A ∪ B = {a, b, c}.

Question 3.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Answer.
If A and B are two sets such that A ⊂ B, then A ∪ B = B.

Question 4.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9,10}; find,
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Answer.
(i) A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= {1, 2, 3, 4, 5, 6}.

(ii) A ∪ C = {1, 2, 3, 4} ∪ {5, 6, 7, 8}
={1, 2, 3, 4, 5, 6, 7, 8}.

(iii) B ∪ C = {3, 4, 5, 6} ∪ {5, 6, 7, 8}
={3, 4, 5, 6, 7, 8}.

(iv) B ∪ D = {3, 4, 5, 6} ∪ {7, 8, 9, 10}
= {3, 4, 5, 6, 7, 8, 9, 10}.

(v) A ∪ B ∪ C = {l, 2, 3, 4} ∪ {3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}.

(vi) A ∪ B ∪ D = {1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪ {7,8, 9, 10}
= {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

(vii) B ∪ C ∪ D ={3, 4, 5, 6} ∪ {5, 6, 7, 8} ∪ {7, 8, 9, 10}
= {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10}
= {3, 4, 5, 6, 7, 8, 9, 10}.

Question 5.
Find the intersection of each pair of sets:
(i) X = {1, 3, 5}
Y = {1, 2, 3}
Answer.
(i) X = {1, 3, 5},
Y = {1, 2, 3}
X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}
B = {a, b, c}
Answer.
A = {a, e, i, o, u}, B = {a, b, c}
A ∩ B = {o}

(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
Answer.
A = {x : x is a natural number and multiple of 3}=(3, 6, 9 …}
B = {x : x is a natural number less than 6}={1, 2, 3, 4, 5}
∴ A ∩ B = {3}

(iv) A = {x : x is a natural number and 1 < x ≤ 6}
B = {x : x is a natural number and 6 < x < 10}
Answer.
(iv) A = {x : x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x <10} = {7, 8, 9}
A ∩ B = Φ

(v) A = {1, 2, 3},
B = Φ
Answer.
A = {1, 2, 3}, B= Φ
A ∩ B = Φ.

Question 6.
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩B
(ii) B ∩ C
(iii)A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Answer.
(i) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13}
= {7, 9, 11}.

(ii) B ∩ C = {7, 9, 11, 13} ∩ {11, 13, 15}
= {11, 13}.

(iii) A ∩ C ∩ D = {3, 5, 7, 9, 11} ∩ {11, 13, 15} ∩ {15, 17}
= {11} ∩ {15, 17}
= Φ.

(iv) A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} =
{11}

(v) B ∩ D = {7, 9, 11, 13} ∩ {15, 17}
= Φ

(vi) A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {11, 13, 15})
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}.

(vii) A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17}
= Φ

(viii) A ∩ (B ∪ D) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {15, 17}
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}
= {7, 9, 11}

(ix) (A ∩ B) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13}
= {7, 9, 11}
B ∪ C = {7, 9, 11, 13} ∪ {11, 13, 15}
= {7, 9, 11, 13, 15}
∴ (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}

(x) (A ∩ D) = {3, 5, 7, 9, 11} ∩ {15, 17}
= {3, 5, 7, 9, 11, 15 ,17}
B ∪ C = {7, 9,11, 13, 15} [From part (ix)]
∴ (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11,15}

Question 7.
If A={x : x. is a natural number}, B = {x : x is an even natural number}, C = {x : x is an odd natural number} and D = {x : x is a prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Answer.
A = {x : x is a natural number} = {1, 2, 3, 4, 5 ……..}
B = {x : x is an even natural number} = {2, 4, 6, 8 ………..}
C = {x : x is an odd natural number} = {1, 3, 5, 7, 9 …………}
D = {x : x is a prime number} = {2, 3, 5, 7 ……….}
(i) A ∩ B = {x : x is a even natural number} = B
(ii) A ∩ C = {x : x is an odd natural number} = C
(iii) A ∩ D = {x : x is a prime number} = D
(iv) B ∩ C = Φ
(v) B ∩ D = {2}
(vi) C ∩ D = {x : x is an odd prime number}.

Question 8.
Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 < x < 6}.
Answer.
{1, 2, 3, 4} and {x : x is a natural number and 4 < x < 6} = {4, 5, 6}
Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u}and {c, d, e, f}
Answer.
{a, e, i, o, u} ∩ (c, d, e, f} = {e}
Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x : x is an even integer} and {x : x is an odd integer}
Answer.
{x : x is an even integer} ∩ {x : x is an odd integer} = Φ.
Therefore, this pair of sets is disjoint.

Question 9.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C
Answer.
(i) A – B = {3, 6, 9,12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}

(ii) A – C= {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10,12, 14, 16}
= {3, 15, 18, 21}.

(iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 12, 18, 21}.

(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}.

(v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}.

(vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21}
= {5, 10, 20}.

(vii) B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
= {20}.

(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}
= {4, 8, 12, 16}.

(ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14}.

(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}
= {5, 10, 15}.

(xi) C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}
= {2, 4, 6, 8, 12, 14, 16}.

(xii) D – C = {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
= {5, 15, 20}.

Question 10.
If X = {a, b, c, d} and Y = {f, b, d, g}, find ;
(i) X – Y
Answer.
X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}.

(ii)Y – X
Answer.
Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}.

(iii) X ∩ Y
Answer.
X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}.

Question 11.
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Answer.
R : set of real numbers
Q : set of rational numbers
Therefore, R – Q is a set of irrational numbers.

Question 12.
State whether each of the following statement is true or false. Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
Answer.
False
As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}
⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
Answer.
False
As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}
⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
Answer.
True
As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) {2, 6, 10} and {3, 7,11} are disjoint sets.
Answer.
True
As {2, 6, 10} ∩ {3, 7, 11} = Φ

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3

Question 1.
Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4} ….. {1, 2, 3, 4, 5}
(ii) {a, b, c} …… {b, c, d}
(iii) {x : x is a student of Class XI of your school} …… {x : x student of your school}
(v) {x : x is a circle in the plane} … {x : x is a circle in the same plane with radius 1 unit}
(v) (x : x is a triangle in a plane} … {x : x is a. rectangle in: the plane} .
(vi) {x : x is an equilateral triangle in a plane} … {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} … {x : x is an integejf}
Answer.
(i)'{2, 3, 4} ⊂ {1, 2, 3, 4, 5}
(ii) {a, b, c} ⊄ {b, c, d}
(iii) {x : x is a student of class XI of your school} ⊂ {x : x is student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} c {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} c {x : x is an integer}.

Question 2.
Examine whether the following statements are true or false :
(i) {a, b} ⊂ {b, c, a}
(ii) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ∈ {a, b, c}
(vi) {x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}
Answer.
(i) False. Each element of {a, b} is also an element of {b, c, a}.
(ii) True, a, e are two vowels of the English alphabet.
(iii) False 2 ∈ {1, 2, 3}; however, 2 ∉ {1, 3, 5}
(iv) True. Each element of {a} is also an element of {a, b, c}
(v) False. The elements of {a, b, c} are a, b, c. Therefore, {a} ⊂ {a, b, c}
(vi) True, {x : x is an even natural number less than 6} = {2, 4} {x : x is a natural number which divides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Question 3.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why ?
(i) {3, 4} ⊂ A
(ii) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) 1 ∈ A
(v) 1 ⊂ A
(vi) {1, 2, 5} ∈ A
(vii) {1, 2, 5} ⊂ A
(viii) {1, 2, 3} ∈ A
(ix) Φ ∈ A
(x) Φ ⊂ A
(xi) {Φ} ⊂ A
Answer.
A = {1, 2, {3, 4}, 5}
(i) The statement {3, 4} ⊂ A is incorrect because 3 e {3, 4}; however, 3g A.
(ii) The statement {3, 4} ∈ A is correct because {3, 4} is an element of A.
(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4}
(iv) The statement 1 ∈ A is correct because 1 is an element of A.
(v) The statement 1 ⊂ A is incorrect because an element of a set can never be a subset of itself.
(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.
(vii) The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.
(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 g A.
(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.
(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.
(xi) The statement {Φ} ⊂ A is incorrect because Φ ∈ {Φ}; however, Φ ∈ A.

Question 4.
Write down all the subsets of the following sets :
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Answer.
(i) The subsets of {a} are Φ and {a}.
(ii) The subsets of {a, b} are Φ, {a}, {b}, and {a, b}.
(iii) The subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}
(iv) The only subset of Φ is Φ.

Question 5.
How many elements has P(A), if A = Φ?
Answer.
We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2^m.
If A = Φ, then n(A) = 0.
n[P(A)] = 2⁰ = 1.
Hence, P(A) has one element.

Question 6.
Write the following as intervals:
(i) {x : x ∈ R, – 4 < x < 6}
(ii) {x : x ∈ R, – 12 < x < – 10}
(iii) {x : x ∈ R, 0 < x < 7}
(iv) {x : x ∈ R, 3 ≤ x ≤ 4}
Answer.
(i) {x : x ∈ R, – 4 < x < 6} = (- 4, 6]
(ii) {x : x ∈ R, – 12 < x < – 10} = (- 12, – 10)
(iii) {x : x ∈ R, 0 < x < 7} = [0, 7)
(iv) {x : x ∈ R, 3 ≤ x ≤ 4} = [3, 4]

Question 7.
Write the following intervals in set-builder form :
(i) (- 3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [- 23, 5)
Answer.
(i) (- 3, 0) = {x : x ∈ R, – 3 < x < 0}
(ii) [6, 12] = {x : x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x : x ∈ R, 6 < x < 12}
(iv) [- 23, 5) = {x : x ∈ R, – 23 ≤ x < 5}.

Question 8.
What universal set (s) would you propose for each of the following:
(i) The set of right triangles
(ii) The set of isosceles triangles
Answer.
(i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.
(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

Question 9.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = (0, 2, 4, 6, 8}, which of the following may be considered as universals set (s) for all the three sets A, B and C
(i) (0, 1, 2, 3, 4, 5, 6}
(ii) 4
(iii) {0, 1, 2, 3, 4, 5, 6, 7,8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Answer.
(i) It can be seen that
A ⊂ {0, 1, 2, 3, 4, 5, 6}
B ⊂ {0, 1, 2, 3, 4, 5, 6}
However, C ⊄ {0, 1, 2, 3, 4, 5, 6}
Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
Therefore, Φ cannot be the universal set for the sets A, B, and C.

(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C ⊂ {0, 1,2, 3, 4, 5, 6, 7, 8, 9, 10}
Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 1 Sets Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

Question 1.
Which of the following are examples of the null set.
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x : x is a natural number, x < 5 and x > 7}
(iv) {y : y is a point common to any two parallel lines}
Answer.
(i) A set of odd natural numbers divisible by 2 is a null set because no odd number is divisible by 2.
(ii) A set of even prime numbers is not a null set because 2 is an even prime number.
(iii) {x : x is a natural number, x < 5 and x > 7} is a null set because a number cannot be simultaneously less than 5 and greater than 7.
(iv) {y : y is a point common to any two parallel lines} is a null set because parallel lines do not intersect.
Hence, they have no common point.

Question 2.
Which of the following sets are finite or infinite
(i) The set of months of a year
(ii) {1, 2, 3,…}
(iii) {1, 2, 3, 99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Answer.
(i) The set of months of a year is a finite set because it has 12 elements.
(ii) {1, 2, 3, …} is an infinite set as it has infinite number of natural numbers.
(iii {1, 2, 3, …………., 99, 100} is a finite set because the numbers from 1 to 100 are finite in number.
(iv) The set of positive integers greater than 100 is an infinite set because positive integers greater than 100 are infinite in number.
(v) The set of prime numbers less than 99 is a finite set because prime numbers less than 99 are finite in number.

Question 3.
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0, 0)
Answer.
(i) The set of lines which are parallel to the x-axis is an infinite set because lines parallel to the x-axis are infinite in number.
(ii) The set of letters in the English alphabet is a finite set because it has 26 elements.
(iii) The set of numbers which are multiple of 5 is an infinite set because multiples of 5 are infinite in number.
(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is quite a big number).
(v) The set of circles passing through the origin (0, 0) is an infinite set because infinite number of circles can pass through the origin.

Question 4.
In the following, state whether A=B or not:
(i) A = {a, b, c, d}; B = {d, c, b, a}
(ii) A = {4, 8, 12, 16}; B = {8, 4, 16,18}
(iii) A = {2, 4, 6, 8,10}; B = {x : x is positive even integer and x < 10}
(iv) A = {x : x is a multiple of 10}; B = {10, 15, 20, 25, 30, ………}
Answer.
(i) A = {a, b, c, d}; B = {d, c, b, a}
The order in which the elements of a set are listed is not significant.
∴ A = B

(ii) A = {4, 8,12, 16}; B = {8, 4,16,18}.
It can be seen that 12 ∈ A but 12 ∉ B and 18 ∈ B but 18 ∉ A.
∴ A ≠ B

(iii) A = {2, 4, 6, 8, 10}
B = {x : x is positive even integer and x < 10}
= {2, 4, 6, 8, 10}
∴ A = B

(iv) A = {x : x is a multiple of 10}
= {10, 20, 30, 40, ……..}
B = {10, 15, 20, 25, 30, …………}
It can be seen that 15 ∈ B but 15 ∉ A.
∴ A ≠ B.

Question 5.
Are the following pair of sets equal? Give reasons.
(i) A = {2, 3}; B ={x : x is solution of x² + 5x + 6 = 0}
(ii) A= {x : x is a letter in the word FOLLOW}; B = {y : y is a letter in the word WOLF}
Answer.
(i) A = {2, 3}; B = {x : x is a solution of x² + 5x + 6 = 0
The equatioh x² + 5x + 6 = 0 can be solved as :
x (x + 3) + 2 (x + 3) = 0
⇒ (x + 2) (x + 3) = 0
⇒ x = – 2 or x = – 3
∴ A = {2, 3}; B = {- 2, – 3}
∴ A ≠ B

(ii) A = {x : x is a letter in the word FOLLOW} = {F, O, L, W}
B = {y : y is a letter in the word WOLF} = {W, O, L, F}
The order in which the elements of a set are listed is hot significant.
∴ A = B

Question 6.
From the sets given helow, select equal sets :
A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2} E = {- 1, 1},F = {0 ,o}, G = {1, – 1}, H = {0, 1}
Answer.
A = {2, 4, 8, 12}; B = {1, 2, 3, 4}; C = {4, 8, 12, 14}
D = {3, 1, 4, 2}; E = {- 1, 1}; F = {0, a}
G = {1, – 1}; A = {0, 1}
It can be seen that
8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H
⇒ A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H
Also, 2 ∈ A, 2 ∉ C
∴ A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H
∴ B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H
∴ C ≠D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H
∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H
Similarly, E≠ F, E ≠ G, E ≠ H F ≠ G, F ≠ H, G ≠ H
The order in which the elements of a set are listed is not significant
∴ B = D and E = G
Hence, among the given sets, B = D and E = G.