Baroda Lions School

Date : 11-07-2024 STD 11 Science Maths Total Marks : 115

SECTION A

* Choose the right answer from the given options. [1 Marks Each] [10]
1. The smallest set A such that A ∪ {1, 2} = {1, 2, 3, 5, 9} is:
(A) {2, 3, 5} (B) {3, 5, 9} (C) {1, 2, 5, 9} (D) {1, 2}
Ans. :
b. {3, 5, 9}
Solution:
A ∪ {1, 2} = {1, 2, 3, 5, 9}
Thus,
A = {1, 2, 3, 5, 9} − {1, 2}

2. Choose the correct answers from the given four option:

If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ....., 18} and N the set of natural numbers
′ ′
is the universal set, then A ∪ (A ∪ B) ∪ B ) is
(A) ϕ (B) N (C) A (D) B

Ans. :
b. N
Solution:
Given that:
A = {1, 3, 5, 7, 9, 11, 13, 15, 17}
B = {2, 4, ...., 18}
U = N = {1, 2, 3, 4, 5, .....}
′ ′ ′ ′ ′
A ∪ (A ∪ B) ∩ B = A [(A ∩ B ) ∪ (B ∩ B )]

′ ′ ′
= A ∪ (A ∩ B ) ∪ ϕ [ ∵ A ∩ A = ϕ]

′ ′
= A ∪ (A ∩ B )
′ ′ ′
= ( A ∪ A) ∩ (A ∪ B )
′ ′ ′
= N ∪ (A ∪ B ) [ ∵ A ∪ A = N]

′ ′
= A ∪B

′ ′
= (A ∪ B ) = (ϕ) = N [ ∵ A ∩ B = ϕ]

Hence, the correct option is (b).

3. Let n(A) = 28,n(A ∩ B) =8, n(A∪B) =52, then n(A ∩ B′):

(A) 30 (B) 32 (C) 20 (D) none of these

Ans. :
c. 20
Solution:
Given n(A) = 28, n(A ∩ B) = 8.
We have A ∩ B′ = A - A ∩ B.

[1]
 This give n(A ∩ B′) = n(A) -n(A ∩ B)
or, n(A ∩ B′) = 28 - 8 = 20.

4. If the set A has 3 elements and the set B = {1, 3, 4, 5}, then the number of elements in
(A × B) is:
(A) 11 (B) 12 (C) 13 (D) 15

Ans. :
b. 12
5. The range of f(x) =
1
is:
1−2 cos x

1 1
(A) [
3
, 1] (B) [ − 1,
3
]

(C) ( − ∞, −1) ∪ [
1
, ∞) (D) [ −
1
, 1]
3 3

Ans. :
b.
1
[ − 1, ]
3

Solution:
We know that −1 ≤ cos x ≤ 1 for all x ∈ R

Now,
−1 ≤ cos x ≤ 1

⇒ −1 ≤ cos x ≤ 1

⇒ −2 ≤ −2 cos x ≤ 2

⇒ −1 ≤ 1 − 2 cos x ≤ 3 (Adding 1 ro each term)

But,
1
cos x ≠
2

⇒ 1 − 2 cos x ∈ [ − 1, 3] − {0}

1 1
⇒ ∈ ( − ∞, −1] ∩ [ , ∞)
1−2 cos x 3

1
∴ Range of f(x) = ( − ∞, −1] ∩ [ , ∞)
3

Disclaimer: The range of the function does not matches with either of the given
1
options. The range matches with option (c) if it is given as ( − ∞, −1] ∩ [ , ∞)
3

6. If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A - B) × (B - C) is:

(A) {(1, 2), (1, 5), (2, 5)} (B) {(1, 4)}
(C) (1, 4) (D) none of these.

Ans. :
b. {(1, 4)}
Solution:
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
(A - B) = {1}
(B - C) = {4}
So, (A - B) × (B - C) = {(1, 4)}

7.

[2]
 Let A = {1, 2, 3, 4, 5} and R be a relation from A to A, R = {(x, y) : y = x + 1}. Find the
range:
(A) {1, 2, 3, 4, 5} (B) {2, 3, 4, 5} (C) {1, 2, 3, 4} (D) {1, 2, 3, 4, 5, 6}

Ans. :
b. {2, 3, 4, 5}
Solution:
Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1, 2), (2, 3), (3, 4), (4, 5)}.
Range = {2, 3, 4, 5}.

8. Choose the correct answers:

2
x +2x+1
The domain of the function f given by f(x) = 2
.
x −x−6

(A) R – {3, –2} (B) R – {–3, 2} (C) R – [3, –2] (D) R – (3, –2)

Ans. :
a. R – {3, –2}
Solution:
2
x +2x+1
Given that: f(x) = 2
x −x−6

2
f(x) is defined if x −x −6 ≠ 0

2
⇒ x − 3x + 2x − 6 ≠ 0

⇒ (x − 3)(x + 2) ≠ 0

⇒ x ≠ −2, x ≠ 3

So, the domain of f(x) = R - {-2, 3}

9. In a function from set A to set B, every element of set A has___________ image in set B:
(A) one and only one (B) different (C) same (D) many

Ans. :
a. One and only one
Solution:
A relation from a set A to a set B is said to be a function if every element of set A has
one and one image in set B.
−−−−−−− −−−−
10. Domain of the function f(x) = √(2 − 2x − x ) is:
−−−−−−− –
(A) −√3 ≤ x ≤ √3 +
−−−−−−−−−
(B) -1- √3 ≤ x ≤ −1
–
(C) +√3

(D) -2 ≤ x ≤ 2

Ans. :
−−−−−−−−−
b. −1 − √3 ≤ x ≤ −1

* State whether the following sentences are True or False. [1 Marks Each] [7]
11. If x ∈ A and A ∈ B , then x ∈ B. If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

[3]
12. If A ⊂ B and B ∈ C , then A ∈ C. If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

13. If A ⊂ B and B ⊂ C , then A ⊂ C. If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

14. If A ⊄ B and B ⊄ C , then A ⊄ C . If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

15. If A ⊄ B and B ⊄ C , then A ⊄ C . If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

16. If x ∈ A and A ⊄ B , then x ∈ B. If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

17. If A ⊂ B and x ∉ B, then x ∉ A. If it is true, prove it. If it is false, give an example.

Ans. : Self Learning

* Answer the following questions in one sentence. [1 Marks Each] [23]

18. Let A = {a, e, i, o, u} and B = {a, b, c, d}. Is A a subset of B? No. (Why?). Is B a subset of A?

Ans. :
i. Is A ⊂ B
According to the given we can state,
For a set to be a subset of another set, it needs to have all element presents in
another set. In the set A = {e, i, o, u} elements are present but these are not present
in set B.
Hence A ⊄ B
ii. Is b ⊂ A
According to the given we can state,
For this condition to be true, are elements of sets B should be element present in
sets A.
In the set B = {b, c, d} elements are present but these elements are not an element in
set A.
Hence B ⊄ A
19. Is x : x is a natural number, x < 5 and, x > 7 null set?

Ans. : {x : x is a natural number, x< 5 and x > 7} is an empty set because there is no
natural number which satisfies simultaneously r < 5 and x > 7.
20. Is the set of positive integers greater than 100 finite or infinite set?

Ans. : The set of positive integers greater than 100 is an infinite set because there
are infinite number of positive integers greater than 100.
21. Is the set of lines which are parallel to the x-axis finite or infinite?

Ans. : The set of lines which are parallel to the x-axis is an infinite set because we
can draw infinite number of lines parallel to x-axis.

[4]
22. Is the set of letters in the English alphabet finite or infinite?

Ans. : The set of letters in the English alphabet is a finite set because there are 26
letters in the English alphabet.
23. Is the pair of set A = {2, 3} and B = {x : x is solution of x2 + 5x + 6 = 0} equal? Give reasons.

Ans. : A = {2, 3} and B = {x : x is solution of x2 + 5x + 6 = 0}

Now x2 + 5x + 6 = 0 ⇒ x2 + 3x + 2x + 6 = 0
⇒ (x + 3)(x + 2) = 0 ⇒ x = -3, -2
∴ B = {-2, -3}
Hence A and B are not equal sets.
24. Make correct statement by filling the symbol ⊂ or ⊄ in the blank space: {x : x is a student
of Class XI of your school} ....... {x : x student of your school}

Ans. : {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}

25. Make correct statement by filling the symbol ⊂ or ⊄ in the blank space: {x : x is a circle in
the plane}.......{x : x is a circle in the same plane with radius 1 unit}

Ans. : {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1
unit}
26. Let A = {1, 2, {3, 4}, 5}. Is the statement 1 ∈ A incorrect and why?

Ans. : 1 is a member of set A.

∴ 1 ∈ A is correct.
27. Let A = {1, 2, {3, 4 }, 5}. Is the statement {1, 2, 5} ∈ A incorrect and why?

Ans. : Here, we can see that 1, 2, 5 is a member of set A

= {1, 2, 5} is a subset of A
Therefore, the given statement is incorrect.
28. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} find: A ∪ B

Ans. : Here A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D ={7, 8, 9, 10}
A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
29. 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: (A ∪ C)′

Ans. : Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9},

A ∪C = {1, 2, 3, 4} ∩ {3, 4, 5, 6}
= {1, 2, 3, 4, 5, 6}
(A ∪ C )
′
= U − (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6}
={7, 8, 9}
30. Taking the set of natural numbers as the universal set, write down the complement of the
set: {x : x is a perfect square}

[5]
 Ans. : Here U = {x : x ∈ N }

Let A = {x : x is a perfect square}

A
′
= U − A = {x : x ∈ N }− {x : x is a perfect square}
={x: x ∈ N , x is not a perfect square}

31. Using properties of sets, show that: A ∩ (A ∪ B) = A

Ans. : We know that if A ⊂ B then

A ∩B = A

Also A ⊂ A ∪B

∴ A ∩ (A ∪ B) = A

32. Is the set {x : x ∈ N and x2 = 4} finite or infinite?

Ans. : Given set = {2}. Thus, it is finite.

33. Consider the sets ϕ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄

between the pair of sets: B . . . C

Ans. : B ⊂ C as each element of B is also an element of C.

34. Is the collection of most dangerous animals of the world set? Justify your answer.

Ans. : A collection of most dangerous animals of the world is not a very clearly
defined set as the ranking of the animals keep on altering and their ranking vary
from countries to countries.
The collection of distnict objects are not well–defined and don't have universal
acceptance as it is.
Therefore,the collection is not set.
35. Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B

Ans. : It is given that A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}

∴ We have A ∪ B = { 2, 4, 6, 8, 10, 12}
36. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Check whether
(a, a) ∈ R for all a ∈ N ? Justify your answer.

Ans. : We have, R = {(a, b): a, b ∈ N and a = b2​}

(a, a) ∈ R, for all a ∈ N
As we can see that 3 ∈ N but 3 ≠ 32 = 9
Hence, the statement is not true.
37. Let R be a relation from Q to Q defined by R = {(a, b): a,b ∈ Q and a – b ∈ Z}. Show that (a,
b) ∈ R implies that (b, a) ∈ R

Ans. : (a, b) ∈ R implies that a – b ∈ Z.

So, b – a ∈ Z. [∵ m ∈ Z ⇒ −m ∈ Z ]
Hence, (b, a) ∈ R
38. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: A × (B ∩ C) = (A × B)
∩ (A × C)

[6]
 Ans. : Given: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

As we see, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = ϕ

By definition if either of the two set P and Q is null set then P ×Q will also be a
null set. i.e. P × Q = ϕ
⇒ A × (B ∩ C) = ϕ ......(i)

Now, (A × B) = {(1, 1),(1, 2),(1, 3),(1, 4),(2, 1),(2, 2),(2, 3),(2, 4)}
And (A × C) = {(1, 5),(1, 6),(2, 5),(2, 6)}
⇒ (A × B) ∩ (A × C) = ϕ ........(ii)

From (i) and (ii), we have

A × (B ∩ C) = (A × B) ∩ (A × C)

Hence proved.
39. 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 and z are distinct elements.

Ans. : Here (x, 1) ∈ A × B ⇒ x ∈ A and 1 ∈ B

(y, 2) ∈ A×B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A × B ⇒ z ∈ A and 1
It is given that n(A) = 3 and n(B) = 2
∴ A = {x, y, z}

and B = {1, 2}
40. 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.

Ans. : Here A = {1, 2, 3, . . . , 14}

We shall consider the ordered pairs which satisfy 3x - y = 0
They are (1, 3), (2, 6), (3, 9) and (4, 12)
Thus R = {(1, 3), (2, 6), (3, 9), (4, 12)}
∴ Domain = {1, 2, 3, 4}

Range = {3, 6, 9, 12}

Codomain = {1, 2,3, . . . , 14}

SECTION B

* Given section consists of questions of 2 marks each. [30]

41. Let A, B and C be the sets such that A ∪ B = A ∪C and A ∩ B = A ∩C Show that B =
C.

Ans. : We know that A = A ∩ (A ∪ B) and A = A ∪ (A ∩ B)

Now A ∩B = A ∩C and A ∪B = A ∪C

∴ B = B ∪ (B ∩ A) = B ∪ (A ∩ B) = B ∪ (A ∩ C) [∵ A ∩ B = A ∩ C]

= (B ∪ A) ∩ (B ∪ C) (By distributive law)

= (A ∪ C) ∩ (B ∪ C)

= (A ∪ C) ∩ (B ∪ C) [∵ A ∪ B = A ∪ C]

[7]
 = (C ∪ A) ∩ (C ∪ B)

= C ∪ (A ∩ B) (by distributive law)

= C ∪ (A ∩ C) [∵ A ∩ B = A ∩ C]

= C ∪ (C ∩ A) = C

Hence B = C.
42. Assume that P(A) = P(B) show that A = B.

Ans. : Let X ∈ A ⇒ {x} ∈ P (A)

⇒ {x} ∈ P (B) [∵ P (A) = P (B)]

⇒ X ∉ B

∴ A ⊂ B . . . (i)
Let X ∈ B ⇒ {x} ∈ P (B)

⇒ {X} ∈ P (A) [∵ P (A) = P (B)]

⇒ X ∈ A . . . (ii)
∴ B ⊂ A

From (i) and (ii) we have A = B

43. 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?

Ans. : Let H be the set of students who know Hindi and E be the set of students
who know English.
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.
44. 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 the number of people who read at least one of the newspaper.

Ans. : Here
n(U) = a + b + c + d + e + f + g + h = 60 ....(i)
n (H) = a + b + c +d = 25 ....(ii)
n(T) = b + c + f + g = 26 .....(iii)
n(I) = c + d + e + f = 26 ....(iv)
n(H ∩ I ) = c + d = 9 .....(v)

n(H ∩ T ) = b + c = 11 .....(vi)
n(T ∩ I ) = c + f = 8 ....(vii)

n(H ∩ T ∩ I ) = c = 3 ....(viii)

Putting value of c in (vii),

[8]
 3+f=8⇒f=5
Putting value of c in (vi),
3 + b = 11 ⇒ b = 8
Putting values of c in (v),
3+d=9⇒d=6
Putting value of c, d, f in (iv),
3 + 6 + e + 5 = 26 ⇒ e = 26 - 14 = 12
Putting value of b, c, f in (iii),
8 + 3 + 5 + g = 26 ⇒ g = 26 - 16 = 10
Putting value of b, c, d in (ii)
a + 8 + 3 + 6 = 25 ⇒ a = 25 - 17 = 8
Number of people who read at least one of the three newspapers
=a+b+c+d+e+f+g
= 8 + 8 + 3 + 6 + 12 + 5 + 10 = 52
45. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and hence show that A
∩ B = B.

Ans. : Here, A ∩ B = { 2, 3, 5, 7 } = B. We know that B ⊂ A and that A ∩ B = B

46. Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence
show that (A ∪ B)′ = A′ ∩ B′

Ans. : Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Therefore, A′ ∩ B′ = { 1, 6 }

Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B )′ = { 1, 6 }
(A ∪ B)′ = {1, 6} = A′ ∩ B′
It can be shown that the above result is true in general. If A and B are any two
subsets of the universal set U, then ,we have
(A ∪ B)′ = A′ ∩ B′. Similarly, (A ∩ B)′ = A′ ∪ B′.
47. If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements, and Y has 32
elements, how many elements does X ∩ Y have?

Ans. : Here,
n (X ∪ Y) = 50, n (X) = 28, n (Y) = 32,
n (X ∩ Y) = ?
By using the formula,we have
n (X ∪ Y) = n (X) + n (Y) – n (X ∩ Y),
we find that
n (X ∩ Y) = n (X) + n (Y) – n (X ∪ Y)
= 28 + 32 – 50 = 10
48. In a school, there are 20 teachers who teach mathematics or physics. Of these, 12 teach
mathematics and 4 teach physics and mathematics. How many teach physics?

Ans. : Let n(P) denote the number of teachers who teach Physics and n(M) denote
the number of teachers who teach mathematics.

[9]
 We have,
n(P∪M) = 20, n(M) = 12 and n(P∩M) = 4
To find : n(P)
We know that
n(P∪M) = n(P) + n(M) - n(P∩M)
⇒ 20 = n(P) + 12 - 4

⇒ 20 = n(P) + 8

⇒ n(P) = 20 - 8
= 12
∴ 12 teachers teach Physics.

49. For sets A and B, show that: P (A ∩ B) = P (A) ∩ P (B)

Ans. : Let x ∈ P (A ∩ B)

⇒ x ⊂ (A ∩ B)

⇒ x ⊂ A and x ⊂ B

⇒ x ∈ P (A) and x ∈ P (B)

⇒ x ∈ P (A) ∩ P (B)

⇒ x ⊂ P (A) ∩ P (B)

∴ P (A ∩ B) ⊂ P (A) ∩ P (B) . . . (i)

Let x ∈ P (A) ∩ P (B)

⇒ x ∈ P (A) and x ∈ P (B)

⇒ x ⊂ A and ⇒ x ⊂ B

⇒ x ⊂ A ∩B

⇒ x ⊂ P (A ∩ B)

∴ P (A) ∩ P (B) ⊂ P (A ∩ B) . . . . (ii)

From (i) and (ii), we have
P (A ∪ B) = P (A) ∩ P (B)

50. If the function t which maps temperature in degree Celcius into temperature in
degree Fahrenheit is defined by t(C) = + 32, then find t(-10).
9C

Ans. : Here it is given that,

9C
t(C) = +32
5

Put C = -10, we get

9×(−10)
t(−10) = +32
5
−9×10
= 5
+32 = −9 × 2 + 32

= -18 + 32 = 14
51. Find the domain and the range of the real function f defined by f(x) = |x - 1|.

Ans. : Here f (x) = |x - 1|

The function f (x) is defined for all values of x
∴Domain of f(x) = R

when x > 1
|x - 1| = x - 1 > 0

[10]
 When x = 1
|x - 1| = 0
When x < 1
|x - 1| = - x + 1 > 0
Range of f(x) = all real numbers ⩾ 0

= [0, ∞)

52. 2

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

Ans. : Here
x
f (x) =
2
1+x

y + yx2 = x2 ⇒ x2(1 - y) = y
2

Put y =
x
2
⇒
1+x
−−−
y y
2
⇒ x = ⇒ x = ±√
1−y 1−y

y
≥ 0
1−y
y
⇒ ≤ 0
y−1

⇒ 0 ≤ y < 1

⇒ y ∈ [0, 1)

∴Range of f(x) = [0, 1)

53. Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find : A × (B ∩ C)

Ans. : Here we have,

A = {1,2,3}, B = {3,4} and C = {4,5,6}
By the definition of the intersection of two sets, (B ∩ C) = {4}
Therefore, A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}
54. Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y)
: y = 2x, x, y ∈ N}. What is the domain, codomain and range of R? Is this relation a
function?

Ans. : Since the relation R is defined on N.Therefore,

The domain of R is the set of natural numbers N. The co-domain is also N. The
range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a
function.
55. Find the domain of the function f(x) =
2
x +3x+5
2
x −5x+4

Ans. : Since x2 –5x + 4 = (x – 4) (x –1), therefore the function f is defined for all real
numbers except at x = 4 and x = 1.
Therefore, the domain of f is R – {1, 4}

SECTION C

* Given section consists of questions of 3 marks each. [45]

56.

[11]
 Decide among the following sets which sets are subsets of each another:
A = {X : X ∈ R} and x satisfies x2 - 8x + 12 = 0}, B = {2, 4, 6} , C = {2, 4, 6, 8, ...}, D = {6}

Ans. : It is given in the question that,

2
A = {x : x ∈ R and x satisfy x − 8x + 12 = 0}

As, 2 and 6 are the only solutions of x

2
− 8x + 12 = 0

∴ A = {2, 6}

Also it is given that,

B = {2, 4, 6}

C = {2, 4, 6, 8, … . }

And, D = {6}
∴ D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

57. Show that the following four conditions are equivalent :

i. A⊂B
ii. A–B=ϕ
iii. A∪B=B
iv. A∩B=A

Ans. : Here, first we will prove (i) ⇔ (ii)

Where, (i) = A ⊂ B and (ii) = A - B ≠ ϕ
Suppose that A ⊂ B
Now, we need to prove A − B ≠ ϕ
If possible, let A − B ≠ ϕ
Thus, there exists X ∈ A, X ≠ B, but this is impossible as A ⊂ B

∴ A −B = ϕ

And A ⊂ B ⇒ A − B ≠ ϕ
Let suppose that A − B ≠ ϕ
Now, we have to prove: A ⊂ B
Let X ∈ A
It can be concluded that X ∈ B( if X ∉ B, then A − B ≠ ϕ)

Thus, A - B = ϕ = A ⊂ B
∴ (i) ⇔ (ii)

Let us assume that A ⊂ B

To prove: A ∪ B = B
⇒ B ⊂ A ∪B

Let us assume that, x ∈ A ∪ B

⇒ x ∈ A or x ∈ B

Taking case I: X ∈ B

A ∪B = B

Taking Case II : X ∈ A

⇒ X ∈ B(A ⊂ B)

⇒ A ∪B ⊂ B

[12]
 Let A ∪ B = B
Let us assume that X ∈ A

⇒ X ∈ A ∪ B(A ⊂ A ∪ B)

⇒ X ∈ B(A ∪ B = B)

∴ A ⊂ B

Thus, (i) ⇔ (iii)

Now, to prove (i) ⇔ (iv)
Suppose that A ⊂ B
It can be observed that A ∩ B ⊂ A

Let X ∈ A
To show: X ∈ A ∩ B
Since, A ⊂ B and X ∈ B
Therefore, X ∈ A ∩ B
⇒ A ⊂ A ∩B

⇒ A = A ∩B

Similarly, let us assume that A ∩B = A

Let X ∈ A
⇒ X ∈ A ∩B

⇒ X ∈ B and X ∈ A

⇒ A ⊂ B

∴ (i) ⇔ (ii)
Therefore, proved that (i) ⇔ (ii) ⇔ (iii) ⇔ (iv)
58. 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 ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law ]

Ans. : Here A ∪X = B∪X for some X

⇒ A ∩ (A ∪ X) = A ∩ (B ∪ X)

⇒ A = (A ∩ B) ∪ (A ∩ X) [∵ A ∩ (A ∪ X) = A]

⇒ A = (A ∩ B) ∪ ϕ [∵ A ∩ X = ϕ]

⇒ A = A ∩B . . . (i)
⇒ A ⊂ B

Also A ∪X = B∪X

⇒ B ∩ (A ∪ X) = B ∩ (B ∪ X)

⇒ (B ∩ A) ∪ (B ∩ X) = B [∵ B ∩ (B ∪ X) = B]

⇒ (B ∩ A) ∪ ϕ = B [∵ B ∩ X = ϕ]

⇒ B∩A = B

⇒ B∩A . . . (ii)
From (i) and (ii), we have
A = B.
59. 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

[13]
 people liked products B and C and 8 liked all the three products. Find how many liked
product C only?

Ans. : Here
n (A) = a + b + c + d = 21 ....(i)
n(B) = b + c + f + g = 26 ....(ii)
n(C) = c + d + e + f = 29 ....(iii)
n(A ∩ B) = b + c = 14 ....(iv)

n(C ∩ A) = c + d = 12 ....(v)
n(B ∩ C) = c + f = 14 ....(vi)

n(A ∩ B ∩ C) = c = 8 ....(vii)

Putting value of c in (iv), (v) and (vi)

b + 8 = 14 ⇒ b = 6
8 + d = 12 ⇒ d = 4
8 + f = 14 ⇒ f = 6
Putting value of c, d, f in (iii),
8 + 4 + e + 6 = 29 ⇒ e = 29 - 18 = 11
Number of people who like product C only = 11
60. Let U be universal set of all the students of Class XI of a coeducational school and A be the
set of all girls in Class XI. Find A′.

Ans. : We know that,In a coeducational school, there can be only boys and girls in a
school.
A is the set of all girls,then
A' = Set of all Students - Set of all girls
A' = Set of all boys in class XI
61. In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each
student likes to play at least one of the two games. How many students like to play both
cricket and football?

Ans. : Let C be the set of students who like to play cricket and F be the set of
students who like to play football.
Here n(C) = 24, n(F) = 16, n(C ∪ F ) = 35.
We know that
n(C ∪ F ) = n(C) + n(F ) − n(C ∩ F )

35 = 24 + 16 - n(C ∩ F )
n(C ∩ F ) = 40 - 35 = 5

62. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking
orange juice and 75 were listed as taking both apple as well as orange juice. Find how

[14]
 many students were taking neither apple juice nor orange juice.

Ans. : Let A denote the set of students taking apple juice and B denote the set of
students taking orange juice
n(U) = 400, n(A) = 100, n (B) = 150 n(A ∩) =75
′ ′ ′
n (A ∩ B ) = n(A ∪ B)

= n (∪) − n (A ∪ B)

= n (∪) − [n(A) + n(B) − n(A ∩ B)]

=400-100-150+75=225
63. A market research group conducted a survey of 1000 consumers and reported that 720
consumers like product A and 450 consumers like product B, what is the least number that
must have liked both products?

Ans. : Let U be the set of consumers questioned, S be the set of consumers who
liked the product A and T be the set of consumers who like the product B.
Given that:
n(U) = 1000, n(S) = 720, n(T) = 450
Therefore, n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
= 720 + 450 – n (S ∩ T) = 1170 – n(S ∪ T)
Thus, n(S ∪ T) is maximum when n(S ∩ T) is least. But S ∪ T ⊂ U implies
n(S ∪ T) ≤ n (∪) = 1000. So, maximum values of n(S ∪ T) is 1000.
Therefore, the least value of n(S ∪ T) is 170.
64. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈A, b is exactly
divisible by a}.
i. Write R in roster form
ii. Find the domain of R
iii. Find the range of R.

Ans. : Here A = {1, 2, 3, 4, 6}

We have to form a set of ordered pairs (a, b) where 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}
65. 2
x ,0 ≤ x ≤ 3
The relation f is defined by f (x) = { and the relation g is defined by
3x, 3 ≤ x ≤ 10
2
x ,0 ≤ x ≤ 2
g(x) = { . Show that f is a function and g is not a function.
3x, 2 ≤ x ≤ 10

2
x ,0 ≤ x ≤ 3
Ans. : f (x) = {
3x, 3 ≤ x ≤ 10

At x = 3, f (x) = x
2

2
∴ f (3) = 3 = 9

Also, at x = 3, f (x) = 3x

⇒ f(3) = 3 × 3 = 9

[15]
 Since, f is defined at x = 3. Hence, f is a function.
2
x ,0 ≤ x ≤ 2
Now, g(x) = {
3x, 2 ≤ x ≤ 10

At x = 2, g(x) = x
2

2
⇒g(2) = 2 = 4

Also, at x = 2, g(x) = 3x ⇒g(2) = 3 × 2 = 6

At x = 2, relation g has two values.

∴ The relation g is not a function.

66. Define the function f: R → R by y = f(x) = x2, x ∈ R. Complete the Table given below by
using this definition. What is the domain and range of this function? Draw the graph of f.

x –4 –3 –2 –1 0 1 2 3 4

y = f(x) = x2

Ans. : The completed table is given below:

x –4 –3 –2 –1 0 1 2 3 4

y = f (x) = x2 16 9 4 1 0 1 4 9 16

Domain of f = {x : x ∈ R}. Range of f = {x2: x ∈ R}. The graph of f is given by the

figure shown below:

67. Draw the graph of the function f : R → R defined by f (x) = x3, x ∈ R.

Ans. : Given function is, f (x) = x3, x ∈ R.

We have f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc.

[16]
 Therefore, f = {(x, x3): x ∈ R}. The graph of f is given in figure showjbelow:

68. Define the real valued function f : R – {0} → R defined by f (x) =

1
, x ∈R – {0}. Complete
x

the Table given below using this definition. What is the domain and range of this function?

x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2

y =
1
... ... ... ... ... ... ... ... ...
x

Ans. :
The completed table is given by:
x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2

y =
1
– 0.5 – 0.67 –1 –2 4 2 1 0.67 0.5
x

The domain is all real numbers except zero and its range is also all real numbers
except zero. The graph of f is given in figure shown below:

69. Let R be the set of real numbers. Define the real function f: R → R by f(x) = x + 10 and
sketch the graph of this function.

Ans. : Here we have f(x) = x + 10

Here f(0) = 10, f(1) = 11, f(2) = 12, ..., f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, ..., f(–10)
 = 0 and so on.
Therefore, the shape of the graph of the given function assumes the form as
shown in figure below:

70. ⎧ 1 − x, x < 0
⎪

The function f is defined by f (x) = ⎨1 ,x = 0

⎩
⎪
x + 1, x > 0

Draw the graph of f(x).

Ans. : Here it is given that, f(x) = 1 – x, x < 0, this gives

f(– 4) = 1 – (– 4) = 5;
f(– 3) =1 – (– 3) = 4,
f(– 2) = 1 – (– 2) = 3
f(–1) = 1 – (–1) = 2; etc, and f(1) = 2, f (2) = 3, f (3) = 4
f(4) = 5 and so on for f(x) = x + 1, x > 0.
Thus, the graph of the given function f(x) is as shown in figure given below.

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