Sets(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

SETS MCQs

Sets(11th And 12th Grade)

Total Questions : 60 | Page 5 of 6 pages

Question 41. Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,
Then n

(A^{c} \cap B^{c})

Discuss Question

Answer: Option C. -> 300
:
C
n(

A^{c}

∩

B^{c}

) = n(U) - n(A∪ B)
= n(U) - [n(A) + n(B) - n(A∩ B)]
= 700 - [200 + 300 - 100] = 300.

Question 42. If the sets A and B are defined as A = {(x, y) : y =

e^{x}

, x ∈ R};
B = {(x, y) : y = x, x ∈ R}, then

B ⊆ A
A ⊆ B
A ∩ B = ∅
A ∪ B = A

Discuss Question

Answer: Option C. -> A ∩ B = ∅
:
C
Since, y =

e^{x}

and y = x do not meet for any x∈ R

∴

A∩ B =∅ .

Question 43. If A = [(

x, y

x^{2} + y^{2} = 25

] And B = [(

x, y

x^{2} + 9 y^{2} = 144],

then

A \cap B

contains

One point
Three points
Two points
Four points

Discuss Question

Answer: Option D. -> Four points
:
D
A = Set of all values (x,y) :

x^{2} + y^{2} = 25 = 5^{2}

If A = [(x,y): X2+y2=25] And B = [(x,y): X2+9y2=144], the...

B =

\frac{x^{2}}{144} + \frac{y^{2}}{16} = 1

i.e.,

\frac{x^{2}}{(12)^{2}} + \frac{y^{2}}{(4)^{2}} = 1

Clearly ,

A \cap B

consists of four points.

Question 44. If the sets A and B are defined as
A = {(x, y) : y =

\frac{1}{x}

, 0 ≠ x ∈ R}
B = {(x, y) : y = -x, x ∈ R}, then

A ∩ B = A
A ∩ B = B
A ∩ B = ∅
None of these

Discuss Question

Answer: Option C. -> A ∩ B = ∅
:
C
Since y =

\frac{1}{x}

, y = -x meet when -x =

\frac{1}{x}

⇒

x^{2}

= -1,
which does not give any real value of x.
Hence, A ∩ B =∅.

Question 45. The group of intelligent students in a class is __________.

a null set
a finite set
a well defined collection
not a well defined collection

Discuss Question

Answer: Option D. -> not a well defined collection
:
D
Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is nota well defined collection.

Question 46.

A = {x : x \in R, x^{2} = 16 and 2 x = 6}

can be represented in the roster form as _________ .

A = {}
A = { 14, 3, 4 }
A = { 3 }
A = { 4 }

Discuss Question

Answer: Option A. -> A = {}
:
A

x^{2} = 16 \Rightarrow x = \pm 4

2 x = 6 \Rightarrow x = 3

There is no value of

x

which satisfies both the givenequations. The set A is an empty set or a null set.
Thus, A = {}.

Question 47. If a set A has n elements, then the total number of subsets of A is

Discuss Question

Answer: Option C. -> 2n
:
C
Number of subsets of A =

^{n} C_{0}

^{n} C_{1}

+ .............+

^{n} C_{n}

2^{n}

Question 48. Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in (A U B )

Discuss Question

Answer: Option B. -> 6
:
B
n(A

\cup

B) = n(A) + n(B) - n(A

\cap B

) = 3 + 6 - 1

(A \cap B)

Since maximum number of elements in A

\cap

B = 3

∴

Minimum number of elements in A

\cup

B = 9 - 3 = 6 .

Question 49.

A = {x : x \in R, x^{2} = 16 and 2 x = 6}

can be represented in the roster form as _________ .

A = {}
A = { 14, 3, 4 }
A = { 3 }
A = { 4 }

Discuss Question

Answer: Option A. -> A = {}
:
A

x^{2} = 16 \Rightarrow x = \pm 4

2 x = 6 \Rightarrow x = 3

There is no value of

x

which satisfies both the givenequations. The set A is an empty set or a null set.
Thus, A = {}.

Question 50. The group of intelligent students in a class is __________.

a null set
a finite set
a well defined collection
not a well defined collection

Discuss Question

Answer: Option D. -> not a well defined collection
:
D
Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is nota well defined collection.