Test: Operations On Sets


10 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Operations On Sets


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QUESTION: 1

Which of the following two sets are disjoint?

Solution:

Two sets are disjoint if the intersection of two sets is the empty set.

QUESTION: 2

The Shaded region in the following figure illustrates

Solution:

First which region is over which region Then We will see that A is on the B so A intersection B and after C is on the A so, A intersection C after that we have to take all intersection part so A intersection B is Union with A intersection C.

The shaded region represents (A ∩ B) ∪ (A ∩ C).

QUESTION: 3

Shaded region in the following figure illustrates

Solution:

P U Q means P Union Q In set theory, the union (denoted by ∪) of a collection of sets is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other.

QUESTION: 4

If the sets A and B are defined as= {(xy) : ex∈ R}; = {(xy) : x,∈ R}, then

Solution:

Since,ex andx do not meet for any∈ R∩ φ.

QUESTION: 5

The intersection of the sets {1, 2, 5} and {1, 2, 6} is the set ______

Solution:

The intersection of the sets A and B, is the set containing those elements that are in both A and B.

QUESTION: 6

From the following Venn diagram, A ∩ (B ∪ C) is

Solution:
QUESTION: 7

If= {x is a multiple of 3} and= {x is a multiple of 5}, then A - B is

Solution:

A - B = A

∴A' ∩ B = A

QUESTION: 8

From the following venn diagram, A ∩ B is

Solution:


A ∩ B = {4, 3}

QUESTION: 9

In probability, the event ‘A or B’ can be associated with set:

Solution:

Probability of event A or B

The probability that Events A and B both occur is the probability of the intersection of A and B. The probability of the intersection of Events A and B is denoted by P(A ∩ B). If Events A and B are mutually exclusive, P(A ∩ B) = 0. The probability that Events A or B occur is the probability of the union of A and B.

QUESTION: 10

If A and B are two given sets, then∩ (∩ B)' is equal to

Solution:

 A ∩ ( A ∩ B )'

=A ∩ (A'  ∪ B' )  (De Morgan's Law)

=(A ∩ A' ) ∪ ( A∩B' )

=ϕ ∪ ( A ∩ ϕB' )

∴A ∩ ( A ∩ B )C

=A∩B'

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