Answer

Verified

381.9k+ views

**Hint:**Here, we are required to find the minimum number of elements that are present in $A \cup B$ when we are given that the set $A$ is having 3 elements and set $B$ is having 6 elements. Hence, we will consider 3 cases, when set $A$ is a subset of set $B$, when one of the elements in $A$ is not in $B$ and when nothing is common in $A$ and $B$. After observing these cases we will find the required minimum possible elements in $A \cup B$.

**Complete step-by-step answer:**

According to the question,

Set $A$ is having 3 elements and Set $B$ is having 6 elements.

Now, let us assume:

Case 1: When $A$ is a subset of $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {a,b,c,d,e,f} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {a,b,c,d,e,f} \right\} = \left\{ {a,b,c,d,e,f} \right\}$

Thus, when $A$ is a subset of $B$, the number of elements in $A \cup B$ are 6.

Case 2: When one of the elements in $A$ is not in $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {a,b,1,2,3,4} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {a,b,1,2,3,4} \right\} = \left\{ {a,b,c,1,2,3,4} \right\}$

Thus, when one of the elements in $A$ is not in $B$ , then the number of elements in $A \cup B$ are 7.

Case 3: When nothing is common in $A$ and $B$

Let us assume,

$A = \left\{ {a,b,c} \right\}$

and $B = \left\{ {1,2,3,4,5,6} \right\}$

Hence, $A \cup B = \left\{ {a,b,c} \right\} \cup \left\{ {1,2,3,4,5,6} \right\} = \left\{ {a,b,c,1,2,3,4,5,6} \right\}$

Thus, when nothing is common in $A$ and $B$, then the number of elements in $A \cup B$ are 9.

Clearly, in every possible case, the number of elements in $A \cup B$ is greater than or equal to the number of elements present in the set containing more elements.

**Hence, we can clearly say that the minimum number of elements that $A \cup B$ can have is 6.**

**Note:**

In mathematics, a set consists of a list of elements or numbers which are enclosed in curly brackets. A set can be written in two forms, i.e. Set-builder form or the roster form.

Set-builder form is used to represent an equation, an inequality or the numbers which have some kind of relation. This is also used to represent an infinite number of elements.

Roster form is the simpler form. In this form, we separate the numbers with the help of commas and they are enclosed again, in brackets.

Usually in a question, we are given set-builder form and to solve it further, we convert it to roster form. This makes the question easier to solve.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE