Answer
Verified
429k+ 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.
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
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE