
The set $\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'$ is equal to
\[\begin{align}
& A.B\cap C' \\
& B.A\cap C \\
& C.B\cap C' \\
& D.A\cap C' \\
\end{align}\]
Answer
569.7k+ views
Hint: In solving these type of questions, we will use various properties of set to reach the final simplified answer. Properties which we will use are:
(i) $\left( A\cup B \right)'=A'\cap B'$ which is called De Morgan's first law.
(ii) $\left( A\cap B \right)'=A'\cup B'$ which is called De Morgan's second law.
(iii) $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)$ which is called association law.
(iv) $\left( A\cup B \right)\cap \left( A\cup C \right)=A\cup \left( B\cap C \right)$ which is called distributive law.
(v) $\left( A\cup B \right)=\left( B\cup A \right)$ which is called commutative law.
(vi) $A\cup \varnothing =A$ which is called law of identity element.
(vii) $A\cap A'=\varnothing $ which is called complement property.
Complete step-by-step answer:
We are given set $\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'$
Using distributive law, we will solve first two terms together and then third term.
\[\left[ \left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)' \right]\cap C'\]
We will use distributive law in both terms to combine $B\cup C\text{ and }B'\cap C'$ we get:
\[\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A\cap \left( B'\cap C' \right) \right)' \right]\cap C'\]
Let us use De Morgan law in second term, so we get:
\[\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A'\cap \left( B\cap C \right) \right) \right]\cap C'\]
Using commutative rule, we get:
\[\left[ \left( \left( B\cup C \right)\cup A \right)\cap \left( \left( B\cup C \right)\cup A' \right) \right]\cap C'\]
Using distributive law, we get:
\[\left[ \left( B\cup C \right)\cup \left( A\cap A' \right) \right]\cap C'\]
As we know, $A\cap A'=\varnothing $ using complement property, therefore, equation becomes
\[\left[ \left( B\cup C \right)\cup \varnothing \right]\cap C'\]
As we know, $A\cup \varnothing =A$ using law of identity element, we get:
\[\left[ B\cup C \right]\cap C'\]
Using distributive law we get:
\[\left( B\cap C' \right)\cup \left( C\cap C' \right)\]
As we know $C\cap C'=\varnothing $ using complement property, therefore, equation becomes
\[\left( B\cap C' \right)\cup \varnothing \]
Since, $A\cup \varnothing =A$ using law of identity element we get:
\[\left( B\cap C' \right)\cup \varnothing =B\cap C'\]
Hence, $\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'$ is equal to $B\cap C'$
So, the correct answer is “Option C”.
Note: Students should remember all the properties of sets. Notation ‘$^{C}$’ can also be used for representing the complement of a set instead of ‘$'$’ only.
Students should not get confused with $\cup ,\cap $ as both are completely different signs.
(i) $\left( A\cup B \right)'=A'\cap B'$ which is called De Morgan's first law.
(ii) $\left( A\cap B \right)'=A'\cup B'$ which is called De Morgan's second law.
(iii) $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)$ which is called association law.
(iv) $\left( A\cup B \right)\cap \left( A\cup C \right)=A\cup \left( B\cap C \right)$ which is called distributive law.
(v) $\left( A\cup B \right)=\left( B\cup A \right)$ which is called commutative law.
(vi) $A\cup \varnothing =A$ which is called law of identity element.
(vii) $A\cap A'=\varnothing $ which is called complement property.
Complete step-by-step answer:
We are given set $\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'$
Using distributive law, we will solve first two terms together and then third term.
\[\left[ \left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)' \right]\cap C'\]
We will use distributive law in both terms to combine $B\cup C\text{ and }B'\cap C'$ we get:
\[\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A\cap \left( B'\cap C' \right) \right)' \right]\cap C'\]
Let us use De Morgan law in second term, so we get:
\[\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A'\cap \left( B\cap C \right) \right) \right]\cap C'\]
Using commutative rule, we get:
\[\left[ \left( \left( B\cup C \right)\cup A \right)\cap \left( \left( B\cup C \right)\cup A' \right) \right]\cap C'\]
Using distributive law, we get:
\[\left[ \left( B\cup C \right)\cup \left( A\cap A' \right) \right]\cap C'\]
As we know, $A\cap A'=\varnothing $ using complement property, therefore, equation becomes
\[\left[ \left( B\cup C \right)\cup \varnothing \right]\cap C'\]
As we know, $A\cup \varnothing =A$ using law of identity element, we get:
\[\left[ B\cup C \right]\cap C'\]
Using distributive law we get:
\[\left( B\cap C' \right)\cup \left( C\cap C' \right)\]
As we know $C\cap C'=\varnothing $ using complement property, therefore, equation becomes
\[\left( B\cap C' \right)\cup \varnothing \]
Since, $A\cup \varnothing =A$ using law of identity element we get:
\[\left( B\cap C' \right)\cup \varnothing =B\cap C'\]
Hence, $\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'$ is equal to $B\cap C'$
So, the correct answer is “Option C”.
Note: Students should remember all the properties of sets. Notation ‘$^{C}$’ can also be used for representing the complement of a set instead of ‘$'$’ only.
Students should not get confused with $\cup ,\cap $ as both are completely different signs.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

