Home / JEE Main Mathematics / Sets

Complement of a Set

Post a Comment

The complement of a set Ais the set of all those elements which are in universal set but not in A. It is denoted by A' or $A^c$. 

If U is a universal set and A $\subset$ U,

then A' = U - A= {x : x $\in$ U but x $\notin A$}
i.e., x $\in$A ⇒ x $\notin$ A'.

The Venn diagram of complement of a set Ais as shown in the figure and shaded portion represents A'.

e.g., If U = {1, 2, 3, 4, 5, …}
and A = {2, 4, 6, 8, …}
So, A' = U - A = {1, 3, 5, 7, …}

By Venn diagram, the operation between three sets can be represented given below. 



From the Venn diagram above:
  1. $A \cap \bar{B} \cap \bar{C}$
  2. $B \cap \bar{C} \cap \bar{A}$
  3. $C \cap \bar{A} \cap \bar{B}$
  4. $A \cap C \cap \bar{B}$
  5. $A \cap B \cap \bar{C}$
  6. $B \cap C \cap \bar{A}$
  7. $A \cap B \cap C$
  8. $\bar{A} \cap \bar{B} \cap \bar{C}$
Note:
  • $\phi$ = U'
  • $\phi'$ = U
  • (A')' = A
  • A $\cup$ A' = U
  • A $\cap$ A' = $\phi$
Sample Problem 1

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} , A = {2, 4, 6, 8} and B = {2, 3, 5, 7} , then (A $\cup$ B)', (A' $\cap$ B'), (A $\Delta$ B), is equal to
(a) {1, 9}, {2, 8}, {3, 4, 5, 6, 7, 8}
(b) {1, 9}, {1, 9} {3, 4, 5, 6, 7, 8}
(c) {1, 9}, {1, 9} {5, 6, 7, 8}
(d) None of the above

Answer: (b)
Given sets are U = {1, 2, 3, 4, 5, 6, 7, 8, 9} , A = {2, 4, 6, 8} and B = {2, 3, 5, 7}

Now, A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9}

Now, B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7} = {1, 4, 6, 8, 9}

(i) A $\cup$ B = {2, 4, 6, 8} - {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}

So, (A $\cup$ B)' = U - (A $\cup$ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5, 6, 7, 8} = {1, 9} 

(ii) (A' $\cap$ B') = {1, 3, 5, 7, 9} - {1, 4, 6, 8, 9} = {1, 9}

(iii) Now, A - B = {2, 4, 6, 8} - {2, 3, 5, 7} = {4, 6, 8}

 and B - A = {2, 3, 5, 7} - {2, 4, 6, 8} = {3, 5, 7} 

A $\Delta$ B = (A - B) $\cup$ (B - A) = {4, 6, 8} - (3, 5, 7) = {3, 4, 5, 6, 7, 8}

Sample Problem 2

If U = {x : $x^5 - 6x^4+ 11x^3 - 6x^2 = 0$}, A = {x: $x^2 -5x+6 = 0$} and B = {x : $x^2 - 3x+2 = 0$} what is (A $\cap$ B)' equal to ?
(a) {1, 3}
(b) {1, 2, 3}
(c) {0, 1, 3}
(d) {0, 1, 2, 3} 

Answer: (c)
U = {x : $x^5 - 6x^4+ 11x^3 - 6x^2 = 0$} = {0, 1, 2, 3}

A = {x: $x^2 -5x+6 = 0$} = {2, 3} and

B = {x : $x^2 - 3x+2 = 0$} = {1, 2}

so, A $\cap$ B = {2}

Hence, (A $\cap$ B)' = U - (A $\cap$ B)

(A $\cap$ B)' = {0, 1, 2, 3} - {2} = {0, 1, 3}

Sample Problem 3
The shaded region in the given figure is
(a) B $\cap$ (A $\cup$ C)
(b) B $\cap$ (A - C)
(c) B $\cup$ (A $\cap$ C)
(d) B - (A $\cup$ C)

Answer: (d) 
It is clear from the figure that set A $\cup$ C is not shading and set B is shading other than A $\cup$ C. i.e., B - (A $\cup$ C).

Post a Comment for "Complement of a Set"