GATE CSE 1996 | Question: 1.1

Question

Let $A$ and $B$ be sets and let $A^c$ and $B^c$ denote the complements of the sets $A$ and $B$. The set $(A-B) \cup (B-A) \cup (A \cap B)$ is equal to

• A. $A \cup B$

• B. $A^c \cup B^c$

• C. $A \cap B$

• D. $A^c \cap B^c$

Comments

$AB'+BA'+AB$

$A(B+B')+A'B$

$A+A'B= (A+A') (A+B)= A+B$

Hence $A \cup B$

---

why my approach is wrong ?

Union taken as +,Intersection taken as *(and)

A-B+B-A+AB=====AB

@gatecse

@Gyanu

@Shashank

---

………………………………………………………………………….

---

$\color{red}{\text{Find Video Solution Below:}}$

Detailed Video Solution

Answer 1

Taking examples of A ={2,3,5,67} and B={1,3,4} and Universal set = {1,2,3,4,5,6,7,8,9,10} we can find out

A-B={2,5,6,7} and B-A={{1,4}

Complements of A={1,4,8,9,10} and that of B ={2,5,6,7,8,9,10}

We find (A-B) U (B-A)U (A meet B) = A U B.

Hence the option A is correct.

Answer 2

A – B) ∪ (B - A) ∪ (A∩B)

(A - B) = 1
(B - A) = 2
(A∩B) = 3
A∪B = (1∪2∪3)
(A – B) ∪ (B - A) ∪ (A∩B) = 1∪2∪3 = (A∪B)

Answer 3

😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊

Answer 4

Solution using set operators.. It is lengthy, but once understood, can be done in less time..