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 $A \cup B$

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why my approach is wrong ?

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

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

@gatecse

@Gyanu

@Shashank

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Detailed Video Solution

Answer 1

$(A - B) \cup (B - A) \cup (A \cap B)$
$A - B$     is A but not B.  i.e. only A
$B - A$    is B but not A.  i.e. only B
$A \cap B$     is A and B both

Union of all is (only A) U (only B) U (both A and B)
$= A \cup B$

Correct Answer: $A$

Answer 2

We can solve it using boolean algebra also

Given expression is :

(A−B)∪(B−A)∪(A∩B)

Which can be written in boolean algebra as

AB' + BA' + AB           { A-B can be expressed as A ∩B' }

Which gives A+B which is nothing but A U B.

Answer 3

Think like OSA:

This is like the equation of HALF ADDER!
(Symmetric difference= XOR)  U (AB)

=A XOR B + AB

AND HALF ADDER GIVES SUM.
SO A + B . which is A U B.
:)

Comments

sir anyways the physical significance of SUM + CARRY is the total SUM of something. and in case of SETS, i dont think a carry would have any physical significance.

am i really too wrong?

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No, you are right

btw I never guessed you can switch in btw boolean algebra and set theory as per requirement :D :/

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lack of knowledge(in me) = creativity of knowledge! :D

Answer 4

hope it might help.....

Comments

Looks like u hav mastered Venn diagram ... (y)