GATE IT 2006 | Question: 23

Question

Let $P$, $Q$ and $R$ be sets let Δ denote the symmetric difference operator defined as $PΔQ=(P \cup Q) - (P ∩ Q).$ Using Venn diagrams, determine which of the following is/are TRUE?

1. $PΔ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)$
2. $P ∩ (Q ∩ R) =  (P ∩ Q) Δ (P Δ R)$

• A. I only
• B. II only
• C. Neither I nor II
• D. Both I and II

Answer 1

Answer C.

Comments

here 

1. PΔ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)

will be same as :

P XOR ( Q.R ) as LHS 

(P XOR Q) . ( P XOR R ) as RHS ... Definitely not equal ( Since the statements should be true for all type of sets ; Considering element only { 0 ,1 } -> it turns out to boolean functions . Hence applied operations of XOR and AND. )

---

i think the question you guys printed here is wrong : ii) P intersection (Q XOR R) = (P intersection Q) XOR (P XOR R)

i think the LHS of the second statement is printed wrong

---

i just take values randomly i take p={1,2},q={3,4},r={5,6} now i just calculate symmetric difference 1st one getting lhs=rhs and second one not satisfying but in answer both are not satisfying so where i am going wrong what's wrong with my example? @Arjun sir

Answer 2

hope it might help.....

Comments

thanks

---

Helped :)

---

all are cleared bro except how you taken that venn diagram i cant understand

Answer 3

………………......………………………………………….

Answer 4

Best Method

 

The symmetric difference using Venn diagram of two subsets A and B is a subset of U, denoted by A △ B and is defined by

A △ B = (A – B) ∪ (B – A) = AB’ + BA’

Let A and B are two sets. The symmetric difference of two sets A and B is the set (A – B) ∪ (B – A) and is denoted by A △ B.

-----------------------------------------------------------------------------------

The correct option is B 

PΔQ=PQ′+P′Q
where Δ is symmetric difference between P and Q
I. LHS =PΔQ∩R=PΔ(QR)=P(QR)′+P′(QR)
=P(Q′+R′)+P′QR
RHS =(PΔQ)∩(PΔR)=(PQ′+P′Q)(PR′+P′R)=PQ′R′+P′QR
LHS ≠ RHS
So statement I is false.
II. LHS =P∩QΔR=P(QR′+Q′R)
=PQR′+PQ′R
RHS =(P∩Q)Δ(P∩R)=PQΔPR=PQ(PR)′+(PQ)′PR=PQ(P′+R′)+(P′+Q′)PR
=PQR′+PQ′R
LHS = RHS
So statement II is true.

 

😊😊😊😊😊😊😊😊😊😊😊😊

Answer 5