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5 Answers

4 votes
4 votes
Answer is b,

AB+AB'+A'C+AC

A(B+B')+C(A'+A)  ......from property of complementary

A+C ,it is independent of B
3 votes
3 votes

We have AB+AB'+A'C+AC

We can rewrite above equation as

AB(C+C')+AB'(C+C')+A'(B+B')C+A(B+B')C ....(As (C+C')=0, (B+B')=0)

We get

ABC+ABC'+AB'C+AB'C'+A'BC+A'B'C+ABC+AB'C

Eliminating Common Terms and rewrite equation as.

ABC+ABC'+AB'C+AB'C'+A'BC+A'B'C

expression in terms of minterms is

m7+m6+m5+m4+m3+m1

K-Map will be

    C'B' C'B CB CB'
    00 01 11 10
A' 0 0 1 1 0
A 1
1
1
1
1

Thus we can write Reduced equation as

A+C

2 votes
2 votes
F =  AB + AB′ + A′C + AC

F = A(B + B') + C(A' + A)

F = A.1 + C.1             ( A + A' = 1 )

F = A + C

Here not present B variable

So Option (B) is correct answer
Answer:

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