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The boolean expression $AB+AB'+A'C+AC$ is independent of the boolean variable

1. A
2. B
3. C
4. None of these
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A(B+B')+C(A+A')=A+C  WHICH  is independent on B hence b is the ans
by Boss (49.3k points)
selected by

AB+AB'+A'C+AC

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

A+C ,it is independent of B
by (309 points)

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

by Boss (18.4k points)
+1 vote
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
by Veteran (59.4k points)