GO Classes Test Series 2024 | Mock GATE | Test 14 | Question: 28

Question

Consider function $\mathbf{G}(\mathbf{A}, \mathbf{B}, \mathbf{C})=\mathbf{A B}+\mathbf{B C}$. Let $\mathbf{F}(\mathbf{A}, \mathbf{B}, \mathbf{C})$ be the dual of $\mathbf{G}(\mathbf{A}, \mathbf{B}, \mathbf{C})$. Then $\mathbf{F}(\mathbf{A}, \mathbf{B}, \mathbf{C})$ can be implemented using $\mathbf{A}$ as the select (control) input to a 2-to-1 multiplexer. The correct implementation of $\mathbf{F}(\mathbf{A}, \mathbf{B}, \mathbf{C})$ is shown in:

Answer 1

dual of AB+BC= (A+B) (B+C)  [we know (A+B) (B+C) = B+AC]

option b) A'(B.B) + A(B+C)

                 A'B + AB + AC

                 B(A + A') + AC

                 B + AC

Comments

if (A+B)(B+C) means to make output 1

either A should be one or if A is zero then c&b should be 1

so A can answer too

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No, because according to you A=1 is enough to get the expression result 1.

But, even if A = 1 , C must be 1 to make (A+B)(B+C) =1

Consider A = 1, B = 0, C = 0 gives (A+B)(B+C) = (1+0)(0+0) = (1)(0) = 0

So if A = 0, B must be 1 and C can be any of {0,1}

And if A = 1, C must be 1 and B can be any of {0,1}