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Minimum number of $2 \times 1$ multiplexers required to realize the following function, $f = \overline{A} \;\overline{B} C + \overline{A}\; \overline{B} \;\overline{C}$

Assume that inputs are available only in true form and Boolean a constant $1$ and $0$ are available.

  1. $1$
  2. $2$
  3. $3$
  4. $7$
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Answer: 2

f= A'B'C + A'B'C' ===> A'B'( C + C' ) ===> A'B' ====> (A+B)' .

The final function represents NOR.So with the help of 2*1 mux we can implement it

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f = A'B'C + A'B'C' => A'B'( C + C' ) => A'B'

 

 

 

 

 

 

Also we cant use A' or B', since it is mentioned in the question that inputs are available in true form only.

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Ans is 2. {i.e. 2(2x1) MUX or 1(2x1) + 1 NOT}

Explanation : f= A'B'C + A'B'C' = A'B'(C+C') = A'B' = (A+B)'   which is NOR so if we take A as a select line then the value of A either 0 or 1,   if is A is 0 then output of nor gate is  B'  and  if A is1 then output of nor gate is 0

Now input for the multiplexer is B' and 0 and select line is A then output is A'B' = (A+B)'
Answer:

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