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.
We cant use B' as input, as it is mentioned in the question that all inputs are available in true form only.
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
shekhar chauhan u meant to say NOR not XOR as (A+B)' is NOR not XOR and in either case for making NOR or XOR we would need 2 2*1 MUX (one mux to complement either A or B and second mux for NOR) because imputs are available in true form only not in their complement forms.
So the answer would be 3 MUX needed.
@ mcjoshi Veteran aren't u writing the reverse
For 1st1st MUX I0=B,I1=B=1,... Similarly inputs for second mux should also be reverse according to me
Answer - B
Source-https://gateoverflow.in/48660/no-of-multipexers-isro-2015
64.3k questions
77.9k answers
244k comments
80.0k users