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Gate2008_8

Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit

                                                         

\(\sum\) m (4,5,6,7,8)
\(\sum\) m (1,6,15)
\(\sum\) m (1,6,8,15)
then f2 is

(A) \(\sum\)m(4,6)            (B) \(\sum\)m(4,8)          (C) \(\sum\)m(6,8)         (D) \(\sum\)m(4,6,8)

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