Q8MCQ1 mark, −0.25Given f1f_1f1, f3f_3f3 and fff in canonical sum of products form (in decimal) for the circuit f1=Σm(4,5,6,7,8)f_1 = \Sigma m(4, 5, 6, 7, 8)f1=Σm(4,5,6,7,8) f3=Σm(1,6,15)f_3 = \Sigma m(1, 6, 15)f3=Σm(1,6,15) f=Σm(1,6,8,15)f = \Sigma m(1, 6, 8, 15)f=Σm(1,6,8,15) then f2f_2f2 is(A)Σm(4,6)\Sigma m(4, 6)Σm(4,6)(B)Σm(4,8)\Sigma m(4, 8)Σm(4,8)(C)Σm(6,8)\Sigma m(6, 8)Σm(6,8)(D)Σm(4,6,8)\Sigma m(4, 6, 8)Σm(4,6,8)Sign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google