Except for option C, all have 4 in f2.

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## 5 Answers

$\begin{matrix} \text{Options} & f_1 \cdot f_2 & f_3 + (f_1 \cdot f_2) & \\ A & \{4,6\} & \{1,4, 6,15\} & {\color{Red}{fails}} \\ B & \{4,8\} & \{1,4,6, 8,15\} & {\color{Red} {fails}}\\ C & \{6,8\} & \{1,6,8,15\} & ok\\ D & \{4,6, 8\} & \{1,4,6, 8, 15\} & {\color{Red} {fails} } \end{matrix}$

answer = **option C**

### 2 Comments

first of all the and term is one only when when both are one. and or is one when any one is one. so the terms in answer wll be there which comes either in f1.f2 or in f3 . as there is an "OR" between them . so the terms 1,6,15 are contributed by f3 while 8 should be contributed by f1.f2. so we want that f1.f2 should atleast result in 8. and we know for that both should contain 8. f1 is having 8 so f2 atleast should contain 8. now what if f2 contain 4 also. then 4 should apper in f. which is not. so 6,8 is the valid option

$f1.f2 + f3 = f$

$(f1 ⋂ f2) U f3 = f$

we use intersection for AND because if a f1 and f2 both are true for a minterm then only output will be true.

we use Union for OR because if one among f1 and f2 is true for a minterm then output will be true.

${4,5,6,7,8}⋂ f2 U {1,6,15} = {1,6,8,15}$

${4,5,6,7,8}⋂ f2 = {8}$

$f2 = {8,any other elements preset in f }$

check for options c contains {6,8} 6 is in f

option (C)

f = f1.f2 + f3

which means what is present in f3 is always present in f because we are taking union with f3.

as f3 is (1,6,15) so f must have these , and instead of these we have one extra i.e. 8( which means 8 is comes from intersection of f1.f2)

and 6 is already be there in f3, so may be it is not present in f2, But 4 is also not present in f2 because , if it is present then f1.f2 = { 4 } which also comes in f & it is not present in f.

That’s why options which have 4 is not correct answer.

and we find answer as {6 , 8}