Test by Bikram | Mock GATE | Test 4 | Question: 36

Question

Which of the following statements is/are CORRECT?

$S_1$: Max term is a sum term which contains all the variables in either direct or complementary form.

$S_2$: Min term is a product term which contains all the variables in either direct or complementary form such that for the combination of variables, the functional output must be $1$.

$S_3$: Min term is a product term which contains all the variables in either direct or complementary form.

• A. only $S_1$ is correct
• B. only $S_3$ is correct
• C. only $S_2$ is correct
• D. All $S_1$, $S_2$ and $S_3$ are correct

Comments

Max term is a sum term which contains all the variables in either direct or complementary form. True.

It doesn't say "A sum term that contains all the variables in direct or complimented form is a maxterm". That would've been False. (Because we need functional output 0 also)

 

By extending this, I think D is the correct answer.

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I don’t think so that S2 is correct, because for each row there will be a minterm, but those minterms which results in 1 for function are called as True minterms, rest of them are false minterms and we express the function as sum of True minterms, so i think its not neccessary that every minterm should result in 1 for the function,

Answer 1

A min term l is a $\text{product}$ (AND) of all variables in the function, in direct or complemented form. A min term has the property that it is equal to $1$ on exactly one row of the truth table.

A max term is a $\text{sum}$ (OR) of all the variables in the function, in direct or complemented form. A max term has the property that it is equal to $0$ on exactly one row of the truth table.

Consider $F = \sum_{w,x,y} (0,2)$

Here, $\bar w \bar x \bar y, \bar w x \bar y$ are max terms and $\bar w \bar x y, w x y$ are min terms.

$\bar w \bar x y$ not being a max terms makes option $S_1$ false and $\bar w \bar x \bar y$ not being a min term makes $S_3$ false.

Correct option $C$

Comments

$S3$ logically follows from $S2$; if you say $S2$ is true, then it means $S3$ has to be true too ($S2 \rightarrow S3$).

Actually, all statements are correct. D should be the correct answer.

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@rawan Thats not true. See the updated answer.