The number of terms in the product-of-sum canonical form of \([ ( x 1 + x 2 )\; ( x 3 x 4 ) ]\) is

Simplify using De Morgan's law: \( (x 1 + x 2)\, x 3x 4 = (x 1 + x 2) + x 3x 4 = x 1 x 2 + x 3x 4\) Count the minterms (\(x 1\) is the MSB): \( x 1 x 2\) gives minterms 0, 1, 2, 3 \(x 3x 4\) gives minterms 3, 7, 11, 15 Together: 0, 1, 2, 3, 7, 11, 15 , which is 7 minterms. The canonical POS form has one maxterm for each row where the function is 0: \(16 - 7 = 9\) terms. Answer: (C) 9

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Question 92

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The number of terms in the product-of-sum canonical form of <span class="math-tex">\([\overline {({x_1} + {x_2})\;\overline {\left( {{x_3}{x_4}} \right)} } ]\)</span> is

(A) 7

(B) 8

(C) 9

(D) 10

✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Simplify using De Morgan's law:

\(\overline{(x_1 + x_2)\,\overline{x_3x_4}} = \overline{(x_1 + x_2)} + x_3x_4 = \bar{x}_1\bar{x}_2 + x_3x_4\)

Count the minterms (\(x_1\) is the MSB):

\(\bar{x}_1\bar{x}_2\) gives minterms 0, 1, 2, 3
\(x_3x_4\) gives minterms 3, 7, 11, 15

Together: {0, 1, 2, 3, 7, 11, 15}, which is 7 minterms.

The canonical POS form has one maxterm for each row where the function is 0: \(16 - 7 = 9\) terms.

Answer: (C) 9

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Question 92

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