← 2022 Paper 2
UPSC 2022 Maths Optional Paper 2 Q5c-ii — Step-by-Step Solution
5 marks · Section B
Boolean algebra · Numerical Analysis · asked 15× in 14 yrs · Read the full method →
Question
Express the Boolean function F(x,y,z)=xy+x′z in product of maxterms form.
Technique
Truth table; maxterm = sum of literals where 0→unprimed, 1→primed; product of maxterms at all F=0 rows.
Solution
Strategy. Truth table; identify rows where F=0; each such row corresponds to a maxterm.
Step 1 — Truth table
| x | y | z | xy | x′z | F=xy+x′z |
|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
Step 2 — Identify rows where F=0
Rows (x,y,z)∈{(0,0,0),(0,1,0),(1,0,0),(1,0,1)}.
Step 3 — Write each maxterm
Maxterm at (x,y,z): sum (OR) of literals where 0→ unprimed, 1→ primed.
- (0,0,0): x+y+z.
- (0,1,0): x+y′+z.
- (1,0,0): x′+y+z.
- (1,0,1): x′+y+z′.
Step 4 — Product of maxterms
F(x,y,z)=(x+y+z)(x+y′+z)(x′+y+z)(x′+y+z′).
Answer
F=(x+y+z)(x+y′+z)(x′+y+z)(x′+y+z′).