← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q5c-ii — Solution

5 marks · Section B

Question

Determine the truth table for the Boolean function $F(x, y, z) = (x + y + z')(x' + y')$. Also derive the full disjunctive normal form of $F(x, y, z)$ from the truth table.

Technique

Evaluate the product-of-sums expression over all $2^3 = 8$ input combinations to build the truth table, then collect the minterms (rows where $F = 1$) to write the full disjunctive normal form (canonical sum of products).

Solution

Here $+$ is OR, juxtaposition is AND, and prime is complement. Evaluate $F = (x + y + z')(x' + y')$. Note that $(x'+y')=0$ only when $x=y=1$, and $(x+y+z')=0$ only when $x=0, y=0, z=1$.

$F = 1$ in rows $0, 2, 3, 4, 5$, i.e. minterms $m_0, m_2, m_3, m_4, m_5$.

Full disjunctive normal form (sum of minterms). Each minterm uses every variable, primed where the variable is $0$:

• $m_0$: $(x,y,z)=(0,0,0) \to x'y'z'$
• $m_2$: $(0,1,0) \to x'yz'$
• $m_3$: $(0,1,1) \to x'yz$
• $m_4$: $(1,0,0) \to xy'z'$
• $m_5$: $(1,0,1) \to xy'z$

$F(x,y,z) = x'y'z' + x'yz' + x'yz + xy'z' + xy'z = \sum m(0,2,3,4,5).$

Answer

$\boxed{F(x,y,z) = x'y'z' + x'yz' + x'yz + xy'z' + xy'z}$ (minterms $0, 2, 3, 4, 5$).