← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 13 yrs · Read the full method →

Question

Simplify the boolean expression $(a+b)\cdot(\bar b+c)+b\cdot(\bar a+\bar c)$ by using the laws of boolean algebra. From its truth table write it in minterm normal form.

Technique

Distribute, apply $x\bar x=0$ and $c+\bar c=1$, then absorption ($a+ac=a$, $\bar a b+b=b$) and $x\bar y+y=x+y$ to collapse to $a+b$; minterm form read off the truth table.

Solution

Setup. Use the standard identities: distributive, complement ($x\bar x=0,\ x+\bar x=1$), absorption ($x+xy=x$), and $x+\bar xy=x+y$.

Step 1 — Expand the first product $(a+b)(\bar b+c)$

Step 2 — Expand the second product $b(\bar a+\bar c)$

Step 3 — Combine and simplify

Group $bc+b\bar c=b(c+\bar c)=b$. So

Now $\bar a b+b=b$ (absorption, $b(\bar a+1)=b$):

Next $a\bar b+b=a+b$ (since $x\bar y+y=x+y$):

Finally $a+ac=a$ (absorption):

Step 4 — Truth table

Step 5 — Minterm (canonical SOP) normal form

$E=1$ for minterms $m_2,m_3,m_4,m_5,m_6,m_7$ (all except $a=b=0$):

Answer