UPSC 2022 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Logic gates and truth tables · Numerical Analysis · asked 2× in 13 yrs · Read the full method →

Question

Find a combinatorial circuit for $f(x,y,z)=[x\cdot(\bar y+z)]+y$ and write its input/output table.

Technique

Truth table from operator hierarchy; Boolean algebra simplification: $x\bar y+y=x+y$ (key identity).

Solution

Step 1 — Analyse the Boolean expression

$f=[x\cdot(\bar y+z)]+y$ .

Hierarchy:

NOT gate on $y$ : produces $\bar y$ .
OR gate combining $\bar y$ and $z$ : produces $\bar y+z$ .
AND gate combining $x$ and $(\bar y+z)$ : produces $x(\bar y+z)$ .
OR gate combining the above with $y$ : produces final $f$ .

Step 2 — Circuit diagram (text representation)

   y ── [NOT] ──┐
                │   ┌── ̄y
                │   │
   z ──────────┼───[OR]───┐  (gives ̄y+z)
                │          │
                            │
   x ──────────────────────[AND]───┐  (gives x·( ̄y+z))
                                    │
                                    │
   y ─────────────────────────────[OR]──── f

Components used: 1 NOT, 2 OR, 1 AND.

Step 3 — Truth table (input/output)

$x$	$y$	$z$	$\bar y$	$\bar y+z$	$x(\bar y+z)$	$f=[\cdot]+y$
0	0	0	1	1	0	0
0	0	1	1	1	0	0
0	1	0	0	0	0	1
0	1	1	0	1	0	1
1	0	0	1	1	1	1
1	0	1	1	1	1	1
1	1	0	0	0	0	1
1	1	1	0	1	1	1

Step 4 — Simplification (bonus)

Looking at the truth table: $f=0$ only when $(x,y,z)\in\{(0,0,0),(0,0,1)\}$ , i.e., when $x=0$ and $y=0$ .

So $f=x+y$ ? Test: $x+y$ at $(0,0,0)=0$ ✓, at $(0,0,1)=0$ ✓, at $(0,1,0)=1$ ✓, at $(1,0,0)=1$ ✓, etc. Yes — $f$ simplifies to $x+y$ .

Simplification: $f=[x(\bar y+z)]+y$ . Use distributive: $f=x\bar y+xz+y$ . Combine $x\bar y+y=x+y$ (absorption-related; verify: $x\bar y+y=(x\bar y+y)\cdot 1=(x+y)(\bar y+y)=(x+y)\cdot 1=x+y$ ). So $f=x+y+xz=x(1+z)+y=x+y$ (since $1+z=1$ ).

So $f=x+y$ simplifies dramatically.

Conclusion

Answer

\boxed{\;f(x,y,z)=[x(\bar y+z)]+y=x+y\text{ (simplified)}.\;}