1. **Problem statement:** Given the Boolean function $F = AC + AB$, we want to deduce its Karnaugh map (K-map). 2. **Recall the variables and function:** The function depends on variables $A$, $B$, and $C$. The function is $F = AC + AB$. 3. **K-map setup:** For three variables $A$, $B$, and $C$, the K-map is a 2x4 grid with rows representing $A$ (0 or 1) and columns representing $BC$ combinations (00, 01, 11, 10). 4. **Fill the K-map:** Evaluate $F$ for each combination of $A$, $B$, and $C$: - $A=0$: $F=0$ because $AC=0$ and $AB=0$. - $A=1$, $B=0$, $C=0$: $F=1\cdot0 + 1\cdot0=0$. - $A=1$, $B=0$, $C=1$: $F=1\cdot1 + 1\cdot0=1$. - $A=1$, $B=1$, $C=0$: $F=1\cdot0 + 1\cdot1=1$. - $A=1$, $B=1$, $C=1$: $F=1\cdot1 + 1\cdot1=1$. 5. **K-map table:** | A \ BC | 00 | 01 | 11 | 10 | |--------|----|----|----|----| | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 1 | 6. **Interpretation:** The K-map shows that $F=1$ for all cases where $A=1$ and either $B=1$ or $C=1$. 7. **Final answer:** The K-map for $F=AC + AB$ is as above, confirming the function's behavior.