The image presents a set of questions from a Grade 11 Information and Communication Technology exam. I will solve question (iv) - Construct the Boolean expression for the logic circuit and draw the truth table as well.

1. Problem Description

The image presents a set of questions from a Grade 11 Information and Communication Technology exam. I will solve question (iv) - Construct the Boolean expression for the logic circuit and draw the truth table as well.

2. Solution Steps

The logic circuit consists of NAND gates and an OR gate. Let's trace the circuit to determine the Boolean expression.

Let $x$ and $y$ be the inputs.

The first NAND gate has inputs $x$ and $x$. The output of the first NAND gate is $\overline{x \cdot x} = \overline{x}$.

The second NAND gate has inputs $y$ and $y$. The output of the second NAND gate is $\overline{y \cdot y} = \overline{y}$.

The third NAND gate has inputs $x$ and $y$. The output of the third NAND gate is $\overline{x \cdot y}$.

The output of the first NAND gate and the third NAND gate are sent as inputs to the lower AND gate which then undergoes a NOT process. Therefore, $\overline{\overline{x} \cdot \overline{x \cdot y}}$. Applying DeMorgan's law to inside the NOT we have $\overline{\overline{x} \cdot \overline{x \cdot y}} = \overline{\overline{x}} + \overline{\overline{x \cdot y}} = x + (x \cdot y)$.

The output of the second NAND gate and the third NAND gate are sent as inputs to the upper AND gate which then undergoes a NOT process. Therefore, $\overline{\overline{y} \cdot \overline{x \cdot y}}$. Applying DeMorgan's law to inside the NOT we have $\overline{\overline{y} \cdot \overline{x \cdot y}} = \overline{\overline{y}} + \overline{\overline{x \cdot y}} = y + (x \cdot y)$.

The last OR gate has inputs $x + (x \cdot y)$ and $y + (x \cdot y)$. Therefore, the output $O$ is $(x + (x \cdot y)) + (y + (x \cdot y))$. This simplifies to $x + y + x \cdot y$.

Boolean expression: $O = x + y + (x \cdot y)$

Truth table:

| x | y | x * y | x + y + (x*y) |

|---|---|-------|---------------|

| 0 | 0 | 0 | 0 |

| 0 | 1 | 0 | 1 |

| 1 | 0 | 0 | 1 |

| 1 | 1 | 1 | 1 |

3. Final Answer

Boolean expression: $O = x + y + (x \cdot y)$

Truth table:

| x | y | x * y | x + y + (x*y) |

|---|---|-------|---------------|

| 0 | 0 | 0 | 0 |

| 0 | 1 | 0 | 1 |

| 1 | 0 | 0 | 1 |

| 1 | 1 | 1 | 1 |