Let `F_1(A,B,C)=(A^^~B)vv[~C^^(AvvB)]vv~A and F_2(A,B)=(AvvB)vv(BrarrA)` be two logical expressions. Then :

Options

A. `F_1 and F_2` both are tautologies
B. `F_1` is tautology but `F_2` is not a tautology
C. `F_1` is not tautology but `F_2` is a tautology
D. Both `F_1 and F_2` are not talutogies

Answer

To determine whether the logical expressions \( F_1(A, B, C) \) and \( F_2(A, B) \) are tautologies, we will evaluate both expressions using truth tables.

### Step 1: Define the expressions
The expressions are given as:
- \( F_1(A, B, C) = (A \land \neg B) \lor [\neg C \land (A \lor B)] \lor \neg A \)
- \( F_2(A, B) = (A \lor B) \lor (B \Rightarrow A) \)

### Step 2: Construct the truth table for \( F_1 \)

We will create a truth table for \( A, B, C \) and evaluate \( F_1 \).

| A | B | C | \( \neg B \) | \( A \land \neg B \) | \( \neg C \) | \( A \lor B \) | \( \neg C \land (A \lor B) \) | \( F_1 \) |
|-------|-------|-------|---------------|-----------------------|---------------|-----------------|-------------------------------|-----------|
| T | T | T | F | F | F | T | F | T |
| T | T | F | F | F | T | T | T | T |
| T | F | T | T | T | F | T | F | T |
| T | F | F | T | T | T | T | T | T |
| F | T | T | F | F | F | T | F | T |
| F | T | F | F | F | T | T | T | T |
| F | F | T | T | F | F | F | F | T |
| F | F | F | T | F | T | F | F | T |

### Step 3: Analyze \( F_1 \)
From the truth table, we can see that \( F_1 \) evaluates to true in all cases except for the last two rows where \( A \) and \( B \) are both false. Therefore, \( F_1 \) is **not a tautology**.

### Step 4: Construct the truth table for \( F_2 \)

Next, we will create a truth table for \( F_2 \).

| A | B | \( B \Rightarrow A \) | \( A \lor B \) | \( F_2 \) |
|-------|-------|-----------------------|-----------------|-----------|
| T | T | T | T | T |
| T | F | T | T | T |
| F | T | F | T | T |
| F | F | T | F | T |

### Step 5: Analyze \( F_2 \)
From the truth table, we can see that \( F_2 \) evaluates to true in all cases. Therefore, \( F_2 \) is a **tautology**.

### Conclusion
- \( F_1 \) is not a tautology.
- \( F_2 \) is a tautology.

Thus, the answer is that \( F_1 \) is not a tautology, but \( F_2 \) is a tautology.