Let `f (x+y)=f(x). f(y)` for all x and y `f(1)=2` If in a triangle `ABC, a =f (3),b=f(1)+f (3), c=f (2)+f (3), then 2A is equal to

Options

A. C
B. 2C
C. 3C
D. 4C

Answer

To solve the given problem step by step, we will follow the logic presented in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Function**: We are given a function \( f \) such that \( f(x+y) = f(x) \cdot f(y) \) for all \( x \) and \( y \), and \( f(1) = 2 \). This suggests that \( f(x) \) might be an exponential function.

2. **Finding \( f(2) \)**:
- Let \( x = 1 \) and \( y = 1 \).
- Then, \( f(2) = f(1 + 1) = f(1) \cdot f(1) = 2 \cdot 2 = 4 \).

3. **Finding \( f(3) \)**:
- Let \( x = 2 \) and \( y = 1 \).
- Then, \( f(3) = f(2 + 1) = f(2) \cdot f(1) = 4 \cdot 2 = 8 \).

4. **Finding \( b \) and \( c \)**:
- We have \( a = f(3) = 8 \).
- \( b = f(1) + f(3) = 2 + 8 = 10 \).
- \( c = f(2) + f(3) = 4 + 8 = 12 \).

5. **Using the Cosine Rule**:
- We will use the cosine rule to find \( \cos A \):
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
- Substituting the values:
\[ \cos A = \frac{10^2 + 12^2 - 8^2}{2 \cdot 10 \cdot 12} = \frac{100 + 144 - 64}{240} = \frac{180}{240} = \frac{3}{4} \]

6. **Finding \( \cos C \)**:
- Now, we will find \( \cos C \):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
- Substituting the values:
\[ \cos C = \frac{8^2 + 10^2 - 12^2}{2 \cdot 8 \cdot 10} = \frac{64 + 100 - 144}{160} = \frac{20}{160} = \frac{1}{8} \]

7. **Using the Cosine Double Angle Formula**:
- We know that \( \cos 2A = 2 \cos^2 A - 1 \).
- Substituting \( \cos A = \frac{3}{4} \):
\[ \cos 2A = 2 \left(\frac{3}{4}\right)^2 - 1 = 2 \cdot \frac{9}{16} - 1 = \frac{18}{16} - 1 = \frac{2}{16} = \frac{1}{8} \]

8. **Equating \( \cos 2A \) and \( \cos C \)**:
- Since \( \cos 2A = \cos C \), we have:
\[ \cos 2A = \frac{1}{8} = \cos C \]
- Therefore, \( 2A = C \).

### Conclusion:
Thus, we find that \( 2A = C \).