Statement 1 : Let `f : r-{3}` be a function given by , `f(x+10)=(f(x)-5)/(f(x)-3)," then " f(10)=f(50)`.
Statement 2 : f (x) is a periodic function.

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to analyze the function defined by the equation:

\[ f(x + 10) = \frac{f(x) - 5}{f(x) - 3} \]

### Step 1: Substitute \( x \) with \( x + 10 \)

We start by substituting \( x \) with \( x + 10 \) in the original equation:

\[ f((x + 10) + 10) = f(x + 20) = \frac{f(x + 10) - 5}{f(x + 10) - 3} \]

### Step 2: Replace \( f(x + 10) \)

Now we replace \( f(x + 10) \) using the original equation:

\[ f(x + 20) = \frac{\frac{f(x) - 5}{f(x) - 3} - 5}{\frac{f(x) - 5}{f(x) - 3} - 3} \]

### Step 3: Simplify the expression

We simplify the right-hand side:

1. For the numerator:
\[ \frac{f(x) - 5}{f(x) - 3} - 5 = \frac{f(x) - 5 - 5(f(x) - 3)}{f(x) - 3} = \frac{f(x) - 5 - 5f(x) + 15}{f(x) - 3} = \frac{-4f(x) + 10}{f(x) - 3} \]

2. For the denominator:
\[ \frac{f(x) - 5}{f(x) - 3} - 3 = \frac{f(x) - 5 - 3(f(x) - 3)}{f(x) - 3} = \frac{f(x) - 5 - 3f(x) + 9}{f(x) - 3} = \frac{-2f(x) + 4}{f(x) - 3} \]

So now we have:

\[ f(x + 20) = \frac{-4f(x) + 10}{-2f(x) + 4} \]

### Step 4: Further simplify \( f(x + 20) \)

This simplifies to:

\[ f(x + 20) = \frac{2(4f(x) - 10)}{2(2f(x) - 4)} = \frac{4f(x) - 10}{2f(x) - 4} \]

### Step 5: Substitute \( x \) with \( x + 20 \)

Next, we substitute \( x \) with \( x + 20 \):

\[ f(x + 40) = \frac{f(x + 20) - 5}{f(x + 20) - 3} \]

### Step 6: Replace \( f(x + 20) \)

Using the expression we derived for \( f(x + 20) \):

\[ f(x + 40) = \frac{\frac{4f(x) - 10}{2f(x) - 4} - 5}{\frac{4f(x) - 10}{2f(x) - 4} - 3} \]

### Step 7: Simplify \( f(x + 40) \)

This will involve similar steps as before, and after simplification, we will find that:

\[ f(x + 40) = f(x) \]

### Conclusion

Since \( f(x + 40) = f(x) \), this shows that \( f(x) \) is periodic with a period of 40. Therefore, we can conclude:

1. \( f(10) = f(50) \) (since \( 50 = 10 + 40 \)).
2. \( f(x) \) is a periodic function.

### Final Statements

- **Statement 1** is true: \( f(10) = f(50) \).
- **Statement 2** is also true: \( f(x) \) is periodic.