"Given a function $f(x)$ defined on $\mathbf{R}$, $f(x) > f(x-1) + f(x-2)$, and when $x < 3$, $f(x) = x$. Which of the following conclusions must be true?

A. $f(10) > 100$
B. $f(20) > 1000$
C. $f(10) < 1000$
D. $f(20) < 10000$"

Given the conditions $ f(x) > f(x-1) + f(x-2) $ and $ f(x) = x $ for $ x < 3 $, let's analyze the function step-by-step for $ x \geq 3 $.

### Initial Values
We start with:
- $ f(1) = 1 $
- $ f(2) = 2 $

### Calculations for $ x \geq 3 $
Now, let's calculate the next few values based on the given recursive inequality $ f(x) > f(x-1) + f(x-2) $:

- $ f(3) > f(2) + f(1) = 2 + 1 = 3 $
- $ f(4) > f(3) + f(2) $
- $ f(5) > f(4) + f(3) $

We can see that each new value of $ f(x) $ grows significantly because it is the sum of the previous two values plus an additional amount.

### Bounding the Growth
We observe that the sequence generated by this inequality grows faster than the Fibonacci sequence. Recall that the Fibonacci sequence is defined as:
\[ F(n) = F(n-1) + F(n-2) \]
where $ F(1) = 1 $ and $ F(2) = 1 $. The Fibonacci sequence grows exponentially, approximately as:
\[ F(n) \sim \frac{1}{\sqrt{5}} \left( \frac{1 + \sqrt{5}}{2} \right)^n \]

Given $ f(x) $ grows faster than the Fibonacci sequence, we know that $ f(x) $ will also grow exponentially but at a rate faster than $ \left( \frac{1 + \sqrt{5}}{2} \right)^x $.

### Estimating $ f(10) $ and $ f(20) $
For $ x = 10 $:
- Since $ f(x) $ grows faster than the Fibonacci sequence, $ f(10) $ will be much larger than the 10th Fibonacci number.
- The 10th Fibonacci number is 55, but since $ f(x) $ grows faster, we expect $ f(10) $ to be much larger, likely in the hundreds.

For $ x = 20 $:
- Similarly, the 20th Fibonacci number is 6765, but $ f(x) $ grows faster.
- Hence, $ f(20) $ will be significantly larger, likely in the thousands.

### Verifying the Options
Given the above reasoning:

- **A. $ f(10) > 100 $**: This must be true because $ f(x) $ grows faster than the Fibonacci sequence, and by $ x = 10 $, $ f(10) $ will exceed 100.
- **B. $\( f(20) > 1000 $**: This must be true because $ f(x) $ grows significantly faster than the Fibonacci sequence, ensuring $ f(20) $ is well above 1000.
- **C. $ f(10) < 1000 $**: This is true because even though $ f(10) $ is significantly large, it won't reach 1000 by $ x = 10 $.
- **D. $ f(20) < 10000 $**: This must be true because although $ f(20) $ is large, it is still likely less than 10000.

### Conclusion
All options A, B, C, and D must be true based on the given conditions and the nature of the function $ f(x) $.

Please let me know if you need more details or have any questions. Here are 5 numbered questions you can ask next:

1. Can you explain how to derive the Fibonacci sequence?
2. What are some examples of functions that grow faster than exponential functions?
3. How would the solution change if $ f(x) \leq f(x-1) + f(x-2) $?
4. Can you provide a proof for the exponential growth of the Fibonacci sequence?
5. What are some applications of recursive sequences in real-life problems?

**Tip:** When dealing with inequalities involving functions, it's helpful to compare them with known sequences or functions to understand their growth behavior.

Explore the analysis of function $f(x)$ defined recursively and by initial conditions. Understand its growth behavior using inequalities and exponential growth comparison. Suitable for advanced high school students.

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