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8. 已知函数 f(x)f(x) 的定义域为 R,f(x)>f(x1)+f(x2)\mathrm{R}, f(x)>f(x-1)+f(x-2), 且当 x<3x<3 时, f(x)=xf(x)=x,则下列...
Aug 14, 2024
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Solution by Steps
step 2
For x=0,1,2x = 0, 1, 2, we have f(0)=0f(0) = 0, f(1)=1f(1) = 1, and f(2)=2f(2) = 2. We can calculate f(3)f(3) using the inequality: f(3) > f(2) + f(1) = 2 + 1 = 3. Thus, f(3) > 3
step 3
Continuing this process, we find f(4) > f(3) + f(2) > 3 + 2 = 5, so f(4) > 5. Similarly, we can derive that f(5) > f(4) + f(3) > 5 + 3 = 8, and so on
step 4
This pattern suggests that f(n)f(n) grows faster than the Fibonacci sequence, which implies that f(10)f(10) and f(20)f(20) will be significantly larger than 100 and 1000, respectively
step 5
Therefore, we can conclude that f(10) > 100 and f(20) > 1000, making options A and B correct. However, we need to determine which statements are definitely true based on the growth pattern
A
Key Concept
Inequalities and function growth rates
Explanation
The function f(x)f(x) grows faster than linear due to the recursive inequality, leading to values exceeding 100 and 1000 for larger inputs.
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