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Let f be a twice differentiable function on (1, 6). If f(2) = 8, f'(2) = 5, f'(x) ≥ 1 and f''(x) ≥ 4, for all x ∈ (1, 6), then:
- f(5) + f'(5) ≥ 28
- f(5) + f'(5) ≤ 26
- f'(5) + f''(5) ≤ 20
- f(5) ≤ 10
Correct answer: f(5) + f'(5) ≥ 28
Solution
The correct option is right because, given that f'(x) is always at least 1 and f''(x) is at least 4, f' is increasing. Starting from f'(2) = 5, we can deduce that f'(5) must be at least 5 + 4*3 = 17. Additionally, using the minimum growth rate of f, we find that f(5) must be at least 8 + 1*3 = 11. Therefore, f(5) + f'(5) is at least 11 + 17 = 28.
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