Question

If \(f(x)\) is a non-negative continuous function for all \(x \ge 1\) such that \(f'(x) \le p f(x)\), where \(p > 0\) and \(f(1) = 0\), then \([f(\sqrt{e}) + f(\sqrt{\pi})]\) is equal to

• A. 0
• B. negative
• C. positive
• D. None of these

Hint:

When \(f'(x) \le k f(x)\) and \(f(a) = 0\) with non-negativity, the function must be identically zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The inequality \(f'(x) \le p f(x)\) suggests considering \(g(x) = e^{-px} f(x)\).

Step 2: Detailed Explanation:
Let \(g(x) = e^{-px} f(x)\). Then \(g'(x) = e^{-px} f'(x) - p e^{-px} f(x) = e^{-px} (f'(x) - p f(x)) \le 0\). So \(g(x)\) is non-increasing on \([1, \infty)\). Since \(f(1) = 0\), \(g(1) = e^{-p} \cdot 0 = 0\). For \(x > 1\), \(g(x) \le g(1) = 0\). But \(g(x) = e^{-px} f(x) \ge 0\) because \(f(x) \ge 0\). Thus \(g(x) = 0\) for all \(x \ge 1\), so \(f(x) = 0\) for all \(x \ge 1\). Hence \(f(\sqrt{e}) = 0\) and \(f(\sqrt{\pi}) = 0\).

Step 3: Final Answer:
The sum is 0, which corresponds to option (A).