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

To solve this problem, we need to analyze the given conditions on the function \( f(x) \). We are given:

• \( f(x) \) is a non-negative continuous function for all \( x \ge 1 \).
• \( f'(x) \leq p f(x) \), where \( p > 0 \).
• \( f(1) = 0 \).

Given that \( f(x) \) is a continuous function and \( f(1) = 0 \), let's check what happens as \( x \) increases from 1:

The inequality \( f'(x) \leq p f(x) \) suggests that the derivative of \( f(x) \) is proportional to the value of \( f(x) \), but possibly less than or equal to it. Let's rewrite this in a form that might resemble differential equations:

Consider that if \( f(x) \) satisfies \( f'(x) = p f(x) \), then the solution would be \( f(x) = C e^{px} \) for some constant \( C \). However, given that \( f(1) = 0 \), this would imply \( C = 0 \) because \( f(1) = C e^{p \cdot 1} = C e^p = 0 \), which gives \( C = 0 \).

This means \(f(x) = 0\) for all \(x \ge 1\), since a continuous and non-negative function that is equal to zero at one point \( x = 1 \) and is less than or equal to zero elsewhere (by the inequality) must be zero everywhere in its domain:

• For \(x \ge 1\), the derivative \(f'(x) \leq 0\), since \(p f(x) = 0\).
• If for some \(x > 1\), \(f(x) > 0\), \(f'(x)\) must increase, but our condition contradicts that, as it does not exceed \(0\).

Therefore, the function \( f(x) \) must remain zero for all \( x \ge 1 \).

Now, let's evaluate the expression \( [f(\sqrt{e}) + f(\sqrt{\pi})] \):

• \( f(\sqrt{e}) = 0 \)
• \( f(\sqrt{\pi}) = 0 \)

Thus, \([f(\sqrt{e}) + f(\sqrt{\pi})] = [0 + 0] = 0\).

Hence, the correct answer is: 0