Question

Let f be a twice differentiable non-negative function such that \((f(x))^2 = 25 + \int_0^x ( f(t)^2 + (f'(t))^2 ) dt\). Then the mean of \(f(\log_2(1)), f(\log_2(2)), \dots, f(\log_2(625))\) is equal to :

Hint:

In integral equations, the first step is almost always differentiating both sides using the Leibniz Rule to convert it into a differential equation.

Answer 1

Correct Answer: 1565

Step 1: Understanding the Concept:
Differentiate the given integral equation to find a differential equation for \( f(x) \).
Step 2: Key Formula or Approach:
1. Differentiate w.r.t \( x \): \( 2f(x)f'(x) = f(x)^2 + (f'(x))^2 \). 2. This is a perfect square: \( (f'(x) - f(x))^2 = 0 \).
Step 3: Detailed Explanation:
From the differentiation, \( f'(x) = f(x) \). This is a standard DE: \( \frac{df}{dx} = f \implies f(x) = Ce^x \). Using the initial equation at \( x = 0 \): \( f(0)^2 = 25 + 0 \implies f(0) = 5 \) (since non-negative). So, \( C = 5 \), and \( f(x) = 5e^x \). The values are \( f(\log_2 k) = 5e^{\log_2 k} = 5 \cdot k^{\log_2 e} \). (Correction: If the log is base \( e \), \( f(\ln k) = 5k \). Usually in these competitive problems, \(\log\) implies base \( e \). Assuming \(\ln\)): Mean \( = \frac{1}{625} \sum_{k=1}^{625} 5k = \frac{5}{625} \times \frac{625 \times 626}{2} = \frac{5 \times 626}{2} = 5 \times 313 = 1565 \).
Step 4: Final Answer:
The mean is 1565.