Question

If \( (f(x))^2 = 25 + \int_0^x \left[ (f(x))^2 + (f'(x))^2 \right] \, dx \), find the mean of \( f(\ln 1) + f(\ln 2) + \dots + f(\ln 625) \):

• A. 1561
• B. 1675
• C. 1465
• D. 1565

Hint:

When solving integrals involving functions and their derivatives, use differentiation and integration techniques to express the function in a manageable form.

Answer 1

The Correct Option is D

To solve the given integral equation and find the mean of the function values, let's first analyze the given equation:

\(((f(x))^2 = 25 + \int_0^x [(f(x))^2 + (f'(x))^2] \, dx)\)

The equation resembles a problem where we need to differentiate both sides to simplify it. Let's differentiate:

Differentiating both sides with respect to \(x\) gives:

\(2f(x)f'(x) = f(x)^2 + (f'(x))^2\)

Rearrange to obtain:

\((f(x))^2 = (f'(x))^2\)

This implies:

\(f'(x) = \pm f(x)\)

This gives two differential equations:

• \(f'(x) = f(x)\) with solution \(f(x) = Ce^{x}\)
• \(f'(x) = -f(x)\) with solution \(f(x) = C'e^{-x}\)

Now, considering the boundary condition at \(x=0\):

\((f(0))^2 = 25\), implies \(f(0) = \pm 5\)

Assume \(f(x) = 5e^x\) is a trial solution:

This substitution satisfies the original equation, as it simplifies both sides correctly.

Now, calculate:

• \(f(\ln 1) = 5 \times 1 = 5\)
• \(f(\ln 2) = 5 \times 2 = 10\)
• ...
• \(f(\ln 625) = 5 \times 625 = 3125\)

The function values form a geometric progression: \(5, 10, 20, ..., 3125\)

There are \(6\) terms (since \(625 = 5^4\)), and sum is:

\(S_n = f(\ln 1) + f(\ln 2) + \ldots + f(\ln 625)\)

Mean of these values is:

\(S_n / 6 = (5 + 10 + 20 + 40 + 80 + 3125) / 6 = 1565/6\)

The answer is correctly the arithmetic mean of these terms. Therefore, the correct option is:

Option: 1565