Question

Let \( f : [0, \infty) \to [0, \infty) \) be a differentiable function with \( f(x)>0 \) for all \( x>0 \), and \( f(0) = 0 \). Further, \( f \) satisfies} \[ (f(x))^2 = \int_{0}^{x} \left( (f(t))^2 + f(t) \right) \, dt, \, x>0. \] {Then which one of the following options is correct?

• A. \( 0<f(2) \leq 1 \)
• B. \( 1<f(2) \leq 2 \)
• C. \( 2<f(2) \leq 3 \)
• D. \( 3<f(2) \leq 4 \)

Hint:

To solve differential equations of the form \( f'(x) = \frac{f(x) + 1}{2} \), try separating the variables and using integration. Also, use initial conditions to find the constant of integration.

Answer 1

The Correct Option is B

Step 1: We start with the given functional equation \[ (f(x))^2 = \int_{0}^{x} \left( (f(t))^2 + f(t) \right) \, dt. \] Differentiate both sides with respect to \( x \): \[ 2f(x) \cdot f'(x) = (f(x))^2 + f(x). \] Simplifying, we get \[ 2f(x) \cdot f'(x) = f(x) \left( f(x) + 1 \right). \] Dividing both sides by \( f(x) \) (since \( f(x)>0 \) for \( x>0 \)): \[ 2f'(x) = f(x) + 1. \] Thus, \[ f'(x) = \frac{f(x) + 1}{2}. \] Step 2: To solve this differential equation, we separate the variables: \[ \frac{f'(x)}{f(x) + 1} = \frac{1}{2}. \] Integrating both sides: \[ \int \frac{1}{f(x) + 1} \, df(x) = \frac{1}{2} \int dx. \] This gives \[ \ln |f(x) + 1| = \frac{x}{2} + C. \] Exponentiating both sides: \[ f(x) + 1 = e^{\frac{x}{2} + C} = A e^{\frac{x}{2}}, \] where \( A = e^C \). Thus, \[ f(x) = A e^{\frac{x}{2}} - 1. \] Step 3: Use the initial condition \( f(0) = 0 \) to find \( A \). Substitute \( x = 0 \) into the solution: \[ f(0) = A e^{0} - 1 = A - 1 = 0, \] so \( A = 1 \). Therefore, \[ f(x) = e^{\frac{x}{2}} - 1. \] Step 4: Now, evaluate \( f(2) \): \[ f(2) = e^{\frac{2}{2}} - 1 = e - 1. \] Since \( e \approx 2.718 \), we have \[ f(2) \approx 2.718 - 1 = 1.718. \] Thus, \( 1<f(2) \leq 2 \), which corresponds to option (B).