Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2\  is :

• A. \( e^2 - 1 \)
• B. \( e^4 + 1 \)
• C. \( e^4 - 1 \)
• D. \( e^2 + 1 \)

Hint:

To find the area under a curve, integrate the function over the given limits.

Answer 1

The Correct Option is A

We are given that \( f(x + y) = f(x) f(y) \), which implies that \( f(x) = e^{\lambda x} \), for some constant \( \lambda \). We also know that \( f'(0) = 4a \), so: \[ f'(x) = \lambda e^{\lambda x}, \quad f'(0) = \lambda = 4a. \] Thus, \( f(x) = e^{4ax} \). Next, we are given the differential equation \( f''(x) - 3a f'(x) - f(x) = 0 \), and we need to solve for \( a \). Substituting \( f(x) = e^{4ax} \) into the equation gives: \[ f''(x) = 16a^2 e^{4ax}, \quad f'(x) = 4a e^{4ax}. \] Substitute into the differential equation: \[ 16a^2 e^{4ax} - 3a (4a e^{4ax}) - e^{4ax} = 0, \] \[ 16a^2 - 12a^2 - 1 = 0 \quad \Rightarrow \quad 4a^2 = 1 \quad \Rightarrow \quad a = \frac{1}{2}. \] Thus, the function becomes: \[ f(x) = e^{2x}. \] The area of the region is given by: \[ \text{Area} = \int_0^2 f(ax) \, dx = \int_0^2 e^{x} \, dx = e^2 - 1. \] Thus, the area of the region is \( \boxed{e^2 - 1} \).