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

Consider the functional equation \( f(x+y)=f(x)f(y) \) for all \( x,y \in \mathbb{R} \). This typically implies an exponential form, so we assume \( f(x)=e^{c x} \). This yields \( f(0)=e^{c \cdot 0}=1 \).

Differentiating \( f(x+y)=f(x)f(y) \) with respect to \( y \) and setting \( y=0 \) results in \( f'(x)=f(x)f'(0) \).

Given \( f'(0)=4a \), we have \( f'(x)=4a f(x) \).

Solving the differential equation \( f'(x)=c f(x) \) yields \( f(x)=e^{4a x} \).

Now, substitute this into the second differential equation \( f''(x)-3a f'(x)-f(x)=0 \).

The second derivative is \( f''(x)=16a^2 e^{4a x} \) and the first derivative is \( f'(x)=4a e^{4a x} \). Substituting these into the equation gives:

\( 16a^2 e^{4a x} - 3a (4a e^{4a x}) - e^{4a x}=0 \)

Simplifying this expression:

\( 16a^2 e^{4a x} - 12a^2 e^{4a x} - e^{4a x}=0 \)

\( (4a^2-1)e^{4a x}=0 \)

For non-trivial solutions, \( 4a^2=1 \), which implies \( a=\frac{1}{2} \). Consequently, \( f(x)=e^{2x} \).

We need to find the area of the region \( R = \{(x,y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2 \}\).

With \( a=\frac{1}{2} \), we have \( f(ax)=e^{2(ax)}=e^{x} \).

The area under the curve \( y=e^{x} \) from \( x=0 \) to \( x=2 \) is calculated as:

\(\int_0^2 e^x \, dx = [e^x]_0^2 = e^2 - e^0 = e^2 - 1 \)

Therefore, the area of region \( R \) is \( e^2 - 1 \).