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

Given the functional equation \( f(x+y)=f(x)f(y) \) for all \( x,y \in \mathbb{R} \), it's known that such an equation often indicates an exponential function. Assume \( f(x)=e^{c x} \). Then:

\( f(0)=e^{c \cdot 0}=1 \)

Taking the derivative of both sides of \( f(x+y)=f(x)f(y) \) with respect to \( y \) and evaluating at \( y=0 \), we get:

\( f'(x)=f(x)f'(0) \)

Substituting \( f'(0)=4a \), we have:

\( f'(x)=4a f(x) \)

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

Now, apply the given second differential equation:

\( f''(x)-3a f'(x)-f(x)=0 \)

Calculate \( f''(x)=16a^2 e^{4a x} \) and \( f'(x)=4a e^{4a x} \). Substitute into the equation:

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

Simplify:

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

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

For nontrivial solutions, \( 4a^2=1 \); thus \( a=\frac{1}{2} \). Hence, \( f(x)=e^{2x} \).

Find the area of region \( R = \{(x,y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2 \}\):

Substitute \( a=\frac{1}{2} \):

\( f(ax)=e^{2(ax)}=e^{x} \)

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

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

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

Answer 2

From $f(x+y)=f(x)f(y)$ and differentiability we get the exponential form. Let $k=f'(0)$. Then $f(x)=e^{kx}$ for all $x$ (standard result for differentiable multiplicative Cauchy equation).

1. Here $f'(0)=4a$, so $k=4a$. Thus $$\boxed{f(x)=e^{4ax}}.$$
2. Use the differential equation $f''(x)-3a f'(x)-f(x)=0$. For $f(x)=e^{4ax}$ compute $$f'(x)=4a\,e^{4ax},\qquad f''(x)=(4a)^2 e^{4ax}=16a^2 e^{4ax}.$$ Substitute: \[ 16a^2 e^{4ax}-3a(4a e^{4ax})-e^{4ax}=e^{4ax}\big(16a^2-12a^2-1\big) =e^{4ax}(4a^2-1). \] This must be $0$ for all $x$, so $4a^2-1=0$.
3. Solve $4a^2-1=0$. Since $a>0$, $$a=\tfrac{1}{2}.$$
4. Now $f(x)=e^{4ax}=e^{4(\tfrac12)x}=e^{2x}$. Therefore $$f(ax)=f\big(\tfrac{1}{2}x\big)=e^{2(\tfrac{1}{2}x)}=e^{x}.$$
5. The area of $R$ is $$\text{Area}=\int_{x=0}^{2}\big(\text{top }y - \text{bottom }y\big)\,dx =\int_{0}^{2} f(ax)\,dx=\int_0^2 e^{x}\,dx.$$ Evaluate: $$\int_0^2 e^{x}\,dx = e^{x}\Big|_0^2 = e^{2}-1.$$

Answer

Area $=e^{2}-1$. (Option 1)