Let $f:\mathbf R o\mathbf R$ be a twice differentiable function such that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbf R$. If $f'(0)=4a$ and $f$ satisfies $f''(x)-3af'(x)-f(x)=0$, $a0$, then the area of the region $R=\{(x,y)\mid0\le y\le f(ax),\ 0\le x\le2\}$ is:

Question

Let $f:\mathbf R	o\mathbf R$ be a twice differentiable function such that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbf R$. If $f'(0)=4a$ and $f$ satisfies $f''(x)-3af'(x)-f(x)=0$, $a>0$, then the area of the region $R=\{(x,y)\mid0\le y\le f(ax),\ 0\le x\le2\}$ is:

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $f(x)=e^{kx}$ with $k=f'(0)=4a$. Then $k^2-3ak-1=0\Rightarrow4a^2=1$, so $a=\frac12$ and $k=2$. Area $=\int_0^2e^{2\cdot x/2}dx=\int_0^2e^x\,dx=e^2-1$. Answer (1).

Q. Let $f:\mathbf R\to\mathbf R$ be a twice differentiable function such that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbf R$. If $f'(0)=4a$ and $f$ satisfies $f''(x)-3af'(x)-f(x)=0$, $a>0$, then the area of the region $R=\{(x,y)\mid0\le y\le f(ax),\ 0\le x\le2\}$ is:

Explanation

Reviewed by Dr. Abubacker

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