Let f:R R be a twice differentiable function such that f(x+y)=f(x)f(y) for all x,y R.

Question

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

Accepted Answer

$\begin{gathered} f(x+y)=f(x)f(y) \ \Rightarrow f(x)={e}^{\lambda x}{f}^{'}(0)=4a \ \Rightarrow {f}^{'}(x)=\lambda {e}^{\lambda x}\Rightarrow \lambda =4a \end{gathered}$

So, $f(x)={e}^{4ax}$

$\begin{gathered} {f}^{''}(x)-3{af}^{'}(x)-f(x)=0 \ \Rightarrow {\lambda }^{2}-3a\lambda -1=0 \ \Rightarrow 16{a}^{2}-12{a}^{2}-1=0\Rightarrow 4{a}^{2}=1\Rightarrow a=\frac{1}{2} \end{gathered}$

$\begin{gathered} F(x)={e}^{2x} \ ={\int }_{2}^{0}{e}^{x}dx={e}^{2}-1 \end{gathered}$

Solutions · 1