Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y,$ where $f\left(0\right)\ne 0.$ If $f^{\prime }\left(0\right)=2,$ then $f\left(x\right)$ is equal to :

• A. $A{e}^x$
• B. $e^{2x}$
• C. $2x$
• D. None of these

Answer 1

The correct option is B $e^{2x}$

If $f\left(x\right)=e^{2x}$
$f(x+y)=e^{2(x+y)}=e^{2x}\cdot e^{2y}=f\left(x\right)\cdot f\left(y\right)$
$f\left(0\right)=1\ne 0$
$f^{\prime }\left(x\right)=2e^{2x}$
$f^{\prime }\left(0\right)=2$
Since $f\left(x\right)=e^{2x}$ satisfy all given conditions
$\therefore f\left(x\right)=e^{2x}$