Question

A function $f:R\to R^{+}$ satisfies $f(x+y)=f\left(x\right)f\left(y\right)\forall x,yϵR.$ If $f^{\prime }\left(0\right)=2$ then$\forall x,yϵR$, $f^{\prime }\left(x\right)=$

• A. $f\left(x\right)$
• B. $2f\left(x\right)$
• C. $\frac{1}{2}\left(f\right(x\left)\right)$
• D. $\frac{1}{f\left(x\right)}$

Answer 1

The correct option is B $2f\left(x\right)$

$f(x+y)=f\left(x\right)\cdot f\left(y\right)$
For $x=y=0$
$f(0+0)=f\left(0\right)\cdot f\left(0\right)\Rightarrow f\left(0\right)=0\text{or}1$
But codomain of $f$ is $R^{+}$
$\therefore f\left(0\right)=1$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)f\left(h\right)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)\left(f\right(h)-1)}{h}=f\left(x\right)\cdot f^{\prime }\left(0\right)=2f\left(x\right)$