Question

A function $f:R\to R$ satisfies the equation $f(x+y)=f\left(x\right),f\left(y\right)$ for all $x,yϵR,f\left(x\right)\ne 0.$ Suppose that the function is differentiable at $x=0$ and $f^{\prime }\left(0\right)=2,$ then $f^{\prime }\left(x\right)=$

• A. $f\left(x\right)$
• B. $2f\left(x\right)$
• C. $-f\left(x\right)$
• D. $-2f\left(x\right)$

Answer 1

Answer: (B)

$2f\left(x\right)$

We have $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,yϵR$
$\therefore f\left(0\right)=f\left(0\right)\Rightarrow f\left(0\right)\left\{f\right(0)-1\}=1$
$\Rightarrow f\left(0\right)=1[\because f(0)\ne 1]$
Now $f^{\prime }\left(0\right)=0\Rightarrow \underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=2$
$\Rightarrow \underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2[\because f(0)=1]$ ...(i)
Now $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)f\left(h\right)-f\left(x\right)}{h}[\because f(x+y)=f(x\left)f\right(y\left)\right]$
$=f\left(x\right)\left(\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}\right)=2f\left(x\right)$ (using (i) only)