Discussion Forum

Let $ f : R \rightarrow R $ be a function such that f(x + y)= f(x) + f(y),$ \forall x,y \epsilon R $. If f(x) is  differentiable at x = 0, then

Answer:

Explanation:

 $f(x+y)=f(x)+f(y)$ as f(x) is differentiable  at x=0

 $\Rightarrow$    $f'(0)=k$ ....(i)

 Now, $f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h \rightarrow 0}\frac{f(x)+f(h)-f(x)}{h}$

 =$\lim_{h \rightarrow 0}\frac{f(h)}{h}$    $[ \frac{0}{0}$ from]

 Given ,$f(x+y)=f(x)+f(y), \forall x,y$

$\therefore$ $f(0)=f(0)+f(0)$

when $x=y=0 \Rightarrow f(0)=0$

 Using L' Hosptial 's rule 

 $\lim_{h\rightarrow 0}\frac{f'(h)}{1}=f'(0)=k$ .....(ii)

 $\Rightarrow$  $f'(x)=k, $on integrating  both sides

$f(x)=kx+C, $ as f(0)=0 $\Rightarrow$ C=0

 So, f(x)=kx

$\therefore$  f(x) is continuous for all $x \epsilon R$ and 

 $f'(x)=k$, i.e constant  for all $x \epsilon R$

 hence , both (b) and (c) are correct.