Answer:
Option B,C
Explanation:
f(x+y)=f(x)+f(y) as f(x) is differentiable at x=0
⇒ f′(0)=k ....(i)
Now, f′(x)=lim
=\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.