Question

Let $f:R \to R$ be a function such that $f(x + y) = f(x) + f(y),\forall x,y \in R.$
If $f(x)$ is differentiable at $x = 0$, then
A) $f(x)$ is differentiable only in a finite interval containing zero
B) $f(x)$ is continuous \[\forall x \in R\]
C) \[{f'}(x)\] is constant \[\forall x \in R\]
D) $f(x)$ is differentiable except at finitely many points

Answer 1

Hint: A function is differentiable at a point means when there is a defined derivative at that point, according to the question the function is differentiable at $x = 0$.
So, we have to use the formula of ${f'}(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h) - f(0)}}{h}$ and by this, we can check every options from the above.

Complete step-by-step answer:
According to the question we have $f(x + y) = f(x) + f(y)$
We have given that $f(x)$ is differentiable at $x = 0$. So let us assume that $x = 0$ and $y = 0$ then, we will have
$f(0 + 0) = f(0) + f(0)$
This equation will be true only if $f(0) = 0$.
Now, we know that:
\[{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\] …(i)
Using the concept $f(x + y) = f(x) + f(y)$ we get:
\[{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x) + f(h) - f(x)}}{h}\]
\[ \Rightarrow {f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h)}}{h}\] …(ii)
The function is differentiable at $x = 0$, so if we replace $x$ with $0$ in eqn (i), then
${f'}(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h)}}{h}$ …(iii)
From equation (ii) and (iii), we can conclude that,
${f'}(x) = {f'}(0)$
Now, integrating both sides we respect to x, we get:
$\int {{f'}(x)} = \int {{f'}(0)} dx$
Using the product integration rule i.e. \[\int {u.vdx = u\int {vdx - \int {{u'}(\int {vdx)dx} } } } \]
$f(x) = x.{f'}(0) + C$ …(iv)
Where C is a constant
At $x = 0$, we have:
$
  f(0) = 0.{f'}(0) + C \\
   \Rightarrow C = 0 \\
 $
Putting this value of C in (iv), we get:
$f(x) = x.{f'}(0)$ \[\forall x \in R\]
Clearly, f(x) is everywhere continuous and \[{f'}(x)\] is constant \[\forall x \in R\].

So, option (B) and option (C) are correct.

Note: Alternate Method: We can also check the continuity of $f(x)$ at $x = 0$, and the other is to check whether $f(x)$ is differentiable there. First, we will check that at $x = 0$, then $f(x)$ is continuous. You can also solve this by using this method
${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x) + f(h) - f(x)}}{h}$
${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h)}}{h}$
So now using L hospital rule formula
${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h)}}{h} = \dfrac{0}{0}$
${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{f'}(h)}}{1}$
So it means that, ${f'}(x) = {f'}(0)$
It will be equal to a constant assume C
THEN, $f(x) = Kx + C$
${f'}(x) = K$
Integrating both sides we get
$f(x) = Kx + C$
So our function is continuous because it is in the linear form we can write this, $y = mx + c$.