Question

If $f(x+y)=f\left(x\right)+f\left(y\right)\forall x,y$ and $f^{\prime }\left(0\right)$ exists, then

• A. $f\left(x\right)=f^{\prime }\left(0\right)x\forall x$
• B. $f\left(x\right)=cx,$ where $c$ is a constant
• C. $f\left(x\right)=xf\left(1\right)$
• D. $f\left(x\right)=\frac{x}{2}f\left(2\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f\left(x\right)=f^{\prime }\left(0\right)x\forall x$
B $f\left(x\right)=cx,$ where $c$ is a constant
C $f\left(x\right)=xf\left(1\right)$
D $f\left(x\right)=\frac{x}{2}f\left(2\right)$

Given, $f(x+y)=f\left(x\right)+f\left(y\right)$

Put $y=0$ $⟹f(x+0)=f\left(x\right)+f\left(0\right)$ $⟹f\left(0\right)=0$

Let us assume $f\left(1\right)=c$

$⟹f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)=2f\left(1\right)=2c$

and $f\left(3\right)=f(2+1)=f\left(2\right)+f\left(1\right)=3f\left(1\right)=3c$

and $f\left(x\right)=cx$, where $c$ is a constant, $c\in R$

It can also be written as $f\left(x\right)=xf\left(1\right)=\frac{x}{2}f\left(2\right)$

Now, since $f^{\prime }\left(0\right)$ exists, $f^{\prime }\left(x\right)=c⟹f^{\prime }\left(0\right)=c$

Therefore, $f\left(x\right)=f^{\prime }\left(0\right)x$ for all $x$.

Therefore, all the above options are correct.