Question

If $f(x+y)=f\left(x\right)+f\left(y\right)+c$, for all real $x$ and $y$ and $f\left(x\right)$ is continuous at $x=0$ and $f^{\prime }\left(0\right)=1$ then $f^{\prime }\left(x\right)$ equals to

• A. $c$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

We have, $f(x+y)=f\left(x\right)+f\left(y\right)+c$
Put $x=0,y=0$
$\Rightarrow f\left(0\right)=2f\left(0\right)+c\Rightarrow f\left(0\right)=-c\Rightarrow \left(1\right)$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+c-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)+c}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$ using (1)
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=f^{\prime }\left(0\right)=1$ (given)
Hence $f^{\prime }\left(x\right)=1$