Question

Let $f(x+y)=f\left(x\right)+f\left(y\right)$ and $f\left(x\right)=x^{2}g\left(x\right)$ for all $x,yϵR$, where $g\left(x\right)$ is continuous function. Then $f^{\prime }\left(x\right)$ is equal to

• A. g'(x)
• B. g(0)
• C. g(0) + g'(x)

Answer 1

We have $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)-f\left(x\right)}{h}$
[$\because f(x+y)=f\left(x\right)+f\left(y\right)$]

$=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=\underset{h\to 0}{lim}\frac{h^{2}g\left(h\right)}{h}=0.g\left(0\right)=0$
[because $g$ is continuous therefore $\underset{h\to 0}{lim}g\left(h\right)=g\left(0\right)].$