Question

Let $f\left(x\right)$ is a differentiable function such that $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+2xy\forall x,y\in R$ and $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=210$ , then $f\left(2\right)$ is equal to

• A. $20$
• B. $105$
• C. $424$
• D. none of these

Answer 1

Answer: (C)

$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{f\left(x\right)+f\left(h\right)+2xh-f\left(x\right)}{h}$
$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\lim }\left(\frac{f\left(h\right)}{h}+2x\right)=210+2x$
$f\left(x\right)=210x+x^2+C$
$f\left(0\right)=0\Rightarrow C=0$
$f\left(x\right)=210x+x^2$
$\therefore f\left(2\right)=424$