Question

Let $f:R\to R$ be a function such that $f(x+y)=f\left(x\right)+f\left(y\right)+x^{2}y+x{y}^2$ $\forall x,y\in R$. If $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=1$, then $f\left(x\right)$ is

• A. $1+\frac{x^2}{2}$
• B. $x-\frac{x^2}{3}$
• C. $x+\frac{x^3}{3}$
• D. $1+\frac{x^2}{3}$

Answer 1

The correct option is C $x+\frac{x^3}{3}$

We know that,
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+x^{2}h+x{h}^2-f\left(x\right)}{h}=\underset{h\to 0}{lim}(\frac{f\left(h\right)}{h}+x^2+xh)=1+x^2+0(\because \underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=1)\Rightarrow f^{\prime }\left(x\right)=1+x^2$

Integrating w.r.t. $x$, we get
$f\left(x\right)=x+\frac{x^3}{3}+c$

Given $f(x+y)=f\left(x\right)+f\left(y\right)+x^{2}y+x{y}^2$
Putting $x=y=0$, we get
$f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0\Rightarrow c=0\therefore f\left(x\right)=x+\frac{x^3}{3}$