Question

Suppose $f:R\to R$ is differentiable function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)+xy(x+y)$ for every $x,y\in R$. If $f^{\prime }\left(0\right)=0$, then which of the following hold(s) good?

• A. $f$ is an odd function.
• B. $f$ is a bijective mapping.
• C. $f$ has a minima but no maxima.
• D. $f$ has an inflection point.

Answer 1

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

The correct options are
A $f$ is a bijective mapping.
B $f$ is an odd function.
D $f$ has an inflection point.

Given that $f(x+y)=f\left(x\right)+f\left(y\right)+xy(x+y)$ ....(1)
and $f^{\prime }\left(0\right)=0$
Put $x=y=0$, we get $f\left(0\right)=0$

Now,

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left(\frac{f(x+h)-f\left(x\right)}{h}\right)=\underset{h\to 0}{lim}\left(\frac{f\left(x\right)+f\left(h\right)+xh(x+h)-f\left(x\right)}{h}\right)$ ...[ from(1) ]
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\left(\frac{f\left(h\right)-f\left(0\right)}{h}\right)+{lim}_{h\to 0}x(x+h)=f^{\prime }\left(0\right)+x^2$
$\Rightarrow f^{\prime }\left(x\right)=x^2$
$\Rightarrow f\left(x\right)=\frac{x^3}{3}+c$

Since, $f\left(0\right)=0\Rightarrow c=0$
Therefore, $\Rightarrow f\left(x\right)=\frac{x^3}{3}$

Thus,

$f\left(x\right)$ is odd function and a bijective mapping
For point of inflection, $f^{\prime \prime }\left(x\right)=0$
$f^{\prime \prime }\left(x\right)=x=0$ at $x=0$ only
Therefore, $f\left(x\right)$ has only one point of inflection.

Hence, options A, B, and D.