Question

If a function satisfies $(x-y)f(x+y)-(x+y)f(x-y)=2(x^{2}y-y^3)\forall x,y\in R$ and $f\left(1\right)=2$, then

• A. $f\left(0\right)=2$
• B. $f\left(x\right)$ must be a quadratic function
• C. If $f:R^{+}\to R^{+}$ then $f\left(x\right)$ is an invertible function.
• D. $f\left(x\right)$ is a periodic function

Answer 1

Answer: (B), (C)

If $f:R^{+}\to R^{+}$ then $f\left(x\right)$ is an invertible function.

$(x-y)f(x+y)-(x+y)f(x-y)=2y(x^2-y^2)\Rightarrow \frac{f(x+y)}{x+y}-\frac{f(x-y)}{x-y}=2y\Rightarrow \frac{f(x+y)}{x+y}-\frac{f(x-y)}{x-y}=(x+y)-(x-y)\Rightarrow \frac{f(x+y)}{x+y}-(x+y)=\frac{f(x-y)}{x-y}-(x-y)=\lambda \text{(say)}\Rightarrow \frac{f\left(x\right)}{x}-x=\lambda \Rightarrow f\left(x\right)=x^2+\lambda x$
Given $f\left(1\right)=2$
$\Rightarrow 1+\lambda =2\lambda =1\therefore f\left(x\right)=x^2+x$
Clearly, $f\left(x\right)$ is a quadratic function, which is non-periodic and we have $f\left(0\right)=0$.
Also, $f$ is invertible in $R^{+}\to R^{+}$