If $f:R \rightarrow (0, \infty)$ be a differentiable function $f(x)$ satisfying $f(x+ y) - f(x - y) = f(x) \{ f(y) - f(y) - y \}, \forall x, y \epsilon R, (f(y) \neq f(-y)$ for all $y \epsilon R)$ and $f' (0) = 2010$.

Now answer the following questions

Which of the following is true for $f(x)$

Here, $2 f' (x) = \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x)}{h} + \frac{f(x x - h) - f(x)}{-h} \right )$

$= \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x - h)}{h} \right )$ (i)

$\therefore 2f'(0) \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(h) - f (-h)}{h} + \frac{f(-h) - f(0)}{-h} \right )$

$= \displaystyle \lim_{h \rightarrow 0}\frac{f(h) - f(-h)}{h}$ ... ..(ii)

Now by given relation, we have

$f(h) - f(-h) = \frac{f(x + h) - f(x - h)}{-h}$ and $f(0) = 1$

From Eqs. (i) and (ii), we have $\frac{f'(x)}{f(x)} = 2010$

$\Rightarrow$ $f(x) = e^{2010e}, f(0) = 1$

$\therefore \{ f(x) \}$ is non-periodic.