# If $f(x) = \sin x+\lfloor \frac{x^2}{a}\rfloor$ be an odd function. Then the set of values of parameter $a$ is/are

Let $f: \left[-10\;,10\right]\rightarrow R\;,$ where $\displaystyle f(x) = \sin x+\lfloor \frac{x^2}{a}\rfloor$ be an odd function. Then the set of values of parameter $a$ is/are

$\bf{options::}$ $(a)\;\; (-10,10)-\left\{ 0\right\}\;\;\;\;\;\;(b)\;\; (0,10)\;\;\;\\;\; (c)\;\;\left[100,\infty \right)\;\;\;\;\;\; (d)\;\;\;\;\;\;\; (100,\infty)$

$\bf{My\; Solution::}$ If function $f(x)$ is an odd function, Then $f(-x) = f(x).$ So Using the formula..

$\Rightarrow \displaystyle \sin x + \lfloor \frac{x^2}{a} \rfloor = \sin x+\lfloor \frac{x^2}{a} \rfloor .$

Now $-10\leq x \leq 10\Rightarrow 0 \leq x \leq 100$.

Now I did not Understand How can I solve it

$$f(-x)=-f(x)\implies-\sin x+\lfloor \frac{x^2}{a}\rfloor=-\sin x-\lfloor \frac{x^2}{a}\rfloor$$ or $$\lfloor \frac{x^2}{a}\rfloor=0\quad\quad\forall x\in [-10,10]$$ Therefore $$0\leq\frac{x^2}{a}\lt 1$$ $a$ must be positive and $$a\gt x^2 \quad\quad \forall x\in [-10,10]$$
hence $$a\gt 100$$
sin(-x) = -sin(x): With this you should be done. sin(x) is an odd function, $\lfloor \frac{x^2}{a}\rfloor$ should also be odd or zero. $\frac{x^2}{a}$ is even and so is $\lfloor \frac{x^2}{a}\rfloor$. So the only value poosible for $\lfloor \frac{x^2}{a}\rfloor$ is $0$, which means $\frac{x^2}{a} \lt 1$ for the defined domain.