A fraction  $f\left(x\right)$ is defined for all real number and satisfy $f(2+x)=f(2-x)$  and  $f(7+x)=f(7-x)$ for all real x. If  $x=0$ is a root of  $f\left(x\right)=0$, then the least number of roots of  $f\left(x\right)=0$ in the interval  $[-1000,1000]$ is

$$\begin{gathered} f(2+x)=f(2-x)\Rightarrow f(4-x)=f\left(x\right)\mathrm{.........}\left(1\right) \\ From f(7+x)=f(7-x)\Rightarrow f(14-x)=f\left(x\right)\mathrm{.........}\left(2\right) \end{gathered}$$

From equation (1) and (2) we get  $f(x+10)=f\left(x\right)$ so period is 10
Now replacing x by  $x+10$ and then  $x-10$ is equations (3) continuity in thus ways
$f(x+10n)=f\left(x\right)\mathrm{....}\left(4\right)$
For $\pm 1,\pm 2,\pm 3,\mathrm{.....}$
Since  $f\left(0\right)=0$ equation (4)
$f(\pm 10)=f(\pm 20)\mathrm{......}=f(\pm 1000)=0$  so total 201
Now  $x=0$ in equation  $\left(1\right)f\left(x\right)=f\left(0\right)=o$ from equation (1) we have 200 more roots