If set $\mathrm{A}=\left\{\mathrm{x}\mid \frac{{\mathrm{x}}^2(5-\mathrm{x})(1-2\mathrm{x})}{(5\mathrm{x}+1)(\mathrm{x}+2)}<0\right\}$ and set $\mathrm{B}=\left\{\mathrm{x}\mid \frac{3\mathrm{x}+1}{6{\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}}>0\right\}$, then $\mathrm{A}\cap \mathrm{B}$ does not contain 

 

We have $\frac{{\mathrm{x}}^2(\mathrm{x}-5)(2\mathrm{x}-1)}{(5\mathrm{x}+1)(\mathrm{x}+2)}<0$

From the sign scheme, solution is

$\left(-2,-\frac{1}{5}\right)\cup \left(\frac{1}{2},5\right)$               (i)

Now $\frac{3\mathrm{x}+1}{\mathrm{x}\left(6{\mathrm{x}}^2+\mathrm{x}-1\right)}>0$

$\Rightarrow \frac{3\mathrm{x}+1}{\mathrm{x}(3\mathrm{x}-1)(2\mathrm{x}+1)}>0$

From the sign scheme solution is

$\left(-\mathrm{\infty },-\frac{1}{2}\right)\cup \left(-\frac{1}{3},0\right)\cup \left(\frac{1}{3},\mathrm{\infty }\right)$               (ii)

Clearly, from (i) and (ii), option(s) (1) and (3) are  elements of $A\cap B$