The straight line $2\mathrm{x}-3\mathrm{y}=1$ divides the circular region ${\mathrm{x}}^2+{\mathrm{y}}^2\le 6$ into two parts. If $\mathrm{S}=\left\{\left(2\cdot \frac{3}{4}\right),\left(\frac{5}{2},\frac{3}{4}\right)\left(\frac{1}{4},-\frac{1}{4}\right)\left(\frac{1}{8},\frac{1}{4}\right)\right\}$ then the number of points in S lying inside the smaller part is:

The smaller region of circle is the region

given by ${\mathrm{x}}^2+{\mathrm{y}}^2\le 6 ......\left(1\right)$

and $2\mathrm{x}-3\mathrm{y}\ge 1 ......\left(2\right)$

We observe that only two points $\left(2,\frac{3}{4}\right)\text{ and }\left(\frac{1}{4},1\right)$

satisfy both the inequations (1) and (2)

$\therefore$   2 points is S lie inside the smaller part.