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

The smaller region of circle is the region 
 
Given by  $x^2+y^2\le \mathrm{6......}\left(1\right)$
And  $2x-3y\ge \mathrm{1.......}\left(2\right)$
We observe that only one point  $(2,\frac{3}{4})$
Satisfy both the inequations (1) and (2) 
$\therefore$  1 point in S lie inside the smaller part.