Number of Points in Set S Lying Inside Circular Region

To determine the number of points in the set $S$ lying inside the smaller part of the circular region $x^2 + y^2 \le 6$ divided by the line $2x - 3y = 1$, we analyze the conditions defining the smaller part:

1. Inside the circular region:
   A point $(x, y)$ must satisfy:
   $x^2 + y^2 < 6$

2. Inside the smaller part:
   The center of the circle is the origin $(0, 0)$.
   Evaluating the line expression $L(x, y) = 2x - 3y - 1$ at the origin:
   $L(0, 0) = 2(0) - 3(0) - 1 = -1 < 0$
   Since the origin lies in the larger part (the major segment), the region on the opposite side of the line from the origin constitutes the smaller part (the minor segment). Therefore, any point $(x, y)$ in the smaller part must satisfy:
   $2x - 3y - 1 > 0$

Now, we test each of the given points in $S$:

1. For $P_1 = \left(2, \frac{3}{4}\right)$:

   • Circle condition:
     $x^2 + y^2 = 2^2 + \left(\frac{3}{4}\right)^2 = 4 + \frac{9}{16} = \frac{73}{16} = 4.5625 < 6 \quad (\text{Satisfied})$
   • Line condition:
     $2x - 3y - 1 = 2(2) - 3\left(\frac{3}{4}\right) - 1 = 3 - \frac{9}{4} = \frac{3}{4} > 0 \quad (\text{Satisfied})$
     Hence, $P_1$ lies inside the smaller part.

2. For $P_2 = \left(\frac{5}{2}, \frac{3}{4}\right)$:

   • Circle condition:
     $x^2 + y^2 = \left(\frac{5}{2}\right)^2 + \left(\frac{3}{4}\right)^2 = \frac{25}{4} + \frac{9}{16} = \frac{109}{16} = 6.8125 > 6 \quad (\text{Not satisfied})$
     Hence, $P_2$ lies outside the circular region.

3. For $P_3 = \left(\frac{1}{4}, -\frac{1}{4}\right)$:

   • Circle condition:
     $x^2 + y^2 = \left(\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right)^2 = \frac{1}{16} + \frac{1}{16} = \frac{1}{8} < 6 \quad (\text{Satisfied})$
   • Line condition:
     $2x - 3y - 1 = 2\left(\frac{1}{4}\right) - 3\left(-\frac{1}{4}\right) - 1 = \frac{1}{2} + \frac{3}{4} - 1 = \frac{1}{4} > 0 \quad (\text{Satisfied})$
     Hence, $P_3$ lies inside the smaller part.

4. For $P_4 = \left(\frac{1}{8}, \frac{1}{4}\right)$:

   • Circle condition:
     $x^2 + y^2 = \left(\frac{1}{8}\right)^2 + \left(\frac{1}{4}\right)^2 = \frac{1}{64} + \frac{1}{16} = \frac{5}{64} < 6 \quad (\text{Satisfied})$
   • Line condition:
     $2x - 3y - 1 = 2\left(\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - 1 = \frac{1}{4} - \frac{3}{4} - 1 = -\frac{3}{2} < 0 \quad (\text{Not satisfied})$
     Hence, $P_4$ lies in the larger part containing the origin.

Thus, exactly $2$ points, $\left(2, \frac{3}{4}\right)$ and $\left(\frac{1}{4}, -\frac{1}{4}\right)$, lie inside the smaller part.