Question

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

Answer 1

$x^2 + y^2 \le 6$ and $2x - 3y = 1$ is shown as

$\therefore$ For the point to lie in the shade part, origin and the point lie on opposite side of straight line $L$.
$\bigg( 2, \frac{3}{4}\bigg) L : 2x - 3_y - 1$ $ L: 4 - \frac {9}{4}- 1 = \frac {3}{4} > 0$
and $ S: x^2 + y^2 - 6, S : 4 + \frac {9}{16}- 6 < 0$
$\Rightarrow \bigg( 2, \frac {3}{4}\bigg)$ lies in shaded part.
For $ \bigg( \frac {5}{2}, \frac {3}{4}\bigg), L : 5 - 9 - 1 < 0 $ [neglect]
For $ \bigg( \frac {1}{4} ,-\frac {1}{4}\bigg), L : \frac {1}{2} + \frac{3}{4}- 1 > 0 $
$ \therefore \bigg( \frac {1}{4} ,-\frac {1}{4}\bigg) $
lies in the shaded part.
For For $\bigg( \frac {1}{8} ,\frac {1}{4} \bigg), L : \frac {1}{4}- \frac {3}{4} - 1 < 0 $ [neglect]
$\Rightarrow $ Only $2$ points lie in the shaded part.