Question

The straight line $2x-3y=1$ divide the circular region $x^2+y^2\le 6$ 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

Answer: 2

$x^2+y^2\le 6$ 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 $L$.
$(2,\frac{3}{4})L:2x-3_y-1$ $L:4-\frac{9}{4}-1=\frac{3}{4}>0$
and $S:x^2+y^2-6,S:4+\frac{9}{16}-6<0$
$\Rightarrow (2,\frac{3}{4})$ lies in shaded part.
For $(\frac{5}{2},\frac{3}{4}),L:5-9-1<0$ [neglect]
For $(\frac{1}{4},-\frac{1}{4}),L:\frac{1}{2}+\frac{3}{4}-1>0$
$\therefore (\frac{1}{4},-\frac{1}{4})$
lies in the shaded part.
For For $(\frac{1}{8},\frac{1}{4}),L:\frac{1}{4}-\frac{3}{4}-1<0$ [neglect]
$\Rightarrow$ Only $2$ points lie in the shaded part.