JEE Advanced 2011 Paper 2Question 1
The straight line $2x-3y=1$ divides the circular region $x^2+y^2\le 6$ into two parts. If $$S=\left\{\left(2,\frac{3}{4}\right),\left(\frac{5}{2},\frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8},\frac{1}{4}\right)\right\},$$ then the number of point(s) in $S$ lying inside the smaller part is
Points and Straight Lines
Answer$2$
Solution
The origin satisfies $2x-3y<1$ and the line is at distance $\frac1{\sqrt{13}}$ from it, so the smaller part is $2x-3y>1$ inside the circle. $(2,\frac34)$: $1.75>1$, $4.56<6$ yes; $(\frac52,\frac34)$: $6.81>6$ outside; $(\frac14,-\frac14)$: $1.25>1$, inside yes; $(\frac18,\frac14)$: $-0.5<1$ no. Count $=2$.