Circle
Region
JEE Advanced 2011 Paper 2
Grade 11
Question:
The line $2x - 3y = 1$ divides the circular region $x^2 + y^2 \leq 6$ into two parts. Consider the set $S = \{ (2, \frac{3}{4}), (\frac{5}{2}, \frac{3}{4}), (\frac{1}{4}, -\frac{1}{4}), (\frac{1}{8}, \frac{1}{4}) \}$. The number of point(s) in $S$ that lie inside the smaller part is _____.
Step-by-Step Solution
Key Concept: The origin satisfies $2(0) - 3(0) - 1 < 0$, so the smaller region is where $2x - 3y - 1 > 0$ and $x^2 + y^2 < 6$. Check each point: $(5/2, 3/4)$ is outside the circle. The remaining inside-circle points on the positive side are $(2, 3/4)$ and $(1/4, -1/4)$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: 2