IIT-JEE 2011 · Paper 2 · Mathematics Q58

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MathematicsInteger+4Question 58 of 60

The straight line $2x-3y=1$ divides the circular region $x^2+y^2\le6$ into two parts. If
$S=\left\{\left(2,\frac34\right),\left(\frac52,\frac34\right),\left(\frac14,-\frac14\right),\left(\frac18,\frac14\right)\right\},$
then the number of point(s) in $S$ lying inside the smaller part is

Your answer

Answer and worked solution

Official answer: 2

Smaller part. The centre gives $2(0)-3(0)-1<0$ , so the smaller segment is where $2x-3y-1>0$ (and inside the circle).

Point	$2x-3y-1$	$x^2+y^2$	In smaller part?
$(2,\frac34)$	0.75 > 0	4.56 < 6	yes
$(\frac52,\frac34)$	1.75 > 0	6.81 > 6	outside circle
$(\frac14,-\frac14)$	0.25 > 0	0.125	yes
$(\frac18,\frac14)$	−1.5 < 0	-	no

Answer: 2

Topics:Equation of a circle Distance of a point from a line

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IIT-JEE 2011 · Paper 2, Mathematics question 58