JEE Advanced 2014 · Paper 1 · Mathematics Q59

← JEE Advanced 2014 · Paper 1

MathematicsInteger+3Question 59 of 60

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2\le d_1(P)+d_2(P)\le4$ , is

Your answer

Answer and worked solution

Official answer: 6

Distances in the first quadrant.
$d_1+d_2=\frac{|x-y|+(x+y)}{\sqrt2}=\frac{2\max(x,y)}{\sqrt2}=\sqrt2\max(x,y).$

Region. $2\le\sqrt2\max(x,y)\le4$ means $\sqrt2\le\max(x,y)\le2\sqrt2$ : the square $[0,2\sqrt2]^2$ with the square $[0,\sqrt2)^2$ removed.

Area. $(2\sqrt2)^2-(\sqrt2)^2=8-2=6$ .

Answer: 6

Topics:Distance of a point from a line Area between curves

← Previous question Next question →

JEE Advanced 2014 · Paper 1, Mathematics question 59