JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Straight Lines

For a point $P(h,k)$ in the first quadrant, $d_1(P)=\frac{|h-k|}{\sqrt{2}}$ and $d_2(P)=\frac{|h+k|}{\sqrt{2}}=\frac{h+k}{\sqrt{2}}$. The condition $2\le d_1(P)+d_2(P)\le 4$ becomes $2\le \frac{|h-k|+h+k}{\sqrt{2}}\le 4$. In the first quadrant, if $h\ge k$, then $|h-k|=h-k$, so $d_1+d_2=\frac{2h}{\sqrt{2}}=h\sqrt{2}$. Thus $2\le h\sqrt{2}\le 4$, giving $\sqrt{2}\le h\le 2\sqrt{2}$. If $h<k$, then $|h-k|=k-h$, so $d_1+d_2=\frac{2k}{\sqrt{2}}=k\sqrt{2}$. Thus $\sqrt{2}\le k\le 2\sqrt{2}$. The region is a square with vertices at $(\sqrt{2},0)$, $(2\sqrt{2},0)$, $(0,\sqrt{2})$, $(0,2\sqrt{2})$, and the diagonals along $x=y$. The area is $(2\sqrt{2}{)}^2-(\sqrt{2}{)}^2=8-2=6$.