For a point P in the plane, let $d_1\left(P\right)$ and $d_2\left(P\right)$ 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 $2 \le d_1\left(P\right)+d_2\left(P\right) \le 4,$ is ________

Solution

$2 \le d_1\left(p\right)+d_2\left(p\right)\le 4$
For $P\left(\alpha , \beta \right), \alpha >\beta$
$\Rightarrow$ $2\sqrt{2} \le 2\alpha \le 4\sqrt{2}$
$\sqrt{2} \le \alpha \le 2\sqrt{2}$
$\Rightarrow$ Area of region $=\left({\left(2\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^2\right)$
$=8-2=6\mathrm{ }\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Hence, the correct answer is 6.