For a point $P$ in the plane, let $d_{1}(P)$ and $d_{2}(P)$ be the distance 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 $2 \le d_{1}(P)+d_{2}(P) \le 4$, is ________.

For a point P in the plane, let d₁(P) and d₂(P) be the distance 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 ≤ d₁(P)+d₂(P) ≤ 4, is ________.

$d_1(P) = \text{distance of } P \text{ from } x-y=0$

$d_2(P) = \text{distance of } P \text{ from } x+y=0$

$R:\ \text{first quadrant, } 2 \le d_1(P)+d_2(P) \le 4$

$\text{To find: area of } R$

$x \ge 0,\quad y \ge 0$ $d = \dfrac{|Ax+By+C|}{\sqrt{A^2+B^2}}$ $d_1(P) = \dfrac{|x-y|}{\sqrt2}$ $d_2(P) = \dfrac{|x+y|}{\sqrt2}$ $1.\ \text{simplify } d_1+d_2 \text{ in the first quadrant}$ $2.\ \text{condition on } \max(x,y) \Rightarrow \text{shape of } R$ $3.\ \text{area of } R$ $x \ge 0,\ y \ge 0 \ \Rightarrow\ |x+y| = x+y$ $|x-y| + (x+y) = 2\max(x,y)$ $d_1+d_2 = \dfrac{2\max(x,y)}{\sqrt2} = \sqrt2\,\max(x,y)$ $2 \le \sqrt2\,\max(x,y) \le 4$ $\Rightarrow\ \sqrt2 \le \max(x,y) \le 2\sqrt2$ $\max(x,y) \le 2\sqrt2 \ \Rightarrow\ x \le 2\sqrt2,\ y \le 2\sqrt2$ $\max(x,y) \ge \sqrt2 \ \Rightarrow\ \text{not both } x < \sqrt2,\ y < \sqrt2$ $R = [0,2\sqrt2]^2 \setminus [0,\sqrt2]^2$ $A_{\text{big}} = (2\sqrt2)^2 = 8$ $A_{\text{small}} = (\sqrt2)^2 = 2$ $\text{Area}(R) = 8 - 2 = 6$ $\boxed{6\ \text{square units}}$