For a point P(x, y) in the plane,let d_1(P) a | vedclass

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(C) Let $P(x, y)$ be a point in the first quadrant,so $x > 0$ and $y > 0$.The distances are $d_1(P) = \frac{|x-y|}{\sqrt{2}}$ and $d_2(P) = \frac{|x+y

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$6$

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(C) Let $P(x, y)$ be a point in the first quadrant,so $x > 0$ and $y > 0$.The distances are $d_1(P) = \frac{|x-y|}{\sqrt{2}}$ and $d_2(P) = \frac{|x+y|}{\sqrt{2}}$.The given condition is $2 \leq \frac{|x-y|}{\sqrt{2}} + \frac{|x+y|}{\sqrt{2}} \leq 4$,which simplifies to $2\sqrt{2} \leq |x-y| + |x+y| \leq 4\sqrt{2}$.Case $1$: $x \geq y$. Then $|x-y| + |x+y| = (x-y) + (x+y) = 2x$.So,$2\sqrt{2} \leq 2x \leq 4\sqrt{2} \implies \sqrt{2} \leq x \leq 2\sqrt{2}$. Since $x \geq y > 0$,this region is a rectangle in the first quadrant.Case $2$: $y > x$. Then $|x-y| + |x+y| = (y-x) + (x+y) = 2y$.So,$2\sqrt{2} \leq 2y \leq 4\sqrt{2} \implies \sqrt{2} \leq y \leq 2\sqrt{2}$. Since $y > x > 0$,this region is also a rectangle.The region $R$ is the union of these two rectangles,which forms an $L$-shaped region.The area is the difference between the area of a large square of side $2\sqrt{2}$ and a small square of side $\sqrt{2}$.Area $= (2\sqrt{2})^2 - (\sqrt{2})^2 = 8 - 2 = 6$.

For a point $P(x, y)$ 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 $2 \leq d_1(P)+d_2(P) \leq 4$ is:

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