For a point P in the plane, let d_1(P) and d_ | vedclass

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Question

let $ p(h, k) $

$2 \leq\left|\frac{h-k}{\sqrt{2}}\right|+\left|\frac{h+k}{\sqrt{2}}\right| \leq 4 $

$\Rightarrow \quad 2 \sqrt{2} \leq|h-k|+|h+k

Vedclass · Accepted Answer

$6$

Vedclass · Answer

let $ p(h, k) $

$2 \leq\left|\frac{h-k}{\sqrt{2}}ight|+\left|\frac{h+k}{\sqrt{2}}ight| \leq 4 $

$\Rightarrow \quad 2 \sqrt{2} \leq|h-k|+|h+k| \leq 4 \sqrt{2} $

$	ext { if } \quad h \geq k$

$\Rightarrow \quad 2 \sqrt{2} \leq x-y+x+y \leq 4 \sqrt{2} \quad$ or $\quad \sqrt{2} \leq x \leq 2 \sqrt{2}$

similarly when $k > h$

we have $\sqrt{2} \leq y \leq 2 \sqrt{2}$

The required area $=(2 \sqrt{2})^2-(\sqrt{2})^2=6$.

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 $2 \leq d_1(P)+d_2(P) \leq 4$, is

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