JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 5)
For a point $$P$$ in the plane, Let $${d_1}\left( P \right)$$ and $${d_2}\left( P \right)$$ 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 $$2 \le {d_1}\left( P \right) + {d_2}\left( P \right) \le 4$$, is
Answer
6
Explanation
Let $P(X, Y)$ is the point in first quadrant.
Now, $2 \leq\left|\frac{x-y}{\sqrt{2}}\right|+\left|\frac{x+y}{\sqrt{2}}\right| \leq 4$
$$2 \sqrt{2} \leq|x-y|+|x+y| \leq 4 \sqrt{2}$$
Case I : $x \geq y$
$\begin{array}{rlrl}2 \sqrt{2} \leq(x-y)+(x+y) \leq 4 \sqrt{2} \\\\ \Rightarrow x \in[\sqrt{2}, 2 \sqrt{2}]\end{array}$
Case II : $x < y$
$\begin{aligned} 2 \sqrt{2} & \leq y-x+(x+y) \leq 4 \sqrt{2} \\\\ y & \in[\sqrt{2}, 2 \sqrt{2}]\end{aligned}$
$\Rightarrow A=(2 \sqrt{2})^2-(\sqrt{2})^2=6$ sq units
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