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

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Therefore, according to the question (Fig.)

Since x, y ≥ 0 in the first quadrant.

When x > y (or y - x < 0),

|x - y| = x - y and |x + y| = x + y

Therefore, Eq. (1) is true given that,

2≤ x - y + x + y ≤ 4⇒ ≤ x ≤ 22

checking with (2, 1) in region x > y, i.e. 2 > 1

Therefore, we shade area below y = x from [2, 22].

Area of this region = 1/2(22 x 22) - (1/2)2 x 2 = 4 - 1 = 3 sq. units

By symmetry about y = x, total area required = 6 sq. units

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