CAT 2019 Slot 1 — QA Question 6
Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals
Answer & solution
Answer: 2
|x| ≥ 1 ⇒ x ≥ 1 and x ≤ −1. This is represented as the shaded region in the figure below.
|x| + |y| ≤ 2 : This will have four subcases depending on which quadrant the point is.
1st quadrant (x > 0; y > 0) : y ≤ 2 − x
2nd quadrant (x < 0; y > 0) : y ≤ 2 + x
3rd quadrant (x < 0; y < 0) : y ≥ −2 − x
4th quadrant (x > 0; y < 0) : y ≥ x − 2
Combining these four, the graph for |x| + |y| ≤ 2 is as shown below.
The intersection of |x| + |y| ≤ 2 and |x| ≥ 1 is as shown as the shaded area in the following image.
We need to find Area âABC + Area âDEF
Area âABC = Area âDEF = (1/2) × BC × AM = (1/2) × 2 × 1 = 1 square unit.
Required area = 1 + 1 = 2 square units.
Hence, 2.