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CAT 2020 Slot 1 — QA Question 26

Domain & RangeMedium

The area of the region satisfying the inequalities |x| - y ≤ 1, y ≥ 0 and y ≤ 1 is

Answer & solution

Answer: 3

Solution

Medium

Rewrite the first inequality as y≥∣x∣−1y\ge |x|-1 and combine with the horizontal strip 0≤y≤10\le y\le 1. The boundary y=∣x∣−1y=|x|-1 is a V-shape; the region between it and the strip is a symmetric trapezium. Find its parallel sides and height, then use the trapezium area formula.

Boundaries: y=∣x∣−1y=|x|-1 (a V with vertex at (0,−1)(0,-1), opening upward), the line y=0y=0 (x-axis) and the line y=1y=1. The required region is where all three of y≥∣x∣−1y\ge|x|-1, y≥0y\ge0, y≤1y\le1 hold.

$y=1$ $y=0$ (-2,1) (2,1) (-1,0) (1,0)
1

Find the bottom edge (y=0y=0). On the x-axis, y≥∣x∣−1y\ge|x|-1 becomes 0≥∣x∣−10\ge|x|-1, i.e. ∣x∣≤1|x|\le1.

−1≤x≤1(at y=0)⇒ bottom side length=1−(−1)=2\begin{aligned} &-1\le x\le 1 \quad\text{(at $y=0$)}\\ &\Rightarrow\ \text{bottom side length}=1-(-1)=2 \end{aligned}
2

Find the top edge (y=1y=1). At y=1y=1, 1≥∣x∣−11\ge|x|-1 gives ∣x∣≤2|x|\le2.

−2≤x≤2(at y=1)⇒ top side length=2−(−2)=4\begin{aligned} &-2\le x\le 2 \quad\text{(at $y=1$)}\\ &\Rightarrow\ \text{top side length}=2-(-2)=4 \end{aligned}
3

Apply the trapezium area formula. Parallel sides 22 and 44, height =1−0=1=1-0=1.

Area=12 (sum of parallel sides)×height⇒ Area=12 (2+4)×1(from steps 1 and 2)⇒ Area=3\begin{aligned} &\text{Area}=\tfrac12\,(\text{sum of parallel sides})\times \text{height}\\ &\Rightarrow\ \text{Area}=\tfrac12\,(2+4)\times 1 \quad\text{(from steps 1 and 2)}\\ &\Rightarrow\ \text{Area}=3 \end{aligned}
Area=3 square units\text{Area}=3\ \text{square units}
CAT 2020 Slot 1 QA Q26: The area of the region satisfying the inequalities |x| - y ≤ 1, y ≥ 0 and y ≤ 1 is — Solution | TheCATExam