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CAT 2020 Slot 3 — QA Question 24

LinesEasy

The area, in sq. units, enclosed by the lines x = 2, y = |x – 2| + 4, the X-axis and the Y-axis is equal to

Answer & solution

  • A

    6

  • 10

  • C

    8

  • D

    12

Solution

Easy

The region is bounded on the right by x=2x=2, on the left by the YY-axis (x=0x=0), below by the XX-axis and above by y=∣x−2∣+4y=|x-2|+4. On 0≤x≤20\le x\le 2 the modulus simplifies, and the area under that line splits neatly into a rectangle plus a triangle.

(2, 4) (0, 6) 2 x y
1

Simplify the curve on the strip. The region lies between x=0x=0 and x=2x=2. For x≤2x\le 2 we have x−2≤0x-2\le 0, so ∣x−2∣=2−x|x-2|=2-x.

y=∣x−2∣+4⇒ y=(2−x)+4(x≤2)⇒ y=6−x\begin{aligned} &y = |x-2|+4\\ &\Rightarrow\ y = (2-x)+4 \quad\text{(}x\le 2\text{)}\\ &\Rightarrow\ y = 6-x \end{aligned}
2

Find the boundary heights. Evaluate the top edge at the two vertical bounds.

x=0: y=6−0=6x=2: y=6−2=4\begin{aligned} &x=0:\ y = 6-0 = 6\\ &x=2:\ y = 6-2 = 4 \end{aligned}
3

Split the area. Under the line, the region (a right trapezium of width 22 with heights 44 and 66) splits into a rectangle of height 44 plus a triangle on top.

rectangle=width×height=2×4=8triangle=12×base×height=12×2×(6−4)=2\begin{aligned} &\text{rectangle} = \text{width}\times\text{height} = 2\times 4 = 8\\ &\text{triangle} = \tfrac12\times\text{base}\times\text{height} = \tfrac12\times 2\times(6-4) = 2 \end{aligned}
4

Add the pieces.

area=8+2(rectangle + triangle)⇒ area=10\begin{aligned} &\text{area} = 8 + 2 \quad\text{(rectangle + triangle)}\\ &\Rightarrow\ \text{area} = 10 \end{aligned}

Use the trapezium area directly: 12(4+6)×2=10\dfrac{1}{2}\big(4+6\big)\times 2 = 10 sq. units.

10 sq. units\boxed{10\ \text{sq. units}}
CAT 2020 Slot 3 QA Q24: The area, in sq. units, enclosed by the lines x = 2, y = |x – 2| + 4, the X-axis and the Y-axis is equal — Solution | TheCATExam