Question 20

Mathematics Definite Integrals Hard

The area of the region bounded by the lines y = |x - 1| and y = 3 - |x| is

(A) 3 sq. units
(B) 4 sq. units
(C) 6 sq. units
(D) 2 sq. units
View Dynamic Solution & Explanation
Correct Solution: Option B

Step-by-step Solution:

Here is the MathJax code for the given solution: \[ y = 3(-x) \Rightarrow y = 3 + x \] \[ y = |x - 1| \Rightarrow y = |-1| \quad \text{[if } x = 0\text{]} \] \[ y = 3 - |x| \] \[ x = 2, \quad 2y = 4 \] \[ 2y = 2y = 2 \] \[ \Rightarrow y = 1 \] Breadth = distance between \( (1,0) \) and \( (2,1) \) or \( (0,3) \) and \( (-1,2) \) \[ = \sqrt{(2 - 1)^2 + (1 - 0)^2} = \sqrt{2} \] or distance between \( (1,0) \) and \( (2,3) \) Length = Distance between \( (0,3) \) and \( (2,1) \) or \( (-1,2) \) and \( (1,0) \) \[ = \sqrt{(0 - 2)^2 + (3 - 1)^2} = \sqrt{8} = 2\sqrt{2} \] \[ \text{Area} = \text{length} \times \text{breadth} = \sqrt{2} \times 2\sqrt{2} = 4 \text{ (units sq)} \]