The area of the region bounded by the curves y = |x - 1| and

Question

What is the area of the region bounded by the curves  y = |x - 1|  and  y = |3| - |x|  ?

(A) 2 units^2
(B) 3 units^2
(C) 4 units^2
(D) 5 units^2
(E) 6 units^2

bv8562 · Answer

y=|x−1| and y= -|x|+|3|

a) when x<0, then

y=-(x-1) => y=-x+1 (Therefore, x=1 and y=1, NOT VALID as x<0)

b) when 0<x<1, then

y=-(x-1) => y=-x+1 (Therefore, x=1 and y=1, NOT VALID as 0<x<1)

c) when x>=1, then

y=x-1 (Therefore, x=1 and y=-1, VALID as x>=1)
and
y=-x+3 (Therefore, x=3 and y=3, VALID as x>=1)

Since, both lines intersect with each other, therefore:
x-1=-x+3 => x=2
y=x-1 => y=2-1 => y=1

Both lines intersect at point (2,1) which is the height of the triangle formed and it's base is 4 units long. Hence, the area would be:

1/2*4*2 = 4 units^2 (C)

This is how I have solved it. Is this the correct approach? Bunuel chetan2u

FPM · Answer

I believe the curves form a quadrilateral.

There are 4 curves:
f(x) = x-1 
f(x) = -x+1 
f(x) = 3 - x
f(x) = 3 +x

The curves intersect at:

-x+1=x-1
-2x=-2
x=1, then (1,0)

3-x=3+x
-2x=0
x=0, then (0,3)

3-x=x-1
-2x=-4
x=2, then (2,1)

3+x=-x+1
2x=-2
x=-1, then (-1,2)

Plotting the points and using Pitagoras we can find the sides of the quadrilateral, which are  \sqrt {2}  and 2 \sqrt {2} .
The area of the quadrilateral is  2^{2} .

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

We need to find what is the area of the region bounded by the curves y = |x - 1| and y = |3| - |x|

Plotting y = |x - 1|

We will have two cases if we open \|x-1\|
-Case 1: x-1 ≥ 0 => x ≥ 1 => \|x-1\| = x-1 => y = x - 1 for x ≥ 1 We can plot the same by getting two points x = 1 , y = 1-1 = 0 => (1,0) x = 2 , y = 2-1 = 1 => (2,1) Line will look as the one shown in purple color below	-Case 2: x-1 ≤ 0 => x ≤ 1 => \|x-1\| = -(x-1) = 1 - x => y = 1 - x for x ≤ 1 We can plot the same by getting two points x = 1 , y = 1-1 = 0 => (1,0) x = 0 , y = 1-0 = 1 => (0,1) Line will look as the one shown in red color below

Attachment:

abs14 img-1.jpg [ 10.62 KiB | Viewed 2353 times ]

Plotting y = |3| - |x|

As 3 > 0
=> |3| = 3
=> y = 3 - |x|

We will have two cases if we open \|x\|
-Case 1: x ≥ 0 => \|x\| = x => y = 3-x We can plot the same by getting two points x = 0 , y = 3-0 = 1 => (0,3) x = 3 , y = 3-3 = 0 => (3,0) Line will look as the one shown in green color below	-Case 2: x ≤ 0 => \|x\| = -x => y = 3-(-x) = 3 + x We can plot the same by getting two points x = 0 , y = 3+0 = 3 => (0,3) x = -3 , y = 3-3 = 0 => (-3,0) Line will look as the one shown in blue color below

Attachment:

abs14 img-2.jpg [ 15.43 KiB | Viewed 2317 times ]

We will find the Point of intersection of both the graphs to get the desired enclosed figure

There will be four points of intersections

1. x=1, y=0 between the two lines of y = |x-1|
2. x=0, y=3 between the two lines of y = |3| - |x|
3. x=2 and y=1 for the lines y = 3 - x and y = x - 1
=> y = 3 - x = x - 1
=> 2x = 4
=> x = 2
=> y = 3 - 2 = 1
4. x=-1 and y=2 for the lines y = 1 - x and y = 3 + x
=> y = 1 - x = 3 + x
=> 2x = -2
=> x = -1
=> y = 3 - 1 = 2

Attachment:

abs14 img-3.jpg [ 15.42 KiB | Viewed 2294 times ]

Enclosed figure is a rectangle with the length of two sides given by

Distance between (0,3) and (2,1) = \(\sqrt{(0-2)^2 + (3-1)^2}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\)
Distance between (2,1) and (1,0) = \(\sqrt{(2-1)^2 + (1-0)^2}\) = \(\sqrt{1 + 1}\) = \(\sqrt{2}\)

=> Area = \(\sqrt{8}\) * \(\sqrt{2}\) = \(\sqrt{16}\) = 4

So, Answer will be C
Hope it helps!

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The area of the region bounded by the curves y = |x - 1| and

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The area of the region bounded by the curves y = |x - 1| and

Prep Toolkit

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