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25 Nov 2021, 04:30
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
51% (02:26) correct
49%
(02:23)
wrong
based on 156
sessions
History
Date
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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
(A) 2 units^2
(B) 3 units^2
(C) 4 units^2
(D) 5 units^2
(E) 6 units^2
11 Dec 2021, 02:15
Kudos
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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
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
20 Dec 2021, 16:09
Kudos
Bookmarks
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 \(\sqrt[2]{2}\) and 2\(\sqrt[2]{2}\).
The area of the quadrilateral is \(2^{2}\).
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 \(\sqrt[2]{2}\) and 2\(\sqrt[2]{2}\).
The area of the quadrilateral is \(2^{2}\).
