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Question

If the area of the curve enclosed by y|y|±x|x|=1 and y=|x| is A, then πA is
(exclude infinite area regions enclosed between the curves)

A
2
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B
1
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C
4
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D
3
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Solution

The correct option is C 4
y|y|±x|x|=1 in the 1st and 2nd quadrant is
y2+x2=1 and y2x2=1 respectively.
In 3rd and 4th quadrant, it is y2+x2=1 and
x2y2=1 respectively


The point of intersection of the curves x2+y2=1and y=x is x=12
Area=2⎜ ⎜1/201x2xdx⎟ ⎟
=2(x1x22+12sin1x)1/20(x22)1/20
=π4
πA=4

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