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Question

The area enclosed by x2+y2=4,y=x2+x+1, y=[sin2x4+cosx4] and x-axis (where [.] denotes the greatest integer function) is:

A
2π3+316
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B
2π3+2316
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C
2313
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D
π3+3
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Solution

The correct option is A 2π3+316
sin2x4+cosx4=1cos2x4+cosx4
=1(cos2x4cosx4)
=54(cosx412)2
Now, for 2x2
12x412
cos2x4cosx4 for all x[2,2]

cos2x4+sin2x4sin2x4+cosx4, x[2,2]
[sin2x4+cosx4]=1 x[2,2]

Now from the figure, to find the enclosed area, we have to find point of intersections of y=1 and the circle.
Solving both equations we get x=3 and x=3
Also we have to find the points of intersection of y=1 and y=x2+x+1.
Solving both equations we get x=1 and x=0
Now dividing the enclosed area into different intervals as per the points of intersections, we get required area
=324x2dx+13dx+01(x2+x+1)dx+30dx+234x2dx
=2π3+316

145904_77059_ans.png

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