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Question

Let y=f(x)=x+3,3x<21+x+2,2x<12+x+1,1x0.
If the area enclosed between |y|=f(|x|) and x2+y2=5 is p+πq sq. units, then the value of (p+q) is

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Solution



Area bounded by the y=f(x) and xaxis from x=3 to x=0
=23x+3 dx+12(1+x+2) dx+01(2+x+1) dx=[23(x+3)3/2]23+[x+23(x+2)3/2]12+[2x+23(x+1)3/2]01=23[10]+[1+2+23(10)]+[0+2+23(10)]=23+53+83=5

Required area
=4(514×πr2)=4(5π4×5)=205πp=20, q=5p+q=15

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