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Consider two circles C1: x^2+y^2-1=0 an...

Consider two circles `C_1: x^2+y^2-1=0` and `C_2: x^2+y^2-2=0`. Let A(1,0) be a fixed point on the circle `C_1` and B be any variable point on the circle `C_2`. The line BA meets the curve `C_2` again at C. Which of the following alternative(s) is/are correct?

Text Solution

Verified by Experts

The correct Answer is:
`f(x)=x^(3)-x^(2), 0 le x le 1`
Refer to the figure given in the question. Let the coordinates of P be `(x, x^(2)), " where " 0 le x le 1.`
Upper boundary: `y=x^(2) and `
lower boundary: `y=f(x)`
Lower limit of `x:0`
Upper limit of `x:x`
`therefore " Area " (OPRO)=int_(0)^(x) t^(2) dt - int_(0)^(x) f(t) dt`
`=[(t^(3))/(3)]_(0)^(x)-int_(0)^(x) f(t) dt `
`=(x^(3))/(3)-int_(0)^(x)f(t) dt`
For the area (OPQO),
The upper curve : `x=sqrt(y)`
and the lower curve : `x = y//2 `
Lower limit of y : 0
and upper limit of `y: x^(2)`
`therefore " Area " (OPQO) =int_(0)^(x^(2))sqrt(t)dt-int_(0)^(x^(2))(t)/(2) dt `
`=(2)/(3)[t^(3//2)]_(0)^(x^(2)) -(1)/(4)[t^(2)]_(0)^(x^(2))`
`=(2)/(3)x^(3)-(x^(4))/(4)`
According to the given condition,
`(x^(3))/(3)-int_(0)^(x)f(t)dt = (2)/(3) x^(3) - (x^(4))/(4)`
On differentiating both sides w.r.t.x, we get
`x^(2) -f(x)*1=2x^(2)-x^(3)`
`rArr f(x)=x^(3)-x^(2), 0 le x le 1`
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