Let P and Q be any points on the curves x - 1 2 + y + 1 2 = 1 and y = x 2 , respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval

Step-by-step solution

Let P x 1 , x 1 2 Minimum distance will be obtained at common normal of the parabola and circle. So for minimum P Q , the distance between centre of circle and P should be minimum. Now, the distance of P from given circle, d = x 1 - 1 2 + x 1 2 + 1 2 - 1 For least value of d , we need to minimize f x 1 = x 1 - 1 2 + x 1 2 + 1 2 i.e. f ' x 1 = 2 x 1 - 1 + 4 x 1 x 1 2 + 1 = 0 From options f ' 1 4 is - ve and f ' 1 2 is + ve So, f ' x 1 = 0 for some x 1 ∈ 1 4 , 1 2 from IMVT