Let P and Q be any points on the curves $(\mathrm{x}-1{)}^2+(\mathrm{y}+1{)}^2=1\text{ and }\mathrm{y}={\mathrm{x}}^2$, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval

$$\begin{gathered} Q is a point on the curve y=x^{2 }i.e Q=\left(x,x^2\right) \\ P is a point on the circle {\left(x-1\right)}^2+{\left(y+1\right)}^2=1 \\ d= PQ=CQ-CP \\ \Rightarrow d=\sqrt{{\left(x-1\right)}^2+{\left(x^2+1\right)}^2}-1 \\ We need to minimise d so it is sufficient to minimise \\ f\left(x\right)={\left(x-1\right)}^2+{\left(x^2+1\right)}^2 \\ \therefore f\text{'}\left(x\right)=2\left(x-1\right)+2\left(x^2+1\right)2x \\ For critical points, f\text{'}\left(x\right)=0 \\ \Rightarrow x-1+2x^3+2x=0 \\ \Rightarrow 2x^3+3x-1=0 \\ f\text{'}\left(\frac{1}{4}\right)<0.f\text{'}\left(\frac{1}{2}\right)>0 \\ \therefore f\text{'}\left(x\right)=0 if\frac{1}{4}<x<\frac{1}{2}\Rightarrow x\in \left(\frac{1}{4},\frac{1}{2}\right) \end{gathered}$$