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

Let $P\left(x_1,{x_1}^2\right)$

Minimum distance will be obtained at common normal of the parabola and circle.

So for minimum $PQ$, the distance between centre of circle and $P$ should be minimum.

Now, the distance of $P$ from given circle,

$d=\sqrt{{\left(x_1-1\right)}^2+{\left({x_1}^2+1\right)}^2}-1$

For least value of $d$, we need to minimize

$f\left(x_1\right)={\left(x_1-1\right)}^2+{\left({x_1}^2+1\right)}^2$

i.e. $f^{\text{'}}\left(x_1\right)=2\left(x_1-1\right)+4x_1\left({x_1}^2+1\right)=0$

From options

$f^{\text{'}}\left(\frac{1}{4}\right)$ is $-\mathrm{ve}$ and $f^{\text{'}}\left(\frac{1}{2}\right)$ is $+\mathrm{ve}$

So, $f^{\text{'}}\left(x_1\right)=0$ for some $x_1\in \left(\frac{1}{4},\frac{1}{2}\right)$ from IMVT