Let P and Q be any points on the curves (x - 1) ^ 2 + (y + 1) ^ 2 =...

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. ${f}^{'}\left({x}_{1}\right) = 2\left({x}_{1} - 1\right) + 4{x}_{1}\left({{x}_{1}}^{2} + 1\right) = 0$

From options

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

So, ${f}^{'}\left({x}_{1}\right) = 0$ for some ${x}_{1} \in \left(\dfrac{1}{4}, \dfrac{1}{2}\right)$ from IMVT