Given that,
$y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4}) ; 0<x<1 \longrightarrow (1)$
Minimum values ,we can get when
$\frac{1}{2} – \frac{3x^{2}}{4} =\frac{5x^{2}}{4}$
$ \Rightarrow \frac{8x^{2}}{4} = \frac{1}{2}$
$ \Rightarrow 2x^{2} = \frac{1}{2}$
$ \Rightarrow x^{2} = \frac{1}{4}$
$ \Rightarrow x = \sqrt \frac{1}{4}$
$ \Rightarrow \boxed{x = \frac{-1}{2}(rejected) \; or \; x= \frac{1}{2}}$
$ \Rightarrow \boxed{ x= \frac{1}{2}}$
Now, $y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4})$
$\Rightarrow y = \text{min} \left[ \frac{1}{2} – \frac{3(\frac{1}{2})^{2}}{4}, \frac{5(\frac{1}{2})^{2}}{4}\right]$
$\Rightarrow y= \text{min} \left[\frac{1}{2} – \frac{3}{16} , \frac{5}{16}\right]$
$\Rightarrow y= \text{min} (\frac{8-3}{16} , \frac{5}{16})$
$\Rightarrow y= \text{min} (\frac{5}{16} , \frac{5}{16})$
$\Rightarrow \boxed{y= \frac{5}{16}}$
$\therefore$ The minimum possible value of $y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4}) $ is $\frac{5}{16}$
Correct Answer $:\text{D}$