If the line $x=\alpha$  divides the area of region $S=\left\{\left(x,y\right)\in R^2,x^5\le y\le x^2,0\le x\le 1\right\}$
  into two equal parts then which of the following is/are CORRECT?
 

A.T.Q
 $$\begin{gathered} \underset{0}{\overset{\alpha }{\int }}\left(x^2-x^5\right)dx=\underset{0}{\overset{1}{\int }}\left(x^2-x^5\right)dx \\ 2\underset{0}{\overset{\alpha }{\int }}\left(x^2-x^5\right)dx=\underset{0}{\overset{1}{\int }}\left(x^2-x^5\right)dx \\ 2{\left[\frac{x^3}{3}-\frac{x^6}{6}\right]}_0^{\alpha }={\left[\frac{x^3}{3}-\frac{x^6}{6}\right]}_0^{\alpha } \\ 2\left[\frac{{\alpha }^3}{3}-\frac{{\alpha }^6}{6}\right]=\left[\frac{1}{3}-\frac{1}{6}\right] \\ 2\frac{\left[2{\alpha }^3-{\alpha }^6\right]}{6}=\frac{1}{6} \\ \Rightarrow 2{\alpha }^6-4{\alpha }^3+1=0 \\ \frac{2{\alpha }^6+1}{4{\alpha }^2}=\alpha \end{gathered}$$
 Let  $f\left(\alpha \right)=2{\alpha }^6-4{\alpha }^3+1$
 $f\left(3/4\right)=2\times \frac{9}{16}-4\times \frac{3}{4}+1=\frac{9}{8}+1-3<0$
$f\left(0\right)>0$