If P(x) is non constant polynomial with real coefficients such that  $P\left({\sin }^{2}x\right)=P\left({\cos }^{2}x\right)forallx\in \left[0,\frac{\pi }{2}\right]$. Consider the following statements : 
I) P(x) is an even function 
II) P(x) can be expressed as a polynomial in ${\left(2x-1\right)}^2$ 
III) P(x) is a polynomial of even degree, then 

Given  $p\left({\sin }^{2}x\right)=p\left({\cos }^{2}x\right);x=\left[0,\frac{\pi }{2}\right]\Rightarrow p\left({\sin }^{2}x\right)=p\left(1-{\sin }^{2}x\right);x\in \left[0,\frac{\pi }{2}\right]$

$\Rightarrow p\left(x\right)=p\left(1-x\right);x\in \left[0,1\right]replace=x\to x+\frac{1}{2}\Rightarrow P\left(\frac{1}{2}+x\right)=p\left(\frac{1}{2}-x\right)$

P(x) is symmetrical about $x=\frac{1}{2}P\left(x\right)$ will have even number of solutions in (0, 1) (or) p(x) is even degree polynomial. Therefore, statements (ii) and (iii) are true clearly, p(x) is nor even function therefore, statement (i) is false