Question

If $P\left(x\right)$ be a polynomial with real coefficients such that $P\left({\sin }^{2}x\right)=P\left({\cos }^{2}x\right)$, for all $x\in [0,\frac{\pi }{2}]$. Consider following statements:

I. $P\left(x\right)$ is an even function.
II. $P\left(x\right)$ can be expreesed as a polynomial in $(2x-1{)}^2$
III. $P\left(x\right)$ is a polynomial of even degree.

• A. all are false
• B. Only I and II are true
• C. Only II and III are true
• D. all are correct

Answer 1

The correct option is C Only II and III are true

$P\left({\sin }^{2}x\right)=P\left({\cos }^{2}x\right),\forall \in [0,\frac{\pi }{2}]\Rightarrow P\left({\sin }^{2}x\right)=P(1-{\sin }^{2}x)\text{Let}{\sin }^{2}x=t\Rightarrow P\left(t\right)=P(1-t),\forall t\in [0,1]$
So, $P\left(x\right)$ is symmetric about $x=\frac{1}{2}$
Therefore, $P\left(x\right)$ can be expreesed as a polynomial in $(2x-1{)}^2$.
Since, $P\left(x\right)$ is symmetric about $x=\frac{1}{2}$, it must be a polynomial of even degree.
We don't have any information about the $P\left(t\right)$ for $t\in [-1,0]$, So, we cannot comment on even/odd.