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(C) Given $P(\sin^2 x) = P(\cos^2 x)$ for $x \in [0, \pi/2)$.Let $t = \sin^2 x$. Then $\cos^2 x = 1 - t$. Since $x \in [0, \pi/2)$,$t \in [0, 1)$.Thus

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only $II$ and $III$ are true

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(C) Given $P(\sin^2 x) = P(\cos^2 x)$ for $x \in [0, \pi/2)$.Let $t = \sin^2 x$. Then $\cos^2 x = 1 - t$. Since $x \in [0, \pi/2)$,$t \in [0, 1)$.Thus,$P(t) = P(1 - t)$ for all $t \in [0, 1)$. Since $P$ is a polynomial,$P(t) = P(1 - t)$ for all $t \in \mathbb{R}$.Let $u = t - 1/2$. Then $t = u + 1/2$ and $1 - t = 1/2 - u$.The condition becomes $P(u + 1/2) = P(1/2 - u)$.Let $Q(u) = P(u + 1/2)$. Then $Q(u) = Q(-u)$,which means $Q(u)$ is an even function of $u$.Since $Q(u)$ is an even polynomial,it can be expressed as a polynomial in $u^2$.Substituting $u = x - 1/2$,we get $P(x) = Q(x - 1/2)$,which is a polynomial in $(x - 1/2)^2$,or equivalently,a polynomial in $(2x - 1)^2$. This confirms statement $II$.Since $Q(u)$ is an even polynomial,its degree must be even. Thus,$P(x)$ must be a polynomial of even degree. This confirms statement $III$.Statement $I$ is false because $P(x) = P(1-x)$ does not imply $P(x) = P(-x)$ (e.g.,$P(x) = x(1-x)$ satisfies the condition but is not even).Therefore,only $II$ and $III$ are true.

Let $P(x)$ be a polynomial with real coefficients such that $P(\sin^2 x) = P(\cos^2 x)$ for all $x \in [0, \pi/2)$. Consider the following statements:
$I.$ $P(x)$ is an even function.
$II.$ $P(x)$ can be expressed as a polynomial in $(2x - 1)^2$.
$III.$ $P(x)$ is a polynomial of even degree.
Then,

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Let $P(x)$ be a polynomial with real coefficients such that $P(\sin^2 x) = P(\cos^2 x)$ for all $x \in [0, \pi/2)$. Consider the following statements:$I.$ $P(x)$ is an even function.$II.$ $P(x)$ can be expressed as a polynomial in $(2x - 1)^2$.$III.$ $P(x)$ is a polynomial of even degree.Then,