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If $\mathrm{P}(\mathrm{x})$ be a polynomial with real coefficients such that $\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(\cos ^{2} \mathrm{x}\right)$, for all $\mathrm{x} \in[0, \pi / 2]$. Consider the following statements :
I. $\mathrm{P}(\mathrm{x})$ is an even function.
II. $\mathrm{P}(\mathrm{x})$ can be expressed as a polynomial in $(2 \mathrm{x}-1)^{2}$
I. $\mathrm{P}(\mathrm{x})$ is a polynomial of even degree
Then,
Options:
I. $\mathrm{P}(\mathrm{x})$ is an even function.
II. $\mathrm{P}(\mathrm{x})$ can be expressed as a polynomial in $(2 \mathrm{x}-1)^{2}$
I. $\mathrm{P}(\mathrm{x})$ is a polynomial of even degree
Then,
Solution:
1834 Upvotes
Verified Answer
The correct answer is:
only II and III are true
$$
\begin{array}{l}
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(\cos ^{2} \mathrm{x}\right) \\
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(1-\sin ^{2} \mathrm{x}\right) \\
\mathrm{P}(\mathrm{x})=\mathrm{P}(1-\mathrm{x}) \forall \mathrm{x} \in[0,1]
\end{array}
$$
Differentiable both sides w.r.t. $x$ $P^{\prime}(x)=-P^{\prime}(1-x)$
So $P^{\prime}(x)$ is symmetric about point $x=\frac{1}{2}$
So $\mathrm{P}^{\prime}(\mathrm{x})$ has highest degree odd
$\Rightarrow \mathrm{P}(\mathrm{x})$ has highest degree even
\begin{array}{l}
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(\cos ^{2} \mathrm{x}\right) \\
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(1-\sin ^{2} \mathrm{x}\right) \\
\mathrm{P}(\mathrm{x})=\mathrm{P}(1-\mathrm{x}) \forall \mathrm{x} \in[0,1]
\end{array}
$$
Differentiable both sides w.r.t. $x$ $P^{\prime}(x)=-P^{\prime}(1-x)$
So $P^{\prime}(x)$ is symmetric about point $x=\frac{1}{2}$
So $\mathrm{P}^{\prime}(\mathrm{x})$ has highest degree odd
$\Rightarrow \mathrm{P}(\mathrm{x})$ has highest degree even
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