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Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:
$I.$ $P(x)$ is an even function.
$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$
$III.$ $P(x)$ is a polynomial of even degree.
Then,
all are false
only $I$ and $II$ are true
only $II$ and $III$ are true
all are true
Solution
(c)
We have,
$P\left(\sin ^2 x\right) =P\left(\cos ^2 x\right), x \in\left[0, \frac{\pi}{2}\right)$
$P\left(\sin ^2 x\right) =P\left(1-\sin ^2 x\right)$
$P(x) =P(1-x), x \in[0,1]$
$P^{\prime}(x) =-P^{\prime}(1-x)$
So, $P^{\prime}(x)$ is symmetric about line $x=\frac{1}{2}$
So, $P^{\prime}(x)$ has highest degree is odd. $\Rightarrow P(x)$ has highest degree is even.
Hence, option $(c)$ is correct.
