Question

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

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

Answer 1

Answer: (C)

Only II and III are true

$P\left({\sin }^{2}x\right)=P\left({\cos }^{2}x\right)$

$⟹P\left({\cos }^2\right(\frac{\pi }{2}-x\left)\right)=P\left({\cos }^{2}x\right)\text{since}\sin x=\cos (\frac{\pi }{2}-x)$

for all $xϵ[0,\pi /2]$.

$⟹P(\frac{\pi }{2}-x)=P\left(x\right)$

Statement (I) is false because for a function to be even it must have the property$P\left(x\right)=P(-x)$

From the above derived equation we can say that our curve is symmetric along the line $x=\frac{\pi }{2}$

$\text{Moreover, if}P\left(x\right)=0\text{for some x=y then}P(\frac{\pi }{2}-y)\text{must also be equal to zero}$

This means that the curve P will always have even number of solution, i.e. even number of roots. Thus its degree must be even. In statement (II), $(2x-1{)}^2$ has even degree.

Thus statement (II) and (III) must be true.