Question

someone please explain with answer

Answer 1

Firstly, we see that setting t=sin2(x) for x∈[0,π/2] then t∈[0,1] and the condition above becomes P(t)=P(1−t)for all t∈[0,1].

False. Let P(x)=(x−12)2.

If we consider the polynomial P(x)−P(1−x) the above condition tells us that it is 0for all x∈[0,1]. Since this polynomial has infinitely many roots it must be that P(x)=P(1−x) for all x. Now let Q be a polynomial such that P(x)=Q(2x−1) (you should think about why we always can do this). Our condition then implies that

Q(x)=P(x+12)=P(1−x2)=Q(−x)

and thus Q is even. Since Qis an even polynomial it must only contain even powers and therefore we can write Q(x)=R(x2)and thus P(x)=R((2x−1)2).

True. This follows from 2