Try this beautiful Subjective Problem from Polynomials appeared in ISI Entrance - 2021.
Let \(a_{0}, a_{1}, \ldots, a_{19} \in \mathbb{R}\) and
\[
P(x)=x^{20}+\sum_{i=0}^{19} a_{i} x^{i} x \in \mathbb{R}
\]
If \(P(x)=P(-x)\) for all \(x \in \mathbb{R}\) and
\(P(k)=k^{2}\), for all \(k=0,1,2, \ldots, 9\)
then find
\[
\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}.\]
Monic Polynomial
Even Polynomial
Degree of a Polynomial
An Excursion in Mathematics (Chapter - 2.1)
ISI Entrance - 2021 , Subjective problem number - 5
\(1-(9 !)^{2}\)
\(\bullet \) Recall the Fundamental Theorem of Algebra i.e. every polynomial \(P(z)\) of degree \(n\) has \(n\) values \(z_{i}\) (some of them possibly degenerate) for which \(P\left(z_{i}\right)=0\).
\(\bullet \) And apply it to construct the polynomial.
\(\bullet \) Observe \(P(0) =0\) i.e. \(0\) is a root of \(P(x) .\)
To construct \(Q(x)\) use followings:
\(Q(x)\) has 19 roots and those are 0 and \(\pm 1, \pm 2, \ldots, \pm 9\).
As \(P(x)\) is monic and of degree 20 , so \(Q(x)\) is also. Hence all factors of \(Q(x)\) are like \((x+a)\).
Therefore,
\[Q(x)=x(x-1)(x+1)(x-2) \]
\[\ldots (x-9)(x+9) \times(x+c)
\]
(Observe extra \((x+c)\) is multiplied to make the degree of \(Q(x)\) to be 20 .)
\(\bullet\) As
\[Q(x)=x(x-1)(x+1)(x-2) \]
\[\ldots (x-9)(x+9) \times(x+c)\]
\(\bullet \)
\[
P(x)=x(x-1)(x+1)(x-2) \]
\[\ldots (x-9)(x+9) \times(x+c)+x^{2} .\]
\[\Rightarrow P(x)=x^{2}(x^{2}-1)(x^{2}-4) \]
\[\ldots (x^{2}-81)+x^{2}\]
\(\bullet \) Have you noticed the coefficients of all odd exponents of \(x\) in \(P(x) \) are \(0?\)
\(\bullet \) Recall we are given that \(P(x)=P(-x)\) for all \(x \in R\) means that \(P(x)\) is an even function and so all odd degree coefficients are 0 . That is, \(a_{i}=0\) for \(i=1,3,5, \ldots, 17,19\).
\(\bullet \) Therefore,
\[P(x)=x^{2}(x^{2}-1)(x^{2}-4) \]
\[\ldots(x^{2}-81)+x^{2} .\]
\[\Rightarrow \frac{P(x)}{x^{2}}=(x^{2}-1)(x^{2}-4)\]
\[ \ldots(x^{2}-81)+1 .\]
\[\Rightarrow \lim _{x \rightarrow 0} \frac{P(x)}{x^{2}}\]
\[=(-1)(-4) \ldots(-81)+1\]
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