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ISI 2021 Subjective Problem 5 I A Problem from Polynomials

Try this beautiful Subjective Problem from Polynomials appeared in ISI Entrance - 2021.

Problem


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
\[
\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}.\]


Key Concepts


Monic Polynomial

Even Polynomial

Degree of a Polynomial

Suggested Book | Source | Answer


An Excursion in Mathematics (Chapter - 2.1)

ISI Entrance - 2021 , Subjective problem number - 5

\(1-(9 !)^{2}\)

Try with Hints


Hint 1

\(\bullet \) 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 \(P(x) .\)

Hint 2

  • Use \(P(x)=P(-x)\) to observe \(P(\pm k)=k^{2}\) for every \(k=1,2, \ldots, 9\).
  • Define a new polynomial \(Q(x)\)
    \[
    Q(x)=P(x)-x^{2}
    \]
  • And see constructing \(Q(x)\) is easier . Hence find \(Q(x)\) and eventually \(P(x)\).

Hint 3

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 .)

Hint 4

\(\bullet\) 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 \(0?\)

Conclusion

\(\bullet \) 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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Try this beautiful Subjective Problem from Polynomials appeared in ISI Entrance - 2021.

Problem


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
\[
\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}.\]


Key Concepts


Monic Polynomial

Even Polynomial

Degree of a Polynomial

Suggested Book | Source | Answer


An Excursion in Mathematics (Chapter - 2.1)

ISI Entrance - 2021 , Subjective problem number - 5

\(1-(9 !)^{2}\)

Try with Hints


Hint 1

\(\bullet \) 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 \(P(x) .\)

Hint 2

  • Use \(P(x)=P(-x)\) to observe \(P(\pm k)=k^{2}\) for every \(k=1,2, \ldots, 9\).
  • Define a new polynomial \(Q(x)\)
    \[
    Q(x)=P(x)-x^{2}
    \]
  • And see constructing \(Q(x)\) is easier . Hence find \(Q(x)\) and eventually \(P(x)\).

Hint 3

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 .)

Hint 4

\(\bullet\) 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 \(0?\)

Conclusion

\(\bullet \) 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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