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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,

$P(x)=x^{2}(x^{2}-1)(x^{2}-4)$

$\ldots(x^{2}-81)+x^{2} .$

$\Rightarrow \frac{P(x)}{x^{2}}=(x^{2}-1)(x^{2}-4)$

$\ldots(x^{2}-81)+1 .$

$\Rightarrow \lim _{x \rightarrow 0} \frac{P(x)}{x^{2}}$

$=(-1)(-4) \ldots(-81)+1$

ISI Entrance Program at Cheenta

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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,

$P(x)=x^{2}(x^{2}-1)(x^{2}-4)$

$\ldots(x^{2}-81)+x^{2} .$

$\Rightarrow \frac{P(x)}{x^{2}}=(x^{2}-1)(x^{2}-4)$

$\ldots(x^{2}-81)+1 .$

$\Rightarrow \lim _{x \rightarrow 0} \frac{P(x)}{x^{2}}$

$=(-1)(-4) \ldots(-81)+1$

ISI Entrance Program at Cheenta

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