Two polynomial functions $f ( x ) , g ( x )$ with integer constant terms and coefficients satisfy the following conditions. What is the maximum value of $f ( 2 )$? [4 points] (가) $\lim _ { x \rightarrow \infty } \frac { f ( x ) g ( x ) } { x ^ { 3 } } = 2$ (나) $\lim _ { x \rightarrow 0 } \frac { f ( x ) g ( x ) } { x ^ { 2 } } = - 4$
(1) 4
(2) 6
(3) 8
(4) 10
(5) 12

We define the sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} P_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad P_n = [X(X-1)]^n \end{array}\right.$$
We denote by $P_n^{(n)}$ the polynomial derived $n$ times of $P_n$.
Determine the degree of $P_n^{(n)}$ and calculate $P_n^{(n)}(1)$.

The Chebyshev polynomials of the first kind satisfy $T_m \cdot T_n = \frac{1}{2}(T_{n+m} + T_{n-m})$ for $0 \leqslant m \leqslant n$, and $T_m \cdot U_{n-1} = \frac{1}{2}(U_{n+m-1} + U_{n-m-1})$ for $0 \leqslant m < n$.
For $m$ and $n$ natural integers such that $m \leqslant n$, we propose to determine the quotient $Q_{n,m}$ and the remainder $R_{n,m}$ of the Euclidean division of $T_n$ by $T_m$.
a) Suppose $m < n < 3m$. Show that $$Q_{n,m} = 2T_{n-m} \quad \text{and} \quad R_{n,m} = -T_{|n-2m|}$$
b) Determine $Q_{n,m}$ and $R_{n,m}$ when $n$ is of the form $(2p+1)m$ with $p \in \mathbb{N}^*$.
c) Suppose that $m > 0$ and that $n$ is not the product of $m$ by an odd integer. Show that there exists a unique integer $p \geqslant 1$ such that $|n - 2pm| < m$ and that $$Q_{n,m} = 2\left(T_{n-m} - T_{n-3m} + \cdots + (-1)^{p-1} T_{n-(2p-1)m}\right) \quad \text{and} \quad R_{n,m} = (-1)^p T_{|n-2pm|}$$

Using the sequence of polynomials $(P_n)$ defined by $f^{(n)}(x) = \frac{P_n(\sin x)}{(\cos x)^{n+1}}$ for $f(x) = \frac{\sin x + 1}{\cos x}$ on $I = ]-\pi/2, \pi/2[$, justify that, for every integer $n \geqslant 1$, the polynomial $P_n$ is monic, of degree $n$ and that its coefficients are natural integers.

Let $I = [ - 1,1 ]$, $\alpha > 0$, and $$f _ { \alpha } : \begin{array}{ccc} {[-1,1]} & \rightarrow & \mathbb{R} \\ x & \mapsto & \dfrac{1}{\alpha^2 + x^2} \end{array}.$$ For $n \in \mathbb{N}^*$, let $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, and let $R_n \in \mathbb{R}_{2n-1}[X]$ be the interpolating polynomial of $f_\alpha$ at the $2n$ real numbers $\{\pm a_{k,n} \mid k \in \llbracket 0, n-1 \rrbracket\}$. Set $Q_n(X) = 1 - (X^2 + \alpha^2) R_n(X)$. Show that $R_n$ is an even polynomial and determine $Q_n(\alpha \mathrm{i})$.

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $Q_n(X) = 1 - (X^2 + \alpha^2)R_n(X)$. Show that there exists $\lambda_n \in \mathbb{R}$ such that $$\forall x \in [ - 1,1 ] , \quad Q _ { n } ( x ) = \lambda _ { n } \prod _ { k = 0 } ^ { n - 1 } \left( x ^ { 2 } - a _ { k , n } ^ { 2 } \right).$$

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $Q_n(X) = \lambda_n \prod_{k=0}^{n-1}(x^2 - a_{k,n}^2)$. Deduce that, for all $x \in [ - 1,1 ]$, $$f _ { \alpha } ( x ) - R _ { n } ( x ) = \frac { ( - 1 ) ^ { n } } { x ^ { 2 } + \alpha ^ { 2 } } \prod _ { k = 0 } ^ { n - 1 } \frac { x ^ { 2 } - a _ { k , n } ^ { 2 } } { \alpha ^ { 2 } + a _ { k , n } ^ { 2 } }.$$

To every $p \in \mathbb{R}[X]$, we associate the function $J(p) = Jp$ from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad J(p)(x) = Jp(x) = \int_x^{x+1} p(t)\,\mathrm{d}t$$
Show that $J$ preserves degree and that $J$ is invertible.

Let $A , B \in \mathscr { D } _ { \rho } \left( \mathbb { R } _ { n } [ X ] \right)$. We assume that $B$ is monic of degree $d \leqslant n$.
9a. Show that there exist elements $Q \in \mathscr { D } _ { \rho } \left( \mathbb { R } _ { n - d } [ X ] \right)$ and $R \in \mathscr { D } _ { \rho } \left( \mathbb { R } _ { d - 1 } [ X ] \right)$ uniquely determined such that $A = B Q + R$.
The elements $Q$ and $R$ are called respectively the quotient and the remainder of the Euclidean division of $A$ by $B$.
9b. Let furthermore $r , s \in \mathbb { R } _ { + } ^ { * }$ with $r < \rho$. Show that, if $\left\| B - X ^ { d } \right\| _ { r , s } < s ^ { d }$, then $$\| Q \| _ { r , s } \leqslant \frac { \| A \| _ { r , s } } { s ^ { d } - \left\| B - X ^ { d } \right\| _ { r , s } } \quad \text { and } \quad \| R \| _ { r , s } \leqslant \frac { s ^ { d } \cdot \| A \| _ { r , s } } { s ^ { d } - \left\| B - X ^ { d } \right\| _ { r , s } }$$ (One may start by treating the case where $B = X ^ { d }$.)

We consider $P = f _ { 0 } + f _ { 1 } X + \cdots + f _ { n } X ^ { n } \in \mathscr { D } _ { \rho } \left( \mathbb { R } _ { n } [ X ] \right)$, with $\lambda = 0$, $f _ { 0 } ( 0 ) = \cdots = f _ { d - 1 } ( 0 ) = 0$ and $f _ { d }$ the constant function equal to 1.
Let $F \in \mathscr { D } _ { \rho } \left( \mathbb { R } _ { d } [ X ] \right)$ be monic and such that $F _ { \mid t = 0 } = X ^ { d }$. Let $R$ be the remainder of the Euclidean division of $P$ by $F$. Show that $F + R$ is monic of degree $d$ and that $( F + R ) _ { \mid t = 0 } = X ^ { d }$.

(More difficult question) Let $f$ be the power series expansion of a rational function $P/Q \in \mathbf{Q}(x)$ all of whose poles are rational numbers. Suppose that the antiderivative $\int_0^x f(t)\,dt$ is globally bounded. Show that $\int_0^x f(t)\,dt$ is then the power series expansion of a rational function in $\mathbf{Q}(x)$.

Show the equality: for all $m \geq 0$ and $\mu \geq 0$, $$\left(\frac{d}{dx}\right)^{\mu} \cdot \left(x^m f\right) = \left(x^m \left(\frac{d}{dx}\right)^{\mu} + \sum_{i=1}^{\min(\mu,m)} \frac{m(m-1)\cdots(m-i+1)\,\mu(\mu-1)\cdots(\mu-i+1)}{i!} x^{m-i} \left(\frac{d}{dx}\right)^{\mu-i}\right) \cdot f.$$

We denote by $\lambda_1, \cdots, \lambda_\ell$ the eigenvalues of $A$, with $\lambda_i \neq \lambda_j$ if $i \neq j$. We denote by $m_1 \geqslant 1, \cdots, m_\ell \geqslant 1$ the multiplicities of $\lambda_1, \cdots, \lambda_\ell$ respectively as roots of $\varphi_A$. Thus we have $$\varphi_A(X) = (X - \lambda_1)^{m_1} \cdots (X - \lambda_\ell)^{m_\ell}$$ with $m = m_1 + \cdots + m_\ell$.
Show that the map $$T : P \in \mathbb{C}_{m-1}[X] \mapsto \left(P(\lambda_1), P'(\lambda_1), \cdots, P^{(m_1-1)}(\lambda_1), \cdots, P(\lambda_\ell), P'(\lambda_\ell), \cdots, P^{(m_\ell-1)}(\lambda_\ell)\right) \in \mathbb{C}^m$$ is an isomorphism and deduce that there exists a unique polynomial $Q \in \mathbb{C}_{m-1}[X]$ such that $$\forall i \in \llbracket 1; \ell \rrbracket, \forall k \in \llbracket 0; m_i - 1 \rrbracket, Q^{(k)}(\lambda_i) = U^{(k)}(\lambda_i)$$

For every polynomial $P = P(X) \in \mathbf{C}_{n-1}[X]$ we set $$g(P) = P(X+1) - P(X)$$
Let $P$ be a non-constant polynomial. Express the degree of the polynomial $g(P)$ in terms of the degree of $P$.

For every polynomial $P = P(X) \in \mathbf{C}_{n-1}[X]$ we set $$g(P) = P(X+1) - P(X)$$ Let $P$ be a non-constant polynomial. Express the degree of the polynomial $g(P)$ in terms of the degree of $P$.

Let $r$ and $s$ be two strictly positive natural integers such that $r > s$. Justify that
$$\frac { 1 } { ( r + k + 1 ) ( s + k + 1 ) } = \frac { 1 } { r - s } \left( \frac { 1 } { s + k + 1 } - \frac { 1 } { r + k + 1 } \right) .$$

Let $\mathbb { R }$ denote the set of all real numbers. Let $a _ { i } , b _ { i } \in \mathbb { R }$ for $i \in \{ 1,2,3 \}$.
Define the functions $f : \mathbb { R } \rightarrow \mathbb { R } , g : \mathbb { R } \rightarrow \mathbb { R }$, and $h : \mathbb { R } \rightarrow \mathbb { R }$ by
$$\begin{aligned} & f ( x ) = a _ { 1 } + 10 x + a _ { 2 } x ^ { 2 } + a _ { 3 } x ^ { 3 } + x ^ { 4 } \\ & g ( x ) = b _ { 1 } + 3 x + b _ { 2 } x ^ { 2 } + b _ { 3 } x ^ { 3 } + x ^ { 4 } \\ & h ( x ) = f ( x + 1 ) - g ( x + 2 ) \end{aligned}$$
If $f ( x ) \neq g ( x )$ for every $x \in \mathbb { R }$, then the coefficient of $x ^ { 3 }$ in $h ( x )$ is

(A)	8
(B)	2
(C)	-4
(D)	-6

If $\Delta _ { r } = \left| \begin{array} { c c c } r & 2 r - 1 & 3 r - 2 \\ \frac { n } { 2 } & n - 1 & a \\ \frac { 1 } { 2 } n ( n - 1 ) & ( n - 1 ) ^ { 2 } & \frac { 1 } { 2 } ( n - 1 ) ( 3 n + 4 ) \end{array} \right|$, then the value of $\sum _ { r = 1 } ^ { n - 1 } \Delta _ { r }$
(1) Is independent of both $a$ and $n$
(2) Depends only on $a$
(3) Depends only on $n$
(4) Depends both on $a$ and $n$

Let the numbers $2 , b , c$ be in an A.P. and $A = \begin{pmatrix} 2 & b & c \\ 4 & b^2 & c^2 \end{pmatrix}$. If $\det ( A ) \in [ 2,16 ]$, then $c$ lies in the interval:
(1) $[2,3]$
(2) $[4,6]$
(3) $\left[3, 2 + 2 ^ { \frac { 3 } { 4 } }\right]$
(4) $\left[2 + 2 ^ { \frac { 3 } { 4 } } , 4\right]$

For some $\mathrm{a}, \mathrm{b}$, let $f(x) = \left| \begin{array}{ccc} \mathrm{a} + \frac{\sin x}{x} & 1 & \mathrm{b} \\ \mathrm{a} & 1 + \frac{\sin x}{x} & \mathrm{b} \\ \mathrm{a} & 1 & \mathrm{b} + \frac{\sin x}{x} \end{array} \right|$, $x \neq 0$, $\lim_{x \rightarrow 0} f(x) = \lambda + \mu\mathrm{a} + \nu\mathrm{b}$. Then $(\lambda + \mu + \nu)^{2}$ is equal to:
(1) 16
(2) 25
(3) 9
(4) 36

Q2 Consider
$$E = P^2 - 4Q^2 - 3P + 6Q$$
where $P$ and $Q$ are the integral expressions
$$P = 2x^2 - x + 2, \quad Q = x^2 - 2x + 1.$$
(1) By factorizing the right side of $E$, we obtain
$$E = (P - \mathbf{LL}Q)(P + \mathbf{M}Q - \mathbf{MN}).$$
(2) When we express $E$ in terms of $x$, we have
$$E = \mathbf{O}x(x - \mathbf{P})(\mathbf{Q})(\mathbf{Q} - \mathbf{Q}).$$
(3) If $x = -\frac{1 - \sqrt{5}}{3 - \sqrt{5}}$, then the value of $E$ is $\mathbf{S} + \mathbf{T}\sqrt{\mathbf{U}}$.

Q2 Consider
$$E = P^2 - 4Q^2 - 3P + 6Q$$
where $P$ and $Q$ are the integral expressions
$$P = 2x^2 - x + 2, \quad Q = x^2 - 2x + 1.$$
(1) By factorizing the right side of $E$, we obtain
$$E = (P - \mathbf{LL}Q)(P + \mathbf{M}Q - \mathbf{MN}).$$
(2) When we express $E$ in terms of $x$, we have
$$E = \mathbf{O}x(x - \mathbf{P})(\mathbf{Q})(\mathbf{Q} - \mathbf{Q}).$$
(3) If $x = -\frac{1 - \sqrt{5}}{3 - \sqrt{5}}$, then the value of $E$ is $\mathbf{S} + \mathbf{T}\sqrt{\mathbf{U}}$.

Set $P = 10a^2 + 14ab - 21bc - 15ca$.
(1) Factorizing $P$, we obtain $$P = (\mathbf{A}a + \mathbf{B}b)(\mathbf{C}a - \mathbf{D}c).$$
(2) If $5a = \sqrt{6}$, $14b = \sqrt{2} + \sqrt{3} - \sqrt{6}$ and $15c = \sqrt{12} - \sqrt{8}$, then $$P = \frac{\mathbf{E}}{\mathbf{E} + \mathbf{F}} \frac{\mathbf{G}}{\mathbf{H}}$$ and hence the greatest integer less than $P$ is $\mathbf{I}$.

We are to find the value of $a$ such that $15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a$ can be factorized as the product of linear expressions in $x$ and $y$.
First of all, the first three terms of the above expression form a quadratic expression in $x$ and $y$ that can be factorized as
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } = ( \mathbf { A } x - \mathbf { B } y ) ( \mathbf { C } x + \mathbf { D } y ) .$$
Hence, when we set
$$\begin{aligned} & 15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a \\ & \quad = ( \mathbf { A } x - \mathbf { B } y + b ) ( \mathbf { C } x + \mathbf { D } y + c ) , \end{aligned}$$
the right-hand side of equation (1) can be expanded into
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } + ( \mathbf { E } b + \mathbf { F } c ) x + ( \mathbf { G } b - \mathbf { H } c ) y + b c .$$
When we compare the coefficients of this expression with the coefficients of the left-hand side of equation (1), we have
$$b = \mathbf { I } , \quad c = - \mathbf { J } ,$$
and hence $a = - \mathbf { K L }$.

Let $n$ be a natural number and $a$ be a real number, where $a \neq 0$. Suppose that the integral expression $x ^ { n } + y ^ { n } + z ^ { n } + a ( x y + y z + z x )$ can be expressed as the product of $x + y + z$ and an integral expression $P$ in $x$, $y$ and $z$, i.e.
$$x ^ { n } + y ^ { n } + z ^ { n } + a ( x y + y z + z x ) = ( x + y + z ) P .$$
We are to find the values of $n$ and $a$.
The equality (1) holds for all $x$, $y$ and $z$. So, consider for example, two triples of $x$, $y$ and $z$ that satisfy $x + y + z = 0$:
$$x = y = 1 , \quad z = - \mathbf { A }$$
and
$$x = y = - \frac { \mathbf { B } } { \mathbf { C } } , \quad z = 1 .$$
By substituting each triple in (1), we obtain the following two equations:
$$\left( - \mathbf { A } \right) ^ { n } = \mathbf { D } \text{ a} - \frac{\mathbf{E}}{\mathbf{F}}$$
$$\left( - \frac { \mathbf { B } } { \mathbf { C } } \right) ^ { n } = \frac { \mathbf{E} }{ \mathbf{F} } \text{ a} - \frac { \mathbf { H } } { \mathbf { I } } .$$
From these two equations, we get an equation in $a$. Solving this, we obtain $a = \mathbf { K }$ and hence by the first equation that $n = \mathbf { L }$.
Conversely, when $a = \mathbf{K}$ and $n = \mathbf{L}$, there exists a $P$ such that (1) holds, and hence these values of $a$ and $n$ are the solution.