9. Let $x$ be a variable that takes real values, and let $f ( x ) , g ( x ) , h ( x )$ be three polynomials with real-valued coefficients. Further, let $f ( x )$ be a polynomial of degree $1 , g ( x )$ a polynomial of degree 2 , and $h ( x )$ a polynomial of degree 3 . Which of the following statements is/are always true for any three polynomials of these types?
(a) The graphs of $f ( x )$ and $g ( x )$ intersect at one or more points.
(b) The graphs of $f ( x )$ and $h ( x )$ intersect at one or more points.
(c) The graphs of $g ( x )$ and $h ( x )$ intersect at one or more points.

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

A function $f(x) = x^{3} + ax^{2} + bx + 4$ satisfies the following condition for two integers $a$ and $b$. What is the maximum value of $f(1)$? [4 points] For all real numbers $\alpha$, the limit $\lim_{x \rightarrow \alpha} \frac{f(2x+1)}{f(x)}$ exists.

For every integer $n \in \mathbb{N}$, we set $F_n(x) = \cos(n \arccos x)$.
Deduce from the above that $F_n$ extends to $\mathbb{R}$ as a unique polynomial function, whose degree and leading coefficient should be specified.

Let $A \in M_p(\mathbb{C})$ and $n \in \mathbb{N}^*$. We denote $$P_n(X) = \left(1 + \frac{X}{n}\right)^n \in \mathbb{C}[X]$$ and $\chi_A$ the characteristic polynomial of $A$ defined by $$\chi_A(X) = \det\left(A - XI_p\right)$$
Show that there exists a unique pair $\left(Q_n, R_n\right) \in \mathbb{C}[X] \times \mathbb{C}_{p-1}[X]$ such that $$P_n = Q_n \chi_A + R_n$$

We choose an even polynomial in $B_{N}$ (see question 2(c)), and we denote it $R_{N}$.
Show that there exist non-negative integers $r, s, t \geqslant 0$, real numbers $c_{1}, \ldots, c_{r}$ different from $\pm 1$, non-zero reals $\rho_{1}, \ldots, \rho_{s}$ and complex numbers $w_{1}, \ldots, w_{t}$ that are neither real nor purely imaginary, such that $$R_{N}(X) = \prod_{j=1}^{r} \frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}} \prod_{k=1}^{s} \frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}} \prod_{\ell=1}^{t} \frac{X^{2} - w_{\ell}^{2}}{1 - w_{\ell}^{2}} \cdot \frac{X^{2} - \overline{w_{\ell}}^{2}}{1 - \overline{w_{\ell}}^{2}}.$$

We choose an even polynomial in $B_{N}$, denoted $R_{N}$, which has the factorisation $$R_{N}(X) = \prod_{j=1}^{r} \frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}} \prod_{k=1}^{s} \frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}} \prod_{\ell=1}^{t} \frac{X^{2} - w_{\ell}^{2}}{1 - w_{\ell}^{2}} \cdot \frac{X^{2} - \overline{w_{\ell}}^{2}}{1 - \overline{w_{\ell}}^{2}}.$$
We decide to replace all $\rho_{k}$ by zeros. We thus replace the corresponding factors of $R_{N}$, $$\frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}},$$ by factors $X^{2}$. We thus obtain a new polynomial $S_{N}$ of the same degree as $R_{N}$.
Show that $0 \leqslant S_{N}(x) \leqslant R_{N}(x)$ for all $x \in [-1,1]$, then that $S_{N} \in B_{N}$.

We choose an even polynomial in $B_{N}$, denoted $R_{N}$, which has the factorisation $$R_{N}(X) = \prod_{j=1}^{r} \frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}} \prod_{k=1}^{s} \frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}} \prod_{\ell=1}^{t} \frac{X^{2} - w_{\ell}^{2}}{1 - w_{\ell}^{2}} \cdot \frac{X^{2} - \overline{w_{\ell}}^{2}}{1 - \overline{w_{\ell}}^{2}}.$$ After replacing all $\rho_k$ by zeros we obtained $S_N$.
Similarly, in the list of $c_{j}$, we decide to replace those that do not belong to $[-1,1]$ by zeros. We thus replace the corresponding factors of $S_{N}$, $$\frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}}$$ by factors $X^{2}$. We thus obtain a new polynomial $T_{N}$.
Show that $0 \leqslant T_{N}(x) \leqslant S_{N}(x)$ for all $x \in [-1,1]$, then that $T_{N} \in B_{N}$.

We choose an even polynomial in $B_{N}$, denoted $R_{N}$, which has the factorisation $$R_{N}(X) = \prod_{j=1}^{r} \frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}} \prod_{k=1}^{s} \frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}} \prod_{\ell=1}^{t} \frac{X^{2} - w_{\ell}^{2}}{1 - w_{\ell}^{2}} \cdot \frac{X^{2} - \overline{w_{\ell}}^{2}}{1 - \overline{w_{\ell}}^{2}}.$$
Using the results of questions 8, 9, and 10, conclude that $R_{N}$ has all its roots in the interval $[-1,1]$.

We denote by $n$ the integer part of $\frac{N}{2}$. We continue the study of the polynomial $R_{N}$ (the even polynomial in $B_N$ minimising $L$).
Show that $\deg R_{N} = 2n$.

We denote by $n$ the integer part of $\frac{N}{2}$. We continue the study of the polynomial $R_{N}$ (the even polynomial in $B_N$ minimising $L$, with all roots in $[-1,1]$).
Show that $R_{N}$ is the square of a polynomial: $R_{N}(X) = U_{N}(X)^{2}$ where $U_{N}(1) = 1$ and $U_{N}(-1) = \pm 1$. What can we say about the parity of $U_{N}$?

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 $n \in \mathbb{N}^*$ and $$T _ { n } ( X ) = \sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } ( - 1 ) ^ { p } \binom { n } { 2 p } X ^ { n - 2 p } \left( 1 - X ^ { 2 } \right) ^ { p }.$$ Show that $T _ { n }$ is a polynomial of degree $n$. Explicitly state the leading coefficient of $T _ { n }$.

108- $P(x)$ is a second-degree polynomial with natural coefficients. If $P(x)$ is divided by $P'(x)$ (the derivative of $P(x)$), the quotient and remainder are $\dfrac{1}{2}x + 1$ and $2-$, respectively. What is the minimum value of the sum of the coefficients of $P(x)$?
(1) $4$ (2) $6$ (3) $7$ (4) $9$
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Let $P ( X ) = X ^ { 4 } + a _ { 3 } X ^ { 3 } + a _ { 2 } X ^ { 2 } + a _ { 1 } X + a _ { 0 }$ be a polynomial in $X$ with real coefficients. Assume that
$$P ( 0 ) = 1 , P ( 1 ) = 2 , P ( 2 ) = 3 , \text { and } P ( 3 ) = 4 .$$
Then, the value of $P ( 4 )$ is
(A) 5
(B) 24
(C) 29
(D) not determinable from the given data.

Define a polynomial $f ( x )$ by $$f ( x ) = \left| \begin{array} { l l l } 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{array} \right|$$ for all $x \in \mathbb { R }$, where the right hand side above is a determinant. Then the roots of $f ( x )$ are of the form
(A) $\alpha , \beta \pm i \gamma$ where $\alpha , \beta , \gamma \in \mathbb { R } , \gamma \neq 0$ and $i$ is a square root of $- 1$.
(B) $\alpha , \alpha , \beta$ where $\alpha , \beta \in \mathbb { R }$ are distinct.
(C) $\alpha , \beta , \gamma$ where $\alpha , \beta , \gamma \in \mathbb { R }$ are all distinct.
(D) $\alpha , \alpha , \alpha$ for some $\alpha \in \mathbb { R }$.

1. With reference to the example, determine the expression of the function $y = f(x)$ and the equation of the curve $\Lambda$, so as to be able to perform a test and verify the operation of the machine.
You are asked to construct a tile with a more elaborate design that, in addition to respecting conditions a), b) and c) described above, has $f'(0) = 0$ and the area of the coloured part equal to 55\% of the area of the entire tile. For this purpose, consider polynomial functions of second and third degree.

2. After verifying that it is not possible to achieve what is required using a second-degree polynomial function, determine the coefficients $a, b, c, d \in \mathbb{R}$ of the function $f(x)$ polynomial of third degree that satisfies the stated conditions. Finally, represent the resulting tile in a Cartesian plane.
Two different types of design are proposed to a customer, derived respectively from the functions $a_n(x) = 1 - x^n$ and $b_n(x) = (1-x)^n$, considered for $x \in [0,1]$, with $n$ a positive integer.

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

$f ( x ) = \left| \begin{array} { c c c } a & - 1 & 0 \\ a x & a & - 1 \\ a x ^ { 2 } & a x & a \end{array} \right| , a \in R$. Then the sum of the squares of all the values of $a$ for $2 f ^ { \prime } ( 10 ) - f ^ { \prime } ( 5 ) + 100 = 0$ is
(1) 117
(2) 106
(3) 125
(4) 136

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 $\mathbf{S} + \mathbf{T}\sqrt{\mathbf{U}}$.

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 $a = - \mathbf { K L }$.

2. For ALL APPLICANTS.

Suppose that the equation
$$x ^ { 4 } + A x ^ { 2 } + B = \left( x ^ { 2 } + a x + b \right) \left( x ^ { 2 } - a x + b \right)$$
holds for all values of $x$.
(i) Find $A$ and $B$ in terms of $a$ and $b$.
(ii) Use this information to find a factorization of the expression
$$x ^ { 4 } - 20 x ^ { 2 } + 16$$
as a product of two quadratics in $x$.
(iii) Show that the four solutions of the equation
$$x ^ { 4 } - 20 x ^ { 2 } + 16 = 0$$
can be written as $\pm \sqrt { 7 } \pm \sqrt { 3 }$.

For each part of the question on Pages 3 and 4, you will be given four possible answers just one of which is correct. Indicate for each part A-J which answer (a), (b), (c), or (d) you think is correct with a tick $( \checkmark )$ in the corresponding column in the table below.
A. The substitution $x = y + t$ transforms the equation $x ^ { 3 } + a x ^ { 2 } + b x + c = 0$ into an equation of the form $y ^ { 3 } + p y + q = 0$ when\ (a) $t = \frac { a } { 3 }$\ (b) $t = - \frac { a } { 3 }$\ (c) $t = a$\ (d) $t = - a$.\
B. The faces of a cube are coloured red or blue. Exactly three are red and three are blue. The number of distinguishable cubes that can be produced (allowing the cube to be turned around) is\ (a) 2\ (b) 4\ (c) 6\ (d) 20 .\
C. The shortest distance from the origin to the line $3 x + 4 y = 25$ is\ (a) 3\ (b) 4\ (c) 5\ (d) 6 .\
D. The numbers 10, 11 and -12 are solutions of the cubic equation\ (a) $x ^ { 3 } - 11 x ^ { 2 } - 122 x + 1320 = 0$\ (b) $x ^ { 3 } - 9 x ^ { 2 } + 122 x - 1320 = 0$\ (c) $x ^ { 3 } - 9 x ^ { 2 } - 142 x + 1320 = 0$\ (d) $x ^ { 3 } + 9 x ^ { 2 } - 58 x - 1320 = 0$.\
E. The maximum gradient of the curve $y = x ^ { 4 } - 4 x ^ { 3 } + 4 x ^ { 2 } + 2$ in the range $0 \leq x \leq 2 \frac { 1 } { 5 }$ occur when\ (a) $x = 0$\ (b) $x = 1 - \frac { 1 } { \sqrt { 3 } }$\ (c) $x = 1 + \frac { 1 } { \sqrt { 3 } }$\ (d) $x = 2 \frac { 1 } { 5 }$.\
F. The expression $x ^ { 2 } y + x y ^ { 2 } + y ^ { 2 } z + y z ^ { 2 } + z ^ { 2 } x + z x ^ { 2 } - x ^ { 3 } - y ^ { 3 } - z ^ { 3 } - 2 x y z$ factorises as\ (a) $( x + y + z ) ( x - y + z ) ( - x + y - z )$\ (b) $( x + y - z ) ( x - y - z ) ( - x + y + z )$\ (c) $( x + y - z ) ( x - y + z ) ( - x + y + z )$\ (d) $( x - y - z ) ( - x - y + z ) ( - x + y - z )$.\
G. The derivative of $x e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$ is\ (a) $- \frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (b) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (c) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) + 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (d) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$.\
H. You are told that the infinite series $1 + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 4 ^ { 2 } } + \ldots$ and $1 + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 5 ^ { 2 } } + \frac { 1 } { 7 ^ { 2 } } + \ldots$ have sums $\frac { \pi ^ { 2 } } { 6 }$ and $\frac { \pi ^ { 2 } } { 8 }$ respectively. The infinite series $1 - \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } - \frac { 1 } { 4 ^ { 2 } } + \ldots + ( - 1 ) ^ { n - 1 } \frac { 1 } { n ^ { 2 } } + \ldots$ has sum equal to\ (a) $\frac { \pi ^ { 2 } } { 9 }$\ (b) $\frac { \pi ^ { 2 } } { 10 }$\ (c) $\frac { \pi ^ { 2 } } { 12 }$\ (d) $\frac { \pi ^ { 2 } } { 16 }$.\
I. A grid of size $3 \mathrm {~cm} \times 5 \mathrm {~cm}$ is drawn, ruled at 1 cm intervals. The number of squares that can be drawn using the grid is\ (a) 15\ (b) 18\ (c) 26\ (d) 37 .\
J. A pack of cards consists of 52 different cards. A malicious dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each. The probability that one player has two identical cards is\ (a) $\frac { 3 } { 13 }$\ (b) $\frac { 12 } { 51 }$\ (c) $\frac { 1 } { 4 }$\ (d) $\frac { 13 } { 51 }$