17. $\int _ { 2 } ^ { 3 } \frac { 3 } { ( x - 1 ) ( x + 2 ) } d x =$
(A) $- \frac { 33 } { 20 }$
(B) $- \frac { 9 } { 20 }$
(C) $\quad \ln \left( \frac { 5 } { 2 } \right)$
(D) $\ln \left( \frac { 8 } { 5 } \right)$
(E) $\ln \left( \frac { 2 } { 5 } \right)$

$\int _ { 0 } ^ { 1 } \frac { 5 x + 8 } { x ^ { 2 } + 3 x + 2 } d x$ is
(A) $\ln ( 8 )$
(B) $\ln \left( \frac { 27 } { 2 } \right)$
(C) $\ln ( 18 )$
(D) $\ln ( 288 )$
(E) divergent

Let $p(x)$ be a polynomial of degree strictly less than 100 and such that it does not have $x^{3} - x$ as a factor. If $$\frac{d^{100}}{dx^{100}} \left( \frac{p(x)}{x^{3} - x} \right) = \frac{f(x)}{g(x)}$$ for some polynomials $f(x)$ and $g(x)$ then find the smallest possible degree of $f(x)$. Here $\frac{d^{100}}{dx^{100}}$ means taking the 100th derivative.

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$. We define the rational fraction $F ( X ) := \dfrac { X ^ { q - 1 } } { 1 + X ^ { p } }$.
Show that there exist constants $\left( a _ { 0 } , b _ { 0 } , \ldots , b _ { \lfloor p / 2 \rfloor - 1 } \right) \in \mathbf { C } ^ { \lfloor p / 2 \rfloor + 1 }$ such that $$F ( X ) = \frac { 1 - ( - 1 ) ^ { p } } { 2 } \cdot \frac { a _ { 0 } } { X + 1 } + \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \left( \frac { b _ { k } } { X - \omega _ { p , k } } + \frac { \overline { b _ { k } } } { X - \overline { \omega _ { p , k } } } \right) ,$$ where the $\omega _ { p , k }$ are constants which one will specify and $F ( X )$ is the rational fraction defined at the beginning of this part.
In the case where $p$ is even, we set $a _ { 0 } = 0$.

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$. We define the rational fraction $F ( X ) := \dfrac { X ^ { q - 1 } } { 1 + X ^ { p } }$, and set $\theta _ { k } := ( 2 k + 1 ) \dfrac { \pi } { p }$.
Calculate $a _ { 0 }$ in the case where $p$ is odd, then show that, for all integers $k \in \llbracket 0 , \lfloor p / 2 \rfloor - 1 \rrbracket$, $b _ { k }$ can be written in the form $$b _ { k } = - \frac { 1 } { p } e ^ { i q \theta _ { k } }$$

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$. We define the rational fraction $F ( X ) := \dfrac { X ^ { q - 1 } } { 1 + X ^ { p } }$, and set $\theta _ { k } := ( 2 k + 1 ) \dfrac { \pi } { p }$.
Deduce the partial fraction decomposition of $F ( X )$ in $\mathbf { R } ( X )$: $$F ( X ) = \frac { 1 - ( - 1 ) ^ { p } } { 2 } \cdot \frac { ( - 1 ) ^ { q - 1 } } { p } \cdot \frac { 1 } { X + 1 } - \frac { 2 } { p } \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } F _ { k } ( X )$$ where, for all $0 \leq k \leq \lfloor p / 2 \rfloor - 1$, $$F _ { k } ( X ) := \frac { \cos \left( q \theta _ { k } \right) X - \cos \left( ( q - 1 ) \theta _ { k } \right) } { X ^ { 2 } - 2 \cos \left( \theta _ { k } \right) X + 1 }$$

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

For $k \in N$, let $\frac { 1 } { \alpha ( \alpha + 1 ) ( \alpha + 2 ) \ldots ( \alpha + 20 ) } = \sum _ { K = 0 } ^ { 20 } \frac { A _ { k } } { \alpha + k }$, where $\alpha > 0$. Then the value of $100 \left( \frac { A _ { 14 } + A _ { 15 } } { A _ { 13 } } \right) ^ { 2 }$ is equal to $\underline{\hspace{1cm}}$.

The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to
(1) $\frac{7}{87}$
(2) $\frac{7}{29}$
(3) $\frac{14}{87}$
(4) $\frac{21}{29}$

If $\frac { 1 } { ( 20 - a ) ( 40 - a ) } + \frac { 1 } { ( 40 - a ) ( 60 - a ) } + \ldots\ldots + \frac { 1 } { ( 180 - a ) ( 200 - a ) } = \frac { 1 } { 256 }$, then the maximum value of $a$ is
(1) 198
(2) 202
(3) 212
(4) 218

If $\left( \frac { 1 } { \alpha + 1 } + \frac { 1 } { \alpha + 2 } + \ldots \ldots + \frac { 1 } { \alpha + 1012 } \right) - \left( \frac { 1 } { 2 \cdot 1 } + \frac { 1 } { 4 \cdot 3 } + \frac { 1 } { 6 \cdot 5 } + \ldots . + \frac { 1 } { 2024 \cdot 2023 } \right) = \frac { 1 } { 2024 }$, then $\alpha$ is equal to $\_\_\_\_$

Q82. If $\left( \frac { 1 } { \alpha + 1 } + \frac { 1 } { \alpha + 2 } + \ldots \ldots + \frac { 1 } { \alpha + 1012 } \right) - \left( \frac { 1 } { 2 \cdot 1 } + \frac { 1 } { 4 \cdot 3 } + \frac { 1 } { 6 \cdot 5 } + \ldots . + \frac { 1 } { 2024 \cdot 2023 } \right) = \frac { 1 } { 2024 }$, then $\alpha$ is equal to $\_\_\_\_$

If $\sum _ { r = 1 } ^ { 25 } \frac { r } { r ^ { 4 } + r ^ { 2 } + 1 } = \frac { p } { q ^ { \prime } }$, where $p$ and $q$ are coprime positive integer, then $\mathrm { p } + \mathrm { q }$ is equal to (A) 841 (B) 984 (C) 976 (D) 890

2. (a) You are given that
$$\frac { 1 } { ( x - 1 ) ( x - 2 ) } = \frac { A } { x - 2 } + \frac { B } { x - 1 }$$
where $A$ and $B$ are constants. Find the values of $A$ and $B$.
(b) Simplify
$$\frac { 1 } { ( x - 1 ) ^ { n + 1 } ( x - 2 ) } - \frac { 1 } { ( x - 1 ) ^ { n } ( x - 2 ) }$$
(c) You are given that
$$\frac { 1 } { ( x - 1 ) ^ { n } ( x - 2 ) } = \frac { A _ { 0 } } { x - 2 } + \sum _ { i = 1 } ^ { n } \frac { A _ { i } } { ( x - 1 ) ^ { i } }$$
where $A _ { 0 } , A _ { 1 } , A _ { 2 } , \ldots$ are constants. Using your answers to (a) and (b), or otherwise, find the values of these constants.

$$f(x) = \frac{\left(1+x+x^{2}+x^{3}\right)(1-x)^{2}}{1-x-x^{2}+x^{3}}$$
Given this, what is the value of $f(\sqrt{2})$?
A) 1
B) 2
C) 3
D) 4
E) 5

$$\frac { a ^ { 4 } - a ^ { 3 } } { a ^ { 4 } + a ^ { 2 } } \cdot \frac { a ^ { 2 } + 1 } { a ^ { 2 } - a }$$
Which of the following is the simplified form of this expression?
A) $a - 1$
B) a
C) 1
D) $a + 1$
E) $a ^ { 2 } + 1$

$$\frac { 2 ( x - y ) } { x - y - 1 } + \frac { x - y - 1 } { x - y - 2 } = 3$$
Given this, what is the difference $x - y$?
A) $\frac { - 1 } { 2 }$
B) $\frac { - 2 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $\frac { 5 } { 3 }$
E) $\frac { 5 } { 4 }$

$$\begin{aligned} & x = \frac { a - b } { a + b } \\ & y = \frac { b - c } { b + c } \end{aligned}$$
Given that, which of the following is the equivalent of the expression $\frac { 1 + y } { 1 - x }$ in terms of $a , b$ and $c$?
A) $\frac { b - c } { a - b }$
B) $\frac { b + c } { a - b }$
C) $\frac { a - b } { a + c }$
D) $\frac { a - c } { b - c }$
E) $\frac { a + b } { b + c }$

$\mathbf { a } , \mathbf { b } , \mathbf { c }$ are non-zero real numbers and $\mathbf { a } + \mathbf { b } + \mathbf { c } = \mathbf { a b }$. Given this,
$$\frac { a b + a c + b c + c ^ { 2 } } { a b c }$$
Which of the following is this expression equal to?
A) $\frac { a + 1 } { a }$
B) $\frac { b + 1 } { b }$
C) $\frac { c + 1 } { c }$
D) $\frac { b } { a }$
E) $\frac { b } { c }$

$$\begin{aligned} & a = \frac { x } { x - y } \\ & b = \frac { y } { x + y } \end{aligned}$$
Given this, what is the value of the expression $\frac { a + b - 1 } { a \cdot b }$?
A) $- 2$
B) $- 1$
C) $0$
D) $1$
E) $2$

$$a \cdot b = \frac { 3 } { 2 }$$
Given that, what is the value of the expression $\left( a + \frac { 1 } { 2 b } \right) \left( b - \frac { 1 } { a } \right)$?
A) $\frac { 1 } { 2 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $\frac { 3 } { 4 }$
E) $\frac { 2 } { 5 }$

For distinct positive real numbers $x$ and $y$,
$$\left( \frac { x } { y } - \frac { y } { x } \right) \cdot \frac { x y } { 4 } = ( x - y ) ^ { 2 }$$
Given that, what is the ratio $\frac { x } { y }$?
A) $\frac { 1 } { 2 }$
B) $\frac { 3 } { 2 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 4 } { 3 }$
E) $\frac { 5 } { 3 }$