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 $$b _ { k } = - \frac { 1 } { p } e ^ { i q \theta _ { k } }$$
Let $r$ and $s$ be two strictly positive natural integers such that $r > s$. Justify that $$\frac { 1 } { ( r + k + 1 ) ( s + k + 1 ) } = \frac { 1 } { r - s } \left( \frac { 1 } { s + k + 1 } - \frac { 1 } { r + k + 1 } \right) .$$
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.
$$\frac { a ^ { 4 } - a ^ { 3 } } { a ^ { 4 } + a ^ { 2 } } \cdot \frac { a ^ { 2 } + 1 } { a ^ { 2 } - a }$$ Which of the following is the simplified form of this expression? A) $a - 1$ B) a C) 1 D) $a + 1$ E) $a ^ { 2 } + 1$
$$\begin{aligned} & x = \frac { a - b } { a + b } \\ & y = \frac { b - c } { b + c } \end{aligned}$$ 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) $2$
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) $\frac { 5 } { 3 }$