As shown in the figure, there is an isosceles triangle ABC with AB as one side of length 4, $\overline { \mathrm { AC } } = \overline { \mathrm { BC } }$, and $\angle \mathrm { ACB } = \theta$. On the extension of segment AB, a point D is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AD } }$, and a point P is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AP } }$ and $\angle \mathrm { PAB } = 2 \theta$. Let $S ( \theta )$ be the area of triangle BDP. Find the value of $\lim _ { \theta \rightarrow + 0 } ( \theta \times S ( \theta ) )$. (Here, $0 < \theta < \frac { \pi } { 6 }$) [4 points]

As shown in the figure, in a right triangle ABC with $\overline { \mathrm { AB } } = 2$ and $\angle \mathrm {~B} = \frac { \pi } { 2 }$, let D and E be the points where the circle with center A and radius 1 meets the two segments $\mathrm { AB }$ and $\mathrm { AC }$ respectively. Let F be the trisection point of arc DE closer to point D, and let G be the point where line AF meets segment BC. Let $\angle \mathrm { BAG } = \theta$. Let $f ( \theta )$ be the area of the common part of the interior of triangle ABG and the exterior of sector ADF, and let $g ( \theta )$ be the area of sector AFE. Find the value of $40 \times \lim _ { \theta \rightarrow 0 + } \frac { f ( \theta ) } { g ( \theta ) }$. (where $0 < \theta < \frac { \pi } { 6 }$) [3 points]

We assume in this part that all roots of $p$ are stable and have multiplicity 1 and we denote by $h = Xp'$ (where $p'$ is the derivative polynomial of $p$) and $h_0$ the reciprocal polynomial of $h$. We recall that, according to question 3, there exists a real number $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$.
For every real number $r > 0$, we denote by $F(r) = J(p(rX))$.
Deduce, using the results of question 4, that $$\frac{n}{2(r-1)} F(r) \underset{r \rightarrow 1}{=} J(h) + o(1)$$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be any twice differentiable function such that its second derivative is continuous and $$\frac { d f ( x ) } { d x } \neq 0 \text { for all } x \neq 0$$ If $$\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x ^ { 2 } } = \pi$$ then
(A) for all $x \neq 0 , \quad f ( x ) > f ( 0 )$.
(B) for all $x \neq 0 , \quad f ( x ) < f ( 0 )$.
(C) for all $x , \frac { d ^ { 2 } f ( x ) } { d x ^ { 2 } } > 0$.
(D) for all $x , \frac { d ^ { 2 } f ( x ) } { d x ^ { 2 } } < 0$.

The value of $\lim _ { x \rightarrow 0 ^ { + } } \frac { \cos ^ { - 1 } \left( x - [ x ] ^ { 2 } \right) \cdot \sin ^ { - 1 } \left( x - [ x ] ^ { 2 } \right) } { x - x ^ { 3 } }$, where $[ x ]$ denotes the greatest integer $\leq x$ is:
(1) $\pi$
(2) 0
(3) $\frac { \pi } { 4 }$
(4) $\frac { \pi } { 2 }$

If $\alpha > \beta > 0$ are the roots of the equation $a x ^ { 2 } + b x + 1 = 0$, and $\lim _ { x \rightarrow \frac { 1 } { \alpha } } \left( \frac { 1 - \cos \left( x ^ { 2 } + b x + a \right) } { 2 ( 1 - \alpha x ) ^ { 2 } } \right) ^ { \frac { 1 } { 2 } } = \frac { 1 } { k } \left( \frac { 1 } { \beta } - \frac { 1 } { \alpha } \right)$, then $k$ is equal to
(1) $2 \beta$
(2) $\alpha$
(3) $2 \alpha$
(4) $\beta$

$$\lim _ { x \rightarrow 1 ^ { + } } ( x - 1 ) \cdot \ln \left( x ^ { 2 } - 1 \right)$$
What is the value of this limit?
A) $\frac { -1 } { 2 }$
B) $-2$
C) 0
D) 1
E) 4

For a function f defined on the set of real numbers
$$\begin{aligned} & \lim _ { x \rightarrow 3 ^ { + } } f ( x ) = 1 \\ & \lim _ { x \rightarrow 3 ^ { - } } f ( x ) = 2 \end{aligned}$$
Given this, what is the value of the limit $\lim _ { x \rightarrow 2 ^ { + } } \frac { f ( 2 x - 1 ) + f ( 5 - x ) } { f \left( x ^ { 2 } - 1 \right) }$?
A) $\frac { -1 } { 2 }$
B) $\frac { 3 } { 2 }$
C) 1
D) 3
E) 4