Let $f$ be the function defined by $f ( x ) = 2 x e ^ { - x }$ for all real numbers $x$. (a) Write an equation of the horizontal asymptote for the graph of $f$. (b) Find the $x$-coordinate of each critical point of $f$. For each such $x$, determine whether $f ( x )$ relative maximum, a relative minimum, or neither. (c) For what values of $x$ is the graph of $f$ concave down? (d) Using the results found in parts (a), (b), and (c), sketch the graph of $y = f ( x )$ in the $x y$ provided below. Note: The $x y$-plane is provided in the pink test booklet only.

The figure above shows the graph of $f$. If $f ( x ) = \int _ { 2 } ^ { x } g ( t ) d t$, which of the following could be the graph of $y = g ( x )$ ?
(A) [graph A]
(B) [graph B]
(C) [graph C]
(D) [graph D]
(E) [graph E]

The function $h$ is defined on $\mathbb{R}$ by $$h : \mathbb{R} \longrightarrow \mathbb{R}, \quad u \longmapsto u - [u] - 1/2$$ where $[u]$ denotes the integer part of $u$.
Carefully draw the graph of the application $h$ on the interval $[-1, 1]$.

Let $z \in \mathbb{C}$. We denote $C_z$ (respectively $\Omega_z$) the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| = 1$ (respectively $|Z(Z-2z)| < 1$). In this question we assume that $z$ is a real number denoted $a$. The curve $C_a$ in polar coordinates $(\rho, \theta)$ in the frame $\mathcal{R}'$ satisfies $$\left(\rho^2 + a^2\right)^2 - 4a^2 \rho^2 \cos^2\theta = 1$$ Simplify this equation when $a = 1$. Study and sketch the shape of the curve $C_1$.

We denote $\mathcal{D}$ the set of functions from $\mathbb{R}$ to $\mathbb{R}$ of class $\mathcal{C}^{\infty}$ and with compact support. We denote $\varphi$ the function defined by: $$\begin{cases} \varphi(x) = 0 & \text{if } |x| \geqslant 1 \\ \varphi(x) = \exp\left(-\frac{x^2}{1-x^2}\right) & \text{if } |x| < 1 \end{cases}$$
a) Study the variations of $\varphi$. b) Sketch the graph of $\varphi$. c) Show that $\varphi$ is $\mathcal{C}^{\infty}$. d) Show that $\mathcal{D}$ is a vector space over $\mathbb{R}$ not reduced to $\{0\}$.

We recall that the function $\phi$ is defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Sketch the graph of $\phi$. Show that $\phi$ defines by restriction to the intervals $] - 1,0 [$ and $] 0 , + \infty [$ respectively

• a bijection $\left. \phi _ { - } : \right] - 1,0 [ \rightarrow ] 0 , + \infty [$,
• a bijection $\left. \phi _ { + } : \right] 0 , + \infty [ \rightarrow ] 0 , + \infty [$.

We denote $\left. \phi _ { - } ^ { - 1 } : \right] 0 , + \infty [ \rightarrow ] - 1,0 \left[ \right.$ and $\left. \phi _ { + } ^ { - 1 } : \right] 0 , + \infty [ \rightarrow ] 0 , + \infty [$ the inverse bijections.

Study the variations of $g_{\sigma}$. Show that the second derivative of $g_{\sigma}$ vanishes and changes sign at exactly two points. Give the shape of the graph of $g_{\sigma}$ and mark the two points mentioned.

Give the shape of the representative curve of $\zeta(x) = \sum_{n=1}^{+\infty} \frac{1}{n^x}$.

For all $s \in [0,1]$, the function $k_s$ is defined by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ Let $s \in ]0,1[$. Sketch the graph of $k_s$ on $[0,1]$.

Let $f(x) = xe^x$ and let $W$ denote the inverse of the bijection $f|_{[-1,+\infty[}$. Sketch, on the same diagram, the curves $\mathcal { C } _ { f }$ and $\mathcal { C } _ { W }$ representing the functions $f$ and $W$. Specify the tangent lines to the two curves at the point with abscissa 0 as well as the tangent line to $\mathcal { C } _ { W }$ at the point with abscissa $- \mathrm { e } ^ { - 1 }$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Sketch the graph of $f$ by making best use of the previous results.

Draw the graph (on plain paper) of $f(x) = \min\{|x|-1, |x-1|-1, |x-2|-1\}$.

Let $h(x) = \frac{x^4}{(1-x)^4}$ and $g(x) = f(h(x)) = h(x) + 1/h(x)$. Sketch the graph of $g(x)$ and show that $g(x)$ has a root between $0$ and $1$.

3. Boccioni's futurist work ``Unique Forms of Continuity in Space'' from 1913, featured on the 20-cent coin, depicts a man advancing rapidly through space. A part of the profile highlighted in the figure, in an appropriate coordinate system, can be approximated by the function
$$f ( x ) = \left\{ \begin{array} { l r } - 4 x ^ { 2 } - 8 x , & - 1 \leq x \leq 0 \\ 1 + \tan \left( x + \frac { 3 } { 4 } \pi \right) , & 0 < x \leq 2 \end{array} \right.$$
Sketch the graph of $f$, after analyzing its continuity and differentiability on the interval $[ - 1 ; 2 ]$. [Figure]

Let $P$ be a variable point on the parabola $y = 4 x ^ { 2 } + 1$. Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is:
(1) $( 3 x - y ) ^ { 2 } + ( x - 3 y ) + 2 = 0$
(2) $2 ( 3 x - y ) ^ { 2 } + ( x - 3 y ) + 2 = 0$
(3) $( 3 x - y ) ^ { 2 } + 2 ( x - 3 y ) + 2 = 0$
(4) $2 ( x - 3 y ) ^ { 2 } + ( 3 x - y ) + 2 = 0$

The combined equation of the two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ can be written as $(ax + by + c)(a'x + b'y + c') = 0$. The equation of the angle bisectors of the lines represented by the equation $2x^2 + xy - 3y^2 = 0$ is
(1) $3x^2 + 5xy + 2y^2 = 0$
(2) $x^2 - y^2 + 10xy = 0$
(3) $3x^2 + xy - 2y^2 = 0$
(4) $x^2 - y^2 - 10xy = 0$

3. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \\ \text { MATHEMATICS \& COMPUTER SCIENCE } \end{array} \right\}$ ONLY.

Computer Science and Computer Science \& Philosophy applicants should turn to page 14.
Let $g ( x )$ be the function defined by
$$g ( x ) = \begin{cases} ( x - 1 ) ^ { 2 } + 1 & \text { if } x \geqslant 0 \\ 3 - ( x + 1 ) ^ { 2 } & \text { if } x \leqslant 0 \end{cases}$$
and for $x \neq 0$ write $m ( x )$ for the gradient of the chord between ( $0 , g ( 0 )$ ) and ( $x , g ( x )$ ).
(i) Sketch the graph $y = g ( x )$ for $- 3 \leqslant x \leqslant 3$.
(ii) Write down expressions for $m ( x )$ in the two cases $x \geqslant 0$ and $x < 0$.
(iii) Show that $m ( x ) + 2 = x$ for $x > 0$. What is the value of $m ( x ) + 2$ when $x < 0$ ?
(iv) Explain why $g$ has derivative - 2 at 0 .
(v) Suppose that $p < q$ and that $h ( x )$ is a cubic with a maximum at $x = p$ and a minimum at $x = q$. Show that $h ^ { \prime } ( x ) < 0$ whenever $p < x < q$.
Suppose that $c$ and $d$ are real numbers and that there is a cubic $h ( x )$ with a maximum at $x = - 1$ and a minimum at $x = 1$ such that $h ^ { \prime } ( 0 ) = - 3 c$ and $h ( 0 ) = d$.
(vi) Show that $c > 0$ and find a formula for $h ( x )$ in terms of $c$ and $d$ (and $x$ ).
(vii) Show that there are no values of $c$ and $d$ such that the graphs of $y = g ( x )$ and $y = h ( x )$ are the same for $- 3 \leqslant x \leqslant 3$.

3. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \\ \text { MATHEMATICS \& COMPUTER SCIENCE } \end{array} \right\}$ ONLY.

Computer Science and Computer Science \& Philosophy applicants should turn to page 20.
(i) Sketch $y = \left( x ^ { 2 } - 1 \right) ^ { n }$ for $n = 2$ and for $n = 3$ on the same axes, labelling any points that lie on both curves, or that lie on either the $x$-axis or the $y$-axis.
(ii) Without calculating the integral explicitly, explain why there is no positive value of $a$ such that $\int _ { 0 } ^ { a } \left( x ^ { 2 } - 1 \right) ^ { n } \mathrm {~d} x = 0$ if $n$ is even.
If $n > 0$ is odd we will write $n = 2 m - 1$ and define $a _ { m } > 0$ to be the positive real number that satisfies
$$\int _ { 0 } ^ { a _ { m } } \left( x ^ { 2 } - 1 \right) ^ { 2 m - 1 } \mathrm {~d} x = 0$$
if such a number exists.
(iii) Explain why such a number $a _ { m }$ exists for each whole number $m \geqslant 1$.
(iv) Find $a _ { 1 }$.
(v) Prove that $\sqrt { 2 } < a _ { 2 } < \sqrt { 3 }$.
(vi) Without calculating further integrals, find the approximate value of $a _ { m }$ when $m$ is a very large positive whole number. You may use without proof the fact that $\int _ { 0 } ^ { \sqrt { 2 } } \left( x ^ { 2 } - 1 \right) ^ { 2 m - 1 } \mathrm {~d} x < 0$ for any sufficiently large whole number $m$.
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3 (Go to problem page)

(1) For $f(x) = \dfrac{x}{x^2+3}$, noting that $f(-x) = -f(x)$,
$$f'(x) = \frac{(x^2+3) - x \cdot 2x}{(x^2+3)^2} = \frac{-x^2+3}{(x^2+3)^2}$$
From this, the increase/decrease of $f(x)$ for $x \geq 0$ is as shown in the table on the right, and from $\lim_{x \to \infty} f(x) = 0$, the rough sketch of $C: y = f(x)$ is as shown in the figure on the right.

$x$	$0$	$\cdots$	$\sqrt{3}$	$\cdots$
$f'(x)$		$+$	$0$	$-$
$f(x)$	$0$	$\nearrow$	$\dfrac{\sqrt{3}}{6}$	$\searrow$

[Figure: Graph of $C: y=f(x)$ with tangent line $l$ at point A, showing the curve passing through the origin and the tangent line intersecting the curve]
Now, the equation of the tangent line $l: y = g(x)$ at $\mathrm{A}(1,\, f(1))$ is, from $f(1) = \dfrac{1}{4}$, $f'(1) = \dfrac{1}{8}$,
$$y - \frac{1}{4} = \frac{1}{8}(x-1), \quad y = \frac{1}{8}(x+1)$$
Then, the common points of $C$ and $l$ are given by $\dfrac{x}{x^2+3} = \dfrac{1}{8}(x+1)$, so $8x = (x^2+3)(x+1)$, giving
$$x^3 + x^2 - 5x + 3 = 0, \quad (x-1)^2(x+3) = 0$$
Therefore, there is one common point other than A, and its $x$-coordinate is $x = -3$.
(2) From (1), $\alpha = -3$. In this case, for $I = \displaystyle\int_{-3}^{1} \{f(x) - g(x)\}^2\,dx$,
$$I = \int_{-3}^{1} \{f(x)\}^2\,dx - 2\int_{-3}^{1} f(x)g(x)\,dx + \int_{-3}^{1} \{g(x)\}^2\,dx$$
Let $I_1 = \displaystyle\int_{-3}^{1} \{f(x)\}^2\,dx$, $\;I_2 = \displaystyle\int_{-3}^{1} f(x)g(x)\,dx$, $\;I_3 = \displaystyle\int_{-3}^{1} \{g(x)\}^2\,dx$.
First, let $x = \sqrt{3}\tan t$ $\left(-\dfrac{\pi}{2} < t < \dfrac{\pi}{2}\right)$, then $dx = \dfrac{\sqrt{3}}{\cos^2 t}\,dt$, and
$$I_1 = \int_{-3}^{1} \frac{x^2}{(x^2+3)^2}\,dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{3\tan^2 t}{9(\tan^2 t+1)^2} \cdot \frac{\sqrt{3}}{\cos^2 t}\,dt = \frac{1}{\sqrt{3}}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{\tan^2 t}{\tan^2 t+1}\,dt$$
$$= \frac{1}{\sqrt{3}}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} \tan^2 t \cos^2 t\,dt = \frac{1}{\sqrt{3}}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} \sin^2 t\,dt = \frac{1}{2\sqrt{3}}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} (1 - \cos 2t)\,dt$$
$$= \frac{1}{2\sqrt{3}}\left[t - \frac{1}{2}\sin 2t\right]_{-\frac{\pi}{3}}^{\frac{\pi}{6}} = \frac{1}{2\sqrt{3}}\!\left(\frac{\pi}{2} - \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{12}\pi - \frac{1}{4}$$
$$I_2 = \int_{-3}^{1} \frac{x}{x^2+3} \cdot \frac{1}{8}(x+1)\,dx = \frac{1}{8}\int_{-3}^{1}\!\left(1 + \frac{x}{x^2+3} - \frac{3}{x^2+3}\right)dx$$
$$= \frac{1}{8}\left[x + \frac{1}{2}\log(x^2+3)\right]_{-3}^{1} - \frac{3}{8}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{1}{3(\tan^2 t+1)} \cdot \frac{\sqrt{3}}{\cos^2 t}\,dt$$
$$= \frac{1}{8}\!\left(4 + \frac{1}{2}\log\frac{4}{12}\right) - \frac{\sqrt{3}}{8}\int_{-\frac{\pi}{3}}^{\frac{\pi}{6}} dt = \frac{1}{2} - \frac{1}{16}\log 3 - \frac{\sqrt{3}}{16}\pi$$
%% Page 10 $$I_3 = \frac{1}{64}\int_{-3}^{1}(x+1)^2dx = \frac{1}{64}\Big[\frac{1}{3}(x+1)^3\Big]_{-3}^{1} = \frac{1}{64\cdot 3}(8+8) = \frac{1}{12}$$
Therefore, from $I = I_1 - 2I_2 + I_3$, $$I = \left(\frac{\sqrt{3}}{12}\pi - \frac{1}{4}\right) - 2\left(\frac{1}{2} - \frac{1}{16}\log 3 - \frac{\sqrt{3}}{16}\pi\right) + \frac{1}{12} = \frac{5}{24}\sqrt{3}\pi + \frac{1}{8}\log 3 - \frac{7}{6}$$
[Commentary]
While there is also a part asking for the equation of the tangent line, the problem is essentially a standard computation problem involving definite integrals.
%% Page 11 4 (Go to problem page)

(1) For positive odd integers $K, L$ and positive integers $A, B$, suppose $KA = LB$ ①
Here, when the remainder of $K$ divided by 4 equals the remainder of $L$ divided by 4, $$K - L = 4n \quad (n \text{ is an integer}) \hfill \textbf{②}$$
From ②, $K = L + 4n$, and substituting into ①: $$(L + 4n)A = LB, \quad L(A - B) = -4nA$$
Since $-4nA$ is a multiple of 4, $L(A-B)$ is also a multiple of 4, but since $L$ is odd, $A - B$ is a multiple of 4.
That is, the remainder of $A$ divided by 4 equals the remainder of $B$ divided by 4.

(2) For positive integers $a, b\ (a > b)$, let $A = {}_{4a+1}\mathrm{C}_{4b+1}$, $B = {}_{a}\mathrm{C}_{b}$. Then, $$A = \frac{{}_{4a+1}\mathrm{P}_{4b+1}}{(4b+1)!} = \frac{(4a+1)\cdot 4a\cdot(4a-1)(4a-2)\cdots(4a-4b+2)(4a-4b+1)}{(4b+1)\cdot 4b\cdot(4b-1)(4b-2)\cdots 2\cdot 1}$$
Here, $r_0 = 4a(4a-4)\cdots(4a-4b+4)$, $\quad r_1 = (4a+1)(4a-3)\cdots(4a-4b+1)$
$r_2 = (4a-2)(4a-6)\cdots(4a-4b+2)$, $\quad r_3 = (4a-1)(4a-5)\cdots(4a-4b+3)$
Also, $s_0 = 4b(4b-4)\cdots 8\cdot 4$, $\quad s_1 = (4b+1)(4b-3)\cdots 5\cdot 1$
$s_2 = (4b-2)(4b-6)\cdots 6\cdot 2$, $\quad s_3 = (4b-1)(4b-5)\cdots 7\cdot 3$
Then $A = \dfrac{r_0\, r_1\, r_2\, r_3}{s_0\, s_1\, s_2\, s_3}$, and $\dfrac{r_0}{s_0} = \dfrac{a(a-1)\cdots(a-b+1)}{b(b-1)\cdots 2\cdot 1} = \dfrac{{}_{a}\mathrm{P}_{b}}{b!} = {}_{a}\mathrm{C}_{b} = B$
$$\frac{r_2}{s_2} = \frac{(2a-1)(2a-3)\cdots(2a-2b+1)}{(2b-1)(2b-3)\cdots 3\cdot 1}$$
From this, $$A = B \cdot \frac{r_1\, r_3\,(2a-1)(2a-3)\cdots(2a-2b+1)}{s_1\, s_3\,(2b-1)(2b-3)\cdots 3\cdot 1} \hfill \textbf{③}$$
Here, let $K = s_1\, s_3\,(2b-1)(2b-3)\cdots 3\cdot 1$, $\quad L = r_1\, r_3\,(2a-1)(2a-3)\cdots(2a-2b+1)$.
Then $K, L$ are positive odd integers, and from ③, $A = B\cdot\dfrac{L}{K}$, so $KA = LB$ holds.

(3) Hereafter, writing in mod 4: since $4a \equiv 4b \equiv 0$, $$r_1 = (4a+1)(4a-3)\cdots(4a-4b+1) \equiv (4b+1)(4b-3)\cdots(4b-4b+1) = s_1$$ $$r_3 = (4a-1)(4a-5)\cdots(4a-4b+3) \equiv (4b-1)(4b-5)\cdots(4b-4b+3) = s_3$$
Also, since $a - b$ is divisible by 2, the parities of $a$ and $b$ agree, and
(i) When $a, b$ are both even: $\quad 2a \equiv 2b \equiv 0$
(ii) When $a, b$ are both odd: $\quad 2a \equiv 2b \equiv 2$
From (i)(ii), regardless of the parity of $a, b$, $2a \equiv 2b$ holds, and $$(2a-1)(2a-3)\cdots(2a-2b+1) \equiv (2b-1)(2b-3)\cdots(2b-2b+1)$$ $$= (2b-1)(2b-3)\cdots 3\cdot 1$$
From the above, $r_1\, r_3\,(2a-1)(2a-3)\cdots(2a-2b+1) \equiv s_1\, s_3\,(2b-1)(2b-3)\cdots 3\cdot 1$ holds, so $L \equiv K$, that is, the remainder of $K$ divided by 4 equals the remainder of $L$ divided by 4.

$-5-$ \copyright\ 電送数学舎 2021
%% Page 12 From the results of (1)(2), $A = {}_{4a+1}C_{4b+1}$ divided by 4 gives the same remainder as $B = {}_{a}C_{b}$ divided by 4.

1. [(4)] Using the result of (3), $${}_{2021}C_{37} = {}_{4\times505+1}C_{4\times9+1} \equiv {}_{505}C_{9} = {}_{4\times126+1}C_{4\times2+1} \equiv {}_{126}C_{2}$$ From this, the remainder when ${}_{2021}C_{37}$ is divided by 4 equals the remainder when ${}_{126}C_{2}$ is divided by 4, and $${}_{126}C_{2} = \frac{126\times125}{2} = 63\times125 \equiv 3\times1 \equiv 3$$
   Therefore, the remainder when ${}_{2021}C_{37}$ is divided by 4 is $\mathbf{3}$.

\subsection*{[Commentary]}
This is a proof problem based on binomial coefficients. The results proved in (1)--(3) connect neatly to the computation in (4). However, the method of writing the proofs is somewhat involved.
%% Page 13 5 (Go to problem page)

(1) For $\alpha > 0$, $0 \leq \theta \leq \pi$, consider two points $\mathrm{A}(-\alpha,\ -3)$, $\mathrm{P}(\theta + \sin\theta,\ \cos\theta)$, and let
$$f(\theta) = \mathrm{AP}^2 = (\theta + \sin\theta + \alpha)^2 + (\cos\theta + 3)^2$$
\begin{align*} f'(\theta) &= 2(\theta + \sin\theta + \alpha)(1 + \cos\theta) + 2(\cos\theta + 3)(-\sin\theta) &= 2(\theta + \sin\theta + \alpha) + 2(\theta + \sin\theta + \alpha)\cos\theta - 2\cos\theta\sin\theta - 6\sin\theta &= 2\theta + 2(\theta + \alpha)\cos\theta - 4\sin\theta + 2\alpha \end{align*}
$$f''(\theta) = 2 + 2\cos\theta - 2(\theta + \alpha)\sin\theta - 4\cos\theta = 2 - 2\cos\theta - 2(\theta + \alpha)\sin\theta$$
$$f'''(\theta) = 2\sin\theta - 2\sin\theta - 2(\theta + \alpha)\cos\theta = -2(\theta + \alpha)\cos\theta$$
Then, the monotonicity of $f''(\theta)$ on $0 \leq \theta \leq \pi$ is as shown in the table on the right, and from $f''\!\left(\dfrac{\pi}{2}\right) < f''(0) = 0$,

\begin{minipage}{0.55\textwidth} there exists exactly one $\theta$ satisfying $f''(\theta) = 0$; denoting it $\theta = \beta$, we have $\dfrac{\pi}{2} < \beta < \pi$. \end{minipage} \begin{minipage}{0.42\textwidth}

$\theta$	$0$	$\cdots$	$\dfrac{\pi}{2}$	$\cdots$	$\pi$
$f'''(\theta)$		$-$	$0$	$+$
$f''(\theta)$	$0$	$\searrow$		$\nearrow$	$4$

\end{minipage}

From this, the monotonicity of $f'(\theta)$ on $0 \leq \theta \leq \pi$ is as shown in the table on the right, and from $f'(\beta) < f'(\pi) = 0$,

\begin{minipage}{0.55\textwidth} $4\alpha > 0$, so there exists exactly one $\theta$ satisfying $f'(\theta) = 0$. \end{minipage} \begin{minipage}{0.42\textwidth}

$\theta$	$0$	$\cdots$	$\beta$	$\cdots$	$\pi$
$f''(\theta)$	$0$	$-$	$0$	$+$
$f'(\theta)$	$4\alpha$	$\searrow$		$\nearrow$	$0$

\end{minipage}

(2) From (1), denote the $\theta$ satisfying $f'(\theta) = 0$ as $\theta = \gamma$,
\begin{minipage}{0.55\textwidth} so $0 < \gamma < \beta$.
From this, the monotonicity of $f(\theta)$ is as shown in the table on the right, and $f(\theta)$ attains its maximum on $0 \leq \theta \leq \pi$ at $\theta = \gamma$. \end{minipage} \begin{minipage}{0.42\textwidth}

$\theta$	$0$	$\cdots$	$\gamma$	$\cdots$	$\pi$
$f'(\theta)$		$+$	$0$	$-$	$0$
$f(\theta)$		$\nearrow$		$\searrow$

\end{minipage}

The condition $0 < \gamma < \dfrac{\pi}{2}$ is equivalent to $f'\!\left(\dfrac{\pi}{2}\right) < 0$, which gives
$$2\cdot\frac{\pi}{2} - 4 + 2\alpha < 0, \quad 2\alpha < 4 - \pi$$
Therefore, the range of $\alpha$ sought is $\displaystyle 0 < \alpha < 2 - \frac{\pi}{2}$.

[Commentary]
This is a basic problem on differentiation and monotonicity. By writing out the sign chart while imagining the graph, the conclusion can be derived smoothly.
%% Page 14 6 (Go to problem page)

(1) For the identity $x^4 + bx + c = (x^2 + px + q)(x^2 - px + r)$, we have $$0 = r - p^2 + q \cdots\textcircled{1}, \quad b = pr - pq \cdots\textcircled{2}, \quad c = qr \cdots\textcircled{3}$$ From \textcircled{1}\textcircled{2}, $b = p(p^2 - q) - pq = p^3 - 2pq$, and since $p \neq 0$, $$q = \frac{1}{2}p^2 - \frac{b}{2p}, \quad r = p^2 - \frac{1}{2}p^2 + \frac{b}{2p} = \frac{1}{2}p^2 + \frac{b}{2p}$$
(2) When $p \neq 0$, $b = (a^2+1)(a+2)$, $c = -\!\left(a + \dfrac{3}{4}\right)(a^2+1)$, from (1), $$q = \frac{1}{2}p^2 - \frac{1}{2p}(a^2+1)(a+2) \cdots\textcircled{4}, \quad r = \frac{1}{2}p^2 + \frac{1}{2p}(a^2+1)(a+2) \cdots\textcircled{5}$$ Substituting \textcircled{4}\textcircled{5} into \textcircled{3}, $$-\!\left(a+\frac{3}{4}\right)(a^2+1) = \frac{1}{4}p^4 - \frac{1}{4p^2}(a^2+1)^2(a+2)^2,$$ $$-(4a+3)(a^2+1)p^2 = p^6 - (a^2+1)^2(a+2)^2$$ $$p^6 + (4a+3)(a^2+1)p^2 - (a^2+1)^2(a+2)^2 = 0 \cdots\cdots\textcircled{6}$$ Dividing the left side of \textcircled{6} by $p^2 - (a^2+1)$, $$\{p^2-(a^2+1)\}\{p^4+(a^2+1)p^2+(a^2+1)(a+2)^2\} = 0 \cdots\cdots\textcircled{7}$$ From \textcircled{7}, letting $f(t)$ and $g(t)$ satisfy $\{p^2-(a^2+1)\}\{p^4+f(a)p^2+g(a)\}=0$, $$f(t) = t^2+1, \quad g(t) = (t^2+1)(t+2)^2$$
(3) When the quartic polynomial $A(x) = x^4 + (a^2+1)(a+2)x - \!\left(a+\dfrac{3}{4}\right)(a^2+1)$ can be factored as a product of two quadratic polynomials with rational coefficients, noting that the coefficient of $x^4$ is 1 and the coefficient of $x^3$ is 0, $$A(x) = (x^2+px+q)(x^2-px+r) \quad (p,\, q,\, r \text{ are rational numbers})$$
(i) When $p = 0$:
$A(x) = (x^2+q)(x^2+r) = x^4+(q+r)x^2+qr$, so $$0 = q+r, \quad (a^2+1)(a+2) = 0, \quad -\!\left(a+\frac{3}{4}\right)(a^2+1) = qr$$ Then $r = -q$, $a = -2$, but $\dfrac{5}{4}\cdot 5 = -q^2$ does not hold.
(ii) When $p \neq 0$:
For integer $a$, from \textcircled{7}, since $p^4+(a^2+1)p^2+(a^2+1)(a+2)^2 > 0$, $$p^2 = a^2+1, \quad p = \pm\sqrt{a^2+1}$$ Also, substituting into \textcircled{4}\textcircled{5}, $$q = \frac{1}{2}(a^2+1) \mp \frac{1}{2\sqrt{a^2+1}}(a^2+1)(a+2) = \frac{1}{2}(a^2+1) \mp \frac{1}{2}\sqrt{a^2+1}\,(a+2)$$ $$r = \frac{1}{2}(a^2+1) \pm \frac{1}{2\sqrt{a^2+1}}(a^2+1)(a+2) = \frac{1}{2}(a^2+1) \pm \frac{1}{2}\sqrt{a^2+1}\,(a+2)$$ Then, the condition for $p, q, r$ to all be rational is that $\sqrt{a^2+1}$ is rational, so $$\sqrt{a^2+1} = \frac{n}{m} \quad (m \text{ and } n \text{ are mutually coprime positive integers})$$
$-8-$ \copyright\ 電送数学舎 2021
%% Page 15 From this, $(a^2+1)m^2 = n^2$, and since $m^2$ and $n^2$ are coprime, $$m^2 = 1, \quad a^2 + 1 = n^2 \cdots\cdots\textcircled{8}$$ From \textcircled{8}, $m=1$ and $(a+n)(a-n) = -1$, and since $a+n > a-n$, $$a+n = 1, \quad a-n = -1$$ Therefore, the integer value of $a$ sought is $a = 0$.

[Commentary]
This is a problem on identities. Parts (1) and (2) serve as guided steps leading to (3).

4

Consider the curve $C: y = x^3 - x$ in the coordinate plane.

1. [(1)] Show that every point $\mathrm{P}$ in the coordinate plane satisfies the following condition (i).
   1. [(i)] There exists a line $l$ passing through $\mathrm{P}$ that intersects the curve $C$ at 3 distinct points.
   
   
2. [(2)] Sketch the region of all possible positions of point $\mathrm{P}$ in the coordinate plane that satisfy the following condition (ii).
   1. [(ii)] There exists a line $l$ passing through $\mathrm{P}$ that intersects the curve $C$ at 3 distinct points, and moreover, the two regions enclosed by the line $l$ and the curve $C$ have equal areas.

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