Let $f ( t )$ be a periodic function of period $T , f ( t + T ) = f ( t ) ( T > 0 )$, and be expanded in the complex Fourier series as follows:
$$f ( t ) = \sum _ { n = - \infty } ^ { \infty } F _ { n } \exp \left( - i \omega _ { n } t \right)$$
Here, $i$ is the imaginary unit, and $t$ is a real number. Answer the following questions.
I. Express $\omega _ { n }$ using $T$ and $n$.
II. Let $M$ be a positive-integer constant and $\delta ( t )$ be the delta function, and define
$$\hat { f } ( t ) = \sum _ { m = 0 } ^ { M - 1 } f ( t ) \delta ( t - m \Delta t ) , \quad \Delta t = \frac { T } { M }$$
Express
$$\lim _ { \varepsilon \rightarrow + 0 } \int _ { - \varepsilon } ^ { T - \varepsilon } \hat { f } ( t ) \exp \left( i \omega _ { k } t \right) d t$$
using $F _ { n } ( n = - \infty , \cdots , - 1,0,1 , \cdots , \infty )$. Here $k$ is an arbitrary integer.
III. $\Delta t$ is given in Question II. Express $F _ { j } ( j = 0,1,2 , \cdots , M - 1 )$ using
$$f ( 0 ) , \quad f ( \Delta t ) , \quad f ( 2 \Delta t ) , \cdots , \quad f ( ( M - 1 ) \Delta t )$$
when $F _ { n } = 0 ( n < 0$ or $n \geq M )$.
IV. Calculate $F _ { j }$ in Question III when $f ( l \Delta t ) = ( - 1 ) ^ { l } \quad ( l = 0,1,2 , \cdots , M - 1 )$.

Problem 5
Consider the continuously differentiable function $f ( x )$ of the real variable $x$. Let $f ( x ) \rightarrow 0$ as $| x | \rightarrow \infty$. $f ( x )$, its derivative $f ^ { \prime } ( x )$, and $x f ( x )$ are absolutely integrable. The Fourier transform of the function $f ( x )$ is denoted by $\mathcal { F } \{ f ( x ) \} ( u )$ or equivalently by $\hat { f } ( u )$, and defined by $$\mathcal { F } \{ f ( x ) \} ( u ) = \hat { f } ( u ) \equiv \frac { 1 } { \sqrt { 2 \pi } } \int _ { - \infty } ^ { \infty } f ( x ) \exp ( - i u x ) \, d x \tag{1}$$ where $u$ is a real variable and $i$ is the imaginary unit. The Fourier transform is defined in the same way for other functions.
I. Express $\mathcal { F } \left\{ f ^ { \prime } ( x ) \right\} ( u )$ in terms of $\hat { f } ( u )$ and $u$.
II. Express $\frac { d \hat { f } ( u ) } { d u }$ in terms of $\mathcal { F } \{ x f ( x ) \} ( u )$.
III. Let the function $f ( x ) = \exp \left( - a x ^ { 2 } \right)$, where $a$ is a positive real constant $( a > 0 )$. The following relation holds for $f ( x )$: $$f ^ { \prime } ( x ) = - 2 a x f ( x ) \tag{2}$$ Apply the Fourier transform on both sides of Eq. (2) to obtain a first-order ordinary differential equation in $\hat { f } ( u )$. Solve this ordinary differential equation to obtain $\hat { f } ( u )$. Note that the integration constant in the solution of this ordinary differential equation can be obtained by calculating $\hat { f } ( 0 )$ with the help of Eq. (1) and the value of the following improper integral: $$\int _ { - \infty } ^ { \infty } \exp \left( - a x ^ { 2 } \right) d x = \sqrt { \frac { \pi } { a } } \tag{3}$$
IV. Consider the function $h ( x , t )$ of the real variables $x$ and $t$. Let $h ( x , t )$ be defined for $- \infty < x < \infty$ and $t \geq 0$, and satisfy the following partial differential equation: $$\frac { \partial h ( x , t ) } { \partial t } = \frac { \partial ^ { 2 } h ( x , t ) } { \partial x ^ { 2 } } \quad ( t > 0 ) \tag{4}$$ given the initial condition $$h ( x , 0 ) = \exp \left( - a x ^ { 2 } \right) \quad ( a > 0 ) \tag{5}$$

1. Apply the Fourier transform with respect to the variable $x$ on both sides of the partial differential equation (4) to obtain an ordinary differential equation with $\hat { h } ( u , t ) \equiv \frac { 1 } { \sqrt { 2 \pi } } \int _ { - \infty } ^ { \infty } h ( x , t ) \exp ( - i u x ) \, d x$ and the independent variable $t$.
2. By solving the ordinary differential equation found in Question IV.1, obtain $\hat { h } ( u , t )$.
3. Use the inverse Fourier transform with respect to the variable $u$ to obtain a solution $h ( x , t )$ satisfying Eq. (4) and Eq. (5).

V. Consider the continuous function $g ( x )$ and its Fourier transform $\hat { g } ( u )$. Let $g ( x ) \rightarrow 0$ as $| x | \rightarrow \infty$ and $g ( x )$ be absolutely integrable. The convolution of the functions $f ( x )$ and $g ( x )$ is defined by $$( f * g ) ( x ) \equiv \int _ { - \infty } ^ { \infty } f ( y ) g ( x - y ) \, d y \tag{6}$$

1. Express $\mathcal { F } \{ ( f * g ) ( x ) \} ( u )$ in terms of $\hat { f } ( u )$ and $\hat { g } ( u )$.
2. Here, the function $h ( x , t )$ satisfies Eq. (4), given the initial condition $h ( x , 0 ) = g ( x )$. Use the result of Question V.1 to find an integral representation of a solution $h ( x , t )$, where $t > 0$.

The Laplace transform of the function $f ( t )$, defined for $t \geq 0$, is denoted by $F ( s ) = \mathcal { L } [ f ( t ) ]$ and its definition is given by
$$F ( s ) = \mathcal { L } [ f ( t ) ] = \int _ { 0 } ^ { \infty } f ( t ) \exp ( - s t ) d t$$
where $s$ is a complex number. In the following, the set of all complex numbers is denoted by $\mathbb { C }$, and the set of the complex numbers with positive real parts is denoted by $\mathbb { C } ^ { + }$.
I. Consider the following function $g ( t )$ defined for $t \geq 0$ :
$$g ( t ) = \int _ { 0 } ^ { \infty } \frac { \sin ^ { 2 } ( t x ) } { x ^ { 2 } } d x$$

1. Find the Laplace transform $G ( s ) = \mathcal { L } [ g ( t ) ] \left( s \in \mathbb { C } ^ { + } \right)$ of the function $g ( t )$.
2. Obtain the value of the following integral using the result of Question I.1: $$\int _ { - \infty } ^ { \infty } \frac { \sin ^ { 2 } ( x ) } { x ^ { 2 } } d x$$

II. Consider the function $u ( x , t )$ that satisfies the following partial differential equation:
$$\frac { \partial u ( x , t ) } { \partial t } = \frac { \partial ^ { 2 } u ( x , t ) } { \partial x ^ { 2 } } ( 0 < x < 1 , t > 0 )$$
under the boundary conditions:
$$\left\{ \begin{array} { l } \left. \frac { \partial u ( x , t ) } { \partial x } \right| _ { x = 0 } = 0 \quad ( t \geq 0 ) \\ u ( 1 , t ) = 1 \quad ( t \geq 0 ) \\ u ( x , 0 ) = \frac { \cosh ( x ) } { \cosh ( 1 ) } \quad ( 0 < x < 1 ) \end{array} \right.$$

1. The Laplace transform of $u ( x , t )$ is denoted by $U ( x , s ) = \mathcal { L } [ u ( x , t ) ]$ ( $s \in \mathbb { C } ^ { + }$). Derive the ordinary differential equation and boundary conditions for $U ( x , s )$ with respect to the independent variable $x$. Here, the function $u ( x , t )$ can be assumed to be bounded. The following relations can also be used: $$\begin{aligned} & \mathcal { L } \left[ \frac { \partial u ( x , t ) } { \partial x } \right] = \frac { \partial U ( x , s ) } { \partial x } \\ & \mathcal { L } \left[ \frac { \partial ^ { 2 } u ( x , t ) } { \partial x ^ { 2 } } \right] = \frac { \partial ^ { 2 } U ( x , s ) } { \partial x ^ { 2 } } \end{aligned}$$
2. Using an analytic function $Q ( s ) ( s \in \mathbb { C } )$, the function $U _ { \mathrm { c } } ( x , s )$ is defined as follows: $$U _ { c } ( x , s ) = \frac { \cosh ( x ) } { ( s - 1 ) \cosh ( 1 ) } - \frac { \cosh ( x \sqrt { s } ) } { Q ( s ) } \quad ( 0 \leq x \leq 1 )$$ When the function $U ( x , s ) = U _ { \mathrm { c } } ( x , s )$ satisfies the differential equation and the boundary conditions derived in Question II.1 for $s \in \mathbb { C } ^ { + }$, find the function $Q ( s )$.
3. Using the function $Q ( s )$ derived in Question II.2, the sequence of complex numbers $\left\{ a _ { r } \right\} ( r = 1,2 , \cdots )$ is defined by arranging all of the roots of $Q ( s ) = 0 ( s \in \mathbb { C } )$ in ascending order of their absolute values. In this case, the following limits $R _ { r } ( x , t )$ are finite for $t \geq 0,0 \leq x \leq 1$, and $r \geq 1$ : $$R _ { r } ( x , t ) = \lim _ { s \rightarrow a _ { r } } \left( s - a _ { r } \right) U _ { \mathrm { c } } ( x , s ) \exp ( s t )$$ and the solution of the partial differential equation is given by $$u ( x , t ) = \sum _ { r = 1 } ^ { \infty } R _ { r } ( x , t )$$ Determine $R _ { 1 } ( x , t ) , R _ { 2 } ( x , t )$, and $R _ { r } ( x , t )$ for $r \geq 3$.

次の文章を読んで、後の設問に答えよ。
第　一　問
いまさらいうまでもなく、仮面はどこにでもあるというものではない。日本の祭に常に仮面が登場するわけではない。世界に視野を広げても、仮面を有する社会は、一部の地域にしか分布しない。オセアニアでは、メラネシアでしか、仮面はつくられていない。南北アメリカなら赤道をはさんで南北に広がる熱帯雨林やウッドランド、サヴァンナ地帯だけで仮面がつくられているというわけではない。アフリカでは広い範囲で仮面の制作と使用が確認できるが、それでもすべての社会に仮面が存在するというわけではない。
いまひとつ、仮面が農耕やシュリョウ・\ruby{漁撈}{ぎょろう}・採集を主たる生業とする社会にはみられるものではけっしてない。牧畜社会にはみられないという点も忘れてはならない。いずれにせよ、仮面は、人類文化に普遍的にみられるものではけっしてない。
ただ、世界の仮面の文化を広くみわたして注目されるのは、仮面の造形や仮面の制作と使用を支える組織のありかたに大きな多様性がみられる一方で、随所に、地域や民族の違いを越えて、驚くほどよく似た慣習や信念がみとめられるという事実である。相互に民族移動や文化の交流がおこったとは考えられない、遠く隔たった場所でよく似た現象がみとめられるというのは、やはり一定の条件のもとでの人類に普遍的な思考や行動のありかたのあらわれだと考えてよい。\textsuperscript{ア}その意味で、仮面の探求は、人間のなかにある普遍的なものの探求につながる可能性をもっている。
地域と時代を問わず、仮面に共通した特性としてあげられるのは、それがいずれも、「異界」の存在を表現したものだという点である。ヨーロッパでいえば、ギリシアのディオニソスの祭典に用いられた仮面から、現代のカーニヴァルに登場する異形の仮面や魔女の仮面まで、日本でいえば、能・狂言や民俗行事のなかで用いられる神がみや死者の仮面から、現代の月光仮面（月からの使者といわれる）やウルトラマン（M78星雲からやって来た人類の味方に至るまで、仮面はつねに、時間の変わり目や危機的な状況において、異界から一時的に来たり、人びとと交わって去っていく存在を可視化するために用いられてきた。それは、アフリカやメラネシアの葬儀や成人儀礼に登場する死者や精霊の仮面についてもあてはまる。たしかに、知識の増大とともに、人間の知識の及ばぬか、月に設定するか、あるいは宇宙の果てに設定するかの違いだけである。たしかに、異界を、山や森に設定するか、世界＝異界は、村をとりまく山や森から、月へ、そして宇宙へと、どんどん遠くへ退いていく。しかし、世界を改変するものとしての異界の力に対する人びとの憧憬、異界からの来訪者への期待が変わることはなかったのである。
ただ、忘れてならないのは、人びとはその仮面のかぶり手を、あるときは歓待し、あるときは慰撫し、またあるときは痛めつけてきたということである。仮面は異界からの来訪者を可視化するものだとはいっても、それはけっして視られるためだけのものではない。それは、あくまでもいったん可視化した対象に人間が積極的にはたらきかけるための装置であった。仮面は、大きな変化や危機に際して、人間がそうした異界の力を一時的に目にみえるかたちにし、それにはたらきかけることで、その力そのものをコントロールしようとして創りだしてきたものの一つである。そして、テレビの画面のなかで繰り広げられる現代の仮面のヒーローたちの活躍もまた、それと同じ欲求に根ざしているのである。
ここでは、仮面が神や霊など、異界の力を可視化し、コントロールする装置であることを強調してきた。しかし、そのような装置は少なくともうひとつある。神霊の憑依、つまり憑霊である。しかも、仮面をかぶった踊り手には、霊が依り憑き、踊り手はその霊になりきることが多かった。いちいち引用の出典を記すまでもない。仮面をかぶった踊り手には、それは神そのものだといった議論は、世界各地の仮面についての民族誌のなかに数多く見いだされる。
たしかに、神や精霊に扮した者は、少なくとも何がしかの神や精霊の属性を帯びることになるという信念が維持されていなければ、彼らとかかわることで福や幸運が享受できるかもしれないという、かすかな期待を人びとが抱くことすら不可能になる。その意味で、儀礼における仮面と憑依との結びつきは、動かしえない事実のようである。
しかし、その一方で神事を脱し芸能化した仮面や子どもたちが好んでかぶる仮面に、憑依という宗教的な体験を想定することはできない。仮面のありかたの歴史的変化が語っているのは、イ仮面は憑依を前提としなくなっても存続しうるという事実である。そしてその点で、仮面は決定的に霊媒と異なる。霊媒は憑依という信念が失われた瞬間、存立しえなくなるからである。
仮面と憑依の相同性を強調した従来の議論に反して、民族誌的事実と歴史的事実は、このように、ともに仮面と憑依との違いを主張している。仮面は憑依とひとつの顔でありつつも、それとは異なる固有の場をもっているのである。では、その固有性とは何か。それを考えるには、顔をもうひとつの顔で覆うという、仮面の定義以外にないであろう。そして、その定義において、仮面が人間の顔ない身体をその存立の与件としている以上、仮面を私たちの身体的経験に照らして考察することにする。
仮面と身体とのかかわり。それはいうまでもなく、仮面が顔、素顔の上につけられるものだという単純な事実に求められる。もちろん、世界を広くみわたしたとき、顔の前につける仮面は、必ずしも一般的だとはいえない。むしろ、顔と体全体を覆ってしまうかぶりものの方が多数を占めるかもしれない。しかし、その場合でも、顔が隠されることが要件であることは間違いない。
変身にとって、顔を変えることが核心的意味をもって明確に示したのは、ロ和辻哲郎であった。私たちは、たとえ未知の他人であっても、その他人の顔を思い浮かべることなしに、その他人とかかわることはできない。また、肖像画や肖像彫刻にみるように、顔だけで人を表象することはできても、顔を除いて特定の人物を表象することははできない。このような経験をもとに、和辻は「人の存在にとっての顔の核心的意義」を指摘し、顔はたんに肉体の一部としてあるのでなく、「肉体を己れに従える主体的なものの座」、すなわち「人格の座」を占めていると述べたのであった。
この和辻の指摘の通り、確かに私たちの他者の認識の方法は顔に集中している。逆にいえば、他者もまた私の顔から私について認知する要\ruby{る}{る}。しかし、他者が私を私として認知するのに、顔だけは絶対に見えるのに、その顔は私にとってもっとも不知な部分として、終生、私につきの言葉を借りていえば、顔は私の人格の座であるはずなのに、自分の身体でも他の部分なら鏡を使わずになんとか見えるのに、その顔は私にとってもっとも不知な部分として、終生、私につきまとうことになる。

次の文章は『沙石集』の一話「耳売りたる事」である。これを読んで、後の設問に答えよ。
南都に、ある寺の僧、耳のびく厚きを、ある貧なる僧ありて、「アたべ。御坊の耳買はん」と云ふ。「どく買ひ給へ」と云ふ。「いかほどに買ひ給はん」と云ふ。「五百文に買はん」と云ふ。「さらば」とて、銭を取りて売りつ。その後、京へ上りて、相者のもとに、耳売りたる僧と同じく行く。相して云はく、「福分おはしまさず」と云ふ時に、「あの御坊の耳、その代銭かくのごとき数にて買ひ候ふ」と云ふ。「さては御耳にして、御心安からん」と相す。さて、耳売りたる数にて買ひ候ふ」と云ふ。「御福分かなひて、世間不階の人なり。「かく耳売る事もあれば、貧窮を売ることもありぬべし」と思ひ、南都を立ち出でて、東の方に住み侍りけるが、学生にて、説法などもする僧なり。
ある上人の云はく、「老僧を仏事に請ずる事あり。身老いて道遠し。予に代はりて、赴き給へかし。ただし三日路なり。想像するに、施物十五貫、文には過ぐべからず。またこれより一日路なる所に、ある神主の有徳なるが、七日逆修をする事あり。これも予を招請すといへどもかんことを欲せず。これは、一日に無下ならば五貫、ようせば十貫づはせんずらん。公、いづれに行き給はんと云ふ。かの僧、「仰すまでもなし。遠路を過ぎて、十五貫文など取り候はより、一日路きて七十貫こそ取り候はれめ」と云ふ。「しからば」とて、一所へは別人をして行かしむ。神主は齢八旬に及びて、病床に臥したり。子息申しけるは、「老体の上、不例日久しく既に海を渡りて、その処に至りぬ。神主は齢八旬に及びて、病床に臥したり。子息申しけるは、「老体の上、不例日久しく仕り候て、安泰頼み難く候へども、もしやと、先づ祈禱に、真読の大般若ありたく候ふ」と申す。「また、逆修は、いかさま用意して、ひて、やがてひきつぎ仕り候はんと云ふ。この僧思ふやう、「先づ大般若を取るべし。また逆修の布施は置き物」と思ひて、

The following passage is the words spoken by Tang Taizong, Li Shimin (r.\ 626--649). Read it and answer the questions that follow. Note that the kana readings added throughout have been omitted in places.
朕聞晋武帝平呉已後、務在驕奢、不復留心治政。何曽退朝、謂其子勉曰、「吾每見主上、不論経国遠図、但説平生常語。ヲ此非下者必遇乱而死。」及 胎厳子孫、爾儞身猶可以免。指サシテ諸孫ヲ曰ハク、「此等ら必ズ遇ヒテ乱ニ死セント。」及ビ 孫ノ綏ニ、果為ニ淫刑、所ヲ謂おもヘラク曽之不忠ハ、其ノ罪大矣。夫リテハ人臣ニ、当シ下ニ進ミテハ思ヒ 朕ガ意ハ不レ然ラ。謂おもヘラク曽之不忠ハ、其ノ罪大矣。夫為リテハ人臣ニ、当シ下ニ進ミテハ思ヒ 蜀サンコトヲ誠ヲ、退キテハ思ヒ補レ過ヲ、将ヲ順ジ其ノ美ヲ、匡救きゃうきうス其ノ悪ヲ上。所ヲ以テ共ニ為スヲ治ヲ也。 曽位極台司ヲ、名器崇重ナリ。当シ$_{\text{c}}$直辞正諫カン、論ジテ道ヲ$_{\text{d}}$佐ク時ニ。今乃チ退キテハ有リ後
${}^{\text{a}}$儞 ${}^{\text{b}}$前史美之
言、進みて無二廷諍一。以テ為スハ明智、不二亦タ謬あやまリナラ一乎。顧たびレテ而不レ扶すケ安クンゾ用二彼ノ相ヲ一。
（『貞観政要』による）

Answer all the following questions.
I. Let a periodic function $f ( x )$ satisfy the condition $f ( x + \pi ) = f ( x - \pi )$. Find the Fourier series expansion of $f ( x )$ for each case, where $f ( x )$ is expressed as follows for the interval $- \pi \leq x \leq \pi$.
1. $f ( x ) = x \quad ( - \pi < x < \pi ) , \quad f ( - \pi ) = f ( \pi ) = 0$ 2. $f ( x ) = x ^ { 2 }$
For the Fourier series expansion, the following equations should be used.
$$\begin{aligned} & f ( x ) = \frac { a _ { 0 } } { 2 } + \sum _ { n = 1 } ^ { \infty } \left( a _ { n } \cos n x + b _ { n } \sin n x \right) \\ & a _ { 0 } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \mathrm { d } x \\ & a _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \cos n x \mathrm {~d} x \\ & b _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \sin n x \mathrm {~d} x \end{aligned}$$
II. Consider the function shown in Figure 5.1 as
$$V ( t ) = A \left| \sin \left( \frac { \omega t } { 2 } \right) \right| \quad \left( \omega = \frac { 2 \pi } { T } \right) .$$
Note that $A$ and $T$ are positive real numbers. The complex Fourier series expansion of $V ( t )$ is given as
$$V ( t ) = - \frac { 2 A } { \pi } \sum _ { n = - \infty } ^ { \infty } \frac { 1 } { 4 n ^ { 2 } - 1 } e ^ { i n \omega t }$$
Let $I ( t )$ be the periodic solution that satisfies the ordinary differential equation
$$L \frac { \mathrm {~d} I ( t ) } { \mathrm { d } t } + R I ( t ) = V ( t )$$
Note that $L$ and $R$ are positive real numbers. Find the coefficient $C _ { n }$, when the complex Fourier series expansion of $I ( t )$ is expressed as
$$I ( t ) = \sum _ { n = - \infty } ^ { \infty } C _ { n } e ^ { i n \omega t }$$

6 (See the solution/explanation page)

In the coordinate space with origin O, consider the cube defined by the inequalities $|x| \leq 1$, $|y| \leq 1$, $|z| \leq 1$. Let $S$ be the part of the surface of this cube satisfying $z < 1$.

Hereafter, when two points A and B in the coordinate space coincide, the line segment AB is defined to represent the point A, and its length is defined to be $0$.

1. [(1)] When a point P in the coordinate space satisfies both of the following conditions (i) and (ii), find the volume $V$ of the region that P can occupy.
   
   • [(i)] $\mathrm{OP} \leq \sqrt{3}$
   • [(ii)] The line segment OP and $S$ have no common point, or have only the point P as a common point.
   
   
2. [(2)] When points N and P in the coordinate space satisfy all of the following conditions (iii), (iv), (v), find the volume $W$ of the region that P can occupy. If necessary, you may use a real number $\alpha$ satisfying $\sin\alpha = \dfrac{1}{\sqrt{3}}$ $\left(0 < \alpha < \dfrac{\pi}{2}\right)$.
   
   • [(iii)] $\mathrm{ON} + \mathrm{NP} \leq \sqrt{3}$
   • [(iv)] The line segment ON and $S$ have no common point.
   • [(v)] The line segment NP and $S$ have no common point, or have only the point P as a common point.
   
   

%% Page 7 1 Go to problem page

(1) For $A_k = \displaystyle\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} |\sin(x^2)|\,dx$, let $x^2 = t$ ($x = \sqrt{t}$), then $2x\,dx = dt$, so
$$A_k = \int_{k\pi}^{(k+1)\pi} |\sin t| \cdot \frac{dt}{2\sqrt{t}} = \int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{2\sqrt{t}}\,dt$$
Now, for $k\pi \leq t \leq (k+1)\pi$, we have $\dfrac{1}{\sqrt{(k+1)\pi}} \leq \dfrac{1}{\sqrt{t}} \leq \dfrac{1}{\sqrt{k\pi}}$, so
$$\frac{|\sin t|}{2\sqrt{(k+1)\pi}} \leq \frac{|\sin t|}{2\sqrt{t}} \leq \frac{|\sin t|}{2\sqrt{k\pi}}$$
Integrating each side from $k\pi$ to $(k+1)\pi$,
$$\frac{1}{2\sqrt{(k+1)\pi}}\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt \leq A_k \leq \frac{1}{2\sqrt{k\pi}}\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt \quad \cdots\cdots\cdots\textcircled{1}$$
Here, since $\sin t$ does not change sign for $k\pi \leq t \leq (k+1)\pi$,
$$\int_{k\pi}^{(k+1)\pi} |\sin t|\,dt = \left|\int_{k\pi}^{(k+1)\pi} \sin t\,dt\right| = \left|-[\cos t]_{k\pi}^{(k+1)\pi}\right|$$ $$= \left|-\cos(k+1)\pi + \cos k\pi\right| = \left|(-1)^{k+1} + (-1)^k\right| = \left|(-1)^k\right||-1+1| = 2$$
Therefore, from \textcircled{1},
$$\frac{1}{\sqrt{(k+1)\pi}} \leq A_k \leq \frac{1}{\sqrt{k\pi}} \quad \cdots\cdots\cdots\textcircled{2}$$
(2) $\displaystyle\int_{\sqrt{n\pi}}^{\sqrt{2n\pi}} |\sin(x^2)|\,dx = \sum_{k=n}^{2n-1}\left(\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} |\sin(x^2)|\,dx\right) = \sum_{k=n}^{2n-1} A_k$, so from \textcircled{2},
$$\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} \leq \sum_{k=n}^{2n-1} A_k \leq \sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}}$$
Since $B_n = \dfrac{1}{\sqrt{n}}\displaystyle\int_{\sqrt{n\pi}}^{\sqrt{2n\pi}} |\sin(x^2)|\,dx$,
$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} \leq B_n \leq \frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}} \quad \cdots\cdots\cdots\textcircled{3}$$
Here, letting $l = k - n$ and considering $n \to \infty$,
$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{k\pi}} = \frac{1}{\sqrt{n}}\sum_{l=0}^{n-1} \frac{1}{\sqrt{(n+l)\pi}} = \frac{1}{n}\sum_{l=0}^{n-1} \frac{1}{\sqrt{\left(1+\dfrac{l}{n}\right)\pi}}$$
$$\to \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx \quad (n\to\infty)$$
Also, letting $m = k+1-n$ and considering $n \to \infty$,
$$\frac{1}{\sqrt{n}}\sum_{k=n}^{2n-1} \frac{1}{\sqrt{(k+1)\pi}} = \frac{1}{\sqrt{n}}\sum_{m=1}^{n} \frac{1}{\sqrt{(n+m)\pi}} = \frac{1}{n}\sum_{m=1}^{n} \frac{1}{\sqrt{\left(1+\dfrac{m}{n}\right)\pi}}$$
$$\to \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx \quad (n\to\infty)$$
%% Page 8 Thus, from \textcircled{3},
$$\lim_{n \to \infty} B_n = \int_0^1 \frac{1}{\sqrt{(1+x)\pi}}\,dx = \frac{1}{\sqrt{\pi}}\bigl[2\sqrt{(1+x)}\bigr]_0^1 = \frac{2}{\sqrt{\pi}}(\sqrt{2}-1)$$
[Commentary]
This is a standard problem combining definite integrals and inequalities with the Riemann sum method. However, it has a somewhat formal atmosphere.
%% Page 9 2 (Go to problem page)

(1) When arranging all 12 balls --- 3 black balls, 4 red balls, and 5 white balls --- in a single row, assume that all $12!$ arrangements are equally likely.

The probability that no two red balls are adjacent is found by first arranging the 8 white and black balls in a row, then choosing 4 of the 9 gaps (between balls or at either end) to insert the 4 red balls one by one. Thus the probability $p$ is, $$p = \frac{8! \times {}_9P_4}{12!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9} = \frac{14}{55}$$
(2) The probability that no two red balls are adjacent and no two black balls are adjacent is found by considering the following cases:

(i) When the arrangement red-black-black-red-black occurs

First arrange a group of 5 white balls and 3 black balls together, then insert 2 of the 4 red balls into the gaps between the black balls, and choose 2 of the 7 positions between the group of 5 white balls and 3 black balls or at either end to insert the remaining 2 red balls one by one. The probability is, $$\frac{(3! \times 6!) \times {}_4P_2 \times {}_7P_2}{12!} = \frac{3 \cdot 2 \cdot 4 \cdot 3 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7} = \frac{1}{2^2 \cdot 5 \cdot 11}$$
(ii) When the arrangement black-red-black occurs but the arrangement red-black-black-red-black does not occur

First arrange 5 white balls, then choose 2 of the 6 gaps (between balls or at either end) to place a group of 2 black balls and 1 single black ball. Next, insert 1 of the 4 red balls into the gap between the black balls, and choose 3 of the 8 positions between the 5 white balls, the group of 2 black balls, and the single black ball or at either end to insert the remaining 3 red balls one by one. The probability is, $$\frac{5! \times ({}_3P_2 \times 1 \times {}_6P_2) \times {}_4P_1 \times {}_8P_3}{12!} = \frac{3 \cdot 2 \cdot 6 \cdot 5 \cdot 4 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} = \frac{2}{3 \cdot 11}$$
(iii) When the arrangement black-red-black does not occur

First arrange 5 white balls, then choose 3 of the 6 gaps (between balls or at either end) to place the 3 black balls one by one. Next, choose 4 of the 9 gaps between the 5 white balls and 3 black balls or at either end to insert the 4 red balls one by one. The probability is, $$\frac{5! \times {}_6P_3 \times {}_9P_4}{12!} = \frac{6 \cdot 5 \cdot 4 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} = \frac{1}{11}$$
From (i)--(iii), letting $r$ denote the probability that no two red balls are adjacent and no two black balls are adjacent,
%% Page 10 $$r = \frac{1}{2^2 \cdot 5 \cdot 11} + \frac{2}{3 \cdot 11} + \frac{1}{11} = \frac{3 + 2 \cdot 20 + 60}{2^2 \cdot 5 \cdot 3 \cdot 11} = \frac{103}{2^2 \cdot 5 \cdot 3 \cdot 11}$$
Therefore, the conditional probability $q$ that no black balls are adjacent, given that no red balls are adjacent, is $$q = \frac{r}{p} = \frac{103}{2^2 \cdot 5 \cdot 3 \cdot 11} \div \frac{14}{55} = \frac{103}{168}$$
[Commentary]
This is a standard probability problem. Part (1) involves routine processing, while part (2) admits various approaches. Here, we perform case analysis by focusing on the arrangement black-red-black.
%% Page 11

\boxed{3

\text{Go to Problem Page}}

1. [(1)] When any point $(\cos\theta,\ a+\sin\theta)$ on the circle $C: x^2+(y-a)^2=1$ is contained in the region represented by the inequality $y>x^2$, we have $a+\sin\theta > \cos^2\theta$, so
   $$a > \cos^2\theta - \sin\theta = 1-\sin^2\theta - \sin\theta = -\!\left(\sin\theta+\frac{1}{2}\right)^2+\frac{5}{4} \quad \cdots\cdots\textcircled{1}$$
   Here, since $\theta$ takes any value, we have $-1\leq\sin\theta\leq 1$, and the maximum value of the right-hand side of \textcircled{1} is $\dfrac{5}{4}$,
   so the range of $a$ for which \textcircled{1} always holds is $a>\dfrac{5}{4}$.
   
2. [(2)] From (1), when $a>\dfrac{5}{4}$, the equation of the tangent line to $C$ at the point $\mathrm{P}(\cos\theta,\ a+\sin\theta)$ on $S$
   $\left(-\dfrac{\pi}{2}\leq\theta<0\right)$ is
   $$x\cos\theta+(a+\sin\theta-a)(y-a)=1$$ $$x\cos\theta+y\sin\theta = a\sin\theta+1 \quad\cdots\cdots\textcircled{2}$$
   The intersection points with the parabola $y=x^2\cdots\cdots\textcircled{3}$ are found by solving \textcircled{2} and \textcircled{3} simultaneously:
   $$x\cos\theta + x^2\sin\theta = a\sin\theta+1$$ $$x^2\sin\theta + x\cos\theta - (a\sin\theta+1)=0 \quad\cdots\cdots\textcircled{3}'$$
   Since $-1\leq\sin\theta<0$, letting $D$ be the discriminant of \textcircled{3}$'$, when $a>\dfrac{5}{4}$,
   $$D = \cos^2\theta + 4\sin\theta(a\sin\theta+1) = (4a-1)\sin^2\theta+4\sin\theta+1$$ $$=(4a-1)\!\left(\sin\theta+\frac{2}{4a-1}\right)^2+\frac{4a-5}{4a-1}>0$$
   So, letting the real solutions of \textcircled{3}$'$ be $x=\alpha,\ \beta\ (\alpha<\beta)$,
   $$\alpha = \frac{-\cos\theta+\sqrt{D}}{2\sin\theta},\quad \beta = \frac{-\cos\theta-\sqrt{D}}{2\sin\theta}$$
   Now, the intersection points are expressed as $(\alpha,\ \alpha^2)$, $(\beta,\ \beta^2)$, and the length $L_{\mathrm{P}}$ of the chord cut off is
   $$L_{\mathrm{P}}^2 = (\alpha-\beta)^2+(\alpha^2-\beta^2)^2 = (\alpha-\beta)^2\{1+(\alpha+\beta)^2\} = \frac{D}{\sin^2\theta}\!\left(1+\frac{\cos^2\theta}{\sin^2\theta}\right)$$
   $$= \frac{D}{\sin^2\theta}\cdot\frac{1}{\sin^2\theta} = \frac{D}{\sin^4\theta} = \frac{(4a-1)\sin^2\theta+4\sin\theta+1}{\sin^4\theta}$$
   Here, letting $u=\sin\theta\ (-1\leq u<0)$, $\theta$ and $u$ correspond one-to-one, and further
   $$f(u) = \frac{(4a-1)u^2+4u+1}{u^4} = \frac{4a-1}{u^2}+\frac{4}{u^3}+\frac{1}{u^4}$$
   gives $L_{\mathrm{P}}^2 = f(u)$, and
   $$f'(u) = \frac{-2(4a-1)}{u^3}-\frac{12}{u^4}-\frac{4}{u^5} = -\frac{2}{u^5}\{(4a-1)u^2+6u+2\}$$
   So, letting $g(u)=(4a-1)u^2+6u+2$, we have $f'(u)=-\dfrac{2}{u^5}g(u)$,
   and noting that $-\dfrac{2}{u^5}>0$, the signs of $f'(u)$ and $g(u)$ agree,
   

$-5-$ \copyright\ 電送数学舎 2023
%% Page 12 $$g(u) = (4a-1)\left(u + \frac{3}{4a-1}\right)^2 + \frac{8a-11}{4a-1}$$
Since $a > \dfrac{5}{4}$, we have $g(-1) = (4a-1) - 6 + 2 = 4a - 5 > 0$, $g(0) = 2 > 0$, and $-\dfrac{3}{4} < -\dfrac{3}{4a-1} < 0$. Based on this, considering the sign of $g(u)$, i.e., $f'(u)$:

• [(i)] When $8a - 11 \geq 0$ $\left(a \geq \dfrac{11}{8}\right)$
  $f'(u) \geq 0$ for $-1 \leq u < 0$, so $f(u)$ is monotonically increasing.
  
• [(ii)] When $8a - 11 < 0$ $\left(\dfrac{5}{4} < a < \dfrac{11}{8}\right)$
  $f'(u) = 0$ has two distinct real roots in $-1 \leq u < 0$; denoting these as $u = u_1,\ u_2\ (u_1 < u_2)$, the increase/decrease of $f(u)$ is as shown in the table on the right. From this, for a certain value of $f(u)$, there exist multiple values of $u$.
  

  $u$	$-1$	$\cdots$	$u_1$	$\cdots$	$u_2$	$\cdots$	$0$
  $f'(u)$		$+$	$0$	$-$	$0$	$+$
  $f(u)$	$4a-4$	$\nearrow$		$\searrow$		$\nearrow$	$\infty$

  
  

From (i) and (ii), the condition for there to exist two distinct points Q, R on $S$ such that $L_{\mathrm{Q}} = L_{\mathrm{R}}$ is: $$\frac{5}{4} < a < \frac{11}{8}$$

[Commentary]
This is an application problem of differentiation dealing with the relationship between a parabola and a line. In part (2), the amount of computation varies depending on the initial setup method, but in any case, it suffices to process things so that the graph of the corresponding function $L_{\mathrm{P}}^2 = f(u)$ is sinuous (has a local maximum and minimum).
%% Page 13 4 (Go to problem page)

(1) For the 4 points $O(0,0,0)$, $A(2,0,0)$, $B(1,1,1)$, $C(1,2,3)$, let $P(x,y,z)$. From $\overrightarrow{OP}\cdot\overrightarrow{OA}=0$, $\overrightarrow{OP}\cdot\overrightarrow{OB}=0$, $\overrightarrow{OP}\cdot\overrightarrow{OC}=1$, $$2x=0,\quad x+y+z=0,\quad x+2y+3z=1$$ From these, $x=0$, $y=-1$, $z=1$, so $P(0,-1,1)$.

(2) Let $\overrightarrow{OH}=(1-t)\overrightarrow{OA}+t\overrightarrow{OB}=(2-t,\,t,\,t)$, then $$\overrightarrow{PH}=(2-t,\,t,\,t)-(0,-1,1)$$ $$=(2-t,\;t+1,\;t-1)$$ Also, $\overrightarrow{AB}=(-1,1,1)$, and from $\overrightarrow{PH}\cdot\overrightarrow{AB}=0$, $$-(2-t)+(t+1)+(t-1)=0$$ Thus $3t-2=0$, so $t=\dfrac{2}{3}$, and $\overrightarrow{OH}=\dfrac{1}{3}\overrightarrow{OA}+\dfrac{2}{3}\overrightarrow{OB}$.
[Figure: Points P, C, B, O, H, A in 3D]

(3) From $\overrightarrow{OP}\perp\overrightarrow{OA}$, $\overrightarrow{OP}\perp\overrightarrow{OB}$, OP is perpendicular to the plane OAB, and furthermore from $\overrightarrow{PH}\perp\overrightarrow{AB}$, $\overrightarrow{OH}\perp\overrightarrow{AB}$.
Now, let $\overrightarrow{OR}=\dfrac{3}{4}\overrightarrow{OA}$, and consider the sphere $S$ of radius $r$ centered at the point $Q$ defined by $\overrightarrow{OQ}=\overrightarrow{OR}+\overrightarrow{OP}$, and find the range of $r$ for which $S$ has a common point with $\triangle OHB$.
[Figure: Points P, Q, O, B, S, H, R, A in 3D]
First, drop a perpendicular from R to OH and let the foot be S; then the minimum value of $r$ is QS.
Here, $\mathrm{QR}=\mathrm{PO}=\sqrt{(-1)^2+1^2}=\sqrt{2}$, $\mathrm{AB}=\sqrt{(-1)^2+1^2+1^2}=\sqrt{3}$, and $$\mathrm{RS}:\mathrm{AH}=\mathrm{OR}:\mathrm{OA}=3:4,\quad \mathrm{AH}:\mathrm{AB}=2:3$$ From these, $\mathrm{RS}=\dfrac{3}{4}\mathrm{AH}=\dfrac{3}{4}\cdot\dfrac{2}{3}\mathrm{AB}=\dfrac{\sqrt{3}}{2}$, and since QR is perpendicular to the plane ABC, $$\mathrm{QS}=\sqrt{\mathrm{QR}^2+\mathrm{RS}^2}=\sqrt{(\sqrt{2})^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{\sqrt{11}}{2}$$
Also, from $R\!\left(\dfrac{3}{2},\,0,\,0\right)$, $\mathrm{RO}=\dfrac{3}{2}$, $\mathrm{RB}=\sqrt{\left(\dfrac{1}{2}\right)^2+(-1)^2+(-1)^2}=\dfrac{3}{2}$,
and since $\mathrm{RH}<\mathrm{RB}$, the maximum value of $r$ is QO or QB, and $$\mathrm{QO}=\sqrt{\mathrm{QR}^2+\mathrm{OR}^2}=\sqrt{(\sqrt{2})^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{17}}{2}$$
From the above, the required range of $r$ is $\dfrac{\sqrt{11}}{2}\leq r\leq\dfrac{\sqrt{17}}{2}$.

[Commentary]
This is a standard problem on applications of vectors in space. By noting that $\triangle OHB$ is a right triangle, the rest reduces to computation.
%% Page 14

\boxed{5

\text{Go to Problem Page}}
(1) Let $q_1(x)$ be the quotient and $r(x)$ be the remainder when $g(x)$ is divided by $f(x)$, so that $$g(x) = f(x)q_1(x) + r(x)$$ Raising both sides to the 7th power, $$g(x)^7 = \{f(x)q_1(x)+r(x)\}^7 = \sum_{k=0}^{6} {}_7C_k\, f(x)^{7-k}\,q_1(x)^{7-k}\,r(x)^k + r(x)^7$$ $$= f(x)\sum_{k=0}^{6} {}_7C_k\, f(x)^{6-k}\,q_1(x)^{7-k}\,r(x)^k + r(x)^7$$ From this, $g(x)^7 - r(x)^7 = f(x)\displaystyle\sum_{k=0}^{6} {}_7C_k\, f(x)^{6-k}\,q_1(x)^{7-k}\,r(x)^k$, so $g(x)^7 - r(x)^7$ is divisible by $f(x)$.
That is, the remainder when $g(x)^7$ is divided by $f(x)$ equals the remainder when $r(x)^7$ is divided by $f(x)$.
(2) Let $q_2(x)$ be the quotient and $h_1(x)$ be the remainder when $h(x)^7$ is divided by $f(x)$, and let $q_3(x)$ be the quotient and $h_2(x)$ be the remainder when $h_1(x)^7$ is divided by $f(x)$. If $h_2(x) = h(x)$, then $$h(x)^7 = f(x)q_2(x) + h_1(x) \cdots\cdots\textcircled{1}, \quad h_1(x)^7 = f(x)q_3(x) + h(x) \cdots\cdots\textcircled{2}$$ From \textcircled{1}\textcircled{2}, $\{h(x)^7 - f(x)q_2(x)\}^7 = f(x)q_3(x) + h(x)$.
By the same argument as (1), $h(x)^{49} - h(x)$ is divisible by $f(x) = (x-1)^2(x-2)$.
Now, let $H(x) = h(x)^{49} - h(x)$. We show that the necessary and sufficient condition for the polynomial $H(x)$ to be divisible by $f(x)$ is that $H(1)=0\cdots\cdots\textcircled{3}$, $H(2)=0\cdots\cdots\textcircled{4}$, $H'(1)=0\cdots\cdots\textcircled{5}$.
(a) Let $Q_1(x)$ be the quotient when $H(x)$ is divided by $f(x)$: $$H(x) = (x-1)^2(x-2)Q_1(x)$$ $$H'(x) = 2(x-1)(x-2)Q_1(x) + (x-1)^2 Q_1(x) + (x-1)^2(x-2)Q_1'(x)$$ From this, \textcircled{3}\textcircled{4}\textcircled{5} hold.
(b) Conversely, when \textcircled{3}\textcircled{4}\textcircled{5} hold, from \textcircled{3}\textcircled{4}, $H(x)$ is divisible by $(x-1)(x-2)$.
Let $Q_2(x)$ be the quotient when $H(x)$ is divided by $(x-1)(x-2)$: $$H(x) = (x-1)(x-2)Q_2(x)$$ $$H'(x) = (x-2)Q_2(x) + (x-1)Q_2(x) + (x-1)(x-2)Q_2'(x)$$ Then from \textcircled{5}, $Q_2(1) = 0$, so $Q_2(x)$ is divisible by $x-1$. Let $Q_3(x)$ be the quotient when $Q_2(x)$ is divided by $x-1$: $$Q_2(x) = (x-1)Q_3(x)$$ Putting it together, $H(x) = (x-1)(x-2)(x-1)Q_3(x) = (x-1)^2(x-2)Q_3(x)$, so $H(x)$ is divisible by $f(x)$.
From (a)(b), the necessary and sufficient condition for $H(x)$ to be divisible by $f(x)$ is \textcircled{3} and \textcircled{4} and \textcircled{5}.
Now, from $h(x) = x^2 + ax + b$, we have $H(x) = (x^2+ax+b)^{49} - (x^2+ax+b)$.
%% Page 15 $$H'(x) = 49(x^2+ax+b)^{48}(2x+a)-(2x+a)$$
From \textcircled{3}, $(1+a+b)^{49}-(1+a+b)=0$, so $$(1+a+b)^{49}=1+a+b \cdots\cdots\textcircled{6}$$
From \textcircled{4}, $(4+2a+b)^{49}-(4+2a+b)=0$, so $$(4+2a+b)^{49}=4+2a+b \cdots\cdots\textcircled{7}$$
From \textcircled{5}, $49(1+a+b)^{48}(2+a)-(2+a)=0$, so $$49(1+a+b)^{48}(2+a)=2+a \cdots\cdots\textcircled{8}$$
(i) When $1+a+b=0$
\textcircled{6} holds, and from \textcircled{8}, $2+a=0$, i.e., $a=-2$, so $b=-1-(-2)=1$.
In this case, $4+2a+b=1$, and \textcircled{7} holds.
(ii) When $1+a+b\neq 0$
From \textcircled{6}, $(1+a+b)^{48}=1$, so $1+a+b=\pm 1$.
(ii-i) When $1+a+b=1$
From \textcircled{8}, $49(2+a)=2+a$, so $2+a=0$, i.e., $a=-2$.
Also, $b=1+2-1=2$, and since $4+2a+b=2$, \textcircled{7} does not hold.
(ii-ii) When $1+a+b=-1$
From \textcircled{8}, $49(2+a)=2+a$, so $2+a=0$, i.e., $a=-2$.
Also, $b=-1+2-1=0$, and since $4+2a+b=0$, \textcircled{7} holds.
From (i)(ii), the pairs $(a,\,b)$ sought are $(a,\,b)=(-2,\,1),\,(-2,\,0)$.
[Commentary]
This is a problem based on polynomial division. In (2), by setting $g(x)=h(x)^7$ and applying the result of (1), the solution becomes somewhat more concise. Also, regarding the description that the necessary and sufficient condition for $H(x)$ to be divisible by $f(x)$ is \textcircled{3} and \textcircled{4} and \textcircled{5}, we were uncertain whether it could be omitted. In the solution above, we have noted it briefly.
%% Page 16 6 Go to problem page

(1) In a coordinate space with O as the origin, let $\mathrm{A}(1,1,1)$, $\mathrm{B}(-1,1,1)$, $\mathrm{C}(-1,-1,1)$, $\mathrm{D}(1,-1,1)$, $\mathrm{E}(1,1,-1)$, $\mathrm{F}(-1,1,-1)$, $\mathrm{G}(-1,-1,-1)$, $\mathrm{H}(1,-1,-1)$.

The region $S$ satisfying $z < 1$ among the surface of the cube defined by $|x|\leq 1$, $|y|\leq 1$, $|z|\leq 1$ represents the five faces of cube ABCD-EFGH other than face ABCD.

Now, the range $V$ of points P satisfying $\mathrm{OP} \leq \sqrt{3}$ and the condition that segment OP and $S$ share no common point or share only point P is:

(a) The interior or surface of cube ABCD-EFGH
In this case, the volume of the range of point P is $2^3 = 8$.

(b) The upper side of plane ABCD
In this case, the range of point P is the upper side of planes OAB, OBC, OCD, ODA, and the interior or surface of a sphere of radius $\sqrt{3}$ centered at O. By symmetry, its volume is $$\frac{1}{6}\left\{\frac{4}{3}\pi(\sqrt{3})^3 - 2^3\right\} = \frac{2}{3}\sqrt{3}\pi - \frac{4}{3}.$$
From (a) and (b), the volume of range $V$ is $$8 + \left(\frac{2}{3}\sqrt{3}\pi - \frac{4}{3}\right) = \frac{2}{3}\sqrt{3}\pi + \frac{20}{3}.$$

(2) The range $W$ of points P satisfying $\mathrm{ON} + \mathrm{NP} \leq \sqrt{3}$, with segment ON and $S$ sharing no common point, and segment NP and $S$ sharing no common point or sharing only point P is:

(c) When O, N, P are collinear
The range of point P in this case is the same as in (1), so its volume is $\dfrac{2}{3}\sqrt{3}\pi + \dfrac{20}{3}$.

(d) When O, N, P are not collinear
Below, we find the range of point P not included in case (c).
In this case, it suffices to consider when point N is on a side of square ABCD, so first let M be the midpoint of side AB and consider the case where point N is on segment AM.
First, as preparation, expressing the range of point P from case (b): $$z \geq 1,\quad z \geq x,\quad z \geq -x,\quad z \geq y,\quad z \geq -y,\quad x^2+y^2+z^2 \leq 3$$
Here, letting $0 \leq t \leq 1$, the inequalities representing the cross-section at the plane $x = t$ passing through point $\mathrm{N}(t, 1, 1)$ are: $$z \geq 1,\quad z \geq t,\quad z \geq -t,\quad z \geq y,\quad z \geq -y$$ $$y^2 + z^2 \leq 3 - t^2$$
This is shown as the shaded region in the figure on the right.
[Figure: cross-section in the $yz$-plane showing shaded region bounded by $z\geq 1$, $z\geq \pm y$, and $y^2+z^2\leq 3-t^2$, with $\sqrt{3-t^2}$ marked on the $z$-axis and N marked]

$-10-$ \copyright\ 電送数学舎 2023
%% Page 17 When O, N, P are not collinear, the range of the newly added point P forms a sector centered at N with radius $r = \sqrt{3-t^2} - \sqrt{2}$ and central angle $\dfrac{3}{4}\pi$. As shown in the figure, this corresponds to the darkly shaded region in the figure on the right.
[Figure: A sector in the $yz$-plane centered at N with radius $\sqrt{3-t^2}$, with shaded region]
Note that when $t = 0 \to 1$, $r = \sqrt{3} - \sqrt{2} \to 0$, and $r$ decreases monotonically as $t$ increases.
Now, this cross-sectional area is $\dfrac{1}{2}\!\left(\sqrt{3-t^2}-\sqrt{2}\right)^2 \cdot \dfrac{3}{4}\pi = \dfrac{3}{8}\pi\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)$,
and considering symmetry, the volume of the range of the newly added point P when O, N, P are not collinear is:
$$4 \times 2\int_0^1 \frac{3}{8}\pi\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)dt = 3\pi\int_0^1\!\left(5 - t^2 - 2\sqrt{2}\sqrt{3-t^2}\right)dt$$
$$= 3\pi\!\left[5t - \frac{t^3}{3}\right]_0^1 - 6\sqrt{2}\pi\int_0^1\sqrt{3-t^2}\,dt = 14\pi - 6\sqrt{2}\pi\int_0^1\sqrt{3-t^2}\,dt$$
Here, let $I = \displaystyle\int_0^1 \sqrt{3-t^2}\,dt$, and substitute $t = \sqrt{3}\sin\theta$ $\left(-\dfrac{\pi}{2} \leqq \theta \leqq \dfrac{\pi}{2}\right)$,
then $dt = \sqrt{3}\cos\theta\,d\theta$, and since $\sin\alpha = \dfrac{1}{\sqrt{3}}$, when $t = 0 \to 1$, $\theta = 0 \to \alpha$.
$$I = \int_0^{\alpha}\sqrt{3 - 3\sin^2\theta}\cdot\sqrt{3}\cos\theta\,d\theta = 3\int_0^{\alpha}\cos^2\theta\,d\theta = \frac{3}{2}\int_0^{\alpha}(1+\cos 2\theta)\,d\theta$$
$$= \frac{3}{2}\!\left[\theta + \frac{1}{2}\sin 2\theta\right]_0^{\alpha} = \frac{3}{2}\alpha + \frac{3}{4}\sin 2\alpha = \frac{3}{2}\alpha + \frac{3}{2}\sin\alpha\cos\alpha$$
$$= \frac{3}{2}\alpha + \frac{3}{2}\cdot\frac{1}{\sqrt{3}}\sqrt{1 - \frac{1}{3}} = \frac{3}{2}\alpha + \frac{\sqrt{2}}{2}$$
Therefore, $14\pi - 6\sqrt{2}\pi I = 14\pi - 6\sqrt{2}\pi\!\left(\dfrac{3}{2}\alpha + \dfrac{\sqrt{2}}{2}\right) = (8 - 9\sqrt{2}\alpha)\pi$.
From (c)(d), the volume of the region $W$ is:
$$\left(\frac{2}{3}\sqrt{3}\pi + \frac{20}{3}\right) + (8 - 9\sqrt{2}\alpha)\pi = \left(8 + \frac{2}{3}\sqrt{3} - 9\sqrt{2}\alpha\right)\pi + \frac{20}{3}$$

[Commentary]
This is a difficult problem asking for the volume of a solid. For (1), I initially tried to compute it by integration but it did not work out well, so I used a method utilizing symmetry. For (2), since the segment OP bends, the problem asks for the volume of the region ``overflowing'' from each side of the square ABCD on the top face of the cube, but the description relies considerably on intuition.

Consider a function $f ( s )$ defined by the following integral for positive real numbers $s$.
$$f ( s ) = \int _ { 0 } ^ { \infty } t ^ { s - 1 } \exp ( - t ) \mathrm { d } t$$
Answer the following questions. You may answer without showing that the above integral converges.
(1) Find the value of $f ( 1 )$.
(2) The inequality $\exp ( t ) > \frac { t ^ { n } } { n ! }$ holds for any positive real number $t$ and non-negative integer $n$.
(a) For positive real numbers $s$, show the following inequality.
$$\int _ { 0 } ^ { 1 } t ^ { s - 1 } \exp ( - t ) \mathrm { d } t < \frac { 1 } { s }$$
(b) When $n > s > 0$, show that the following inequality holds for any real number $c$ that satisfies $c > 1$.
$$\int _ { 1 } ^ { c } t ^ { s - 1 } \exp ( - t ) \mathrm { d } t < \frac { n ! } { n - s }$$
(3) When the second-order derivative of $f ( s )$ is expressed as
$$\frac { \mathrm { d } ^ { 2 } f ( s ) } { \mathrm { d } s ^ { 2 } } = \int _ { 0 } ^ { \infty } g ( t , s ) \exp ( - t ) \mathrm { d } t$$
find a function $g ( t , s )$. You may answer without showing that the order of differentiation and integration can be exchanged.
(4) Find the value of $D$ defined as
$$D = \int _ { 0 } ^ { \infty } ( \log t ) ^ { 2 } \exp ( - t ) \mathrm { d } t - \left( \int _ { 0 } ^ { \infty } ( \log t ) \exp ( - t ) \mathrm { d } t \right) ^ { 2 }$$
Here, you may use the fact that the following relation holds.
$$\left. \frac { \mathrm { d } ^ { 2 } } { \mathrm {~d} s ^ { 2 } } \log f ( s ) \right| _ { s = 1 } = \frac { \pi ^ { 2 } } { 6 }$$
(5) Define a function $p ( r )$ for positive real numbers $r$ and $\alpha$ as
$$p ( r ) = \frac { r } { \alpha } \exp \left( - \frac { r ^ { 2 } } { 2 \alpha } \right)$$
Find the value of $S$ defined as
$$S = \int _ { 0 } ^ { \infty } ( \log r ) ^ { 2 } p ( r ) \mathrm { d } r - \left( \int _ { 0 } ^ { \infty } ( \log r ) p ( r ) \mathrm { d } r \right) ^ { 2 }$$

Consider a particle moving on the coordinate plane, and denote the location of the particle at time $t \in \{ 0,1,2 , \ldots \}$ by $\left( X _ { t } , Y _ { t } \right)$. The initial location of the particle is $\left( X _ { 0 } , Y _ { 0 } \right) = ( 0,0 )$. Also, if $\left( X _ { t } , Y _ { t } \right) = ( a , b )$, then $\left( X _ { t + 1 } , Y _ { t + 1 } \right) = ( a + 1 , b )$ with probability $p , \left( X _ { t + 1 } , Y _ { t + 1 } \right) = ( a , b + 1 )$ with probability $q$, and $\left( X _ { t + 1 } , Y _ { t + 1 } \right) = ( a , b )$ with probability $1 - p - q$. Here, it is assumed that $p , q > 0 , p + q < 1$, and the movements of the particle at different time points are independent. Let $( X , Y )$ denote the location of the particle such that $\left( X _ { t + 1 } , Y _ { t + 1 } \right) = \left( X _ { t } , Y _ { t } \right)$ for the first time. Answer the following questions.
(1) Show that the probability that $( X , Y ) = ( 1,2 )$ is $3 p q ^ { 2 } ( 1 - p - q )$.
(2) For non-negative integers $n$, find the probability that $X + Y = n$.
(3) For non-negative integers $n$, let $f _ { n }$ denote the probability that $X = n$.
(a) Find $f _ { 0 }$.
(b) Express the probability that $X \geq n + 1$ given the condition $X \geq n$, using $f _ { 0 }$.
(c) Show that $f _ { n } = \left( 1 - f _ { 0 } \right) ^ { n } f _ { 0 }$.
(4) Express the expectation of $X$ using $p$ and $q$.
(5) Express the correlation coefficient between $X$ and $Y$ $$\frac { E \left[ \left( X - \mu _ { X } \right) \left( Y - \mu _ { Y } \right) \right] } { \sqrt { E \left[ \left( X - \mu _ { X } \right) ^ { 2 } \right] E \left[ \left( Y - \mu _ { Y } \right) ^ { 2 } \right] } }$$ using $p$ and $q$, where $\mu _ { X } = E [ X ]$ denotes the expectation of $X$ and $\mu _ { Y } = E [ Y ]$ denotes the expectation of $Y$.

Problem 5

I. We consider a continuous and absolutely integrable function $f ( t )$ of a real variable $t$ and denote the Fourier transform of the function $f ( t )$ as $\mathcal { F } \{ f ( t ) \}$. We define a function $F ( \omega )$ by the following formula:
$$F ( \omega ) = \mathcal { F } \{ f ( t ) \} = \int _ { - \infty } ^ { \infty } f ( t ) \exp ( - i \omega t ) \mathrm { d } t$$
where $\omega$ is a real variable and $i$ is the imaginary unit.

1. We define $g ( t ) = f ( a t )$ for a constant $a$ satisfying $a > 0$ and $$G ( \omega ) = \mathcal { F } \{ g ( t ) \} = \mathcal { F } \{ f ( a t ) \}$$ Express $G ( \omega )$ using the function $F$.
2. When $f ( t ) = \exp \left( - t ^ { 2 } \right)$ and $a = 2$, sketch the graph of the Fourier transformed functions $F ( \omega )$ and $G ( \omega )$ defined by Equation (2) as a function of $\omega$ to show the difference between them.
3. We define $h ( t ) = f ( t ) \exp ( - i b t )$ for a constant $b$ satisfying $b > 0$ and $$H ( \omega ) = \mathcal { F } \{ h ( t ) \} = \mathcal { F } \{ f ( t ) \exp ( - i b t ) \}$$ Express $H ( \omega )$ using the function $F$.
4. When $f ( t ) = \exp \left( - t ^ { 2 } \right)$ and $b = 2$, sketch the graph of the Fourier transformed functions $F ( \omega )$ and $H ( \omega )$ defined by Equation (3) as a function of $\omega$ to show the difference between them.

II. Let $N$ be a positive integer. We define a discrete Fourier transform $D _ { 1 } , \cdots , D _ { N }$ by the following formula:
$$D _ { m } = \frac { 1 } { \sqrt { N } } \sum _ { n = 1 } ^ { N } c _ { n } \exp \left( - i \frac { 2 \pi } { N } n m \right)$$
for a complex sequence $c _ { 1 } , \cdots , c _ { N }$. Here, $m$ is an integer satisfying $1 \leq m \leq N$.

1. Calculate $S \left( n , n ^ { \prime } \right)$ : $$S \left( n , n ^ { \prime } \right) = \frac { 1 } { N } \sum _ { m = 1 } ^ { N } \exp \left\{ i \frac { 2 \pi } { N } \left( n - n ^ { \prime } \right) m \right\}$$ Here, $n$ is an integer satisfying $1 \leq n \leq N$, and $n ^ { \prime }$ is an integer satisfying $1 \leq n ^ { \prime } \leq N$.
2. Let $U _ { m n }$ be a complex number satisfying $$D _ { m } = \sum _ { n = 1 } ^ { N } U _ { m n } c _ { n }$$ Show that the matrix $\mathbf { U } = \left[ U _ { m n } \right] _ { 1 \leq m \leq N , 1 \leq n \leq N }$ is a unitary matrix.
3. Derive an equation for the inverse discrete Fourier transform $c _ { n }$ from $D _ { 1 } , \cdots , D _ { N }$. Here, $n$ is an integer satisfying $1 \leq n \leq N$.
4. For any complex value $z , \bar { z }$ is the complex conjugate of $z$. We define $Q$ by $$Q = \sum _ { n = 1 } ^ { N } \left( \overline { c _ { n } } c _ { n + 1 } + \overline { c _ { n + 1 } } c _ { n } \right)$$ Express $Q$ in terms of $D _ { m }$ and $\overline { D _ { m } }$. Here, we impose the condition $c _ { N + 1 } = c _ { 1 }$.

Problem 6

Consider an electric vehicle charging station with a single charger installed and let us observe the number of vehicles at the station at regular time intervals.
Arriving vehicles at the station are lined up in the queue in the order of arrival, and only the first vehicle in the queue can be charged. In the interval between one observation and the next observation, assume that one new vehicle arrives with probability $p ( 0 < p < 1 )$, and that the vehicle charging at the head of the queue completes charging with probability $q ( 0 < q < 1 )$. Here, assume that $p$ and $q$ are constants and $p + q < 1$.
The queue can accommodate $N ( N \geq 2 )$ vehicles, including the vehicle being charged at the head of the queue, and the $( N + 1 )$-th vehicle shall give up and leave the station without queuing up. The vehicle which completes charging leaves the station immediately.
In the interval between one observation and the next observation, either only one or no vehicles arrive at the station and either only one or no vehicles complete charging. Moreover, assume that both arrival of new vehicle and completion of charging for the first vehicle do not occur together in any one interval.
I. When there are $i ( 0 < i < N )$ vehicles in the queue, find the probability for the following condition: no new vehicle arrives and the first vehicle does not leave in the interval between one observation and the next observation.
Let $\pi _ { i } ( 0 \leq i \leq N )$ be the probability that $i$ vehicles are in the queue in the steady state.
II. Express the relationship between $\pi _ { i }$ and $\pi _ { i + 1 }$. Here, $i \leq N - 1$.
III. Express $\pi _ { i }$ using $p , q$ and $N$.
IV. Find the expected value of the number of vehicles at the station in the steady state using $p , q$ and $N$. Here, $p < q$.

Problem 5

Consider a function $f ( t )$ of a real number $t$, where $| f ( t ) |$ and $| f ( t ) | ^ { 2 }$ are integrable. Let $F ( \omega ) = \mathcal { F } [ f ( t ) ]$ denote the Fourier transform of $f ( t )$. It is defined as
$$F ( \omega ) = \mathcal { F } [ f ( t ) ] = \int _ { - \infty } ^ { \infty } f ( t ) \exp ( - i \omega t ) \mathrm { d } t$$
where $\omega$ is a real number and $i$ is the imaginary unit. Then, the following equation is satisfied:
$$\int _ { - \infty } ^ { \infty } | F ( \omega ) | ^ { 2 } \mathrm {~d} \omega = 2 \pi \int _ { - \infty } ^ { \infty } | f ( t ) | ^ { 2 } \mathrm {~d} t$$
Also, let $R _ { f } ( \tau )$ denote the autocorrelation function of $f ( t )$. It is defined as
$$R _ { f } ( \tau ) = \int _ { - \infty } ^ { \infty } f ( t ) f ( t - \tau ) \mathrm { d } t$$
where $\tau$ is a real number.
I. Consider a case where $f ( t )$ is defined as follows:
$$f ( t ) = \begin{cases} \cos ( a t ) & \left( | t | \leq \frac { \pi } { 2 a } \right) \\ 0 & \left( | t | > \frac { \pi } { 2 a } \right) \end{cases}$$
Here, $a$ is a positive real constant. Find the followings:

1. $F ( \omega )$,
2. $R _ { f } ( \tau )$,
3. $\mathcal { F } \left[ R _ { f } ( \tau ) \right]$.

II. Find the values of the following integrals. Here, you may use the results of I.

1. $\int _ { - \infty } ^ { \infty } \frac { \cos ^ { 2 } \frac { \pi x } { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \mathrm {~d} x$,
2. $\int _ { - \infty } ^ { \infty } \frac { \cos ^ { 4 } \frac { \pi x } { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 4 } } \mathrm {~d} x$.

$$\frac { 0,2 - 0,025 } { 0,5 }$$
What is the result of this operation?
A) $\frac { 3 } { 5 }$
B) $\frac { 4 } { 5 }$
C) $\frac { 7 } { 20 }$
D) $\frac { 8 } { 25 }$
E) $\frac { 12 } { 25 }$

$$\frac { 5 \left( 2 - \frac { 3 } { 5 } \right) } { 2 \left( 3 - \frac { 5 } { 2 } \right) }$$
What is the result of this operation?
A) $\frac { 5 } { 2 }$
B) $\frac { 7 } { 2 }$
C) 3
D) 5
E) 7

For natural numbers $x$ and $y$
$$\begin{array} { r | r | r } x & \frac { 10 } { m } \\ { } ^ { - } & = ^ { y } \frac { 15 } { 3 } \end{array}$$
Given this, what is the remainder when the product $x \cdot y$ is divided by 5?
A) 0
B) 1
C) 2
D) 3
E) 4

Let $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ be propositions with their negations denoted by $\mathbf { p } ^ { \prime } , \mathbf { q } ^ { \prime } , \mathbf { r } ^ { \prime }$ respectively. Which of the following is equivalent to the proposition
$$p \vee q \Rightarrow q \wedge r$$
?
A) $\mathrm { p } ^ { \prime } \wedge \mathrm { q } ^ { \prime } \Rightarrow \mathrm { q } ^ { \prime } \vee \mathrm { r } ^ { \prime }$
B) $\mathrm { p } ^ { \prime } \wedge \mathrm { q } ^ { \prime } \Rightarrow \mathrm { q } ^ { \prime } \wedge \mathrm { r } ^ { \prime }$
C) $p ^ { \prime } \vee q ^ { \prime } \Rightarrow q ^ { \prime } \wedge r ^ { \prime }$
D) $q ^ { \prime } \wedge r ^ { \prime } \Rightarrow p ^ { \prime } \vee q ^ { \prime }$
E) $q ^ { \prime } \vee r ^ { \prime } \Rightarrow p ^ { \prime } \wedge q ^ { \prime }$

$$\begin{aligned} & A = \{ a , b , e \} \\ & B = \{ a , b , c , d \} \end{aligned}$$
Given this, how many sets $K$ satisfy the condition $( A \cap B ) \subseteq K \subseteq ( A \cup B )$?
A) 3
B) 4
C) 5
D) 8
E) 9

On the set of positive integers, the operations $\oplus$ and $\otimes$ are defined using the greatest common divisor and least common multiple as follows:
$$\begin{aligned} & a \oplus b = \operatorname { GCD } ( a , b ) \\ & a \otimes b = \operatorname { LCM } ( a , b ) \end{aligned}$$
Accordingly, what is the result of the operation $18 \oplus ( 12 \otimes 4 )$?
A) 2
B) 3
C) 6
D) 8
E) 9

The sum of a three-digit number $ABC$ and a two-digit number $AB$ is 392.
Accordingly, what is the sum $\mathrm { A } + \mathrm { B } + \mathrm { C }$?
A) 7
B) 9
C) 11
D) 15
E) 19

A two-digit number $AB$ is called a symmetric prime if both $AB$ and $BA$ are prime numbers.
For a symmetric prime number $AB$, which of the following cannot be the product A.B?
A) 7
B) 9
C) 15
D) 21
E) 63

A greengrocer bought lemons in bags containing 12 lemons each and sold them in groups of three. The greengrocer bought one bag of lemons for 5 TL and sold 3 lemons for 2 TL.
How much profit did this greengrocer make from the sale of 4 bags of lemons?
A) 6
B) 8
C) 9
D) 10
E) 12

A car tire seller observes that when he applies a 25\% end-of-season discount on tires, the number of tires sold in a day increases by 40\%.
Accordingly, by what percentage has the money entering the seller's cash register increased in a day?
A) 5
B) 10
C) 15
D) 20
E) 25

A farmer filled five of six containers with capacities of 5, 9, 12, 15, 23, and 45 liters with sunflower oil and olive oil. The amount of sunflower oil the farmer put in the containers is 4 times the amount of olive oil.
Accordingly, how many liters is the empty container?
A) 5
B) 9
C) 12
D) 15
E) 23

The weights of four wrestlers participating in a wrestling competition were measured at one-week intervals. The change in the wrestlers' weights in the second measurement compared to the first measurement is shown in the graph below.
If the average weight of the wrestlers in the first measurement was 56 kilograms, what is it in the second measurement?
A) 53
B) 54
C) 55
D) 57
E) 58