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.
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4 (Go to problem page)
(1) For positive odd integers $K, L$ and positive integers $A, B$, suppose $KA = LB$
①Here, 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.
- [(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.
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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).