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大阪大学 前期理系 1980年度 問5
解答の著作権は解答作成者に帰属します。無断転載、複製を禁止します。
解答作成者: 森 宏征
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入試情報
| 大学名 | 大阪大学 |
|---|---|
| 学科・方式 | 前期理系 |
| 年度 | 1980年度 |
| 問No | 問5 |
| 学部 | 理学部 ・ 医学部 ・ 歯学部 ・ 薬学部 ・ 工学部 ・ 基礎工学部 |
| カテゴリ | |
| 状態 |    |
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次の条件を満たす関数 $f(x)\,\,\,(x > 0)$ を求めよ．
曲線 $C : y = f(x)\,\,\,(x > 0)$ 上の任意の点Pにおける接線と$x$軸との交点をQとすると，線分PQは$y$軸によって2等分される．
また，
点$(0,\,\,3)$と $C$ 上の点との距離の最小値は$\sqrt{\vphantom{b} 2}$である．
\end{FRAME}
\noindent{\color[named]{BurntOrange}\bfseries \fbox{解答}}
\vskip 2mm
$\P(t,\,\,f(t))\,\,\,(t > 0)$とおく．
Pにおける $C$ の接線は，
\begin{align*}
y = f'(t)(x - t) + f(t)
\tag*{$\cdott\MARU{1}$}
\end{align*}
$y$軸が線分PQを二等分することから，
$\Q(-t,\,\,0)$とおける．
Qは\MARU{1}上の点だから，
\begin{align*}
0 = f'(t)(-t - t) + f(t) = 0 \qquad
\therefore \,\,\,
f'(t) - \frac{1}{2t}f(t) = 0
\tag*{$\cdott\MARU{2}$}
\end{align*}
$\displaystyle\int \left({-\dfrac{1}{2t}}\right)dt
= -\dfrac{1}{2}\log t + A\,\,\,(Aは定数)$ に留意して，\smallskip
\MARU{2}の両辺に $e^{-\frac{1}{2}\log t} = \dfrac{1}{\sqrt{\vphantom{b} t}}$ %
を掛ければ，\\
\begin{minipage}{240pt}
\begin{gather*}
\frac{1}{\sqrt{\vphantom{b} t}}f'(t)
- \frac{2}{t\sqrt{\vphantom{b} t}}f(t) = 0 \\[1mm]
\left\{\frac{1}{\sqrt{\vphantom{b} t}}f(t) \right\}^{\! \prime} = 0
\quad
\left(\because\,\,\,
\left(\frac{1}{\sqrt{\vphantom{b} t}} \right)^{\!\! \prime}
= -\frac{2}{t\sqrt{\vphantom{b} t}}
\right)\\[1mm]
\frac{1}{\sqrt{\vphantom{b} t}}f(t) = B \quad (Bは定数) \\
\therefore \,\,\,
f(x) = B\sqrt{\vphantom{b} x}
\quad\mbox{\footnotesize$(tをxにとり換えた)$}
\end{gather*}
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\noindent%
$B \leqq 0$ ならば点$(0,\,\,3)$と $C$ 上の点との距離の最小値は
$3\,\,(> \sqrt{\vphantom{b} 2})$となり矛盾．
ゆえに $B > 0$ である．
$(0,\,\,3)$との距離の最小値を与える $C$ 上の点を$\R(u^2,\,\,Bu)\,\,\,
(u > 0)$とおく．
Rは中心$\I(0,\,\,3)$，半径$\sqrt{\vphantom{b} 2}$の円 $K$ と $C$ の接点である．
Rが $K$ 上にあることから，\\
\begin{minipage}{230pt}
\begin{align*}
u^4 + (Bu - 3)^2 = 2
\tag*{$\cdott\MARU{3}$}
\end{align*}
\smallskip$\bekutoru{$\I\R$}$ はRにおける $C$ の接線に垂直であり，
この接線はベクトル \smallskip$(1,\,\,f'(u^2)) = \dfrac{1}{2u}(2u,\,\,B)$ に
平行だから，
\begin{gather*}
(u^2,\,\,Bu - 3) \cdot (2u,\,\,B) = 0 \\
\therefore \,\,\,
2u^3 + B(Bu - 3) = 0
\tag*{$\cdots\cdotssp\MARU{4}$}
\end{gather*}
$\MARU{3} \times 2 - \MARU{4} \times u$より $Bu = v$ とおけば，
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\begin{gather*}
2(Bu - 3)^2 - Bu(Bu - 3) = 4 \\
v^2 - 9v + 14 = (v - 2)(v - 7) = 0 \\
\therefore \,\,\,
v = 2\,\,\,または\,\,\,7
\end{gather*}
$v = 2$ ならば\MARU{3}から，
\begin{gather*}
u^4 + (2 - 3)^2 = 2 \qquad
\therefore \,\,\,
u = 1 \quad(\,\because\,\,\,u > 0)
\end{gather*}
よって $2 = v = B \cdot 1$ から $B = 2$ を得る．
$v = 7$ ならば\MARU{3}から，
\begin{align*}
u^4 + (7 - 3)^2 = 2 \qquad
\therefore \,\,\,
u^4 = -14
\end{align*}
となり不適．
ゆえに，
\begin{align*}
f(x)
= \textcolor{red}{\boldsymbol{2\sqrt{\vphantom{b} x}}}
\tag*{$\Ans$}
\end{align*}
\end{document}
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