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(1)

\(C_1 , C_2\) の式より \[\begin{align} x^2 = -x^2 +2kx +1 & -k^2 \\ \text{∴} \quad 2x^2 -2kx +k^2 -1 & = 0 \quad ... [1] \ . \end{align}\] [1] の \(2\) つの解が P , Q の \(x\) 座標なので, 判別式 \(D\) について \[\begin{align} \dfrac{D}{4} & = k^2 -2 ( k^2 -1 ) \\ & = 2 -k^2 \gt 0 \\ \text{∴} \quad -\sqrt{2} & \lt k \lt \sqrt{2} \quad ... [2] \ . \end{align}\] また, 解と係数の関係から \[\begin{align} \dfrac{2k}{2} \gt 0 \ & , \ \dfrac{k^2 -1}{2} \gt 0 \\ \text{∴} \quad k & \gt 1 \quad ... [3] \ . \end{align}\] [2] [3] より, 求める範囲は \[ \underline{1 \lt k \lt \sqrt{2}} \ . \]

(2)

P , Q の \(x\) 座標を \(p , q \ ( p \gt q )\) とおくと \[ p+q = k \ , \ pq = \dfrac{k^2 -1}{2} \quad ... [4] \ . \] P \(( p , p^2 )\) , Q \(( q , q^2 )\) なので \[ \text{G} \left( \dfrac{p+q}{3} , \dfrac{p^2 +q^2}{3} \right) \ . \] [4] を用いて \[\begin{align} p^2 +q^2 & = ( p+q )^2 -2pq \\ & = k^2 -2 \cdot \dfrac{k^2 -1}{2} = 1 \ . \end{align}\] ゆえに, G \(\left( \dfrac{k}{3} , 1 \right)\) .
(1) の結果より, \(\dfrac{1}{3} \lt \dfrac{k}{3} \lt \dfrac{\sqrt{2}}{3}\) なので, 求める軌跡は \[ \underline{\text{線分} \ : \ y = \dfrac{1}{3} \ \left( \dfrac{1}{3} \lt x \lt \dfrac{\sqrt{2}}{3} \right)} \ . \]

(3)

\[\begin{align} S & = \dfrac{1}{2} \left| p q^2 -p^2 q \right| \\ & = \dfrac{1}{2} pq | p-q | \ . \end{align}\] なので, [4] も用いて \[\begin{align} S^2 & = \dfrac{1}{4} (pq)^2 \left\{ (p+q)^2 -4pq \right\} \\ & = \dfrac{1}{4} \cdot \dfrac{( k^2 -1 )^2}{4} \left\{ k^2 -2 ( k^2 -1 ) \right\} \\ & = \underline{\dfrac{1}{16} ( k^2 -1 )^2 ( 2 -k^2 )} \ . \end{align}\]

(4)

\(t =k^2\) とおくと, \(1 \lt t \lt 2\) ... [5] .
\(f(t) = ( t-1 )^2 ( 2-t )\) とおけば, [5] において \(f(t)\) が最大になるとき, \(S\) も最大となる. \[\begin{align} f'(t) & = 2 ( t-1 ) ( 2-t ) -( t-1 )^2 \\ & = ( t-1 ) ( 5 -3t ) \ . \end{align}\] したがって, [5] における \(f(t)\) の増減は下表の通り. \[ \begin{array}{c|ccccc} t & ( 1 ) & \cdots & \dfrac{5}{3} & \cdots & ( 2 ) \\ \hline f'(t) & (0) & + & 0 & - & \\ \hline f(t) & & \nearrow & \text{最大} & \searrow & \end{array} \] したがって, \(f(t)\) の最大値は \[ f \left( \dfrac{5}{3} \right) = \left( \dfrac{2}{3} \right)^2 \cdot \dfrac{1}{3} = \dfrac{4}{27} \ . \] よって, \(S\) が最大となるのは \[ k = \underline{\dfrac{\sqrt{15}}{3}} \] のときで, このとき, G の座標は \[ \underline{\left(\dfrac{\sqrt{15}}{9} , \dfrac{1}{3} \right)} \ . \]

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(1)

円 \(C_1 , C_2\) の半径をそれぞれ \(r_1 , r_2\) とおく.
円 \(C_1 , C_2\) の接点を R , それぞれの中心を \(\text{O} {} _ 1 , \text{O} {} _ 2\) , \(x\) 軸との接点を \(\text{H} {} _ 1 , \text{H} {} _ 2\) とおく.
\(d_1 = \sin 2 \theta\) , \(\angle \text{O} {} _ 1 \text{OH} {} _ 1 = \theta\) なので \[ r_1 = \sin 2 \theta \cdot \sin \theta = \underline{2 \sin^2 \theta \cos \theta} \ . \] \(d_2 = \dfrac{r_2}{\sin \theta}\) であり, \(d_1 = d_2 +r_1 +r_2\) なので \[\begin{gather} \dfrac{r_2}{\sin \theta} +r_2 +2 \sin^2 \theta \cos \theta = 2 \sin \theta \cos \theta \\ \text{∴} \quad r_2 = \underline{\dfrac{2 \sin^2 \theta \cos \theta ( 1 -\sin \theta )}{1 +\sin \theta}} \ . \end{gather}\]

(2)

\[\begin{align} \text{OR} & = \dfrac{( 1 +\sin \theta ) r_2}{\sin \theta} \\ & = 2 \sin \theta \cos \theta ( 1 -\sin \theta ) \ . \end{align}\] R は PQ の中点なので \[\begin{align} \text{PQ} & = 2 \tan \theta \cdot \text{OR} \\ & = 4 \sin ^2 \theta ( 1 -\sin \theta ) \ . \end{align}\] \(x = \sin \theta\) とおくと, \(0 \lt x \lt \dfrac{1}{\sqrt{2}}\) ... [1] .
\(f(x) = x^2 ( 1-x )\) とおくと \[ f'(x) = 2x -3x^2 = x ( 2 -3x ) \ . \] したがって, [1] における \(f(x)\) の増減は下表の通り. \[ \begin{array}{c|ccccc} x & ( 0 ) & \cdots & \dfrac{2}{3} & \cdots & \left( \dfrac{1}{\sqrt{2}} \right) \\ \hline f'(x) & & + & 0 & - & \\ \hline f(x) & & \nearrow & \text{最大} & \searrow & \end{array} \] ゆえに, \(f(x)\) の最大値は \[ f \left( \dfrac{2}{3} \right) = \dfrac{4}{9} \cdot \dfrac{1}{3} = \dfrac{4}{27} \ . \] よって, PQ の長さの最大値は \[ 4 \cdot \dfrac{4}{27} = \underline{\dfrac{16}{27}} \ . \]

(3)

\(\sin \theta = \dfrac{2}{3}\) のとき \[ \cos \theta = \dfrac{\sqrt{5}}{3} \ , \ \tan \theta = \dfrac{2}{\sqrt{5}} \ . \] 直線 \(m\) は OR と垂直なので, 傾きは \[ \tan \left( \theta +\dfrac{\pi}{2} \right) = -\dfrac{1}{\tan \theta} = -\dfrac{\sqrt{5}}{2} \ . \] また \[\begin{align} \text{OQ} & = \dfrac{\text{OR}}{\cos \theta} = 2 \sin \theta ( 1 -\sin \theta ) \\ & = 2 \cdot \dfrac{2}{3} \cdot \dfrac{1}{3} = \dfrac{4}{9} \ . \end{align}\] よって, このときの \(m\) の式は \[ \underline{y = -\dfrac{\sqrt{5}}{2} \left( x -\dfrac{4}{9} \right)} \ . \]

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(1)

\(\overrightarrow{\text{OP}} = t \overrightarrow{a}\) , \(\overrightarrow{\text{OQ}} = ( 1-t ) \overrightarrow{b} +t \overrightarrow{c}\) なので \[\begin{align} \overrightarrow{\text{OM}} & = \dfrac{1}{2} \left( \overrightarrow{\text{OP}} +\overrightarrow{\text{OQ}} \right) \\ & = \underline{\dfrac{t}{2} \overrightarrow{a} +\dfrac{1-t}{2} \overrightarrow{b} +\dfrac{t}{2} \overrightarrow{c}} \ . \end{align}\]

(2)

\(\text{OM} = \text{BM}\) より \[\begin{align} \left| \overrightarrow{\text{OM}} \right|^2 -\left| \overrightarrow{\text{BM}} \right|^2 & = \left( \overrightarrow{\text{OM}} +\overrightarrow{\text{BM}} \right) \cdot \left( \overrightarrow{\text{OM}} +\overrightarrow{\text{BM}} \right) \\ & = \left( 2 \overrightarrow{\text{OM}} -\overrightarrow{\text{OB}} \right) \cdot \overrightarrow{\text{OB}} \\ & = t \left( \overrightarrow{a} -\overrightarrow{b} +\overrightarrow{c} \right) \cdot \overrightarrow{b} \\ & = t \left( 2 -\left| \overrightarrow{b} \right|^2 \right) = 0 \ . \end{align}\] \(t \neq 0\) なので \[ \left| \overrightarrow{b} \right| = \text{OB} = \underline{\sqrt{2}} \ . \]

(3)

\(\text{OM} = \text{AM}\) より \[\begin{align} \left| \overrightarrow{\text{OM}} \right|^2 -\left| \overrightarrow{\text{AM}} \right|^2 & = \left( \overrightarrow{\text{OM}} +\overrightarrow{\text{AM}} \right) \cdot \left( \overrightarrow{\text{OM}} +\overrightarrow{\text{AM}} \right) \\ & = \left( 2 \overrightarrow{\text{OM}} -\overrightarrow{\text{OA}} \right) \cdot \overrightarrow{\text{OA}} \\ & = \left\{ -(1-t) \overrightarrow{a} +(1-t) \overrightarrow{b} +t \overrightarrow{c} \right\} \cdot \overrightarrow{a} \\ & = 1 -(1-t) \left| \overrightarrow{a} \right|^2 = 0 \ . \end{align}\] \(t \neq 0\) なので \[ \left| \overrightarrow{a} \right| = \text{OA} = \dfrac{1}{\sqrt{1-t}} \ . \] \(\text{OM} = \text{CM}\) より, 同様にすれば \[ \left| \overrightarrow{c} \right| = \text{OC} = \dfrac{1}{\sqrt{1-t}} \ . \] ゆえに, \(\text{OA} = \text{OC}\) ... [1] .
条件より \[\begin{gather} \overrightarrow{a} \cdot \overrightarrow{b} = \dfrac{\sqrt{2}}{\sqrt{1-t}} \cos \angle \text{AOB} = 1 \\ \text{∴} \quad \cos \angle \text{AOB} = \dfrac{\sqrt{1-t}}{\sqrt{2}} \ . \end{gather}\] \(\overrightarrow{a} \cdot \overrightarrow{c} = 1\) より, 同様にすれば \[ \cos \angle \text{COB} = \dfrac{\sqrt{1-t}}{\sqrt{2}} \ . \] ゆえに, \(\angle \text{AOB} = \angle \text{COB}\) ... [2] .
△OAB と △OCB は, OB を共有し, [1] [2] より, \(2\) 辺とその間の角がそれぞれ等しいので \[ \triangle \text{OAB} \equiv \triangle \text{OCB} \ . \]

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(1)

\[\begin{align} f'(x) & = \dfrac{1}{\sqrt{x}} \cdot e^{-x} -2 \sqrt{x} e^{-x} \\ & = \dfrac{( 1 -2x ) e^{-x}}{\sqrt{x}} \ . \end{align}\] \(f'(x) = 0\) をとくと \[ x = a = \underline{\dfrac{1}{2}} \ . \] さらに \[\begin{align} f''(x) & = -\dfrac{1}{2 x^{\frac{3}{2}}} \cdot e^{-x} -\dfrac{2 ( 1 -2x ) e^{-x}}{\sqrt{x}} \\ & = \dfrac{( 4x^2 -4x +1 ) e^{-x}}{2 x^{\frac{3}{2}}} \\ & = \dfrac{\left( x -\frac{1 -\sqrt{2}}{2} \right) \left( x -\frac{1 +\sqrt{2}}{2} \right) e^{-x}}{2 x^{\frac{3}{2}}} \ . \end{align}\] \(x \geqq 0\) において, \(f'(x) = 0\) をとくと \[ x = b = \underline{\dfrac{1 +\sqrt{2}}{2}} \ . \] また \[\begin{align} & f(0) = 0 \ , \quad \displaystyle\lim _ {x \rightarrow \infty} f(x) = 0 \ , \\ & \displaystyle\lim _ {x \rightarrow +0} f'(x) = \displaystyle\lim _ {x \rightarrow +0} \left( \dfrac{1}{\sqrt{x}} -2 \sqrt{x} \right) e^{-x} = \infty \end{align}\]

したがって, \(f(x)\) の増減, 凹凸は下表の通り. \[ \begin{array}{c|ccccc} x & 0 & \cdots & \dfrac{1}{2} & \cdots & \dfrac{1 +\sqrt{2}}{2} & \cdots & ( \infty ) \\ \hline f'(x) & ( \infty ) & + & 0 & - & & - & \\ \hline f''(x) & & - & & - & 0 & + & \\ \hline f(x) & 0 & \nearrow ( \cap ) & & \searrow ( \cap ) & & \searrow ( \cup ) & (0) \end{array} \] \[ f \left( \dfrac{1}{2} \right) = \sqrt{\dfrac{2}{e}} \ , \ f \left( \dfrac{1 +\sqrt{2}}{2} \right) = \sqrt{2 +2 \sqrt{2}} e^{\frac{1 +\sqrt{2}}{2}} \] なので, \(y = f(x)\) のグラフの概形は下図の通り.

(2)

\(G(x) = \displaystyle\int x e^{-2x} \, dx\) とおくと \[\begin{align} G(x) & = -\dfrac{1}{2} x e^{-2x} +\dfrac{1}{2} \displaystyle\int e^{-2x} \, dx \\ & = -\dfrac{1}{2} x e^{-2x} -\dfrac{1}{4} e^{-2x} \\ & = -\dfrac{2x +1}{4} e^{-2x} +C \quad ( \ C \ \text{は積分定数} \ ) \ . \end{align}\] なので \[\begin{align} V(k) & = G(k) -G(0) \\ & = -\dfrac{2k +1}{4} e^{-2k} +\dfrac{1}{4} \\ & = \underline{\dfrac{1}{4} \left\{ 1 -( 2k +1 ) e^{-2k} \right\}} \ . \end{align}\]

(3)

求める体積 \(W\) は \[\begin{align} W & = 4 \pi \displaystyle\int _ {0}^{b} x e^{-2x} \, dx -4 \pi \displaystyle\int _ {0}^{a} x e^{-2x} \, dx \\ & = 4 \pi \left\{ V(b) -V(a) \right\} \\ & = \pi \left\{ 1 -( 2b+1 ) e^{-2b} -1 +( 2a+1 ) e^{-2a}\right\} \\ & = \underline{\pi \left\{ 2 e^{-1} -( 2 +\sqrt{2} ) e^{-1 -\sqrt{2}} \right\}} \ . \end{align}\]

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(1)

\[\begin{align} \theta _ 2 & = \dfrac{\pi}{2} -\dfrac{\pi -\theta}{2} = \underline{\dfrac{\theta}{2}} \ , \\ \theta _ 3 & = \pi -\dfrac{\pi -\theta}{2} -\dfrac{\pi -\frac{\theta}{2}}{2}= \underline{\dfrac{3 \theta}{4}} \ . \end{align}\]

(2)

「 \(\theta _ {n+1} +\dfrac{\theta _ n}{2} = \theta\) 」 ... [A] とする.

1. 1*　\(n = 1\) のとき \[ \theta _ 2 +\dfrac{\theta _ 1}{2} = \dfrac{\theta}{2} +\dfrac{\theta}{2} = \theta \] で, [A] が成立する.

2. 2*　\(n = k\) のとき [A] が成立すると仮定すると \[\begin{align} \theta _ {k+2} & = \angle \text{P} {} _ {k+4} \text{P} {} _ {k+2} \text{P} {} _ {k+3} \\ & = \pi -\angle \text{P} {} _ {k+1} \text{P} {} _ {k+2} \text{P} {} _ {k+3} -\angle \text{P} {} _ {k} \text{P} {} _ {k+2} \text{P} {} _ {k+1} \\ & = \pi -\dfrac{\pi -\theta _ {k}}{2} -\dfrac{\pi -\theta _ {k+1}}{2} \\ & = \dfrac{\theta _ {k}}{2} +\dfrac{\theta _ {k+1}}{2} \ . \end{align}\] 両辺に \(\dfrac{\theta _ {k+1}}{2}\) を加えれば \[ \theta _ {k+2} +\dfrac{\theta _ {k+1}}{2} = \theta _ {k+1} +\dfrac{\theta _ {k}}{2} = \theta \ . \] ゆえに, \(n = k+1\) のときも, [A] が成立する.

1* 2* から, 数学的帰納法により, すべての自然数 \(n\) について, [A] が成立することが示され, 題意も示された.

(3)

[A] を変形すると \[ \theta _ {n+1} -\dfrac{2 \theta}{3} = -\dfrac{1}{2} \left( \theta _ {n} -\dfrac{2 \theta}{3} \right) \ . \] つまり, 数列 \(\left\{ \theta _ {n} -\dfrac{2 \theta}{3} \right\}\) は, 初項 \(\theta _ 1 -\dfrac{2 \theta}{3} = \dfrac{\theta}{3}\) , 公比 \(-\dfrac{1}{2}\) の等比数列なので \[\begin{align} \theta _ {n} -\dfrac{2 \theta}{3} & = \dfrac{\theta}{3} \left( -\dfrac{1}{2} \right)^{n-1} \\ \text{∴} \quad \theta _ {n} = \dfrac{2 \theta}{3} & +\dfrac{\theta}{3} \left( -\dfrac{1}{2} \right)^{n-1} \ . \end{align}\] よって \[ \displaystyle\lim _ {n \rightarrow \infty} \theta _ n = \underline{\dfrac{2 \theta}{3}} \ . \]

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