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(1)

\[\begin{align} \dfrac{1}{1 +\sin x} & = \dfrac{1 -\sin x}{\cos^2 x} \\ & = \left( \tan x +\dfrac{1}{\cos x} \right)' \end{align}\] よって \[\begin{align} \displaystyle\int _ 0^{\frac{\pi}{4}} \dfrac{dx}{1 +\sin x} & = \left[ \tan x +\dfrac{1}{\cos x} \right] _ 0^{\frac{\pi}{4}} \\ & = 1 +\sqrt{2} -1 \\ & = \underline{\sqrt{2}} \end{align}\]

(2)

求める積分値を \(I\) とおく.
\(t = \sqrt{x-1}\) とおくと \[\begin{align} x & = t^2 +1 \\ \text{∴} \quad dx & = 2t \, dt \end{align}\] また \[ \begin{array}{c|ccc} x & \dfrac{4}{3} & \rightarrow & 2 \\ \hline t & \dfrac{1}{\sqrt{3}} & \rightarrow & 1 \end{array} \] なので \[\begin{align} I & = \displaystyle\int _ \frac{1}{\sqrt{3}}^1 \dfrac{2t}{(t^2+1)^2 t} \, dt \\ & = 2 \displaystyle\int _ \frac{1}{\sqrt{3}}^1 \dfrac{dt}{(t^2+1)^2} \end{align}\] さらに, \(t = \tan \theta \ \left( -\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2} \right)\) とおくと \[ dt = \dfrac{d \theta}{\cos^2 \theta} \] また \[ \begin{array}{c|ccc} t & \dfrac{1}{\sqrt{3}} & \rightarrow & 1 \\ \hline \theta & \dfrac{\pi}{6} & \rightarrow & \dfrac{\pi}{4} \end{array} \] よって \[\begin{align} I & = 2 \displaystyle\int _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta \, d \theta \\ & = \displaystyle\int _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \left( 1 +\cos 2 \theta \right) \, d \theta \\ & = \left[ \theta +\dfrac{\sin 2 \theta}{2} \right] _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \\ & = \left( \dfrac{\pi}{4} +\dfrac{1}{2} \right) -\left( \dfrac{\pi}{6} +\dfrac{\sqrt{3}}{4} \right) \\ & = \underline{\dfrac{\pi}{12} +\dfrac{2 -\sqrt{3}}{4}} \end{align}\]

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(1)

\(C\) 上の点 P \((p,q)\) における接線の式は \[ \dfrac{px}{v^2} +\dfrac{qy}{u^2} = 1 \] これと \(\ell\) の式と比較して \[\begin{align} \dfrac{p}{v^2} = \dfrac{1}{a} & , \ \dfrac{q}{u^2} = \dfrac{1}{b} \\ \text{∴} \quad p = \dfrac{v^2}{a} & , \ q = \dfrac{u^2}{b} \quad ... [1] \end{align}\] また, 点 P は \(C\) 上にあるので, [1] を用いて \[\begin{gather} \dfrac{\left( \frac{v^2}{a} \right)^2}{u^2} +\dfrac{\left( \frac{u^2}{b} \right)}{v^2} = 1 \\ \dfrac{v^2}{a^2} +\dfrac{u^2}{b^2} = 1 \\ \text{∴} \quad v^2 = \underline{b^2 \left( 1 -\dfrac{u^2}{a^2} \right)} \end{gather}\]

(2)

\(u\) の取りうる値の範囲は, \(0 \lt u \lt a\) .
回転体は, 半径 \(v\) の球を \(x\) 軸方向に \(\dfrac{u}{v}\) 倍した立体なので \[\begin{align} V & = \dfrac{4 \pi v^3}{3} \cdot \dfrac{u}{v} \\ & = \dfrac{4 \pi}{3} u \cdot b^2 \left( 1 -\dfrac{u^2}{a^2} \right) \\ & = \dfrac{4 \pi b^2}{3 a^2} \underline{u ( a^2-u^2 )} _ {[ \text{A} ]} \end{align}\] [A] を \(f(u)\) とおいて, \(f(u)\) が最大のときを考えればよい. \[\begin{align} f'(u) & = a^2 -3u^2 \\ & = \left( a +\sqrt{3} u \right) \left( a -\sqrt{3} u \right) \end{align}\] \(f'(u) = 0\) をとくと \[ u = \dfrac{a}{\sqrt{3}} \] したがって, \(f(u)\) の増減は下表のとおりで \[ \begin{array}{c|ccccc} u & (0) & \cdots & \frac{a}{\sqrt{3}} & \cdots & (a) \\ \hline f'(u) & & + & 0 & - & \\ \hline f(u) & & \nearrow & \text{最大} & \searrow & \end{array} \] 最大値は \[\begin{align} f \left( \dfrac{a}{\sqrt{3}} \right) & = \dfrac{a}{\sqrt{3}} \cdot \dfrac{2 a^2}{3} \\ & = \dfrac{2 \sqrt{3} a^3}{9} \end{align}\] よって, 求める最大値は \[ \dfrac{4 \pi b^2}{3 a^2} f \left( \dfrac{a}{\sqrt{3}} \right) = \underline{\dfrac{8 \sqrt{3} \pi ab^2}{27}} \] このとき, (1) の結果より \[\begin{align} v^2 & = b^2 \left( 1-\dfrac{1}{a^2} \cdot \dfrac{a^2}{3} \right) \\ & = \dfrac{2 b^2}{3} \end{align}\] よって, \(C\) の方程式は \[ \underline{\dfrac{3 x^2}{a^2} +\dfrac{3 y^2}{2 b^2} = 1} \]

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(1)

P \(( t , t^2 )\) , Q \(( X , Y )\) とおく.
条件より, \(\overrightarrow{\text{AQ}} = k \overrightarrow{\text{AP}}\) なので \[\begin{gather} \left( \begin{array}{c} X-a \\ Y-b \end{array} \right) = k \left( \begin{array}{c} t-a \\ t^2-b \end{array} \right) \\ \text{∴} \quad \left\{ \begin{array}{ll} X = k(t-a)+a & ... [1] \\ Y = k(t^2-b)+b & ... [2] \end{array} \right. \end{gather}\] [1] より \[ kt = X +(k-1) a \] これを [2] に代入して \[\begin{align} kY & = \left\{ X +(k-1)a \right\}^2 -k^2b +kb \\ \text{∴} \quad Y & = \dfrac{1}{k} \left\{ X +(k-1)a \right\}^2 -(k-1)b \end{align}\] よって, \(C'\) の方程式は \[ \underline{y = \dfrac{1}{k} \left\{ x +(k-1)a \right\}^2 -(k-1)b} \]

(2)

\(C , C'\) の式から \(y\) を消去すると \[\begin{gather} kx^2 = x^2 +2(k-1)ax +(k-1)^2a^2 -k(k-1)b \\ \text{∴} \quad x^2 -2ax -(k-1)a^2 +kb = 0 \quad ... [3] \end{gather}\] [3] の判別式を \(D\) とおくと \[\begin{align} \dfrac{D}{4} & = a^2 +(k-1)a^2 -kb\ \\ & = k ( a^2-b ) \gt 0 \quad ( \ \text{∵} \ k>1 , \ b \lt a^2 ) \end{align}\] したがって, \(C\) と \(C'\) は \(2\) つの点で交わる.
[3] の \(2\) つの解を \(\alpha , \beta \ ( \alpha \lt \beta )\) とおくと, 解と係数の関係より \[ \alpha +\beta = 2a , \ \alpha \beta = -(k-1)a^2 +kb \] これを用いれば, 求める面積 \(S\) は \[\begin{align} S & = \displaystyle\int _ {\alpha}^{\beta} \left[ x^2 -\dfrac{1}{k} \left\{ x +(k-1)a \right\}^2 +(k-1)b \right] \, dx \\ & = -\dfrac{k-1}{k} \displaystyle\int _ {\alpha}^{\beta} ( x-\alpha )( x-\beta ) \, dx \\ & = \dfrac{k-1}{6k} ( \beta -\alpha )^3 \\ & = \dfrac{k-1}{6k} \left\{ ( \alpha +\beta )^2 -4 \alpha \beta \right\}^{\frac{3}{2}} \\ & = \dfrac{k-1}{6k} \left\{ 4a^2 +4(k-1)a^2 -4kb \right\}^{\frac{3}{2}} \\ & = \underline{\dfrac{4 \sqrt{k} (k-1) (a^2-b)^{\frac{3}{2}}}{3}} \end{align}\]

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(1)

\[ f'(x) = 3x^2+6ax+3b = 3 (x^2+2ax+b) \] \(f'(x) = 0\) が異なる \(2\) つの実数解をもつので, 判別式 \(D\) について \[\begin{gather} \dfrac{D}{4} = a^2 -b \gt 0 \\ \text{∴} \quad b \lt a^2 \quad ... [1] \end{gather}\] このとき, \(2\) つの解を \(\alpha , \beta \ ( \alpha \lt \beta )\) とおくと \[\begin{align} \alpha = -a -\sqrt{a^2-b} & , \ \beta = -a+\sqrt{a^2-b} \\ \text{∴} \quad \alpha +\beta = -2a & , \ \alpha \beta = b \quad ... [2] \end{align}\] 条件より \[\begin{align} f( \alpha ) f( \beta ) & \lt 0 \quad ... [3] , \\ f( \alpha ) +f( \beta ) & \lt 0 \quad ... [4] \end{align}\] が成り立つ条件を求めればよい.
ここで \[ f(x) = \dfrac{x+a}{3} f'(x) -2(a^2-b)x-ab \] なので \[\begin{align} f( \alpha ) & = -2(a^2-b) \alpha -ab , \\ f( \beta ) & = -2(a^2-b) \beta -ab \end{align}\] これと [2] を用いれば, [3] より \[\begin{align} f( \alpha ) f( \beta ) = 4(a^2-b)^2 b -4a^2b(a^2-b) +a^2b^2 & \lt 0 \\ 4a^4b -8a^2b^2 +4b^3 -4a^4b +4a^2b^2 +a^2b^2 & \lt 0 \\ b^2 ( 4b -3a^2 ) & \lt 0 \\ \text{∴} \quad b \lt \dfrac{3 a^2}{4} \quad ... [5] & \end{align}\] また, [4] より \[\begin{gather} f( \alpha ) +f( \beta ) = 4a(a^2-b) -2ab \lt 0 \\ a ( 2a^2-3b ) \lt 0 \\ \text{∴} \quad \left\{ \begin{array}{ll} b \lt \dfrac{2a^2}{3} & \ ( \ a<0 \text{のとき} ) \\ b \gt \dfrac{2a^2}{3} & \ ( \ a>0 \text{のとき} ) \end{array} \right. \quad ... [6] \end{gather}\] よって, 求める条件は, [1] かつ [5] かつ [6] なので, 点 \((a,b)\) の範囲は下図斜線部（ただし, ○と境界は除く）

(2)

条件と (1) の途中経過より \[\begin{align} f( \alpha ) +f( \beta ) & = 2a ( 2a^2-3b ) = 0 \quad ... [7] , \\ f( \alpha ) f( \beta ) & = b^2 ( 4b -3a^2 ) = -1\quad ... [8] \end{align}\] [7] をとくと \[ a=0 \ \text{または} \ a^2 = \dfrac{3b}{2} \]

1. 1*　\(a = 0\) のとき
   [8] に代入して \[\begin{align} b^3 & = -\dfrac{1}{4} \\ \text{∴} \quad b & = -\dfrac{1}{\sqrt[3]{4}} \end{align}\]

2. 2*　\(a^2 = \dfrac{3b}{2}\) のとき
   [8] に代入すると \[\begin{gather} b^2 \left( 4b -3 \cdot \dfrac{3b}{2} \right) = -1 \\ b^3 = 2 \\ \text{∴} \quad b = \sqrt[3]{2} \end{gather}\] したがって \[ a = \pm \sqrt{\dfrac{3 \sqrt[3]{2}}{2}} = \pm \dfrac{\sqrt{3}}{\sqrt[3]{2}} \]

よって, 求める値は \[ (a,b) = \underline{\left( 0 , -\dfrac{1}{\sqrt[3]{4}} \right) , \ \left( \pm \dfrac{\sqrt{3}}{\sqrt[3]{2}} , \sqrt[3]{2} \right)} \]

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(1)

\[\begin{align} \textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{p} -\overrightarrow{u _ k} \right|^2 & = \textstyle\sum\limits _ {k=1}^n k \left( \left| \overrightarrow{p} \right|^2 -2 \overrightarrow{p} \cdot \overrightarrow{u _ k} + \left| \overrightarrow{u _ k} \right|^2 \right) \\ & = \dfrac{n(n+1)}{2} \left| \overrightarrow{p} \right|^2 -2 \left( \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right) \cdot \overrightarrow{p} +\textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 \\ & = \dfrac{n(n+1)}{2} \left| \overrightarrow{p} -\dfrac{2}{n(n+1)} \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \\ & \qquad +\textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 -\dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \quad ... [1] \end{align}\] よって \[ \overrightarrow{v} = \underline{\dfrac{2}{n(n+1)} \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k}} \]

(2)

(1) の結果と条件より \[\begin{align} \left| \overrightarrow{v} \right|^2 & = \dfrac{4}{n^2 (n+1)^2} \left\{ \left( \textstyle\sum\limits _ {k=1}^n k \cos \dfrac{k \alpha}{n} \right)^2 +\left( \textstyle\sum\limits _ {k=1}^n k \sin \dfrac{k \alpha}{n} \right)^2 \right\} \\ & = \underline{\dfrac{4}{\left( 1 +\frac{1}{n} \right)^2}} _ {[2]} \left\{ \left( \underline{\dfrac{1}{n} \textstyle\sum\limits _ {k=1}^n \dfrac{k}{n} \cos \dfrac{k \alpha}{n}} _ {[3]} \right)^2 +\left( \underline{\dfrac{1}{n} \textstyle\sum\limits _ {k=1}^n \dfrac{k}{n} \sin \dfrac{k \alpha}{n}} _ {[4]} \right)^2 \right\} \end{align}\] ここで, [2] ～ [4] について, \(n \rightarrow \infty\) とすれば \[\begin{align} [2] & \rightarrow 4 , \\ [3] & \rightarrow \displaystyle\int _ 0^1 x \cos \alpha x \, dx \\ & = \left[ \dfrac{x \sin \alpha x}{\alpha} \right] _ 0^1 -\dfrac{1}{\alpha} \displaystyle\int _ 0^1 \sin \alpha x \, dx \\ & = \dfrac{\sin \alpha}{\alpha} +\dfrac{1}{\alpha} \left[ \dfrac{\cos \alpha x}{\alpha} \right] _ 0^1 \\ & = \dfrac{\alpha \sin \alpha +\cos \alpha -1}{\alpha^2} , \\ [4] & \rightarrow \displaystyle\int _ 0^1 x \sin \alpha x \, dx \\ & = \left[ -\dfrac{x \cos \alpha x}{\alpha} \right] _ 0^1 +\dfrac{1}{\alpha} \displaystyle\int _ 0^1 \cos \alpha x \, dx \\ & = -\dfrac{\cos \alpha}{\alpha} +\dfrac{1}{\alpha} \left[ \dfrac{\sin \alpha x}{\alpha} \right] _ 0^1 \\ & = \dfrac{-\alpha \cos \alpha +\sin \alpha}{\alpha^2} \end{align}\] よって \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} \left| \overrightarrow{v} \right|^2 & = 4 \left\{ \left( \dfrac{\alpha \sin \alpha +\cos \alpha -1}{\alpha^2} \right)^2 +\left( \dfrac{-\alpha \cos \alpha +\sin \alpha}{\alpha^2} \right)^2 \right\} \\ & = \underline{\dfrac{4}{\alpha^4} \left( \alpha^2 -2 \alpha \sin \alpha -2 \cos \alpha +2 \right)} \end{align}\]

(3)

(1) の結果より \[\begin{align} \left| \overrightarrow{v} \right| & = \dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right| \\ \text{∴} \quad \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right| & = \dfrac{n(n+1)}{2} \left| \overrightarrow{v} \right| \end{align}\] これを用いれば, [1] より \[\begin{align} m & = \textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 -\dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \\ & = \dfrac{n(n+1)}{2} \left( 1 -\left| \overrightarrow{v} \right|^2 \right) \end{align}\] よって, (2) の結果も用いて \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} \dfrac{m}{n^2} & = \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1 +\frac{1}{n}}{2} \left( 1 -\left| \overrightarrow{v} \right|^2 \right) \\ & = \underline{\dfrac{\alpha^4 -4 \alpha^2 +8 \alpha \sin \alpha +8 \cos \alpha -8}{2 \alpha^4}} \end{align}\]

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