関数 \[ f(x) = \dfrac{x}{x^2 +3} \] に対して, \(y = f(x)\) のグラフを \(C\) とする. 点 A \(( 1 , f(1) )\) における \(C\) の接線を \[ \ell \ : \ y = g(x) \] とする.
(1) \(C\) と \(\ell\) の共有点で A と異なるものがただ \(1\) つ存在することを示し, その点の \(x\) 座標を求めよ.
(2) (1) で求めた共有点の \(x\) 座標を \(\alpha\) とする. 定積分 \[ \displaystyle\int _ {\alpha}^{1} \left\{ f(x) -g(x) \right\}^2 \, dx \] を計算せよ.
【 解 答 】
(1)
\[\begin{align} f'(x) & = \dfrac{1 \cdot ( x^2 +3 ) -x \cdot 2x}{( x^2 +3 )^2} \\ & = \dfrac{3 -x^2}{( x^2 +3 )^2} \end{align}\] なので, \(\ell\) の式は \[\begin{align} y & = g(x) = f'(1) ( x -t ) +f(t) \\ & = \dfrac{2}{16} ( x -t ) +\dfrac{1}{4} = \dfrac{x+1}{8} \end{align}\] これと \(C\) の式から \(y\) を消去すれば \[\begin{align} \dfrac{x}{x^2 +3} & = \dfrac{x+1}{8} \\ 8x & = ( x^2 +3 ) ( x+1 ) \\ x^3 +x^2 -5x +3 & = 0 \\ ( x+3 ) ( x-1 )^2 & = 0 \\ \text{∴} \quad x & = 1 , -3 \end{align}\] よって, \(C\) と \(\ell\) の共有点は A の他に \(1\) つだけ存在し, その \(x\) 座標は \(\underline{3}\) .
(2)
\[\begin{align} & \left\{ f(x) -g(x) \right\}^2 = \left\{ \dfrac{x}{x^2 +3} -\dfrac{x+1}{8} \right\}^2 \\ & \qquad = \dfrac{x^2}{( x^2 +3 )^2} -\dfrac{1}{4} \cdot \dfrac{x^2 +x}{x^2 +3} +\dfrac{1}{64} (x+1)^2 \\ & \qquad = \dfrac{1}{x^2 +3} -\dfrac{3}{( x^2 +3 )^2} -\dfrac{1}{4} -\dfrac{1}{4} \cdot \dfrac{x}{x^2 +3} \\ & \qquad \qquad +\dfrac{3}{4} \cdot \dfrac{1}{x^2 +3} +\dfrac{1}{64} (x+1)^2 \\ & \qquad = \dfrac{7}{4} \cdot \underline{\dfrac{1}{x^2 +3}} _ {[1]} -3 \cdot \underline{\dfrac{1}{( x^2 +3 )^2}} _ {[2]} \\ & \qquad \qquad -\dfrac{1}{4} \cdot \underline{\dfrac{x}{x^2 +3}} _ {[3]} +\underline{\dfrac{1}{64} ( x+1 )^2 -\dfrac{1}{4}} _ {[4]} \end{align}\] [1] ～ [4] について, それぞれ積分すると
- [1] について
\(x = \sqrt{3} \tan \theta \ \left( -\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2} \right)\) とおくと \[ dx = \dfrac{\sqrt{3} \, d \theta}{\cos^2 \theta} \ , \ \begin{array}{c|ccc} x & -3 & \rightarrow & 1 \\ \hline \theta & -\dfrac{\pi}{6} & \rightarrow & \dfrac{\pi}{3} \end{array} \] なので \[\begin{align} \displaystyle\int _ {-3}^{1} [1] \, dx & = \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{3 ( \tan^2 +1 )} \cdot \dfrac{\sqrt{3} \, d \theta}{\cos^2 \theta} \\ & = \dfrac{1}{\sqrt{3}} \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} d \theta = \dfrac{1}{\sqrt{3}} [ \theta ] _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} = \dfrac{\sqrt{3} \pi}{6} \end{align}\] - [2] について
[1] と同様に置換して \[\begin{align} \displaystyle\int _ {-3}^{1} [2] \, dx & = \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{9 ( \tan^2 +1 )^2} \cdot \dfrac{\sqrt{3} \, d \theta}{\cos^2 \theta} \\ & = \dfrac{\sqrt{3}}{9} \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 \theta \, d \theta = \dfrac{\sqrt{3}}{9} \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1 +\cos 2 \theta}{2} \, d \theta \\ & = \dfrac{\sqrt{3}}{18} \left[ \theta +\dfrac{1}{2} \sin \theta \right] _ {-\frac{\pi}{6}}^{\frac{\pi}{3}} \\ & = \dfrac{\sqrt{3} \pi}{36} +\dfrac{1}{12} \end{align}\] - [3] について \[\begin{align} \displaystyle\int _ {-3}^{1} [3] \, dx & = \dfrac{1}{2} \displaystyle\int _ {-3}^{1} \dfrac{( x^2 +3 )'}{x^2 +3} \, dx \\ & = \dfrac{1}{2} \left[ \log ( x^2 +3 ) \right] _ {-3}^{1} \\ & = \dfrac{1}{2} \left( \log 4 -\log 12 \right) = -\dfrac{1}{2} \log 3 \end{align}\]
- [4] について \[\begin{align} \displaystyle\int _ {-3}^{1} [4] \, dx & = \dfrac{1}{64} \left[ \dfrac{1}{3} ( x +1 )^3 \right] _ {-3}^{1} -\dfrac{1}{4} \cdot 4 \\ & = \dfrac{1}{64} \cdot \dfrac{1}{3} ( 8 +8 ) -1 = -\dfrac{11}{12} \end{align}\]
よって, 求める積分値 \(I\) は \[\begin{align} I & = \dfrac{7}{4} \cdot \dfrac{\sqrt{3} \pi}{6} -3 \left( \dfrac{\sqrt{3} \pi}{36} +\dfrac{1}{12} \right) +\dfrac{1}{4} \cdot \dfrac{1}{2} \log 3 -\dfrac{11}{12} \\ & = \underline{\dfrac{5 \sqrt{3} \pi}{24} +\dfrac{1}{8} \log 3 -\dfrac{7}{6}} \end{align}\]