関数 \(f(x) = \langle \! \langle x \rangle \! \rangle -2 \langle \! \langle x-1 \rangle \! \rangle +\langle \! \langle x-2 \rangle \! \rangle\) を考える. ここで, 実数 \(u\) に対して \(\langle \! \langle u \rangle \! \rangle = \dfrac{u +|u|}{2}\) とする. このとき以下の各問いに答えよ.
(1) \(f(x)\) のグラフをかけ.
(2) \(g(x) = \displaystyle\int _ {0}^{1} f(x-t) \, dt\) とおくとき, \(g(x)\) の最大値を求めよ.
(3) (2) の \(g(x)\) に対して, \(p(s) = \displaystyle\int _ {0}^{3} (x-s)^2 g(x) \, dx\) とおくとき, \(p(s)\) の最小値を求めよ.
【 解 答 】
(1)
\[\begin{align} f(x) & = \dfrac{x +|x|}{2} -2 \cdot \dfrac{x-1 +| x-1 |}{2} +\dfrac{x-2 +| x-2 |}{2} \\ & = \dfrac{1}{2} \left( |x| -2 | x-1 | +| x-2 | \right) \ . \end{align}\]
\(x \lt 0\) のとき \[ f(x) = \dfrac{1}{2} \left\{ -x +2 (x-1) -(x-2) \right\} = 0 \ . \]
\(0 \leqq x \lt 1\) のとき \[ f(x) = \dfrac{1}{2} \left\{ x +2 (x-1) -(x-2) \right\} = x \ . \]
\(1 \leqq x \lt 2\) のとき \[ f(x) = \dfrac{1}{2} \left\{ x -2 (x-1) -(x-2) \right\} = 2-x \ . \]
\(x \geqq 2\) のとき \[ f(x) = \dfrac{1}{2} \left\{ x -2 (x-1) +(x-2) \right\} = 0 \ . \]
よって, \(f(x)\) のグラフは, 下図の通り.
(2)
\(u = x-t\) とおくと
\[
dt = -du , \quad \begin{array}{c|c} t & 0 \rightarrow 1 \\ \hline u & x \rightarrow x-1\end{array} \ .
\]
ゆえに
\[
g(x) = \displaystyle\int _ {x-1}^{x} f(u) \, du \quad ... [1] \ .
\]
(1) で求めたグラフの形状より, \(1 \leqq x \lt 2\) で \(g(x)\) は最大となる.
この区間では
\[\begin{align}
g(x) & = \displaystyle\int _ {x-1}^{1} u \, du +\displaystyle\int_{1}^{x} ( 2-u ) \, du \\
& = \left[ \dfrac{u^2}{2} \right] _ {x-1}^{1} +\left[ 2u -\dfrac{u^2}{2} \right]_{1}^{x} \\
& = \dfrac{1}{2} -\dfrac{(x-1)^2}{2} +2x -\dfrac{x^2}{2} -\dfrac{3}{2} \\
& = -x^2 +3x -\dfrac{3}{2} \\
& = -\left( x -\dfrac{3}{2} \right)^2 +\dfrac{3}{4} \ .
\end{align}\]
よって, \(g(x)\) の最大値は
\[
g \left( \dfrac{3}{2} \right) = \underline{\dfrac{3}{4}} \ .
\]
(3)
[1] より\(0 \leqq x \lt 1\) のとき \[\begin{align} g(x) & = \displaystyle\int _ {0}^{x} u \, du \\ & = \left[ \dfrac{u^2}{2} \right] _ {0}^{x} = \dfrac{x^2}{2} \ . \end{align}\]
\(2 \leqq x \lt 3\) のとき \[\begin{align} g(x) & = \displaystyle\int _ {x-1}^{2} (2-u) \, du \\ & = \left[ 2u -\dfrac{u^2}{2} \right] _ {x-1}^{2} \\ & = \dfrac{1}{2} ( x-3 )^2 \ . \end{align}\]
\(A = \displaystyle\int_{0}^{3} g(x) \, dx\) , \(B = \displaystyle\int_{0}^{3} x g(x) \, dx\) , \(C = \displaystyle\int_{0}^{3} x^2 g(x) \, dx\) とおくと \[\begin{align} p(s) & = A s^2 -2B s +C \\ & = A \left( s -\dfrac{B}{A} \right)^2 -\dfrac{B^2}{A} +C \end{align}\] なので, \(p(s)\) の最小値は, \(-\dfrac{B^2}{A} +C\) と表せる. \[\begin{align} A & = \displaystyle\int _ {0}^{1} \dfrac{x^2}{2} \, dx +\displaystyle\int _ {1}^{2} \left( -x^2 +3x -\dfrac{3}{2} \right) \, dx +\displaystyle\int _ {2}^{3} \dfrac{1}{2} (x-3)^2 \, dx \\ & = \left[ \dfrac{x^3}{6} \right] _ {0}^{1} +\left[ -\dfrac{x^3}{3} +\dfrac{3x^2}{2} -\dfrac{3x}{2} \right] _ {1}^{2} +\left[ \dfrac{(x-3)^3}{6} \right] _ {2}^{3} \\ & = \dfrac{1}{6} +\dfrac{1}{3} -\left( -\dfrac{1}{3} \right) -\left( -\dfrac{1}{6} \right) = 1 \ . \\ B & = \displaystyle\int _ {0}^{1} \dfrac{x^3}{2} \, dx +\displaystyle\int_ {1}^{2} \left( -x^3 +3x^2 -\dfrac{3x}{2} \right) \, dx +\displaystyle\int_ {2}^{3} \dfrac{1}{2} x (x-3)^2 \, dx \\ & = \left[ \dfrac{x^4}{8} \right] _ {0}^{1} +\left[ -\dfrac{x^4}{4} +x^3 -\dfrac{3x^2}{4} \right] _ {1}^{2} +\left[ \dfrac{(x-3)^4}{8} +\dfrac{(x-3)^3}{2} \right] _ {2}^{3} \\ & = \dfrac{1}{8} +1 -0 -\left( -\dfrac{3}{8} \right) = \dfrac{3}{2} \ . \\ C & = \displaystyle\int _ {0}^{1} \dfrac{x^4}{2} \, dx +\displaystyle\int _ {1}^{2} \left( -x^4 +3x^3 -\dfrac{3x^2}{2} \right) \, dx +\displaystyle\int _ {2}^{3} \dfrac{1}{2} x^2 (x-3)^2 \, dx \\ & = \left[ \dfrac{x^5}{10} \right] _ {0}^{1} +\left[ -\dfrac{x^5}{5} +\dfrac{3x^4}{4} -\dfrac{x^3}{2} \right] _ {1}^{2} +\left[ \dfrac{(x-3)^5}{10} +\dfrac{3 (x-3)^4}{4} +\dfrac{3 (x-3)^3}{2} \right] _ {2}^{3} \\ & = \dfrac{1}{10} +\dfrac{8}{5} -\dfrac{1}{20} -\left( -\dfrac{17}{20} \right) = \dfrac{5}{2} \ . \end{align}\] よって, 求める最小値は \[ -\left( \dfrac{3}{2} \right)^2 +\dfrac{5}{2} = \underline{\dfrac{1}{4}} \ . \]