\(2\) 次関数 \(f(x) = x^2+ax+b\) に対して \[ f(x+1) = c \displaystyle\int _ 0^1 ( 3x^2 +4xt ) f'(t) \, dt \] が \(x\) についての恒等式になるような定数 \(a , b , c\) の組をすべて求めよ.
【 解 答 】
\[\begin{align}
( \text{左辺} ) & = (x+1)^2 +a(x+1) +b \\
& = x^2 +(a+2)x +a+b+1
\end{align}\]
また, \(f'(x) = 2x+a\) なので
\[\begin{align}
( \text{右辺} ) & = c \displaystyle\int _ 0^1 ( 3x^2 +4xt )( 2t+a ) \, dt \\
& = c \displaystyle\int _ 0^1 \left\{ 8xt^2 +2\left( 3x^2+2ax \right) t +3ax^2 \right\} \, dt \\
& = c \left[ \dfrac{8x}{3} t^3 +\left( 3x^2+2ax \right) t^2 +3ax^2 t \right] _ 0^1 \\
& = c \left( \dfrac{8x}{3} +3x^2 +2ax +3ax^2 \right) \\
& = 3(a+1)c x^2 +\left( 2a +\dfrac{8}{3} \right) cx
\end{align}\]
よって, 係数を比較して
\[
\left\{ \begin{array}{ll} 3(a+1)c = 1 & ... [1] \\ \left( 2a +\dfrac{8}{3} \right) c = a+2 & ... [2] \\ a+b+1=0 & ... [3] \end{array} \right.
\]
[3] より, \(b=-a-1\) .
[1] より, \(a \neq -1\) なので
\[
c =\dfrac{1}{3(a+1)}
\]
これを [2] に代入して
\[\begin{align}
2a +\dfrac{8}{3} & = 3 (a+1)(a+2) \\
6a+8 & = 9a^2+27a+18 \\
9a^2+21a+10 & = 0 \\
(3a+2)(3a+5) & = 0 \\
\text{∴} \quad a & = -\dfrac{2}{3} , -\dfrac{5}{3}
\end{align}\]
\(a = -\dfrac{2}{3}\) のとき \[\begin{align} b & = \dfrac{2}{3} -1 = -\dfrac{1}{3} , \\ c & = \dfrac{1}{3 \cdot \dfrac{1}{3}} = 1 \end{align}\]
\(a = -\dfrac{5}{3}\) のとき \[\begin{align} b & = \dfrac{5}{3} -1 = \dfrac{2}{3} \\ c & = \dfrac{1}{3 \cdot \left( -\dfrac{2}{3} \right)} = -\dfrac{1}{2} \end{align}\]
よって, 求めるすべての組は \[ (a,b,c) = \underline{\left( -\dfrac{2}{3} , -\dfrac{1}{3} , 1 \right) , \left( -\dfrac{5}{3} , \dfrac{2}{3} , \dfrac{1}{2} \right)} \]