\\(2\\) \u6b21\u95a2\u6570 \\(f(x) = x^2+ax+b\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\nf(x+1) = c \\displaystyle\\int _ 0^1 ( 3x^2 +4xt ) f'(t) \\, dt\r\n\\]\r\n\u304c \\(x\\) \u306b\u3064\u3044\u3066\u306e\u6052\u7b49\u5f0f\u306b\u306a\u308b\u3088\u3046\u306a\u5b9a\u6570 \\(a , b , c\\) \u306e\u7d44\u3092\u3059\u3079\u3066\u6c42\u3081\u3088.<\/p>\r\n
\\[\\begin{align}\r\n( \\text{\u5de6\u8fba} ) & = (x+1)^2 +a(x+1) +b \\\\\r\n& = x^2 +(a+2)x +a+b+1\r\n\\end{align}\\]\r\n\u307e\u305f, \\(f'(x) = 2x+a\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n( \\text{\u53f3\u8fba} ) & = c \\displaystyle\\int _ 0^1 ( 3x^2 +4xt )( 2t+a ) \\, dt \\\\\r\n& = c \\displaystyle\\int _ 0^1 \\left\\{ 8xt^2 +2\\left( 3x^2+2ax \\right) t +3ax^2 \\right\\} \\, dt \\\\\r\n& = c \\left[ \\dfrac{8x}{3} t^3 +\\left( 3x^2+2ax \\right) t^2 +3ax^2 t \\right] _ 0^1 \\\\\r\n& = c \\left( \\dfrac{8x}{3} +3x^2 +2ax +3ax^2 \\right) \\\\\r\n& = 3(a+1)c x^2 +\\left( 2a +\\dfrac{8}{3} \\right) cx\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u4fc2\u6570\u3092\u6bd4\u8f03\u3057\u3066\r\n\\[\r\n\\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.\r\n\\]\r\n[3] \u3088\u308a, \\(b=-a-1\\) .
\r\n[1] \u3088\u308a, \\(a \\neq -1\\) \u306a\u306e\u3067\r\n\\[\r\nc =\\dfrac{1}{3(a+1)}\r\n\\]\r\n\u3053\u308c\u3092 [2] \u306b\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\n2a +\\dfrac{8}{3} & = 3 (a+1)(a+2) \\\\\r\n6a+8 & = 9a^2+27a+18 \\\\\r\n9a^2+21a+10 & = 0 \\\\\r\n(3a+2)(3a+5) & = 0 \\\\\r\n\\text{\u2234} \\quad a & = -\\dfrac{2}{3} , -\\dfrac{5}{3}\r\n\\end{align}\\]\r\n
\\(a = -\\dfrac{2}{3}\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nb & = \\dfrac{2}{3} -1 = -\\dfrac{1}{3} , \\\\\r\nc & = \\dfrac{1}{3 \\cdot \\dfrac{1}{3}} = 1\r\n\\end{align}\\]<\/li>\r\n
\\(a = -\\dfrac{5}{3}\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nb & = \\dfrac{5}{3} -1 = \\dfrac{2}{3} \\\\\r\nc & = \\dfrac{1}{3 \\cdot \\left( -\\dfrac{2}{3} \\right)} = -\\dfrac{1}{2}\r\n\\end{align}\\]<\/li>\r\n<\/ul>\r\n
\u3088\u3063\u3066, \u6c42\u3081\u308b\u3059\u3079\u3066\u306e\u7d44\u306f\r\n\\[\r\n(a,b,c) = \\underline{\\left( -\\dfrac{2}{3} , -\\dfrac{1}{3} , 1 \\right) , \\left( -\\dfrac{5}{3} , \\dfrac{2}{3} , \\dfrac{1}{2} \\right)}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(2\\) \u6b21\u95a2\u6570 \\(f(x) = x^2+ax+b\\) \u306b\u5bfe\u3057\u3066 \\[ f(x+1) = c \\displaystyle\\int _ 0^1 ( 3x^2 +4xt ) f'(t) \\, dt \\] \u304c \\(x\\) … \u7d9a\u304d\u3092\u8aad\u3080