\\(a\\) \u3092\u5b9f\u6570\u3068\u3059\u308b.\r\n\u50be\u304d\u304c \\(m\\) \u3067\u3042\u308b \\(2\\) \u3064\u306e\u76f4\u7dda\u304c, \u66f2\u7dda \\(y = x^3 -3ax^2\\) \u3068\u305d\u308c\u305e\u308c\u70b9 A , \u70b9 B \u3067\u63a5\u3057\u3066\u3044\u308b.<\/p>\r\n
(1)<\/strong>\u3000\u7dda\u5206 AB \u306e\u4e2d\u70b9\u3092 C \u3068\u3059\u308b. C \u306f\u66f2\u7dda \\(y = x^3 -3ax^2\\) \u4e0a\u306b\u3042\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\u76f4\u7dda AB \u306e\u65b9\u7a0b\u5f0f\u304c \\(y = -x-1\\) \u3067\u3042\u308b\u3068\u304d, \\(a , m\\) \u306e\u5024\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n A , B \u306e \\(x\\) \u5ea7\u6a19\u3092\u305d\u308c\u305e\u308c \\(p , q\\) \u3068\u3057, \\(f(x) = x^3 -3ax^2\\) \u3068\u304a\u304f. (2)<\/strong><\/p>\r\n (1)<\/strong> \u3088\u308a, C \\(\\left( a , -2a^3 \\right)\\) .
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n\u6761\u4ef6\u3088\u308a, \\(p , q\\) \u306f\r\n\\[\r\nf'(x) = 3x^2 -6ax = m \\ \\text{\u3059\u306a\u308f\u3061} \\ 3x^2 -6ax -m = 0\r\n\\]\r\n\u306e\u7570\u306a\u308b \\(2\\) \u89e3\u306a\u306e\u3067, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a\r\n\\[\r\np+q = 2a , \\ pq = -\\dfrac{m}{3} \\quad ... [1]\r\n\\]\r\nC \u306e\u5ea7\u6a19\u306f \\(\\left( \\dfrac{p+q}{2} , \\dfrac{f(p)+f(q)}{2} \\right)\\) \u3067\u3042\u308a\r\n\\[\\begin{align}\r\n\\dfrac{f(p)+f(q)}{2} & = \\dfrac{1}{2} \\left\\{ p^3 +q^3 -3a \\left( p^2+q^2 \\right) \\right\\} \\\\\r\n& = \\dfrac{1}{2} \\left[ ( p+q )^3 -3pq( p+q ) -3a \\left\\{ ( p+q )^2 -2pq \\right\\} \\right] \\\\\r\n& = \\dfrac{1}{2} \\left\\{ 8a^3 -3 \\cdot \\left( -\\dfrac{m}{3} \\right) \\cdot 2a -3a \\left( 4a^2 +2 \\cdot \\dfrac{m}{3} \\right) \\right\\} \\\\\r\n& = \\dfrac{1}{2} \\cdot \\left( -4a^2 \\right) = -2a^3\r\n\\end{align}\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nf \\left( \\dfrac{p+q}{2} \\right) & = \\left( \\dfrac{p+q}{2} \\right)^3 -3a \\left( \\dfrac{p+q}{2} \\right)^2 \\\\\r\n& = \\dfrac{1}{8} \\cdot 8a^3 -3a \\cdot \\dfrac{1}{4} \\cdot 4a^2 = -2a^3\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\r\n\\dfrac{f(p)+f(q)}{2} = f \\left( \\dfrac{p+q}{2} \\right)\r\n\\]\r\n\u3086\u3048\u306b, \u70b9C\u306f \\(y = f(x)\\) \u4e0a\u306b\u5b58\u5728\u3059\u308b.<\/p>\r\n
\r\nC \u306f\u76f4\u7dda AB \u4e0a\u306b\u3042\u308b\u306e\u3067\r\n\\[\\begin{align}\r\n-2a^3 & = -a-1 \\\\\r\n2a^3 -a -1 & = 0 \\\\\r\n( a-1 )( 2a^2 +2a +1 ) & = 0\r\n\\end{align}\\]\r\n\\(2a^2 +2a +1 = 2 \\left( a +\\dfrac{1}{2} \\right)^2 +\\dfrac{1}{2} \\gt 0\\) \u306a\u306e\u3067\r\n\\[\r\na = \\underline{1}\r\n\\]\r\n\u3053\u306e\u3068\u304d, \\(y = f(x) = x^3 -3x^2\\) \u3068 \\(y = -x -1\\) \u3088\u308a, \\(y\\) \u3092\u6d88\u53bb\u3057\u3066\r\n\\[\\begin{align}\r\nx^3 -3x^2 +x +1 & = 0 \\\\\r\n( x-1 )( x^2 -2x -1 ) & = 0\r\n\\end{align}\\]\r\n\\(p , q\\) \u306f \\(x^2 -2x -1 = 0\\) \u306e \\(2\\) \u89e3\u306a\u306e\u3067, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a\r\n\\[\r\npq = -1 \\quad ... [2]\r\n\\]\r\n[1] [2] \u3088\u308a\r\n\\[\\begin{align}\r\n-\\dfrac{m}{3} & = -1 \\\\\r\n\\text{\u2234} \\quad m & = \\underline{3}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a\\) \u3092\u5b9f\u6570\u3068\u3059\u308b. \u50be\u304d\u304c \\(m\\) \u3067\u3042\u308b \\(2\\) \u3064\u306e\u76f4\u7dda\u304c, \u66f2\u7dda \\(y = x^3 -3ax^2\\) \u3068\u305d\u308c\u305e\u308c\u70b9 A , \u70b9 B \u3067\u63a5\u3057\u3066\u3044\u308b. (1)\u3000\u7dda\u5206 AB \u306e\u4e2d\u70b9\u3092 C \u3068\u3059\u308b. C … \u7d9a\u304d\u3092\u8aad\u3080