{"id":1963,"date":"2021-11-14T09:46:36","date_gmt":"2021-11-14T00:46:36","guid":{"rendered":"https:\/\/www.roundown.net\/nyushi\/?p=1963"},"modified":"2021-11-14T14:06:55","modified_gmt":"2021-11-14T05:06:55","slug":"tkr202103","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/tkr202103\/","title":{"rendered":"\u6771\u5927\u7406\u7cfb2021\uff1a\u7b2c3\u554f"},"content":{"rendered":"<hr \/>\n<p>\u95a2\u6570\r\n\\[\r\nf(x) = \\dfrac{x}{x^2 +3}\r\n\\]\r\n\u306b\u5bfe\u3057\u3066, \\(y = f(x)\\) \u306e\u30b0\u30e9\u30d5\u3092 \\(C\\) \u3068\u3059\u308b.\r\n\u70b9 A \\(( 1 , f(1) )\\) \u306b\u304a\u3051\u308b \\(C\\) \u306e\u63a5\u7dda\u3092\r\n\\[\r\n\\ell \\ : \\ y = g(x)\r\n\\]\r\n\u3068\u3059\u308b.<\/p>\r\n<ol>\r\n<li><p><strong>(1)<\/strong>\u3000\\(C\\) \u3068 \\(\\ell\\) \u306e\u5171\u6709\u70b9\u3067 A \u3068\u7570\u306a\u308b\u3082\u306e\u304c\u305f\u3060 \\(1\\) \u3064\u5b58\u5728\u3059\u308b\u3053\u3068\u3092\u793a\u3057,\r\n\u305d\u306e\u70b9\u306e \\(x\\) \u5ea7\u6a19\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<li><p><strong>(2)<\/strong>\u3000<strong>(1)<\/strong> \u3067\u6c42\u3081\u305f\u5171\u6709\u70b9\u306e \\(x\\) \u5ea7\u6a19\u3092 \\(\\alpha\\) \u3068\u3059\u308b. \u5b9a\u7a4d\u5206\r\n\\[\r\n\\displaystyle\\int _ {\\alpha}^{1} \\left\\{ f(x) -g(x) \\right\\}^2 \\, dx\r\n\\]\r\n\u3092\u8a08\u7b97\u305b\u3088.<\/p><\/li>\r\n<\/ol>\r\n<hr \/>\r\n<!--more-->\r\n<h4>\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n<p><strong>(1)<\/strong><\/p>\r\n<p>\\[\\begin{align}\r\nf'(x) & = \\dfrac{1 \\cdot ( x^2 +3 ) -x \\cdot 2x}{( x^2 +3 )^2} \\\\\r\n& = \\dfrac{3 -x^2}{( x^2 +3 )^2}\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \\(\\ell\\) \u306e\u5f0f\u306f\r\n\\[\\begin{align}\r\ny & = g(x) = f'(1) ( x -t ) +f(t) \\\\\r\n& = \\dfrac{2}{16} ( x -t ) +\\dfrac{1}{4} = \\dfrac{x+1}{8}\r\n\\end{align}\\]\r\n\u3053\u308c\u3068 \\(C\\) \u306e\u5f0f\u304b\u3089 \\(y\\) \u3092\u6d88\u53bb\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n\\dfrac{x}{x^2 +3} & = \\dfrac{x+1}{8} \\\\\r\n8x & = ( x^2 +3 ) ( x+1 ) \\\\\r\nx^3 +x^2 -5x +3 & = 0 \\\\\r\n( x+3 ) ( x-1 )^2 & = 0 \\\\\r\n\\text{\u2234} \\quad x & = 1 , -3\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(C\\) \u3068 \\(\\ell\\) \u306e\u5171\u6709\u70b9\u306f A \u306e\u4ed6\u306b \\(1\\) \u3064\u3060\u3051\u5b58\u5728\u3057, \u305d\u306e \\(x\\) \u5ea7\u6a19\u306f \\(\\underline{3}\\) .<\/p>\r\n<p><strong>(2)<\/strong><\/p>\r\n<p>\\[\\begin{align}\r\n& \\left\\{ f(x) -g(x) \\right\\}^2 = \\left\\{ \\dfrac{x}{x^2 +3} -\\dfrac{x+1}{8} \\right\\}^2 \\\\\r\n& \\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 \\\\\r\n& \\qquad = \\dfrac{1}{x^2 +3} -\\dfrac{3}{( x^2 +3 )^2} -\\dfrac{1}{4} -\\dfrac{1}{4} \\cdot \\dfrac{x}{x^2 +3} \\\\\r\n& \\qquad \\qquad +\\dfrac{3}{4} \\cdot \\dfrac{1}{x^2 +3} +\\dfrac{1}{64} (x+1)^2 \\\\\r\n& \\qquad = \\dfrac{7}{4} \\cdot \\underline{\\dfrac{1}{x^2 +3}} _ {[1]} -3 \\cdot \\underline{\\dfrac{1}{( x^2 +3 )^2}} _ {[2]} \\\\\r\n& \\qquad \\qquad -\\dfrac{1}{4} \\cdot \\underline{\\dfrac{x}{x^2 +3}} _ {[3]} +\\underline{\\dfrac{1}{64} ( x+1 )^2 -\\dfrac{1}{4}} _ {[4]}\r\n\\end{align}\\]\r\n[1] \uff5e [4] \u306b\u3064\u3044\u3066, \u305d\u308c\u305e\u308c\u7a4d\u5206\u3059\u308b\u3068<\/p>\r\n<ul>\r\n<li>[1] \u306b\u3064\u3044\u3066<br \/>\r\n\\(x = \\sqrt{3} \\tan \\theta \\ \\left( -\\dfrac{\\pi}{2} \\lt \\theta \\lt \\dfrac{\\pi}{2} \\right)\\) \u3068\u304a\u304f\u3068\r\n\\[\r\ndx = \\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}\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\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} \\\\\r\n& = \\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}\r\n\\end{align}\\]<\/li>\r\n<li>[2] \u306b\u3064\u3044\u3066<br \/>\r\n[1] \u3068\u540c\u69d8\u306b\u7f6e\u63db\u3057\u3066\r\n\\[\\begin{align}\r\n\\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} \\\\\r\n& = \\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 \\\\\r\n& = \\dfrac{\\sqrt{3}}{18} \\left[ \\theta +\\dfrac{1}{2} \\sin \\theta \\right] _ {-\\frac{\\pi}{6}}^{\\frac{\\pi}{3}} \\\\\r\n& = \\dfrac{\\sqrt{3} \\pi}{36} +\\dfrac{1}{12}\r\n\\end{align}\\]<\/li>\r\n<li>[3] \u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {-3}^{1} [3] \\, dx & = \\dfrac{1}{2} \\displaystyle\\int _ {-3}^{1} \\dfrac{( x^2 +3 )'}{x^2 +3} \\, dx \\\\\r\n& = \\dfrac{1}{2} \\left[ \\log ( x^2 +3 ) \\right] _ {-3}^{1} \\\\\r\n& = \\dfrac{1}{2} \\left( \\log 4 -\\log 12 \\right) = -\\dfrac{1}{2} \\log 3\r\n\\end{align}\\]<\/li>\r\n<li>[4] \u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {-3}^{1} [4] \\, dx & = \\dfrac{1}{64} \\left[ \\dfrac{1}{3} ( x +1 )^3 \\right] _ {-3}^{1} -\\dfrac{1}{4} \\cdot 4 \\\\\r\n& = \\dfrac{1}{64} \\cdot \\dfrac{1}{3} ( 8 +8 ) -1 = -\\dfrac{11}{12}\r\n\\end{align}\\]<\/li>\r\n<\/ul>\r\n<p>\u3088\u3063\u3066, \u6c42\u3081\u308b\u7a4d\u5206\u5024 \\(I\\) \u306f\r\n\\[\\begin{align}\r\nI & = \\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} \\\\\r\n& = \\underline{\\dfrac{5 \\sqrt{3} \\pi}{24} +\\dfrac{1}{8} \\log 3 -\\dfrac{7}{6}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u95a2\u6570 \\[ f(x) = \\dfrac{x}{x^2 +3} \\] \u306b\u5bfe\u3057\u3066, \\(y = f(x)\\) \u306e\u30b0\u30e9\u30d5\u3092 \\(C\\) \u3068\u3059\u308b. \u70b9 A \\(( 1 , f(1) )\\) \u306b\u304a\u3051\u308b \\(C\\) \u306e\u63a5\u7dda\u3092 \\[  &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/tkr202103\/\">\u7d9a\u304d\u3092\u8aad\u3080 <span class=\"meta-nav\">&rarr;<\/span><\/a>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[164],"tags":[139,165],"class_list":["post-1963","post","type-post","status-publish","format-standard","hentry","category-tokyo_r_2021","tag-tokyo_r","tag-165"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1963","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=1963"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1963\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=1963"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=1963"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=1963"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}