{"id":211,"date":"2011-12-02T23:25:34","date_gmt":"2011-12-02T14:25:34","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=211"},"modified":"2021-09-14T12:58:07","modified_gmt":"2021-09-14T03:58:07","slug":"iks200903","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/iks200903\/","title":{"rendered":"\u533b\u79d1\u6b6f\u79d1\u59272009\uff1a\u7b2c3\u554f"},"content":{"rendered":"<hr \/>\n<p>\u95a2\u6570 \\(f(x) = \\sin 2x +a \\cos x\\) \u306b\u3064\u3044\u3066, \u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n<ol>\r\n<li><p><strong>(1)<\/strong>\u3000\\(f(x)\\) \u304c\u533a\u9593 \\(-\\dfrac{\\pi}{2} \\lt x \\lt \\dfrac{\\pi}{2}\\) \u306e\u76f8\u7570\u306a\u308b \\(2\\) \u70b9\u3067\u6975\u5024\u3092\u53d6\u308b\u3088\u3046\u306a, \\(a\\) \u306e\u5024\u306e\u7bc4\u56f2\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<li><p><strong>(2)<\/strong>\u3000\\(a\\) \u304c <strong>(1)<\/strong> \u3067\u6c42\u3081\u305f\u7bc4\u56f2\u306b\u3042\u308b\u3068\u304d, \\(\\displaystyle\\int _ {-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\big| f(x) \\big| \\, dx\\) \u3092 \\(a\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n<li><p><strong>(3)<\/strong>\u3000\\(a\\) \u304c <strong>(1)<\/strong> \u3067\u6c42\u3081\u305f\u7bc4\u56f2\u306b\u3042\u308b\u3068\u304d, \\(f(x)\\) \u304c\u6975\u5024\u3092\u53d6\u308b \\(x\\) \u306e\u5024\u3092 \\(x = \\alpha , \\beta \\ \\left( \\text{\u305f\u3060\u3057} -\\dfrac{\\pi}{2} \\lt \\alpha \\lt \\beta \\lt \\dfrac{\\pi}{2} \\right) \\quad\\) \u3068\u3059\u308b. \\(\\displaystyle\\int _ {\\alpha}^{\\beta} \\big| f(x) \\big| \\, dx\\) \u3092 \\(a\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/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) & = 2 \\cos 2x -a \\sin x \\\\\r\n& = -\\left( 4 \\sin^2 x +a \\sin x -2 \\right)\n\\end{align}\\]\r\n\\(t = \\sin x\\) \u3068\u304a\u304f\u3068, \\(-\\dfrac{\\pi}{2} \\lt x \\lt \\dfrac{\\pi}{2}\\) \u306e\u3068\u304d, \\(t\\) \u306f\u5358\u8abf\u5897\u52a0\u3067, \\(-1 \\lt t \\lt 1\\) .<br \/>\r\n\u3057\u305f\u304c\u3063\u3066, \\(g(t) = 4t^2 +at -2 \\quad ... [1]\\) \u3068\u304a\u3044\u3066, \\(g(t) = 0\\) \u304c \\(-1 \\lt t \\lt 1\\) \u306b\u7570\u306a\u308b \\(2\\) \u3064\u306e\u5b9f\u6570\u89e3\u3092\u3082\u3064\u6761\u4ef6\u3092\u6c42\u3081\u308c\u3070\u3088\u3044.<br \/>\r\n\u5224\u5225\u5f0f\u3092 \\(D\\) \u3068\u3059\u308b\u3068\r\n\\[\r\nD = a^2+8 \\gt 0\n\\]\r\n\u306a\u306e\u3067, \u6c42\u3081\u308b\u6761\u4ef6\u306f<\/p>\r\n<ul>\r\n<li><p>\\(g(-1) = -a+2 \\gt 0\\)<\/p><\/li>\r\n<li><p>\\(g(1) =a+2 >0\\)<\/p><\/li>\r\n<li><p>\u8ef8\u306b\u3064\u3044\u3066\uff1a \\(-1 \\lt -\\dfrac{a}{8} \\lt 1\\)<\/p><\/li>\r\n<\/ul>\r\n<p>\u3088\u3063\u3066\r\n\\[\r\n\\underline{-2 \\lt a \\lt 2}\n\\]\r\n<p><strong>(2)<\/strong><\/p>\r\n<p>\\[\r\nf \\left( -\\dfrac{\\pi}{2} \\right) = -1 , \\ f(0) = a , \\ f \\left( \\dfrac{\\pi}{2} \\right) = 1\n\\]\r\n\u306a\u306e\u3067, <strong>(1)<\/strong> \u306e\u3068\u304d, \\(y = f(x)\\) \u306e\u6982\u5f62\u306f\u4e0b\u56f3\u306e\u3088\u3046\u306b\u306a\u308b.<\/p>\r\n<img decoding=\"async\" src=\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/ikashika_200903_01.png\" alt=\"\" title=\"ikashika_200903_01\" class=\"aligncenter size-full\" \/>\r\n<p>\\(f( \\gamma ) = 0\\) \u306a\u308b \\(\\gamma \\ \\left( -\\dfrac{\\pi}{2} \\lt \\gamma \\lt \\dfrac{\\pi}{2} \\right)\\) \u3092\u304a\u304f\u3068\r\n\\[\\begin{align}\r\nf( \\gamma ) & = \\sin 2\\gamma +a\\cos \\gamma \\\\\r\n& = \\left( 2\\sin \\gamma +a \\right) \\cos \\gamma = 0 \\\\\r\n\\text{\u2234} \\quad & \\sin \\gamma = -\\dfrac{a}{2} \\quad ... [2]\n\\end{align}\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nF(x) & = \\displaystyle\\int f(x) \\, dx \\\\\r\n& = -\\dfrac{\\cos 2x}{2} +a\\sin x +C \\quad ( \\ C \\text{\u306f\u7a4d\u5206\u5b9a\u6570} ) \\\\\r\n& = \\sin^2 x +a \\sin x -\\dfrac{1}{2} +C\n\\end{align}\\]\r\n\u3068\u304a\u304f\u3068, \u6c42\u3081\u308b\u9762\u7a4d\u306f\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {-\\frac{\\pi}{2}}^{\\gamma} & f(x) \\, dx +\\displaystyle\\int _ {\\gamma}^{\\frac{\\pi}{2}} f(x) \\, dx \\\\\r\n& = F \\left( -\\dfrac{\\pi}{2} \\right) +F \\left( \\dfrac{\\pi}{2} \\right) -2 F\\left( \\gamma \\right) \\\\\r\n& = \\left( -a+\\dfrac{1}{2} \\right) +\\left( a+\\dfrac{1}{2} \\right) -2 \\left( \\dfrac{a^2}{4} -a \\cdot \\dfrac{a}{2} -\\dfrac{1}{2} \\right) \\\\\r\n& = 1 +\\dfrac{a^2}{2} +1 \\\\\r\n& = \\underline{\\dfrac{a^2}{2} +2}\n\\end{align}\\]\r\n<p><strong>(3)<\/strong><\/p>\r\n<p>\\(\\sin \\alpha , \\ \\sin \\beta\\) \u306f\u65b9\u7a0b\u5f0f [1] \u306e\u89e3\u306a\u306e\u3067, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a\r\n\\[\r\n\\sin \\alpha +\\sin \\beta = -\\dfrac{a}{4} , \\ \\sin \\alpha \\sin \\beta = -\\dfrac{1}{2}\n\\]\r\n\u3053\u308c\u3092\u7528\u3044\u308c\u3070, \u6c42\u3081\u308b\u9762\u7a4d\u306f\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {\\alpha}^{\\gamma} & f(x) \\, dx +\\displaystyle\\int _ {\\gamma}^{\\beta} f(x) \\, dx \\\\\r\n& = F \\left( \\alpha \\right) +F \\left( \\beta \\right) -2 F\\left( \\gamma \\right) \\\\\r\n& = \\sin^2 \\alpha +a\\sin \\alpha -\\dfrac{1}{2} +\\sin^2 \\beta +a\\sin \\beta \\\\\r\n& \\qquad -\\dfrac{1}{2} -2 \\left( \\dfrac{a^2}{4} -a \\cdot \\dfrac{a}{2} -\\dfrac{1}{2} \\right) \\\\\r\n& = \\left( \\sin \\alpha +\\sin \\beta \\right)^2 -2\\sin \\alpha \\sin \\beta +a \\left( \\sin \\alpha +\\sin \\beta \\right) +\\dfrac{a^2}{2} \\\\\r\n& = \\dfrac{a^2}{16} +1 +\\dfrac{a^2}{4} \\\\\r\n& = \\underline{\\dfrac{5a^2}{16} +1}\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u95a2\u6570 \\(f(x) = \\sin 2x +a \\cos x\\) \u306b\u3064\u3044\u3066, \u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088. (1)\u3000\\(f(x)\\) \u304c\u533a\u9593 \\(-\\dfrac{\\pi}{2} \\lt x \\lt \\dfrac{\\pi}{2}\\ &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/iks200903\/\">\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":[27],"tags":[145,15],"class_list":["post-211","post","type-post","status-publish","format-standard","hentry","category-ikashika_2009","tag-ikashika","tag-15"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/211","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=211"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/211\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=211"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=211"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=211"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}