设f为抛物线y2 16x(x)=x2+ax+b,A=﹛x︱...

来源：蜘蛛抓取(WebSpider) 时间：2011-10-05 03:25 标签： 设x1 x2是m的方程
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Henri Darmon and
Andrew Granville
On the Equations zm = F(x, y) and Axp + Byq = Czr
Bull. London Math. Soc. 1995 27 (6): 513-543
doi:10.1112/blms/27.6.513
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Bull. London Math. Soc.
10.1112/blms/27.6.513
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Very Short Introductions
World's Classics设f(x)=-1/3x^2+1/2x^2+2ax.(1)若f(x)在(2/3,+∞)上存在单调递增区间，
欢迎百度一句话一类题QQ
(转发者2013考上清华哦！)
十八年高考教学告诉你：高考还有30多天20多天10多天孩子怎么复习？高考前什么东西是最重要的
问：这一道题，为什么一定要最等号？我的解法为什么不满分？
问：我平时考90分左右，现在我想考上120分，“一句话一类题”这本书短时间能够帮到我么？怎么购？是不是要保密？
答：因为区间(2/3,+∞)的左端点是开的。如果是闭的，就能够取等号。
这一个问题有一个规律。当且仅当，全部取等号时，才取等号。如“a&=b&c”不能取等号。
“a&=b&=c”取等号。得a&=c
欢迎有所有高考考的有问题，向我的信箱发。一定会解答。但要提供的东西如上面这么详细哦。
欢迎百度一句话一类题QQ
(转发者2012考上清华哦！)
第二个问题。你对一句话一类题的提问，全部是肯定的答案。欢迎你尽快使用一句话一类题。高考的时间紧了，宜早不宜迟！
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（2015安徽）设函数f（x）=x2-ax+b．
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（2015安徽）设函数f（x）=x2-ax+b．
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（2015安徽）设函数f（x）=x2-ax+b．&（Ⅰ）讨论函数f（sinx）在（-
2）内的单调性并判断有无极值，有极值时求出最值；&（Ⅱ）记fn（x）=x2-a0x+b0，求函数{f（sinx）-f0（sinx）}在[-
2]上的最大值D2&（Ⅲ）在（Ⅱ）中，取an=bn=0，求s=b-
满足条件D≤1时的最大值．&
解：（Ⅰ）设t=sinx，在x∈（-π2，π2）递增， 即有f（t）=t2-at+b（-1＜t＜1），f′（t）=2t-a， ①当a≥2时，f′（t）≤0，f（t）递减，即f（sinx）递减； 当a≤-2时，f′（t）≥0，f（t）递增，即f（sinx）递增． 即有a≥2或a≤-2时，不存在极值． 时，取得最大值，设g（t）=|-t（a-a0）+（b-b0）|， 而g（1）=|-（a-a0）+（b-b0）|，g（-1）=|（a-a0）+（b-b0）|， 则当（a-a0）（b-b0）≥0时，D=g（t）max=g（-1）=|（a-a0）+（b-b0）|； 当（a-a0）（b-b0）≤0时，D=g ②当-2＜a＜2时，-1＜t＜a2，f′（t）＜0，f（sinx）递减； a2＜t＜1，f′（t）＞0，f（sinx）递增． f（sinx）有极小值f（a2）=b-a24； （Ⅱ）设t=sinx，t∈[-1，1]，|f（t）-f0（t）|=|-t（a-a0）+（b-b0）|， 易知t=±1t）max=g（1）=|-（a-a0）+（b-b0）|． （Ⅲ）由（Ⅱ）得ab≥0时，D=|a+b|，当ab≤0时，D=|a-b|． 即有ab≥0-1≤a+b≤1或ab≤0-1≤a-b≤1， 点（a，b）在如图所示的区域内， 则有s=b-a24，当b取最大值1时，a24取最小值0时， smax=1．
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