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[数列] 数列不等式 $|a_n-a_{n+1}/2|\le1$

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hjfmhh Posted at 2016-6-7 18:03:04 |Read mode
NSF)_R1ZLXT{}]@S29(LDGQ.jpg

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realnumber Posted at 2016-6-7 20:56:58

QQ图片20160605141125--1.jpg
QQ图片20160607205416-2.png

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游客 Posted at 2016-6-8 02:41:54
未命名.PNG

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Tesla35 Posted at 2016-6-9 12:18:15
回复 2# realnumber


    浙江那个向量题咋做

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realnumber Posted at 2016-6-9 17:46:45
Last edited by realnumber at 2016-6-10 10:14:00回复 4# Tesla35
\[ \vv{a}=(\cos{x},\sin{x}),\vv{b}=(2\cos y,2\sin y),\vv{e}=(\cos z,\sin z).\]
\[\sqrt{6}\ge \abs{\cos{(x-z)}}+\abs{2\cos{(y-z)}}\ge \abs{\cos{(x-z)}+2\cos{(y-z)}}=\abs{(\cos x +2\cos y)\cos z +(\sin x +2\sin y)\sin z}.\]
\[6\ge (\cos x +2\cos y)^2+(\sin x +2\sin y)^2.\]
\[\cos{(x-y)}\le \frac{1}{4}.\]
\[\vv{a}·\vv{b}   最大为\frac{1}{2}.\]

上面当平面向量处理了,不好.
有人这么解比较好
\[\abs{\vv{a}·\vv{e}+\vv{b}·\vv{e}}\le \abs{\vv{a}·\vv{e}}+\abs{\vv{b}·\vv{e}}\le \sqrt{6}\]
再两边平方下.

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游客 Posted at 2016-6-9 22:36:31
Last edited by 游客 at 2016-6-10 01:01:00 未命名.PNG


未命名.PNG

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isee Posted at 2016-6-10 08:45:51
完整的浙江卷理科,丢一套上来呀。

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realnumber Posted at 2016-6-10 09:55:32
回复 7# isee

11.JPG

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isee Posted at 2016-6-10 10:25:40

    太好了,已经下载。。。。。

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isee Posted at 2016-6-10 10:27:10
回复  Tesla35
\[ \vv{a}=(\cos{x},\sin{x}),\vv{b}=(2\cos y,2\sin y),\vv{e}=(\cos z,\sin z).\]
\[\sqr ...
realnumber 发表于 2016-6-9 17:46
$\abs{\vv a \vv e+\vv b \vv e}\leqslant \abs{\vv a \vv e}+\abs{\vv b \vv e}\leqslant \sqrt 6$两边平方,具体怎么算?

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游客 Posted at 2016-6-10 10:32:38
未命名.PNG

第一小题的答案为什么还带着参数,这样真的可以吗?
那样很多不等式恒成立问题直接变形就算解决了?

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realnumber Posted at 2016-6-10 10:50:49
回复 10# isee


    进一步得到$\abs{(\vv{a}+\vv{b})·\vv{e}}\le \sqrt {6}$
上式对任意的$\vv{e}$均成立,即得左边最大值小于等于$\sqrt {6}$,
即$\abs{\vv{a}+\vv{b}}\le \sqrt {6}$,再两边平方下.

--就是你说的啊,

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isee Posted at 2016-6-10 10:57:35
回复  isee


    进一步得到$\abs{(\vv{a}+\vv{b})·\vv{e}}\le \sqrt {6}$
上式对任意的$\vv{e}$均成立 ...
realnumber 发表于 2016-6-10 10:50
哦~~~~~~~~~~~~~

这正是任意单位向量的意思啊~~

对对对,赞一个!

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游客 Posted at 2016-6-10 11:02:37
Last edited by 游客 at 2016-6-10 11:11:00回复 13# isee


    $\abs{m}+\abs{n}=\abs{m+n}(mn>0)或\abs{m-n}(mn<0)$
跟去年18题(2)考得一样。




14题考两球面交线,考生万年啊!

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isee Posted at 2016-6-10 11:27:27
回复  isee


    $\abs{m}+\abs{n}=\abs{m+n}(mn>0)或\abs{m-n}(mn
游客 发表于 2016-6-10 11:02
原来常套数啊

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游客 Posted at 2016-6-10 12:21:38
第5题搞脑子,跟b是否为0有关,坑的不浅;
第8题前3个选项跟15题考一样的东西,是否重复考查?
第8题第四个正确选项什么鬼?难啊!

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realnumber Posted at 2016-6-10 14:44:31
回复 16# 游客


    8.觉得考察数感啊,很不错问题
A,a=b=100,c=-10100
B.a=100,b=-10000,c=0
C.a=200,b=-200,c=0
D.可得$\abs{a^2+a+b^2+b}\le1$,这样$\abs{a},\abs{b}$,都不会太大,进一步c也一样.

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阿鲁 Posted at 2016-6-10 14:48:20
回复 17# realnumber

正是此意

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阿鲁 Posted at 2016-6-10 15:05:29
回复 17# realnumber

我估计在 $\abs{a^2+b+c}+\abs{a+b^2-c}\le1$ 之下 $a^2+b^2+c^2$ 可能连 1 都达不到

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realnumber Posted at 2016-6-10 15:25:54
回复 19# 阿鲁


    要不改为求$a^2+b^2+c^2$的上界?看看谁求得小

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