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[函数] $k=6$是不是这个选择题的反例?

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isee Posted at 2014-5-15 10:10:49 |Read mode
我觉得这四个选项都不对,好奇怪,不知道思维陷到哪个误区了。



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$k=6$ 是不是这个选择题的反例?

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realnumber Posted at 2014-5-15 11:33:12
分段函数符号找起来麻烦,写成分类的样子.
当$x\ge3 或x\le-2$,$f(x)=x+4+k$.
当$-2< x<3,f(x)=x^2-1+k$,
当k=6时,第一类解得$x=-10$,第2类无解.

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 Author| isee Posted at 2014-5-15 11:55:19
分段函数符号找起来麻烦,写成分类的样子.
当$x\ge3 或x\le-2$,$f(x)=x+4+k$.
当$-2< x ...
realnumber 发表于 2014-5-15 11:33

    fxck

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 Author| isee Posted at 2014-5-15 11:56:10
这个+k原来是这么回事!

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 Author| isee Posted at 2014-5-15 11:56:26
回复 2# realnumber


    THX!

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realnumber Posted at 2014-5-15 13:06:37
en ,我也明白了,看来这题目应该出得更严谨些,
记$g(x)=(x^2-1)⊙(4+x),f(x)=g(x)+k$

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kuing Posted at 2014-5-15 13:09:24
还可以加括号,或者先规定运算符优先级别

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realnumber Posted at 2014-5-15 21:58:15
其实应该没误会的,原题中已经特别加了括号(4+x)+k,而不是4+x+k.也算暗示吧.

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 Author| isee Posted at 2014-5-15 22:34:23
回复 8# realnumber

嗯,属于先入为主了。

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kuing Posted at 2014-5-16 00:26:03
好一个暗示……

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