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[函数] 零点之和

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isee Posted 2013-11-13 22:16 |Read mode
题目:定义在$(0,+\infty)$上的函数$f(x)$满足:
(1)当$x\in [1,3)$时,$f(x)=1-\abs{x-2}$;
(2)$f(3x)=3f(x)$。
设关于$x$的函数$F(x)=f(x)-a$的零点从小到大依次为$x_1,x_2,\cdots,x_n,\cdots$。
若$a=1$,则$x_1+x_2+x_3=$______;
若$a\in (1,3)$,则$x_1+x_2+\cdots +x_{2n}=$_________.

PS:2013年11月海淀高三上理科期中第14题。

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kuing Posted 2013-11-13 22:41
画图看交点,两两配对和为定值,最后变成数列求和

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明春 Posted 2013-11-13 23:33
对称轴?
还有就是相似形?相似比为3?

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爪机专用 Posted 2013-11-13 23:39
回复 3# 明春

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乌贼 Posted 2013-11-14 00:30
Last edited by 乌贼 2013-11-14 00:36不明白,根据函数性质(1)有$f(\frac32)=\frac12$,而根据性质(2)有$f(\frac32)=f(3\times\frac12)=3f(\frac12)=-\frac32$
没看清$x\in[1,3)$,没问题了

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kuing Posted 2013-11-14 01:46
QQ截图20131114014739.gif

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 Author| isee Posted 2013-11-15 12:53
第二空,真做起来,找对称轴,还是麻烦的。需要力强的想像能力。

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