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isee Posted at 2014-1-13 23:17:43 |Read mode
Last edited by isee at 2014-1-14 11:24:002014年1月北京东城高三上数学理科期末8


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核心题干:
设$\left\{\begin{aligned}\abs x &\leqslant 2,\\ \abs y &\leqslant 2\end{aligned}\right.$则$\max\{4x+y,3x-y\}$的取值范围是(     )
A.$[-6,10]$       B.$[-7,10]$         C.$[-6,8]$        D. $[-7,8]$


个人,笨解法,Key:以$4x+7=3x-y$分界,结果 B

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其妙 Posted at 2014-1-13 23:42:02
回复 1# isee
记$z=\max\{4x+y,3x-y\}$,则$z\geqslant 4x+y,z\geqslant 3x-y$,

两式相加得,$2z\geqslant(4x+y)+(3x-y)=7x\geqslant-14$(因为$x\geqslant-2$),

故$z\geqslant -7$。

又$4x+y\leqslant4|x|+|y|\leqslant4\times2+2=10$,

$3x-y\leqslant3|x|+|y|\leqslant3\times2+2=8$,
故$z=\max\{4x+y,3x-y\}\leqslant10$,

选$B$.

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其妙 Posted at 2014-1-13 23:48:58
回复 2# 其妙
顺便说一下取等条件在可行域里:
当$x=-2,y=1$时,$4x+y=-7,3x-y=-7$,故$z=\max\{4x+y,3x-y\}$取最小值-7。

当$x=2,y=2$时,$4x+y=10,3x-y=4$,故此时$z=\max\{4x+y,3x-y\}$可取最大值10。

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乌贼 Posted at 2014-1-13 23:57:43
回复 1# isee
$4x+y=3x-y$还是$4x+7=3x-y$

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乌贼 Posted at 2014-1-14 00:20:09
$z$的取值范围是不是连续的?

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爪机专用 Posted at 2014-1-14 01:43:47
回复 5# 乌贼
显然连续
I am majia of kuing

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 Author| isee Posted at 2014-1-14 11:20:23
回复 4# 乌贼

是$y$不是7,这个意思:

$\max\{4x+y,3x-y\}=\left\{\begin{aligned}4x+y,&&4x+y&\geqslant 3x-y\\ 3x-y,&&4x+y&<3x-y\end{aligned}\right.$

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kuing Posted at 2014-1-14 16:50:59
还可以试试换元。

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