题目内容
如图,在正四棱柱ABCD-A1B1C1D1中,AA1=2,AB=1,M、N分别在AD1,BC上移动,并始终保持MN∥平面DCC1D1,设BN=x,MN=y,则函数y=f(x)的图象大致是( )
A.

B.

C.
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D.
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【答案】分析:由MN∥平面DCC1D1,我们过M点向AD做垂线,垂足为E,则ME=2AE=BN,由此易得到函数y=f(x)的解析式,分析函数的性质,并逐一比照四个答案中的图象,我们易得到函数的图象.
解答:解:若MN∥平面DCC1D1,
则|MN|=
=
即函数y=f(x)的解析式为
f(x)=
(0≤x≤1)
其图象过(0,1)点,在区间[0,1]上呈凹状单调递增
故选C
点评:本题考查的知识点是线面平行的性质,函数的图象与性质等,根据已知列出函数的解析式是解答本题的关键.
解答:解:若MN∥平面DCC1D1,
则|MN|=


即函数y=f(x)的解析式为
f(x)=

其图象过(0,1)点,在区间[0,1]上呈凹状单调递增
故选C
点评:本题考查的知识点是线面平行的性质,函数的图象与性质等,根据已知列出函数的解析式是解答本题的关键.

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