军队文职数学2模拟题及答案解析B.?
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C.?
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a11?xa12?xa13?xa14?x22.设f?x??a21?2xa22?2xa23?2xa24?2xa,则多项式f?x?可能的最高次数是31?3xa32?3xa33?3xa34?3xa41?4xa41?4xa43?4xa44?4x()A.0B.1C.2D.3??11
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A.1B.3C.1或3D.无法确定6军队文职数学2模拟题及答案解析答案解析
一、单项选择题(共9题,每小题1分,共9分。)1.设f?x?????1,x?1
,则??
0,x?1f?f??f?x????等于()A.0B.1C.???1,x?1??0,x?1
D.?
??0,x?1??1,x?1
1.【答案】B【解析】由f?x?的定义知f?x??1,故f??f?x????1,从而f?f??f?x?????1,故选B。2.极限lim?n???1??1?)?1?21?2?3??1?1?2?3????n?1???的值是(?A.?1B.1C.32D.342.【答案】B【解析】11?2?11?2?3???11?2?3????n?1??12?1?2??13?1?3????221?n?1??1?n?1??22?3?23?4???2?n?1???n?2??2??1?2?3?13?4???1??n?1??n?2??2?????2??111?2?13?13?14???n?1??n?2???2??1?2?1?n?2???nn?2,∴极限lim?n???11?1?2?1?2?3???1?1?2?3????n?1?????limnn??n?2?1,故选B。7军队文职数学2模拟题及答案解析3.若函数f?x????e2x?1x?0lnax?0在x?0处连续,则a?()?A.e2B.eC.2D.13.【答案】D【解析】左极限:xlim?lna?lna;右极限:lim2x?0?e?1?0,该点函数值f?0??0。根据?0x函数连续的条件可知:左极限=右极限=该点函数值;lna?0,a?1。故选D。4.设函数f?x?可导,且曲线y?f?x?在点?x0,f?x0??处的切线与直线y?2?x垂直,则当?x?0时,该函数在x?x0处的微分dy是()A.与?x同阶但非等价无穷小B.与?x等价无穷小C.比?x高阶的无穷小D.比?x低阶的无穷小4.【答案】B【解析】由题设可知f??xdy
0??1,而dy|x?x0?f??x0??x??x,因而?limx?0?x
|x?x0?1,即在x?x0处dy与?x是等价无穷小,故选B。5.计算?
xf???x?dx?()A.xf??x??f(x)B.xf??x??CC.xf??x??f(x)?CD.xf??x??f(x)5.【答案】C【解析】根据不定积分的分部积分法,可得?xf???x?dx??xdf??x??xf?(x)?f(x)?C。故选C。8军队文职数学2模拟题及答案解析6.函数z?ln(1?x2
?y2
)当x?1,y?2时的全微分dz是()A.23dx?23
dyB.dx?dy
C.12
3dx?3dy
D.113dx?3dy6.【答案】C【解析】?z?x?2x1?x2?y2,?z?y?2y1?x2?y2,?z
?(1,2)
?1?zx3,?y(1,2)?
2
3
,因此dz?13dx?2
3
dy。故选C。7.求曲线x?t1?t1?t,y?t,z?t2在对应于t?1的点处的法平面方程是()A.x?
12?y?2?z?11?48B.x?4y?8z?2?0C.x?1y?2z?12?4?8D.2x?8y?16z?1?07.【答案】D【解析】曲线在对应于t?1的点为(1
2
,2,1),该点处的切向量为(x?(1),y?(1),z?(1))?(
1(1?t)2,?1
t,2t)t?1
?(124
,?1,2),所以所求法平面方程为2x?8y?16z?1?0。故选D。8.已知A,B均为n阶方阵,且AB=O,B?O,则必有()A.?A?B?2
?A2
?B
2
B.B?0C.B*
?0
9军队文职数学2模拟题及答案解析D.A*
?08.【答案】D【解析】AB=O不一定有BA=O,选项A错误;B?O,但是B可以为零,也可以不为零,B*
可以为零,也可以不为零,故选项B和C错误;AB=O,B?O,所以方程AX=0有非零解,故A?0,从而A*
?A
n?1
?0,故选D。9.若A是m?n矩阵,且A的n个列向量线性无关,则A的秩为()A.大于nB.大于mC.等于mD.等于n9.【答案】D【解析】由于矩阵A的秩等于行向量组的秩或是列向量组的秩,又因为A的n个列向量线性无关,故矩阵A中列向量组的秩为n,即矩阵A的秩为n。故选D。二、单项选择题(共14题,每小题1.5分,共21分。)10.已知f?x?是定义在R上的偶函数,其导函数f'?x?,若f'?x?小于f?x?,且f?x?1?=f?3?x?,f?2015?=2,则不等式f?x??2ex?1的解集为()A.?1,???B.?e,???C.???,0?D.????,1???e?10.【答案】A【解析】因为函数f?x?是定义在R上的偶函数,所以f?x?1??f?3?x??f?x?3?,即f?x??f?x?4?,则函数f?x?是周期为4的周期函数,?f?2015??f?2015?4?504??f??1??f?1??2,?f?1??2。设g?x??f?x??x??f?x?ex,则g'?x??f'ex?0,故g?x?在R上是减函数,10