高等数学模拟试题（一）
一、选择题（本大题共5个小题，每小题3分，共15分）
1.B1排除法,当x?0时,??.排除A.当x???时,ln?1?x???,xx2???,排除CD.2.D由函数在某一点导数定义可知：f?x??f?a?x?a???x??0??lim?lim??x?x?ax?ax?ax?ax?a进一步利用函数连续性可知，lim??x?=??a?.故应选D.f??a??limx?a3.A有函数的表达式可知，只需考虑在分界点x?0处连续性即可.1因为f?0?=1，limxcos?0,所以f?x?在x?0处不连续.故函数x?0x的连续区间为???，0?U?0,+??.故选A4.B因为1?11?????,24n?12?2n?12n?1?1?1??1??11?1???11?1所以limxn?lim???L?2??lim??1???????L?????n??n??3154n?1?n??2??3??35???2n?12n?1??1?1?1?lim?1???,故选B.n??2?2n?1?25.Aexex整理方程得:y?=，分离变量可得:ydy?,xx?1?e??y?1?e?故y2?2ln?1?ey??c,将yx?0代入得c??2ln2.故应选A.
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二、填空题（本大题共5个小题，每小题3分，共15分）
6.原点或?0,0?解析:因为偶函数关于y轴对称，奇函数关于原点对称，由f??x??ln?1???x????x??ln2??1??x??x2???1???f?x?可知:?ln??1?x2?x?????f?x?=ln7.?18?1??x??x为奇函数.2?
cosx11cosx1cosx?0?sinx解析式??lim??limglim??lim4x??sinxx?????2x?4x?????2x??0?x??2???2x???2?22221?sinx1??glim??.4x????2?82
8.2ex?1?x1在x?0处,由于limf?x??0,f?0??0,lim?f?x??lim??.x?0?x?0x?02x2故x?0为间断点.ex?1?xe?2x在x?1处,由于limfx?lim?,limfx?lime?1??e?1.???????x?1?x?1?x?1x?12x2故x?1为间断点.9.x?lnx?1?解析:由题意可知f??x??lnx,又?lnxdx?xlnx??xdlnx?xlnx?x?c.10.x?1yz?2??4?13uvuuv解在直线上选取一点?1，0，?2?,因两平面法向量n1??1,1,1?,n2??2,?1,3?vvvijkvuvuuvvvv 所以，s?n1?n2?111?4i?j?3k.是直线方程的方向向量,因此,
2?13所给直线方程为
2
x?1yz?2??.4?13三、计算题（本大题共7个小题，每小题7分，共49分）
11.?1??x2?1???x2?解:lim??ax?b??lim??ax?b?x??x?1x????x?1????1?lim?x?1?ax?b?x??x?1x???lim?1?a?x?lim?1?b??0,?lim
x??x??
所以，lim?1?a?x=lim?1?b?；x??x???1?a?0由于a,b为常数，故当且仅当?时上式成立.?1?b?0因此，a?1,b??1.12.1???1?解:设an??n?1,2,L?,则数列?an?为n2222,0,,0,,0,L,,0,L352n?1
显然，liman?0,即当n??时，an为无穷小量.n??n?1
1但是在这一变化过程中,有许多项为0,无意义.an1当n??时，不是无穷大量.an13.解析:根据隐函数显化思想.假设函数y?y?x?为方程的解,代入方程,即得恒等式:xy?x??ex?ey?x??0.由复合函数求导法则可知:y?x??xy??x??ex?ey?x?
上式两边同时对x求导?注意y?y?x?是x的函数?，y??x??0.ex?y由上式解得:y??x??y?y?x??.y?x?e 3
14.ehlna?1解:左边=lim?lna.使用分部积分公式，h?0h???0xedx????2?2?x??0??xd?e2?x???xe??0???x2?x??0??e?xdx20??
0??2xe所以,a?e2.15.?xe?xdx??2??x??00xde?x??edx??2e??x??0?2.解:F?x,y,z??lnx?lny?3lnz???x2?y2?z2?5R2?.?x?0,y?0,z?0,? ?'1?Fx?x?2?x?0??x?R?F'?1?2?y?0??yy则?,解得?y?R,故lnxyz3?ln33R.即Fmax?ln33R.?'1??z?3R?Fz??2?z?0z?2222??x?y?z?5R16.解:因为?edx不能用初等函数表示，所以先交换积分顺序再求解.I??dy?1edy??1dy?221214yyx1yyyx31edy??1dy?2edy??1x?e?ex?dx?e?e.x82221x1yxyx
17.解因为n?2?！n?n?1??un?1lim?limn?2n??un???n?1??n?1?!nnn?2?n??lim??n???n?1?n?1n?1??1?1.e所以原级数收敛.?n
四、应用与证明题（本大题共3个小题，每小题7分，共21分）
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18.解:y?x3?x2?2x,x
y?y??1????,??3???凸13y??3x2?2x?2,?1?,?????3??1y???6x?2,令y???0得,x?.3
?116?拐点?，?凹327??1??116???1?故曲线的拐点为?，?；凸区间为???,?,凹区间为?,???.3??327???3?19.先求出曲线 y?lnx与直线 y??e?1??x的交点坐标；再利用积分求面积.??y?lnx由?可知交点坐标为?e,1?.故所求区域D的面积??y??e?1??x1e?1?y13S???dxdy??dy?ydx???e?1?y?ey?dy?.0e02D
20.证明:令f?x??x3?9x?1.因为f??3???1?0,f??2??9?0,f?0???1?0,f?4??27?0.又f?x?在??3,4?上连续，所以可知f?x?在??3，?2?，??2,0?，?0,4?各区间所以方程有3个实根.
内至少有一个零点.又因为它是一元三次方程，即f?x??0至多有三个根； 5
专升本高等数学模拟试题(一)
