《高等数学(二)》密押试卷
一、选择题(1~10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的)
x2?1?() 1. limx?1x?1A.0 B.1 C.2 D.3
?x?1??x?1??limx?1?2x2?1?limC lim??.
x?1x?1x?1x?1x?12. 设函数f?x?在x?1处可导,且f??1??2,则limx?0f?1?x??f?1??()
xA.-2
1B.? 21C. 2D.2 A limx?0f?1?x??f?1?f?1?x??f?1???lim??f??1???2.
x?0x?x?π?3. 设函数f?x??cosx,则f???=()
?2?A.-1
1B.-
2C.0 D.1
π???A 因为f?x??cosx,f??x???sinx,所以f?????sin??1.
2?2?4. 设函数f?x?在区间?a,b?连续且不恒为零,则下列各式中不恒为常数的是() A.f?a?
第 1 页 共 8 页
B.
?f?x?dx
abf?x? C.lim?x?bxD.
?f?t?dta
??0;B项,fa?D 设f?x?在?a,b?上的原函数为F?x?.A项,??????bf?x?dx????F?b??F?a????0;C项,?limf?x?????F?b????0;D项,
???????x?b?????a???xf?t?dt???f?x?.故A、B、C项恒为常数,D项不恒为常数.
??a???5.
2x?dx?()
3A. 3x?C 3B. x?C
x3?C3C. x?CD. 2
x3x2dx?3?C?C .
6. 设函数f?x?在区间?a,b?连续,且I?u???f?x?dx??f?t?dt,a?u?b,则
aauuI?u?()
A.恒大于零 B.恒小于零 C.恒等于零 D.可正,可负
C 因定积分与积分变量所用字母无关,故
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I?u???f?x?dx??f?t?dt??f?x?dx??f?x?dx??f?x?dx?0.
aaauauuuaa7. 设函数z?ln?x?y?,则A.0
1B.
2C.ln2 D.1
B 因为z?ln?x?y?,
?z?x?1,1??().
?z1?z?,所以?xx?y?x?1,1??1. 28. 设函数z?x3?y3,则A. 3x2 B. 3x2?3y2
y4C.
4?z=(). ?yD. 3y2
D 因为z?x3?y3,所以
??
?2z
?z=3y2. ?y9. 设函数z=x??,则?x?y=(). A. ???? B.???? C.x???? D.y????
B 因为z=x??,则?x=e, ?x?y=????.
10. 设事件A,B相互独立,A,B发生的概率分别为0.6,0.9,则A,B都不发生的概率为(). A.0.54 B.0.04 C.0.1 D.0.4
第 3 页 共 8 页
??
?z
y
?2z
B 事件A,B相互独立,则A,B也相互独立,故P(AB)=P(A)P(B)=(1-0.6)×(1-0.9)=0.04. 二、填空题(11~20小题,每小题4分,共40分)
511.函数f?x??的间断点为x=________.
x?1
1 f?x?在x=1处无定义,故f?x?在x=1处不连续,则x=1是函数f?x?的间断点.
12.设函数f(x)={lnx,??≥1,在x?1处连续,则a=________.
a?x,??<1
f?x??lim1 lim?a?x??a?1,因为函数f?x?在x?1处连续,故??x?1x?1x?1?limf?x??f?1??ln1?0,即a-1=0,故a=1.
sin2x=________.
x?03x13.lim
2sin2x2cos2x2 lim ?lim? 3.x?0x?033x314. 当x→0时,f?x?与sin2x是等价无穷小量,则lim
1 由等价无穷小量定义知,limf?x??1
x?0sin2x.
f?x?=________.
x?0sin2x15. 设函数y?sinx,则y???=________.
?cosx
因为y?sinx,故y??cosx,y????sinx,y?????cosx.
16.设曲线y=ax2+2x在点(1,a+2)处的切线与直线y=4x平行,则a=________.
1 因为该切线与直线y=4x平行,故切线的斜率k=4,而曲线斜率y′(1)=2a+2,故2a+2=4,即a=1. 17. ?2xedx?________.
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x2
e?C ?2xedx??edx2?ex?C.
x2x2x2218.
?π20esinxcosxdx? ________.
π20π20πsinx20e-1 ?e19.
π2
sinxcosxdx??esinxd?sinx??e? =e-1.
???01dx?________. 21?x??aa11π dx?limdx?limarctanx?limarctana?a???01?x2a??0a??1?x22.
?020. 设函数z?ex?y,则dz=________.
exdx?dy dz?
?z?zdx?dy?exdx?dy. ?x?y三、解答题(21~28题,共70分.解答应写出推理、演算步骤)
21.(本题满分8分) 计算lim?1?x?.
x?02x
1??lim?1?x?=lim??1+x?x??e2x?0??. 解: x?02x222.(本题满分8分)
设函数y=sinx2+2x,求dy.
解:因为y???x2??cosx2?2?2xcosx2?2, 故dy??2xcosx2?2?dx. 23.(本题满分8分) 计算?lnxdx.1e
eee解:?lnxdx?xlnx??xd?lnx?1 11第 5 页 共 8 页
成考专升本高等数学(二)考前密押试卷
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