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则 ?lnt?ln(1?t)?dt??lnttndt?n?1,2,001n1?.
(II)?lnttndt???lnt?tndt??0011111n?1lntdt ,故由 ????20n?1?n?1?10?un??lnttndt?01?n?1?2,
根据夹逼定理得0?limun?limn??n??1?n?1?2?0,所以limun?0.
n?? (17)【解析】根据题意得
????t??d??????t??2t?2??2???t?2t?2??222t?2???t?dy??3dt??? ?,2dx2t?24?1?t?2t?2dxdt22dydydt?dxdxdt即????t??2t?2??2???t??6?t?1?,整理有????t??t?1?????t??3?t?1?,解
???t??????t???3?t?1??1?t?1??y?y?3?1?t?. ,令,即y??t???1?t???1??5,???1??6??211dt???dt??1?t1?t所以y?edt?C???1?t??3t?C?,t??1.因为y?1?????1??6,所以C?0,??3?1?t?e??故y?3t?t?1?,即???t??3t?t?1?,
3故??t???3t?t?1?dt?t2?t3?C1．
253又由??1??,所以C1?0,故??t??t2?t3,(t??1).
22(18)【解析】油罐放平,截面如图建立坐标系之后,边界椭圆的方程为：
x2y2??1 a2b2阴影部分的面积
S??b2?b2ab2222xdy?b?ydy ??bbwww.lookwell.com.cn ;免费考研辅导视频 乐考无忧官方考研交流群：341384403
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令y?bsint,y??b时t????2;y?2b?时t?. 26?1123S?2ab?6?costdt?2ab?6?(?cos2t)dt?(??)ab
??23422223所以油的质量m?(??)abl?.
34(19)【解析】由复合函数链式法则得
?u?u???u???u?u, ???????x???x?y?x?????u?u???u???u?u?????a?b?, ?y???y???y?????2u???u?u??2u???2u???2u???2u?? ??????2?????2?2?x?x?????????x?????x???x?????x?2u?2u?2u?2?2?2, ?????????2u???u?u??2u???2u???2u???2u?? ??????2?????2??x?y?y?????????y?????y???y?????y?2u?2u?2u ?a2?b2?(a?b),
?????????2u???u?u??2u?2u?2u?2u??a?b)?b(a2?a) ??a(a2?b2?y?y??????????????????2?2u?2u2?u ?a?b?2ab, 22????????2?2u?u2?2u故42?12?52
?x?x?y?y?2u?2u?2u2?(5a?12a?4)2?(5b?12b?4)2??12(a?b)?10ab?8??0,
????????2?5a2?1a2??4??0?2所以 ?5b?1b2??40,
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则a??2222或?2,b??或?2.又因为当(a,b)为(?2,?2),(?,?)时方程(3)不满足,所以当555522(a,b)为(?,?2)?,(?2,?)满足题意.
55D(20)【解析】I???r2sin?1?r2cos2?drd?
???rsin?1?r2?cos2??sin2???rdrd?
D???y1?x2?y2dxdy
D??dx?011x031?22?y1?x?ydy???1??1?x??dx 03??22111??dx???1?x03301322?1?13dx???2cos4?d????.
303161?1?(21)【解析】令F?x??f?x??x3,对于F?x?在?0,?上利用拉格朗日中值定理,得存在
3?2????0,?, 使得
2?1?1?F???F?0??F????．
2?2??1??1?对于F?x?在?,1?上利用拉格朗日中值定理,得存在???,1?,使得
?2??2??1?1F?1??F???F????,
?2?2??1?两式相加得 f?????f??????2??2.
?1??1?所以存在???0,?,???,1?,使f?????f??????2??2.
?2??2?(22) 【解析】因为方程组有两个不同的解,所以可以判断方程组增广矩阵的秩小于3,进而可
以通过秩的关系求解方程组中未知参数,有以下两种方法.
方法1：( I )已知Ax?b有2个不同的解,故r(A)?r(A)?3,对增广矩阵进行初等行变换,得
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11a??11?1???????A??0??101???0??101?
?1?1?1?11a??????1??1???0??10?01??1??2?1??11???1???0??102?0a???01??????1? a???1??1?1111??1111?????当??1时,A??0001???0001?,此时,r(A)?r(A),故Ax?b无解(舍去)．
?000a??0000??????11?11???1?,由于r(A)?r(A)?3,所以a??2,故???1 ,a??2. 当???1时,A??0?20?000a?2???方法2：已知Ax?b有2个不同的解,故r(A)?r(A)?3,因此A?0,即
?11A?0??10?(??1)2(??1)?0,
11?知??1或-1.
当??1时,r(A)?1?r(A)?2,此时,Ax?b无解,因此???1.由r(A)?r(A),得a??2. ( II ) 对增广矩阵做初等行变换
3??10?1?2???111?2??1?1?12???1? ?????A??0?201???020?1??010??2??11?11??0000???????0000???????3??2?3?x1?1?????x1?x3??1?????2可知原方程组等价为?,写成向量的形式,即?x2??x3?0?????.
???x??1??2??x??12?3???0???2?????www.lookwell.com.cn ;免费考研辅导视频 乐考无忧官方考研交流群：341384403
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?3??2?1????1??因此Ax?b的通解为x?k?0????? ,其中k为任意常数.
???1??2???0???????0?14???(23)【解析】由于A???13a?,存在正交矩阵Q,使得QTAQ为对角阵,且Q的第一列为
?4a0???11(1,2,1)T,故A对应于?1的特征向量为?1?(1,2,1)T.
66???根据特征值和特征向量的定义,有?A?????1????6??2?????1?6??1????6??1??6?2?,即 ?6?1??6??0?14??1??1??0?14??????????13a2??2A??13?1,由此可得.故a??1,??211????????. ?4a0??1??1??4?10??????????1?41?(??4)(??2)(??5)?0,
由?E?A?1??3?41?可得A的特征值为?1?2,?2??4,?3?5.
??41?4??x1?????由(?2E?A)x?0,即?1?71??x2??0,可解得对应于?2??4的线性无关的特征向量
??41?4??x????3?为?2?(?1,0,1)T.
?51?4??x1?????由(?3E?A)x?0,即?121??x2??0,可解得对应于?3?5的特征向量为
??415??x????3??3?(1,?1,1)T.
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由于A为实对称矩阵,?1,?2,?3为对应于不同特征值的特征向量,所以?1,?2,?3相互正交,只需单位化：
?1???1?111?(1,2,1)T,?2?2?(?1,0,1)T,?3?3?(1,?1,1)T, ?16??1?6取Q???,?22,?3???1??6??1?6
?22?33?11?23??2?0?1??,则QTAQ?????4?3????.
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