2014-2015年考研数学二真题及答案解析 下载本文

0??g(t)dt?x?a,x?[a,b],

ax(II)

?a?a?ag(t)dtf(x)dx?bf(x)g(x)dx. ?ab【解析】(I)由积分中值定理

?g?t?dt?g????x?a?,??[a,x]

ax?0?g?x??1,?0?g????x?a???x?a? ?0??g?t?dt??x?a?

ax(II)直接由0?g?x??1,得到

0??g?t?dt??1dt=?x?a?

aaxx(II)令F?u??'?f?x?g?x?dx??aua?a?ag?t?dtfxdx??

uF?u??f?u?g?u??fa??g?t?dt?g?u?a?u??g?u??f?u??fa??g?t?dt???a??u??

由(I)知0??g?t?dt??u?a? ?a?a??g??taauudt? u又由于f?x?单增,所以f?u??fa???g?t?dt??0

ua?F'?u??0,?F?u?单调不减,?F?u??F?a??0

取u?b,得F?b??0,即(II)成立. (20)(本题满分11分)

设函数f(x)?x,x??0,1?,定义函数列 1?x记Sn是由曲线y?fn(x),直线x?1f1(x)?f(x),f2(x)?f(f1(x)),?,fn(x)?f(fn?1(x)),?,及x轴所围成平面图形的面积,求极限limnSn.

n??【解析】f1(x)?xxxx,f2(x)?,f3(x)?,?,fn(x)?, 1?x1?2x1?3x1?nx11x??111xnndx ?Sn??fn(x)dx??dx??001?nx01?nx13

1111111??1dx??dx??2ln(1?nx)10 n0n01?nxnn11??2ln(1?n) nnln(1?n)ln(1?x)1?1?0?1 ?limnSn?1?lim?1?lim?1?limn??n??x??x??1?xnx(21)(本题满分11分) 已知函数f(x,y)满足

?f?2(y?1,)且f(y,y?)?y(y?21?)(?2y曲线)求yln,f(x,y)?0所围成的图形绕直线y??1旋转所成的旋转体的体积.

【解析】因为

?f?2(y?1),所以f(x,y)?y2?2y??(x),其中?(x)为待定函数. ?y又因为f(y,y)?(y?1)2??2?y?lny,则?(y)?1??2?y?lny,从而

f(x,y)?y2?2y?1??2?x?lnx?(y?1)2??2?x?lnx.

令f(x,y)?0,可得(y?1)??2?x?lnx,当y??1时,x?1或x?2,从而所求的体积为

2V??????2121?y?1?dx???1?2?x?lnxdx22?x2?lnxd?2x??2??22??x2?x????lnx(2x?)?????2??dx12?12???

??2ln2??(2x?(22)(本题满分11分)

x)422155????2ln2??????2ln2??.44???1?23?4??? 设矩阵A??01?11?,E为三阶单位矩阵.

?120?3???(I)求方程组Ax?0的一个基础解系; (II)求满足AB?E的所有矩阵B.

【解析】

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?1?23?4100??1?23?4100????01?11010?01?11010?AE??????? ?120?3001??04?31?101?????6?1??1?23?4100??10012????10???010?2?1?31?, ??01?110?001?3?1?41??001?3?1?41????? (I)Ax?0的基础解系为????1,2,3,1? (II)e1??1,0,0?,e2??0,1,0?,e3??0,0,1?

TTTTAx?e1的通解为x?k1???2,?1,?1,0???2?k1,?1?2k1,?1?3k1,k1? Ax?e2的通解为x?k2???6,?3,?4,0???6?k2,?3?2k2,?4?3k2,k2? Ax?e3的通解为x?k3????1,1,1,0????1?k3,1?2k3,1?3k3,k3?

6?k2?1?k3??2?k1???1?2k?3?2k1?2k123??B????1?3k1?4?3k21?3k3????k?kk123??(23)(本题满分11分)

TTTTTT(k1,k2,k3为任意常数)

?1?1 证明n阶矩阵?????11?1??0???1?1??0?与

?????????1?1??0?01??02?相似. ????0n??1??1??????2【解析】已知A????1??1?,B=???00?1?,

??????????1???n?则A的特征值为n,0(n?1重).

A属于??n的特征向量为(1,1,?,1)T;r(A)?1,故Ax?0基础解系有n?1个线性无关

的解向量,即A属于??0有n?1个线性无关的特征向量;故A相似于对角阵

?n???0?. ?=??????0??

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B的特征值为n,0(n?1重),同理B属于??0有n?1个线性无关的特征向量,故B相似于对角阵?.

由相似关系的传递性,A相似于B.

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