?2u??u?u?2u???2u???2u???2u???(?)?????x?y?y??????2?y?????y??2?y?????y?2u?2u?2u?a2?b2?(a?b)?????????u??u?u?u?u?u?u?(a?b)?a(a?b)?b(b?a)?y2?y??????2??????2????2?2u?2u2?u?a?b?2ab??2??2????222222
?2u?2u?2u?2u?2u22故42?12?52?(5a?12a?4)2?(5b?12b?4)2?x?x?y?y?x???(12(a?b)?10ab?8)?0?5a2?12a?4?0(1)?当?5b2?12b?4?0(2)满足等式?12(a?b)?10ab?8?0(3)?22解得a??或a??2,b??或b??2552?a????5?a??2又因为?或?不满足(3)式?b??2?b??2?5?2?a??2??a???故?5或?2b???5?b??2??
解:I???r2sin?1?r2cos2?drd?D???rsin?1?r2(cos2??sin2?)rdrd?D???y1?(rcos?)2?(rcos?)2
dxdyD???y1?x2?y2dxdyD??1dx?x00y1?x2?y2dy??110dx?x021?x2?y2d?1?x2?y2?3??11? 03??1?(1?x2)2???dx??11132203dx??0(1?x)dx?1?3??20cos4?d??133?16?
(21)(本题满分10分)设函数f(x)在闭区间?0,连续,在开区间1??0,1?内可导,且f(0)?1,111f(1)?,证明存在??(0,),??(,1),使得f?(?)?f?(?)??2??2322
1证明:令F(x)?f(x)?x331?1?对于F(x)在?0,?上利用L?中值定理,得???(0,),2?2?11F()?F(0)?F?(?)(1) 221?1?对于F(x)在?,1?上利用L?中值定理,得???(,1),2?2?11F(1)?F()?F?(?)(2)22两式相加得f?(?)?f?(?)??2??2(22)(本题满分11分)11????a?????设A??0??10?,b??1??1?1?1??????已知线性方程组Ax?b存在两个不同的解 (I)求?,a(II)求方程组Ax?b的通解解:
(I)因为线性方程组AX?b存在两个不同解,所以r(A)?3,即|A|?0,解得
???1或??1。
1a??11?11??11?11???11???????1???020?1?, 当???1时,A??0?201???020?1????1?11????020a?1??000a?2?因为r(A)?r(A)?3,所以a??2;
1??1111??111a??111??????1???0001?, 当??1时,A??0001???000?1111??000a?1??0000???????显然r(A)?r(A),所以??1,故???1,a??2。
3???10?1?2??11?11?????1(II)由A???020?1??010??,得方程组AX?b的通解为
2??0000??0????000?????3???21??????1?。 X?k?1????(其中k为任意常数)
??2?0??0???????(23)(本题满分11分)?0?14???设A???13a?,正交矩阵Q使得QTAQ为对角阵,若Q的第一列为
?4a0???1(1,2,1)T,求a,Q6解:?0?14???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的特征值为?1?(1,2,1)T, 66?1??1?????66?????0?14??1??1??2??2???????故A???1???,即??13a??2???1?2?,由此可得?1??6??6??4a0??1????????1??1??????6??6?
2019年整理[版]10年考研政治英语数学真题及精析(下)
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