2020年全国硕士研究生招生考试(数学一)参考答案及解析
1.D
解析:A选项可知(B选项(C选项(?x0(et?1)dt)'?ex?1~x2;
333222??x0ln(1?t)dt)'?ln(1?x)~x; sint2dt)'?sinx2cosx~x2;
sin3tdt)'?sinxsin3(1?cosx)~14x. 2sinx0D选项(2.C
?1-cosx0f(x)=0,且解析:当f(x)在x?0处可导时,有f(x)在x=0处连续,f(0)=limx?0f(x)?f?0?f(x)存在设为a,则有,lim=limx?0x?0xxlimx0f(x)x=limx0f(x)x?xxlimx0f(x)x?limxx0xa?00.
3.A
函数f(x,y)在点(0,0)处可微,,则有
f(x,y)-f(0,0)- (x,y)?(0,0)lim抖fx-抖x(0,0)fyy(0,0)fyy(0,0)
x2+y2抖fx-抖x(0,0)x+y22f(x,y)-=lim()()x,y?0,0=0即有(x,y)?(0,0)lim|ng(x,y,f(x,y))|x?y22=0
4.A
5.B
解析:矩阵A经初等列变换化成B,根据左行右列,应该选B. 6.C
解析:由于两直线相交,故两直线的方向向量无关,即?1,?2无关,由因为两直线上有两点
a1c1a2c2a2?a3b2?b3?0,故选C. c2?c3组成的向量与两直线的方向向量共面,故b1b27.D
P(ABC)?P(ABC)?P(ABC)?p(AB)?p(ABC)?p(AB)?p(ABC)?p(BC)?p(ABC)?p(A)?p(AB)?[p(AC)?p(ABC)]?p(B)?p(AB)?[p(BC)?p(ABC)]?p(C)?p(BC)?[p(AC)?p(ABC)]11111?1?1?0???0???0??0???41241212?12?45?12?8.B
E?Xi??EXi?100?i?1i?11001001001001?502
11D?Xi??DXi?100???2522i?1i?1
?100??100?X?50x?50?????55?50??i?1i??i?1i?P?剟1????1???P?55?5???????????
9.-1
limln(1?x)?(e?1)x2xx??x?0121x?(x?x2)22??1
x210.?2
t2?1d2ydy1d2y?2??2解析:?,2??3dxtdxtdxt?111.am?n 解析:12.4e
???0f(x)dx??[?af?(x)?f??(x)]dx?am?n.
0??f(x,y)=òxy0extdt;32fyⅱ(x,y)=exxyx,fy(x,1)=exx;
22ⅱfyx(x,1)=ex3x2x+ex;解析:
33ⅱfyx(1,1)=4e.a13.
0a1?1?11a01?10a=a00aa00a0aa0?111?1a=a?01?1aaa00a0aa?(?1)4?1a1?1
1?10?1001?1+a2a00a?(?1)4?1a1?1 =a?0a?10a0aa =?4a2?a4. 14.
2 ?Cov(X,sinX) ?EXsinX?EXEsinX
?1?f(x)????0?EXsinX??11?x???,??22?其他1?2?
???xsinxdx???22?20???22?22xsinxdx???xcosx0??cosxdx?=0????
EX?EsinX?0
Cov(X,sinX) ?EXsinX?EXEsinX?2?
?fx'?3x2?y?0?15.解:对函数关于x,y分别求导,令并两偏导数同时为零,得?',解得2??fx?24y?x?01?x??x?0??6''''''.又fxx?6x,fxy??1,fyy?48y,或在?0,0?处,AC?B2??1?0,从而??1?y?0?y???12函数在此处不取极值;在?,?11?2?处,AC?B?3?0,A?1?0,从而函数在此处取极小
?612?1?11?f?,???. 612216??值,且f?
1?11?,???.综上函数的极值为612216??16.解:由条件知
P?4x?yx?y,Q?4x2?y24x2?y2,可得
?Q?4x2?8xy?y2?P??222?x?y?4x?y? .令
l:4x2?y2??2,其中?为充分小的正数,取顺时针方向.则
2020年全国硕士研究生招生考试(数学一)--答案解析
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