2014年全国硕士研究生入学统一考试数学一
真题及答案解析
1、C
y?x?sin1xx?sin1x?1 x1lim?y?x??limsin?0x??x??x1?y?x?sin存在斜渐近线y?xxx??k?limy?limxx??2、D
解：令F(x)?f(x)?g(x)?f(x)?f(0)(1?x)?f(1)x
有F(0)?F(1)?0，F?(x)?f?(x)?f(0)?f(1)，F??(x)?f??(x) 当f??(x)?0时，F(x)在[0,1]上是凹的，所以F(x)?0，从而f(x)?g(x) 3、D
区域如图： 选择极坐标：
y -1 -1 0 x
???20d??1cos??sin?0f?rcos?,rsin??rdr???d??f?rcos?,rsin??rdr20?1若为直角坐标0?1
11?x00dx?1?x20f?x,y?dy??dx?f?x,y?dy
4、A 解析：
I??(x?acosx?bsinx)2dx?????(x2?a2cos2x?b2sin2x?2axcosx?2bxsinx?2absinxcosx)dx????2?(x2?a2cos2x?b2sin2x?2bxsinx)dx0?
322当a?0,b?2时，I最小5、B 解析：
?2(?3??a2??b2?2?b)故 a, cosx+b, sinx=2sinx0aba000cdc000b0dab0ab00?c?(?1)4?100bdcd0?a?(?1)2?1cd00??a?d?(?1)3?3??adabab?c?b?(?1)2?3cdcdabab?bccdcdabcd
?(bc?ad)??(ad?bc)2
6、A 解析：
已知?1，?2，?3无关设?（1?1+k?3)??2(?2?l?3)?0即?1?1+?2?2+（k?1?l?2)?3?0??1??2?k?1?l?2?0从而?1+k?3，?2+l?3无关反之，若?1+k?3，?2+l?3无关，不一定有?1，?2，?3无关?1??0??0???????例如，?1=?0?，?2=?1?，?3=?0??0??0??0???????7、B
P(AB)?P(A)P(B)?[P(A?B)?P(AB)]P(B)?(0.3?P(AB))0.5?P(AB)?0.3P(B?A)?P(B)?P(AB)?0.5?0.3?0.2 8、（D） 解：Y2?
111(X1?X2)，EY2?E[(X1?X2)]?(EX1?EX2)， 22211DY2?D[(X1?X2)]?(DX1?DX2)。
24??y11fY1(y)?[f1(y)?f2(y)]，EY1??[f1(y)?f2(y)]dy?(EX1?EX2)?EY2，
??222EY??21????y212[f1(y)?f2(y)]dy?(EX12?EX2)， 22DY1?EY12?(EY1)2??112(EX12?EX2)?(EX1?EX2)22412222?2EX?2EX?(EX)?(EX)?2EX1?EX2?1212??4122??DX?DX?EX?EX1212?2EX1?EX2???4122??DX?DX?(EX)?(EX)?2EX1?EX2?1212??412???DX?DX?(EX?EX)?DY21212??4
9、
F（x,y,z）=x2(1?siny)?y2(1?sinx)?zF'x?2x(1?siny)?y2cosxF'y?x2(?cosy)?2y(1?sinx)Fz??1'Fx'(1,0,1)?2Fy??1'
?2(x?1)?(?1)(y?0)?(?1)(z?1)?0?2x?y?z?1?010.
f'?x??2(x?1)x?[0,2]?f(x)?x2?2x?c又f(x)是奇函数?f(0)?0?c?0?f(x)?x?2xx?[0,2]f(x)的周期为4?f(7)?f(3)?f(?1)??f(1)??(1?2)?111、
2
xy'?y(lnx?lny)?0yxln?0xy yu??y'?u,x?ux?u,x?u?ulnuy'?u,1?ulnu?uxu,1dx??ulnu?u?xdxlnlnu?1?lnx?C1lnu?1?cx即y?xecx?1又y(1)=e3?y?xe2x?112、
答案：?
由斯托克斯公示：
?e3?ec?1?c?2
cos?cos?cos??ds?z?Lzdx?ydz????0??xz12??y0??y01y??????x2?ds??zzy???12ds?12?2??
?y+z?0(上侧)其中?为?22?x?y?113、
[?2,2]
f(x1,x2,x3)?x12?x22?2ax1x3?4x2x3?x12?2ax1x3?a2x32?x22?4x2x3?a2x32?(x1?ax3)?(x2?2x3)2?(4?a2)x32?y12?y22?(4?a2)y32若负惯性指数为1，则4?a2?0，a?[?2,2]14、
22 5nE(c?Xi?)?c?E(Xi)?ncE(X)?nc?222nn2?i?1i?1?2x3dx23?? 15. 解：
2nc142?5nc222?x??=??C?25n3?24?
?limx??x1(t(e?1)?t)dt1x2ln(1?)xt21t??limx??x1(t(e?1)?t)dtx2?1x21t1x(e?1)?x12?lim?limx(ex?1?)x??x??1x21x
11?t?t2??(t2)?1?t1e?1?t12令?tlim?lim?x??x??xt2t2216.
解：
y3?xy2?x2y?6?0(1)(1)式两端对x求导得：(3y2?2xy?x2)y'?y2?2xy?0(2)当y'?0时,解得y?（舍）或0y??2x把y??2x代入(1)式，得x?1,y??2（2）式两端对x求导得(6yy'?2y?2xy'?2x)y'?(3y2?2xy?x2)y''?2yy'?2y?2xy'?0代入x?1,y??2,y'x?1?0,得y''x?1?
4?09所以当x?1时,f(x)有极小值f(1)??217、 解
：
?z?f'?ex?cosy,?x?2z?cosy?(f''?ex?cosy?ex?f'?ex)?f''(?ex?cosy)2?f'?ex?cosy2?x?z?f'?ex?(?siny),?y?2z??ex[f''?ex?(?siny)?f'?cosy]?(ex)2siny2f''?f'?cosy?ex2?y?2z?2z2xx2x??f''?e?(4z?e?cosy)e?x2?y2?f''(ex?cosy)?4f(ex?cosy)?ex?cosy令t?ex?cosy,?f''(t)?4f(t)?t?y''?4y?x求特征值：
?2?4?0y??(ax?b)?x??2?y(x)?C1e2x?C2e?2x1x41?x4再求非其次特征值。代入?y??-?y?y(x)?y??C1e2x?C2e?2xy(0)=0=C1?C2y'(0)=0=x1?x2?14?1?C1?C2?0C????116????1?2C1?2C2??C??14??16?212?1?2?1?f(?)?e?e??16164
?x2?y2?118、设曲面，?1:?，方向向上．
z?1?
??1??1??(x?1)3dydz?(y?1)3dzdx?(z?1)dxdy
????(3(x?1)2?3(y?1)2?1)dxdydz????(3x2?6x?3?3y2?6y?3?1)dxdydz
??????(3x2?3y2?7)dxdydz??dz??(3x2?3y2?7)dxdy
?110D(z)??dz?d??(3r2?7)rdr?4?
0002?z其中
???(6x?6y)dxdydz?0，因为积分区域关于xoz,yoz对称，积分函数
?f(x,y)?6x?6y分别是y,x的奇函数．
在曲面?1上，
33(x?1)dydz?(y?1)dzdx?(z?1)dxdy?0 ???1故
33(x?1)dydz?(y?1)dzdx?(z?1)dxdy??4?． ??? 19、
（1）证明：
cosan?an?cosbn?an?cosan?cosbn又0?an?
? 2?cosan?cosbn?0?an?bn
an?0,bn?0,又?bn收敛
n?1???an收敛
n?1?故liman?0
n??（2）
ancosan?cosbn?bnbnan?bna?bsinnn2222?an?bn?an?bnan?bn4bn? 2222bn?an?2bn?2sinbn2bn??2bn20?an????2,0?bn??2
且
?b收敛
nn?1?? 20.
an收敛 bn?1n??1?23?4??1?23?4??1?23?4????r1?r3???4r2?r3??(A)=?01?11??????01?11??????01?11?
?120?3??04?31??001?3???????
?1?205??1001???2r2?r1??r3?r2010?2010?2?????????3r3?r1???? ?001?3??001?3?????x1??x4?x1???1?????x2?2x4?x2?2???c c为任意常数
?3?x3?3x4?x3?????x4?x4?x4??1??x1?设 B=?x2?x?3y1y2y3z1??z2? z3??1??0? 0??0??1? 0??0??0? 1??100??010?001??412?3???1?31? ?1?41???x1??1??1?23?4?????A?x2???0???01?11?x??0??120?3?3?????y1??0??1?23?4?????A?y2???1???01?11?y??0??120?3?3?????z1??0??1?23?4?????A?z2???0???01?11?z??1??120?3?3????即
100??1?23?4?1?23?4???01?11010????01?11?120?3?001????04?31100??1?205?1?23?4?????01?11010???010?2?001?3?001????0013?1001???010-2?001-3?-1??-1-31? -1-41??26?x1???1??2???????x2???c?2????1??x3?1?3???1???????x?1??0??4??y1???1??6???????y2???c?2????3??y3?2?3???4???????y?1??0??4??z1???1???1???????z2???c?2???1? ?z3?3?3??1???????z?1??0??4?
??c1?2?c2?6?c3?1???2c?12c?32c?1123? ?B???3c1?13c2?43c3?1???ccc23?1?c1,c2,c3为任意常数
21、 解：
?1?1设A?????1111?01???1?0B??????1??0?1000?1?101??02? ??0n???1?E?A??1?1??1?1?(??n)?n?1
??1??n=0
所以A的n个特征值为?1=n,?2=又因为A是一个实对称矩阵，所以A可以相似对角化，且
A?n?????0????，?E?B?0??0?0’’000?1?2?(??n)?n?1
?00??N所以B的n个特征值为?1=n,?2=??n’=0
0又0E?B?0000-10-20-n
00所以r(0E?B)?1
故B的n-1重特征值0有n-1个线性无关的特征向量
所以B也可以相似对角化，且B?n?????0??? ??0?所以A与B相似。
22、（同数三22题） 解：（1）
Fy(y)?P(Y?y)?P(Y?y,X?1)?P(Y?y,X?2)?P(Y?yX?1)P(X?1)?P(Y?yX?2)P(X?2) ?11P(Y?yX?1)?P(Y?yX?2)22① 当y?0时，FY(y)?0 ② 当0?y?1时，FY(y)?1113y??y?y 22241111y??y?? 22224③ 当1?y?2时，FY(y)?④ 当y?2时，FY(y)?11??1 22?0??3y?4综上：FY(y)???1?y?24??1（2）
y?00?y?1
1?y?2y?2?3?4??1fY(y)?F'Y(y)???4?0??0?y?11?y?2 其他EY=??+?-?yfY(y)dy?34?10ydy?12ydy?1431133????42424
23、
解：（1）
x2???F（x;?）=?1?e???0x?0
其他?2X?x2?e?'（x;?）=??所以f（x;?）=F?0?x?0其他x2?
EX=?2+?-?xf(x;?)dx???0??0x2x?2xe?dx???2??0xde???x2???xe??x2????0???0e?x2?dx???EX=?+?-?xf(x;?)dx?2??x2?e?x2?dx令x?t?t02t?et?12tdt????0t1?edt???t?(2)设x1,x2,,xn为样本的观测值
n?nn1??xi22?xn??i?1ie?i?1L(?)??f(xij?)??ni?1????0xi?0(i?1,2,其他,n)
xi?（0i=1，2，，n）时，当L(?)=2?xinni?1n?e?1?i?1?xi2nn
lnL(?)?nln2?ln?xi?nln??i?11?xi?i?1n2dlnL(?)令：
d?n1???2??n2?1xi?i?1n2?0
??1??n?x为?的最大似然估计值nii,xn2独立同分布（3）
x1,x2,??1limP?n???n?,xn独立同分布，?x12,x22,,n)1nn又EXi2??(i?1,2,由辛傾大数定理?????xi?E2i?1n22x?i????1i?1E1n?xi2?i?1n?Exini?11n
???存在实数a=?,??a??limP?nn??st.对于???0,???1
2014年全国硕士研究生入学统一考试数学一真题及答案解析
