西北师范大学2022年研究生入学数学分析试题硕士研究生入学考试数学分析试题七西北师范大学硕士研究生2022年入学考试《数学分析》试题1.设f(x)?x3?x2?1?x??23?2??的反函数为f3????1(x),求极限limfx?3?1(x).解:由于当x?时,f?(x)?3x2?2x?3x?x?2??2?,因此在?0,??f(x)??上严格增加?33???且连续,所以f?131?31??2?上严格增加且连续.其中.(x)在?,??f??????2727???3?令x3?x2?1?3,即(x?2)(x2?x?2)?0,得x?2.从而f(2)?3,f?1(3)?2.又因f?1(x)在x?3处连续,故limfx?3?1(x)?f?1(3)?2.2.求?sinxcosx1?sinx323dx.解:??sinxcosx1?sinx1?sinx222dx??sinxcosx1?sinx121222d(sinx)??sinx(1?sinx)1?sinx222d(sinx)?121?sinxd(sinx)?2?2?(1?sinx)1?sinx222d(sinx)?12??2?1??d(sinx)2?2?1?sinx?12sinx?C.2??1?sin12xd(1?sinx)?222?1d(sinx)?ln(1?sinx)?22223.计算??Dxydxdy,其中D是由曲线y?x,x?y?1,x?y?4围成,且在x轴上方.2解:令x?rcos?,y?rsin?,那么J?r,D:xy223?2?4???3?23?4,0?r?1,23???Ddxdy???44d??rcos?rsin?22221rdr???444cos?sin?2d???rdr?1??441?sin?sin?22d??122r213????4433?1??1d???(?cot???)??222?sin??3???43???32????(4??).2?2?44.计算??(x?y)dx?(x?4y)dyx?4y22L,其中L是正方形x?y?2的边界,取逆时针方向.22解法一:设L0表示椭圆x?4y?1,D表示由封闭曲线L和L0所围成的区域.那么L0的参数174方程为x?cost,y?122sint,t?[0,2?].记P(x,y)?x?yx?4y22,Q(x,y)?x?4yx?4y22,那么?Q?x??x?8xy?4y(x?4y)2222??P?y,?(x,y)?D.且P,Q,?Q?P都在D上连续,因此根据Green公式得:,?x?y(x?y)dx?(x?4y)dyx?4y22??(x?y)dx?(x?4y)dyx?4y22L????L0???0dxdy?0,D从而??2?0(x?y)dx?(x?4y)dyx?4y22L??(x?y)dx?(x?4y)dyx?4y22L0????11??cost?sint(?sint)?cost?2sintcost??????dt?22?????2?012dt??.解法二:L?L1?L2?L3?L4,其中L1:y?2?x,0?x?2;L2:y?2?x,?2?x?0;L3:y??2?x,?2?x?0;L4:y?x?2,0?x?2.??(x?y)dx?(x?4y)dyL1x?4y22??0225x?105x?16x?165x?6x?4(x?2)22dx??2022?5x5x?16x?160?222dx,(x?y)dx?(x?4y)dyL2x?4y20222???20dx???5x?65x?16x?16dx??5x?65x?16x?16dx,0?22??(x?y)dx?(x?4y)dyL3x?4y22?5?x?25x?16x?165x?6dx??20210?5x5x?16x?16dx,(x?y)dx?(x?4y)dyL4x?4y22??2025x?16x?16dx,???(x?y)dx?(x?4y)dyx?4y822L??L1??????L2L3(x?y)dx?(x?4y)dyL4x?4y82022?2025x?16x?16dx?8?5201x?2165x?165dx??518?16?x????5?25?2dx175?2arctan5x?84201???2?arctan?arctan2???.2??注:解法二实际计算较繁.5.设?,???0,????2?x?,证明:sin(???)1?sin(???)?sin?1?sin??sin?1?sin?.证明:令f(x)?1?x,x?(0,??).那么当x?(0,??)时,f?(x)?1(1?x)2?0,因此f(x)在???(0,??)内严格增加.由于当?,???0,?时,有?2?0?sin(???)?sin?cos??cos?sin??sin??sin?,从而有f(sin?(??sin?(??)?))f(?s?in,i?s即?1?sin?(??)?1sin?1?sin??sin?sin(???)1?sin(???)si??ns?i?nsin?s?in?sinsin?1?sin?sin?1?sin???1?sin??sin?sin?sin???,故?1?sin??1?sin?.?x2y,?6.证明函数f(x,y)??x2?y2?0,?x?y?0,x?y?0,2222在点(0,0)连续且偏导数存在,但在此点不成微.证明:ⅰ>由于???0,?????0,使得当x?0??,y?0??时,有xyx?y222?0?x222x?y2y?y?0??,所以lim(x,y)?(0,0)xyx?y22?0?f(0,0),从而函数f在原点(0,0)连续;f(0??x,0)?f(0,0)?x0?0?xⅱ>由于fx?(0,0)?lim?xz?x?x?0?lim?x?0?lim?x?0?0,fy?(0,0)?lim?yz?y?y?0?limf(0,0??y)?f(0,0)?y?y?0?lim0?0?y?y?0?0,176即函数f在原点(0,0)的两个偏导数都存在;ⅲ>若函数f在原点(0,0)可微,那么?x?y?x??y222?z?dz?f(0??x,0??y)?f(0,0)?fx?(0,0)?x?fx?(0,0)?y?应是较??lim?x??y高阶的无穷小量.为此考察极限22?z?dz??0??lim?x?y32,??0(?x??y)222此极限不存在,事实上由于当?x??y时??1?1??lim??3??3.??x?0?222?2?lim?x?0?x?x?23?lim?x?01?3?13,lim?x?0?x?x?23(?x??x)2222222(?x??x)222或当?y?k?x,?x?0时,?z?dzk?x23223?x?0?0lim???x?0?0lim?k(1?k)23与k有关.(?x?k?x)因而函数f在原点(0,0)不成微.7.设无穷积分???af(x)dx收敛,证明函数F(x)??xaf(t)dt在区间[a,??)上一致连续.证明:由于F(x)在[a,??)上连续，且limF(x)存在,所以F(x)在[a,??)上一致连续.x???证明:???0,ⅰ>由于...