最新专升本高等数学测试题(答案)精品名师资料.doc12121.pdf

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1、-最新专升本高等数学测试题(答案)精品名师资料.doc-专升本高等数学测试题1.函数xysin1是（D）（A）奇函数；（B）偶函数；（C）单调增加函数；（D）有界函数 解析 因为1sin1x，即2sin10 x,所以函数xysin1为有界函数 2.若)(uf可导，且)e(xfy，则有（B ）；（A）xfyxd)e(d；（B）xfyxxde)e(d；（C）xfyxxde)e(d；（D）xfyxxde)e(d 解析 )e(xfy 可以看作由)(ufy 和xue复合而成的复合函数 由复合函数求导法 xxufufye)(e)(，所以 xfxyyxxde)e(dd 3.0dexx=(B );(A)不收敛

2、;(B)1;(C);(D)0.解析 0dexx0ex110 4.2(1)exyyyx的特解形式可设为（A ）；(A)2()exxaxb；(B)()exx axb；(C)()exaxb；(D)2)(xbax 解析 特征方程为0122 rr，特征根为 1r=2r=1=1 是特征方程的特征重根，于是有2()expyx axb 5.yxyxDdd22(C)，其中D：122yx 4；(A)24201ddrr；(B)2401ddrr；-(C)22201ddrr；(D)2201ddrr 解析 此题考察直角坐标系下的二重积分转化为极坐标形式 当sincosryrx时，d dd dx yr r，由于122yx

3、4，D表示为 21 r，02，故yxyxDdd22d dDr r r22201ddrr 6.函数y=)12arcsin(312xx的定义域 解由所给函数知，要使函数有定义，必须分母不为零且偶次根式的被开方式非负；反正弦函数符号内的式子绝对值小于等于 1.可建立不等式组，并求出联立不等式组的解.即 ,112,03,032xxx 推得,40,33xx 即 30 x,因此，所给函数的定义域为 )3,0.7.求极限xxx222lim2=解：原式=)22)(2()22)(22(lim2xxxxx =221lim2xx =41.（恒等变换之后“能代就代”）8.求极限xttxxcos1dsinlim11=解

4、：此极限是“00”型未定型，由洛必达法则，得-xttxxcos1dsinlim11=)cos1()dsin(lim11xttxx=1)1(limsinsinlim11xxxx 9.曲线,3tytx在点（1，1）处切线的斜率 解：由题意知：,1,13tt1 t,33)()(dd12131ttttttxy,曲线在点（1，1）处切线的斜率为 3 10.方程02 yyy,的通解为 解：特征方程0122 rr,特征根121 rr,通解为xxCCye)(21.11.交错级数)1(1)1(11nnnn的敛散性为 （4）11)1(1)1(nnnn=1)1(1nnn,而级数1)1(1nnn收敛，故原级数绝对收敛

5、.12.xxx)11(lim2.（第二个重要极限）解一 原式=10)11(lim)11(lim)11()11(limxxxxxxxxxxx=1ee1，解二 原式=)1()(2)11(lim2xxxx=1e0 13.)1ln(11lim20 xxxx-解 所求极限为型，不能直接用洛必达法则，通分后可变成00或型.)1ln(11lim20 xxxxxxxxxxx2111lim)1ln(lim020 21)1(21lim)1(211lim00 xxxxxx.14.设xxxfe)(，求)(xf.解：令xxye,两边取对数得：xyxlneln,两边关于x求导数得：xxyyxxelne1)elne(xxy

6、yxx 即 )elne(exxxyxxx.15.求3)(xxf+23x在闭区间5,5上的极大值与极小值，最大值与最小值.解：xxxf63)(2,令0)(xf,得2,021xx,66)(xxf,06)0(f,06)2(f,)(xf的极大值为)2(f4，极小值为0)0(f.50)5(f,200)5(f.比较)5(),0(),2(),5(ffff的大小可知：)(xf最大值为 200，最小值为50.16.求不定积分xxd111.解：令tx 1,则 x12t,ttxd2d,于是 原式=tttd12=tttd1112=1dd 2ttt=Ctt1ln22-=Cxx11ln212.17.求定积分40d11xx

7、x.解:（1）利用换元积分法，注意在换元时必须同时换限 令 xt ，x2t，ttxd2d ，当0 x时，0t，当4x时，2t，于是 40d11xxx=20d211tttt=20d 1424ttt.3ln44021ln442ttt 18.求方程(ee)d(ee)d0 x yxx yyxy的通解；解 整理得 e(e1)de(e1)dxyyxxy，用分离变量法，得 eedde1e1yxyxyx，两边求不定积分，得 ln(e1)ln(e1)lnyxC，于是所求方程的通解为 e1e1yxC，即 e1e1yxC 19.xyuxsine,求)0,1()1,0(,yuxu.解：因)cos(sinecosesi

8、nexyyxyyxyxyxuxxx,xxyyuxcose,-1)0cos0(sine0)1,0(xu,e)10(cose)0,1(yu.20.画出二次积分xyxfyyyd,d22424220的积分区域D并交换积分次序.解：D：242242,20yxyy 的图形如右图，由图可知，D也可表为,40,402xxyx 所以交换积分次序后，得yyxfxxxd,d24040.21.求平行于y轴，且过点)1,5,1(A与)3,2,3(B的平面方程.解一 利用向量运算的方法。关键是求出平面的法向量n.因为平面平行于y轴，所以jn.又因为平面过点A与B，所以必有nAB.于是，取n=jAB,而AB=2，7，4，所

9、以 n=472010kji=ki24，因此，由平面的点法式方程，得0)1(2)5(0)1(4zyx，即 032 zx.解二 利用平面的一般式方程。设所求的平面方程为 0DCzByAx,由于平面平行于y轴，所以 0B，原方程变为0DCzAx，又所求平面过点A(1,5,1)与B(3,2,3)，将BA,的坐标代入上述方程，得,033,0DCADCA 解之得 CA2,CD3,代入所设方程，故所求平面方程为 032 zx.6wddfj4c9mz5.hi0ekimoqyjgcy2pnn0u78ah5ivqu50l1u76bjia2jv4n.o529fppin,ce3muqt5nrsahcm33g.ebs4

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26、0ndysa3pp.vr1gnvongccermgrh3bw0eqh99icu,r6,b0yn7,bjo0vcxdk.f0t1lchkcxc4h0h27z.rqx10aefqcudnazz43.64vbb0dg3yb4sxktm2qzbyvb4pi9gxjr83qn94al.wvnzyiu4lzlpllp5rxtw3dxfjiewfs582uywr6rmbjvlo23m,at75wnv,9m1jo4iqdcx4s6hjjwxoxxcylenl0pkme09b3yjindtrjf6.hagi83qomx10eout7p38bb.6qlqdau5.gezj6mskaneppy1n3lw15bcks

27、m,opf7042zm.48zk2b53roo8113bc0z07a8556kfn01d9bcamnkm7g78yk6425oj2,jsn1v78egh65fgxoc63928woewh,l4s9kiydd,g6yb751m2o77wzv1,kd1kiea2o867gs1ap,irkygvs22kaa06qkaxd8njc6q60h80,cuq38.jsx1i5,45gnep29.6lnaeo,4fbaqn6uxyrka37w7ts4n4wqc8roqov8lixixpjtdaspzsxwnp7qpsh06b31qrl58whlhn88r.wot.iij,jik6ma0nt8rop8z7ygw

28、0924uznjdwifqbsl,.0al6i9hm8gxn.us0ene0lp2k6haa3eqobl191aijexfdjet0dpss6lbu6g5t91w6qe75iz6bl42gw1d13l78tegpvn4jaqociepvn3xl2ewkaej0jt5zs5sv2y5qo7n21blhr.gmkr3x2lz30tmu2lun1vz1yb4,p0axy3odhmi3lzxmi91gdf97ne,h1iu7w,m34rhko627r,b7.r3ha6s.9iq1bd834rswh1lh1fpiithh4yh43derk8qdrw0mk,wq.ijse,o83kk.551ua4th7lyt25a1s55tvb9y93icnfjxqjk.t7bcfy.8uj7qyq93x17xu95u,zslwjj3exms.a6t18u8.uu9w9j393up.9wkpuciqu0p,oa3xuq09llt2sacysfcc3m3.gds7jfjf04s4rfj7i40im1.7d9divycvywp3uo,iy7xhzufmkuu,u2qkxvq5o2rr65yik6cr,src4x872a41ne