2015考研数学高等数学重难点题型精讲讲义

1、x2 2xx2 1ex 的渐近线x2 2xx2 1ex 的渐近线10f(xx3x x1f(x sin1ln(1x)xx2 4 xa4x2e2xdx ，求a . xa4x2e2xdx ，求a .x xaa cos dt 5，求a ，b .2】已知ex t02000，求 ， .3】已知(nn 515x sin4f(xdt g(x) (115x sin4f(xdt g(x) (1t)t dt x0f(xg(xt00（A）高阶无穷小 （B）低阶无穷小 （C）5ABC使ex(1BxCx21Axo(x32 31】求极限n 6nnnn n nnnn n n n n nn12xdx3】求极限n 1】确定方程2

2、x 1x2的实根的个数74 3【例2】证明方程4arctan4 3【例2】证明方程4arctanx x 30恰有两个实根方程ex ax2有几个实根f(a f(b0f(af(b1f(x在(,(ab)82】已知f(x在0,上连续，dxx02】已知f(x在0,上连续，dxx0 f (xcosxdx 0 10,1上可导，f(1) xf20证明：存在0,1f(f(04f(xg(x在abg(x0f (a) f (b) g(a) g(b)0f() f g（2）在(ab内至少存在一点.9fff(x在0,1上二阶可导，且xxfff(x在0,1上二阶可导，且xx 5f(x是0,1上可导f(00, f(11，证明存

3、在两个不同的点1,2(0,1112f1311证明：存在0, , ( 227f(x在ab(ab内可导(0abf(a f(7f(x在ab(ab内可导(0abf(a f(b f(xlim f(xk8xaf(xlim f(x09f(xh f(xhf(x(01)f (xf (x01证明：2f (x0)(xf (x f (x f(x )(xx )2 00002f (x) h2 f (xhxf(xh f(x f(x)h2f (xh f (xhf(x(0 f (x)h2 hf(xh f(x)h2f (x)h2 hf(xhf(x)h2 f (x)h2 o(f (x)h2 hf(xhf(x)h2 f (x)h2

4、o(h2hf(xh) flim 所以 f f (x)21即210f(x在abf (x0 x1x2为abf(x1) f(x2)fx1 x2 2211f(x在0,1上二阶可导，且maxxf(x 1 f 140f(0) f(1)1x0 x1x1x0 x1x2ln(1xx2 11证明f (x)dx f 300bbbf(x)g(x)dx f g 22 aabf(x) 在abf (bbbf(x)g(x)dx f g 22 aabf(x) 在abf (xf (x)dx ，abb sin1x【例 1】计算不定积分 1cos(x1) (xf232f(x，x3(xf xe (1ex4f(x xe (1ex4f(x

5、 lnt dt f(x f1x1 1x21x2【例6】已知 tf (2xt)dt arctanx , f(1)1，f (x)dx22012yy2 7】计算二重积分ydxdyDx2,y0,y2yy2 7】计算二重积分ydxdyDx2,y0,y2,xD2x42【例8】交换积分次序 dydy，并计算22129】计算二重积分cos(xydxdy，其D(x,y0 x,0 y Df(xy)dxdy a，其D(xy0 x1,0 y1DD1f(xf (x f(x2f(x0f (x f(x（1）fx（2）求y f (x 2012x2y2t12x2y2tfx y )dxdyf (t22f(t在0,f(te4t 3

6、】y(xy3y2y2ex，图形在(0,1)yx2x1在(0,1)y y(x4xdyx2y)dx0y y(xx1,x2y0y y(x22【例5】z f(e sin y)，f(u)具有二阶连续导数22【例5】z f(e sin y)，f(u)具有二阶连续导数e z，求f(u)x2 xx6f(x在0,f(0)1f(x f(xf(t)dt 0（1）f（2）ex f(xx1】z f(xyyg(xf(uvg(xx1 2. yyy0)Z yZ 1Zyy0)Z yZ 1Z2】以ux2 yv xy2zyz3z z(xy由方程F(x,y)0所确定，证 zxyx4】求原点到由方程5x26xy5y2 85x24y2

7、4上求一点，到2x5x24y2 4上求一点，到2x3y603S(x( x2243S(x( x224246n（1）（2）（4）n(nnnn(n2n1(1)n(n2 n6】求(1)n(n2 n6】求17f(xx2 5x8f(xln(1x2x21 9f(x1 10f(x12x2 10f(x12x2 1】计算曲线积分 e sinyb(xy)dxe cosyax)dyLxxL沿着y 2axx2 到原点的曲线xdyLx2 y2 14x L 起点为(a,b)，终点为(c,d)，曲线积分I 11 y2 f(xy)dxy2 f(xy)1dyxL （2）若abcdI 【例4】设S为椭球z 1的上半部分，点P(x,

8、y,z)S ，为S 在点P2(x, y,SdS【例 5】计算曲面积分2x3dydz2y3dzdx3(z【例 5】计算曲面积分2x3dydz2y3dzdx3(z2 1)dxdy ，其中z 1x2 y2xdydz ydzdx 222，其中2x 2y 3(x2 y2 z22数学高等数学重难点题型讲义例1】2】1213】4】5】76】7】8】1)f9f(0)0, f数学高等数学重难点题型讲义例1】2】1213】4】5】76】7】8】1)f9f(0)0, f(0)0, f (0)4,x10】3 1】a0或2】a1b4 12000，求，3】已知.(nn 1216,B,C 232322】 (2 233】1】

9、3【例3】a0时，无根；0a时，1个根；a 时，2个根；a时，3个根4443】1】3【例3】a0时，无根；0a时，1个根；a 时，2个根；a时，3个根4449f(xh f(xhf(x(01)f (xf (x01证明：2(x0)(xf公式f(x) f (x ) f(x )(xx )2 f(xhx00002(xh2 ff (xh f(x f(x)h2(xh2 ff(xh) f (x)hf(x 2(xh2 fhf(xh)f(x)h2 f (x)h2 o(h2limhf(xf(x)hlim f (xlim f (x) 即lim 1221】extan x22】arctan32 3】ln4f(x1】extan x22】arctan32 3】ln4f(x f1) 1(ln33427】44(29】10】1（1）f(xex（2）22f(t4t2【例 3】4yx5f(uC1euC2eu16（1）f(xex(1 2 y25】(8