2001-2017年江苏专转本高等数学真题(及答案)

1、 f5dxf=2f2fl-丄11+Jx-101+t0k1+1丿=2(2-ln3)16、flnxf1lnxf1lnxf111+x1+x1+x1+x1+xxf1-丄kx1+x丿dx=2t-In(1+1)(1+x)2lnx-41+Xff1dx=-应+InX-In(1+X)+C=-应+1亠+C14x14x14x17、原方程可化为y+-y=沁,于是有P(x)=q(x)=沁,则方程的通解为xy=eP(x)dx卩q(x)efp(x加=丄(fsinxdx+Cx218、x2、dx+C丿 HYPERLINK l bookmark94 o Current Document xx2-f2dx仃sinxf2=e-xdx

2、IJAxdx+C HYPERLINK l bookmark259 o Current Document kx2丿)=(-cosx+C)，由y(兀)=0得C=-1,因而所求解为y=-Cosx+lx2x2ijk依题意所求平面经过点(1,1,1),法向量n=131123所求平面方程为7(x-1)-2(y-1)-(z-1)=0,即7x-2yZ-4=0.二7i-2j-k,因而19、设“=x2-y,v=y2-x则=f(u,v),于是有1=f舟+f舟=2xf:-fd2兀dfQudfQvdfQudffdv=2x1+12+2QxQyQuQyQvQyQuQyQvQy=2x(-厂+2yf)-(-厂+2yf)=-2x

3、f+(4xy+1)f-2yf111221221120、1222dy=*f2(4-y2)-(y-2)2dy=y-20ffxdxdy=f2dyf4-y2xdx=f2x2D0y-220f2(2y-y2)dy=.03或ffxdxdy=f4d0f4cos0r(rcos0-2)dr=f4cos0r3-r2D00Qi丿 HYPERLINK l bookmark265 o Current Document 兀丫64A=J41一cos40-16cos20d0=I3丿04-y24cos0d90f4(4cos40-3cos20)d030ry+cos20)2-(1+cos20)|d0=21624301624301+2

4、cos20+1+cos402丿-2(1+cos20)d0=f4(cos20-cos40)d0=30匸匸44=308f叮1.m1山)-J41sm20+sin403124丿0四、证明题21、由limf(x)=lim|x|=0=f(0)，因而函数f(x)=x|在x=0连续。x0 x0因W(0)=limfx-f0=lim凶=lim二=-1,xT0-xxT0-xxT0-x广(0)=lim=lim=lim=1,所以函数/(x)=|x|在x=0处不可导。+xT0+xxT0+xxT0+x22、设f(x)=2x3+1-3x2,f(x)=6x2-6x=6x(x-1),令f(x)=0得x=0,x=1,12因为(0)=1,f(1)=0,f(丄丿=0,而limf(x)=lim(2x3+1-3x2)=+,2丿xT+gxT+g当-时,所以当x-时，函数f(x)有最小值/丿=f(1)=0，因而,有/(x)0,即有2x3+1当-时,五、综合题24、(1)设a=J2f(x)dx,则f(x)=1+2axx丿丄+2a,于是a=J2f(x)dx=J2丄+2adxx2111x2丿2=丄+2a,解得a=-丄.因而f(x)=丄-1.122x2(2)J+af(x)dx=J+W1dx,由于J+a丄dx=limJdx=limJdx1x2bT+a1x2bT+a1x21Ix2丿=lim1：=1收敛，而J+a3dx=limJcdx=