2022年 湖南农业大学 高等数学期末复习题

l2cosl2cos2x1、下列极限中，正确的是((A))x喻0x喻0x 1x喻0n喻父(A)f(1)>f(0)>f(1)-f(0)(B)f(1)>f(1)-f(0)>f(0)(C)f(1)-f(0)>f(1)>f(0)(D)f(1)>f(0)-f(1)>f(0)(A)f[g(x)](B)g[g(x)](C)x喻父xx喻父xx喻0x喻父xaxx分析：由f(x)在点x=0处连续，得x喻0+x喻0+limf(x)=x喻0-limx喻0-limx喻0-(A)f(x)=xsinxEQ\*jc3\*hps24\o\al(\s\up5(x),,)(x2-1x<0(C)f(x)=x((x2-1x<012x+1(A)可去间断点(B)跳跃间断点(C)无穷间断点(D)振荡间断点8．设f(x)=ax3-6ax2+b在区间[-1,2]上的最大值为3,最小值为-29,又a>0,则（C） 1-A.只有铅直渐近线D.没有渐近线EQ\*jc3\*hps24\o\al(\s\up2147483645(m),喻0)13.设函数y=sec2x，求y,=2sec2xtan2x。14.lim(1-x)1/x=1/ex喻0330.x喻m(2x50250x喻4=-=-31-5-x。EQ\*jc3\*hps23\o\al(\s\up2(l),h)EQ\*jc3\*hps23\o\al(\s\up2(im),喻0)EQ\*jc3\*hps23\o\al(\s\up10(一),h)EQ\*jc3\*hps24\o\al(\s\up8(2),2)EQ\*jc3\*hps24\o\al(\s\up8(x),x)EQ\*jc3\*hps24\o\al(\s\up8(3),1)EQ\*jc3\*hps24\o\al(\s\up0(im),喻2)EQ\*jc3\*hps24\o\al(\s\up6(一),x)x3EQ\*jc3\*hps24\o\al(\s\up7(ax),x)EQ\*jc3\*hps24\o\al(\s\up7(一),1)x+4x喻1x喻1有喻EQ\*jc3\*hps24\o\al(\s\up4(x),x)EQ\*jc3\*hps24\o\al(\s\up4(一),1)喻EQ\*jc3\*hps24\o\al(\s\up4(x),x)EQ\*jc3\*hps24\o\al(\s\up4(1),1))dyd2y)dyd2y1EQ\*jc3\*hps23\o\al(\s\up9(d),d)EQ\*jc3\*hps23\o\al(\s\up9(y),x)EQ\*jc3\*hps23\o\al(\s\up6(1),1)EQ\*jc3\*hps13\o\al(\s\up11(2),2)EQ\*jc3\*hps24\o\al(\s\up9(d),d)EQ\*jc3\*hps14\o\al(\s\up15(2),x)EQ\*jc3\*hps24\o\al(\s\up9(y),2)EQ\*jc3\*hps24\o\al(\s\up9(1),t)2EQ\*jc3\*hps24\o\al(\s\up9(+),2)EQ\*jc3\*hps24\o\al(\s\up9(+t),4t3)2。EQ\*jc3\*hps23\o\al(\s\up0(im),喻0)x喻0（xe1)x喻0x(e1)x喻0xx喻02x2x喻0（xe1)x喻0x(e1)x喻0xx喻02x2EQ\*jc3\*hps23\o\al(\s\up0(m),喻1)EQ\*jc3\*hps23\o\al(\s\up5(一),x)解1EQ\*jc3\*hps24\o\al(\s\up11(m),喻1)EQ\*jc3\*hps24\o\al(\s\up11(m),喻1))EQ\*jc3\*hps24\o\al(\s\up4(m),喻1)EQ\*jc3\*hps24\o\al(\s\up4(m),喻1)2)dy2)dyEQ\*jc3\*hps23\o\al(\s\up8(d),d)EQ\*jc3\*hps23\o\al(\s\up8(y),x)EQ\*jc3\*hps23\o\al(\s\up8(y),x)EQ\*jc3\*hps23\o\al(\s\up14(一),1)EQ\*jc3\*hps23\o\al(\s\up8(t),2)EQ\*jc3\*hps24\o\al(\s\up8(d),d)EQ\*jc3\*hps13\o\al(\s\up13(2),x)EQ\*jc3\*hps24\o\al(\s\up8(y),2)EQ\*jc3\*hps24\o\al(\s\up8(d),dx)EQ\*jc3\*hps24\o\al(\s\up8(t),2)EQ\*jc3\*hps24\o\al(\s\up8(d),dt)EQ\*jc3\*hps24\o\al(\s\up8(t),2)EQ\*jc3\*hps24\o\al(\s\up8(d),d)EQ\*jc3\*hps24\o\al(\s\up8(t),x)EQ\*jc3\*hps24\o\al(\s\up8(1),2)EQ\*jc3\*hps24\o\al(\s\up8(+),2)2x喻0xex一1d2ydx222x-EQ\*jc3\*hps26\o\al(\s\up0(m),喻0)xxx喻0x喻02x2x=e22xxcot22xxcot3EQ\*jc3\*hps24\o\al(\s\up7(ay),y2)EQ\*jc3\*hps14\o\al(\s\up10(x2),ax)EQ\*jc3\*hps24\o\al(\s\up7(dy),dx)3dy38.limx(-).x+x2-12=122x22x2exx2;因为limf,(x)=1x喻0EQ\*jc3\*hps24\o\al(\s\up0(im),喻0)40、已知f(x)=x3+ax2+bx在x=1处有极值-2,试确定系数a、b的值,求函数的单调区间、极值、凹凸y=f(x)=x3-3x,y'=3x2-3,y"=6x.由y'=3x2-3=0,解得x1=-1,x2=1.由y"=6x=0,解得x=0.x01y’+0 --0+y’’-- 0+++y极大值y(-1)=2,极小值y(1)=-2,曲线与x轴交点(-,0),(0,0),(,0).2EQ\*jc3\*hps24\o\al(\s\up0(im),喻0)EQ\*jc3\*hps24\o\al(\s\up0(im),喻0)EQ\*jc3\*hps24\o\al(\s\up7(3x-),5x+)EQ\*jc3\*hps24\o\al(\s\up7(si),t)EQ\*jc3\*hps24\o\al(\s\up7(n2x),anx)EQ\*jc3\*hps24\o\al(\s\up0(im),喻0)EQ\*jc3\*hps24\o\al(\s\up7(2),s)EQ\*jc3\*