东南大学2013-14-2高数(B,I)期中试卷分析.pdf

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1、2013pêÆ(B,I)Ï¥Áò©ÛNovember18,2013Öœeuy©¥†Ø§ž˜½Ø‡s1µ¿„§Ï–s•Ð)K•{"˜.W˜K(K
7K§11K8©§Ù{ˆKzK4©§÷©32©)11.¼êf(x)=mä:´0Ú1;x1ex1§‚©O´111˜˜˜aaa£££aaa¤¤¤mä:Ú111aaa£££ÃÃá¡¡¤¤¤mä:;8>>2<(1+3x)ln(1+x);x>0;62.®•f(x)=3x=0?ëY§Ka=e6;>>:acosx;x0.0rq1pBBBBppCCCC23.4•limBBB@x+2x+3xx

2、CCCA=;x!+122e2x4.y=arcsine2x§Kdy=pdx;p1e4x15.x!0ž§exesinx†xn´dá§Kn=3§a=;66.f(x)=x2ln(1+x)§Kf(5)(0)=40;7.y=y(x)´d•§y=yex+2eysinx7x+2¤(½Û¼ê§Ky0(0)=e4;.OŽeˆK(K
4K§zK8©§÷©32©)pxx2+21.¦¼êy=êy0(x).(x+1)2p!xx2+21x22(x2x+1))))y0(x)=+=pp(x+1)2xx2+2x+1(x+1)3x2+22sin2x+x3cos1x

3、2.OŽ4•limx!0(1+xsinx)(1cosx)
4•11•2sin2x+x3cos12sin2x+x3cos1x1x)))lim=lim
limx!0(1+xsinx)(1cosx)x!01+xsinxx!01cosx2sin2x+x3cos1sin2x1xsinx1=lim=4lim+2limxcos=4x!0x2x!0x2x!0x2!n+1n+2n+n3.OŽ4•lim++


+n!1n2+nsin1n2+nsin2n2+nsinnn+1n+2n+n)))Pxn=++


+,n2+nsin1n2+nsin2n2+nsinnn2+n(n+1)n2+

4、n(n+1)223Kyn=n2+nxnn2n=zn;nlim!1yn=nlim!1zn=2;!n+1n+2n+n3dY%½nlim++


+=n!1n2+nsin1n2+nsin2n2+nsinn28>>11>>><x;x,0;4.?ؼêf(x)=>>xe13x=0?ëY5ÚŒ5¶eŒ>>1>:;x=0.2§¦f0(0).))))Ñ"„P99SKl18K)‰"8>>>:y=sint+tcost;.d2y(1)¦•‚L3t=¤éA:?ƒ‚•§¶(2)OŽ4dx20pp1dyBBBB22C

5、CCC)))=(t2cott)jt==2§ƒ:@;(1+)A§ƒ‚•§dx44224t=43py=2x+242d2y12=dx2sintsin3to.(K÷©7©)Ѽêf(x)=xsinx‘kLagrange{‘4Maclaurinúª.x46cosxxsinx)))f(x)=x2+x6(0<<1)66!5sinx+xcosx5{‘•Œ±¤x(0<<1)5!Ê.(K÷©8©)(1)QãCauchy¥Š½n¶(2)y²Cauchy¥Š½n"))))Ñ"
4•12•8.(K÷©7©)a1a2anxn=++


+(a

6、i>0;i=1;2;


;n)1+a1(1+a1)(1+a2)(1+a1)


(1+an)y²êfxngÂñ"an+1yyyÏ•xn+1xn=>0§¤±êfxngüN4O.(1+a1)


(1+an+1)qϕx=11+11n1+a11+a1(1+a1)(1+a2)+11(1+a1)(1+a2)(1+a1)(1+a2)(1+a3)+


+11(1+a1)


(1+an1)(1+a1)


(1+an)=11<1(1+a1)


(1+an)¤±êfxngkþ..êfxngÂñ.Ô.K÷©6©f(x)3[a;b]þëY

7、§3(a;b)SŒ§…f(a)f(b)<0;f0(c)=0;a0.yBacxb0xACFigure1:1ÔK«¿ãyyy{{{˜˜˜Ø”f(a)<0;f(b)>0.Xã1§Ï•f(x)3[a;b]þëY…f(c)f(b)<0§•3x02(c;b)¦f(x0)=0.dLagrange¥Š½n§•32(x0;b)¦0f(b)f(x0)f()=>0bx0dLagrange¥Š½n§•32(c;)(a;b)¦f0()f0(c)00f()=