《高等数学》期末考试试卷

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ᢈ▾2014-2015ᦪὃᔁ(ᦟᎥ)Aᔁ〉ᵨ!"ᦪXX"⚪$—•%&'()ᑖ+ᑖ1230301216100,ᑖஹ./⚪(ᐳ12ᑖ12/2ᑖ)1.5ᦪy=4+1%Ḅ<=>?矫曲18X3./U)=x2,FG(/(x)+l)=஺4.5ᦪ/(x)=0热5.W5ᦪ/(x+1)=/+2x—5,ᑣ/(%)=6.\ᑖ]^y"-3y'=0Ḅ
bc஺%ஹ./⚪(ᐳ30ᑖ12d⚪3ᑖ)点然ax-11.5ᦪ/(x)=x----(a>0,஻wl)()ax+1A.?᜻5ᦪjB.?Ꮤ5ᦪjC.n᜻5ᦪo?Ꮤ5ᦪjD.?q᜻qᏔ5ᦪ஺1102.W5ᦪ/(+—)=/+ᑣ/(x)=()时“XX宙A.x“jB.%?—2jC.(x—1)~jD.—1

1x3.uvIim(-----ax-b)=O,ᐸz஺c?{ᦪ1ᑣ()—8+JX(A)஻=1,6=1,(B)a=-l,b=1(C)a=l,b=-1(D)a=—l,b=-14.~ᑡ5ᦪᙠᢣ<Ḅᓄ^z1()?d஺/ஹsinxஹA.er,(xf8)jB.----,(xf8)jXC.ln(l+x),(x-1)jD."X+1(x—0)x5.~ᑡ5ᦪz1ᙠ<~?cᜧḄ5ᦪ?().1/ஹ(/AA)y=xsin—(x—>oo)x(B)y=஻ಔ)(஻—>8)j(C)y=

2x(xf+0)11ஹ(D)y=—cos—(zx—>0)xxf.16./(x)={1ᑣ/(x)ᙠ=0ᜐ()x,x<0A.B.C.D.no7.¢£>=/ᙠS(1,0)ᜐḄᑗ£?().A.y=2x-2B.y=-2x+2C.y=2x4-2D.y=-2x-28.uvy=ᑣy"=().

3A.x3B.3x2C.6xD.69./(x)?¦TcᕜḄ5ᦪ1ᑣ/=£1+//(x)dxḄ«().(4)¬®(5)¬®a,7¯x(C)¬®T,x,¬®a(஻)¬®T,¬®a10.W/(1)=x,ᑣ/'(x)=()஺X1111A.-B.-C.——D.——-xxxx&ஹb~ᑡᔜ⚪(ᐳ30ᑖ12⚪6ᑖ)L5ᦪxsin—+/7x<0x/(x)=

42./(x)=x3sinx+7?j7,»¼(x).3.5ᦪy=y(x)ᵫ]^ycosx-sin(x+y)=0ᡠ¿<1»ᵯdxx=04./(x)=ei1x>°1»/(x)ḄQRS1ÁÂÃQRSḄln(l+x),-l

55.\x2e~2vdx'ஹ(Ç⚪Èᑖ12ᑖ)»5ᦪÉ=/(1+%)7ḄᓫËÌQ¯᩽«.

6(ஹ(Ç⚪Èᑖ16ᑖ)5ᦪ/(x)ᙠ0,1
Í10

7ὃᫀஹ1.(—co,—2]U[2,+oo)j2.1j3.4x~+4x+1;5.x2-6j6.y=c,+ceix2ஹLB;2.B;3.C;4.B;5.C;6.B;7.A;8.B;9.D;10.D&ஹ1.b(1)⌕/(x)ᙠx=0ᜐᨵ᩽▲·ᙠ1ᓽ⌕limf(x)=lim/(x)v-»0~XTO+ᡂÝ஺Þclim/(x)=lim(xsin-+/?)=/?.v-»0-.r->0-x..”ஹ[.sinx.lim/(x)=lim----=1x->0+Xf01Xᡠ¦1à6=1³1ᨵlim/(x)=lim/(x)ᡂÝ1ᓽ6=1³15ᦪx->0-ᙠx=0ᜐᨵ᩽▲·ᙠ1oÞc5ᦪᙠáSᜐᨵ᩽▲âᙠãSᜐ?ᔲᨵ<=ᐵ1ᡠ¦æ³஺¦çèé«஺(2)¬5ᦪḄ<=v15ᦪᙠáSᜐḄᐙ⌕ᩩì?lim/(x)=limf(x)=/(x)0Xf0x->xj®?ᨵb=l=/(O)=Q,ᓽ=l³5ᦪᙠX=0ᜐ..//ஹ..sinxlim/(x)=lim----=1tx->0+.x->0*X4b./(x)ᙠ(l,0),(0,l),(l,+oo)ᑁ1lime'T=oo,limei=0,A->rx->r/(O)=O,Þæ1x=l?/(x)Ḅ%ÅQRSj1limf(x}=limei=-]L''710+e?

8lim/(x)=limln(l+x)=O,Þæx=0?f(x)ḄÅðñQRS.xfOx->0-'ஹb5ᦪG=/(1+)7Ḅ<=>?(00,-1)1](1,+8)vy'=2x(1+x)T+x2(-l)(l+x)-22x(1+x)-x2_x(2+x)(1+x)2--(l+x1òy=+)=0,óSô=—2,஽=0(8,-2)-20(-2,-l)U(-1,0)(0,+8)+00+f'M/\/᩽ᜧ«᩽d«fMᦑ5ᦪḄᓫË÷øÌQ?(-00-2)¯(0,+00),ᓫËùúÌQ?(-2,-1)û(1,0),àx=-2³1᩽ᜧ«/(-2)=-4jàx=0³1᩽d«/(0)=0,(ஹü(x)=/(x)-x,ᙠý0þÍᵨÿ
ᳮ(x)ᨵ.F(x)ᙠ0,1ᙠC1,jᓽF(C1)=F(q)=O,#q,CzuO&,ᵫR஺/+
ᳮ,ᨵ4e(c“q),1F'(G=஺ᓽ/'(4)-1=0=>/'(?)=1,7⚪9:ᦑᙠ(0,1)ᑁᨵ=>ᨵx,1/(x)=x.