成人高考專升本高數(shù)二真題及答案
2015年成人高考專升本高數(shù)二真題及答案1.X+1口日?2+T=(A.01B.2C.1D.22 .當(dāng)x0時,sin3x是2x的()B.等階無窮小量D.高階無窮小量A.低階無窮小量C.同階但不等價無窮小量3 .函數(shù)f(x)=x+1,xv0,在x=0處()x12,x>0A.有定義且有極 限B.有定義但無極 限C.無定義但有極限D(zhuǎn).無定義且無極4 .設(shè)函數(shù)f(x)=x/則f(x)=()ntn1A.(1+x) e2B.(2+x) e2C. (1+2)e2D.(1+2x) e25 .下列區(qū)間為函數(shù)f(x)=x4.4x的單調(diào)增區(qū)間的是()A (汽B.(-g,0)C. (-1,1 )D.(1 , + g)A.O1 3B.3 / 3f(t) dt1 1c / f(t) dt3D.3 A 4t) dt16.已知函數(shù)f(x)在區(qū)間-3,3上連續(xù),則ZU(3x)B. -2x - + cos x + cA.-2x"+cosx+cdx=()C丁cosx+c1D.-V-cosx+c(x)=()A.-1B.OC.1D.29.設(shè)二元函數(shù)z=xyJiJ?z=(A.yxZB.yxy+iC.yxlnxD.xy10.設(shè)二元函數(shù)z=cos(xy),左二()2A.ysin(xy)2B.ycos(xy)2C.-ysin(xy)D.-ycos(xy)11.xmosin?=12.lim(1,)3=X13.設(shè)函數(shù)14.設(shè)函數(shù)y=x+sinx,貝Udy=(1+cosx)dx15.設(shè)函數(shù)ix2+er16.若/f(x)dx=cos(lnx)+C,則f(x)=sin(Inx)117/x|x|dx=18./d(xInx)=xlnx+C19.由曲線y=x2,直線x=1及x軸所圍成的平面有界圖形的面積匕?z20.設(shè)二兀函數(shù)z=ex則-eex-e21.計算limxFInx22.設(shè)函數(shù)y=COS(X34+1),求y1y'=cos(x2+1)=-sin(x2+1)?(x2+1)'2=-2xsin(x+1)23.計算科£X42dx=/4+x211/2d(4+x2)2”4+x2=2ln(4+x2)+C424.計算/of(x)dx淇中X,?<1*x)=£,?,14111/of(x)dx=/oxdx+x21=7Io+ln(1+x)|i32*ln25.f(x)dx.已知已)是連續(xù)函數(shù),且/f(t)edt=x,求/。等式兩邊對x求導(dǎo),得f(x)ex=1f(x)=etof(x)dx=JOexdx=exlu=e-126 .已知函數(shù)發(fā)f(x)=Inx-x.求f(x)的單調(diào)區(qū)間和極值;什僅)的定義域為(0,十八),f(x)=x-1.令f(x)=0得駐點x=1.當(dāng)Ovxv1時,f'(x)>0;當(dāng)x>1時,f'(x)<0.f(x)的單調(diào)增區(qū)間是(07),單調(diào)減區(qū)間是(f(x)在x=1處取得極大值f(1)=-1判斷曲線y=f(x)的凹凸性1,f”yy(x ' y)=2因為f"(x)二八2<0,所以曲線y=f(x)是凸的.f'xx(X'y)=1,f-xy(x,y)=-1A=f"xx(-6,-3)=1,B=fwxy(-6,-3)=-1,C=fwyy(-6,-3)=2B2-AC=-1<0,A>0,故f(x,y)在卜6,-3)處取得極小值,極小值為f(-6,-3)=9.28.從裝有2個白球,3個黑球的袋中任取3個球,記取出白球的個數(shù)為X.求X的概率分布;03PX=0=非=0.1,C5C2?C3PX=1=芋=0.6,C521c?cPX=2=0.3,C5因此X的概率分布為0.10.60.3求X的數(shù)學(xué)期望E(x).E(X)=0X0.1+1XO.6+2X0.3=1.2227 .求二元函數(shù)f(x,y)=y-xy+y2+3x/極值.f*=x-y+3,f,y=-x+2y由_L口解得x=-6,y=-3-x+2y=0