高職升本高等數(shù)學(xué)模擬試卷

市高等院校“高職升本科”招生統(tǒng)一考高等數(shù)學(xué)標(biāo)準(zhǔn)模擬試卷（一I卷（選擇題，共40注意事項(xiàng)：一、選擇題：本大題共10小題,每小題4分，共40分。1e

e

cos xf(xg(x在點(diǎn)x0處可導(dǎo)f(x0g(x00,f(x0g(x00g(x),f(x不是f(x)g(x)的駐 B．是f(x)g(x)的駐點(diǎn)，但不是極值C．是f(x)g(x)的極大 D．是f(x)g(x)的極小f(ex)xexf(1)0f(x)(lnf(x) B．ln2x3y2z1

lnxC．f(x) 2

ln2L2xy10z3

及平面4x2yz20L平行于 B．在 C．垂直于 D．與斜 f(x在區(qū)間a,bf(x)0,則方程xf(t)dt

dt0 bfa,b上根的個數(shù) B. C. xf(x)0sin(xu)duf(xsin B． 2設(shè)f(x,y)是連續(xù)函數(shù)，則二重積分0 f(x,y)dy等 1A.

dy

2f(x, B.dy

f(x,4 4

44C.0dy1f(x, D.0dy1y2f(x,44D

x2y22dxdy，D(xyx2y24I42A.42

y2 2d2(r2 d(2r2)rdr+ d(r2 2 (x2

2)dy2dx1x2(xyf(x在1,2有二階導(dǎo)數(shù)，f(1f(2)0F(x)x1)2f(xF(x 已知yx是微分方程y ()的解，則(x)的表達(dá)式 ln B.x

y y市高等院?！案呗毶究啤闭猩y(tǒng)一考試高等數(shù)學(xué)標(biāo)準(zhǔn)模擬試卷（一）第Ⅱ卷（非選擇題，共110分二三注意事項(xiàng)：x0時，無窮小量（1cosx）與asin2x為等階無窮小量，則2f(x的一個原函數(shù)為sinax，則xf(x)dxsinxcos3sin3sinxcos

dx

已知a,bc都是單位向量，且滿足abc0，則abbcca

f(x)y2yy0三、解答題：本大題共8小題，共86分。解答應(yīng)寫出文字說明，證明過程或演算步驟。x2x確定常數(shù)x2x

(axb)]（exex

dx f(x為區(qū)間0,22xxf(t)dt20tf(2t)dt2x3x f(x)在0,2上的最大值及最小（22

dyd2設(shè)ytcos(t

t)

cos

，t0，求 , dxdx p(1,0,4)，平行于平面：3x4yz10，且與直線lx1y3z2（設(shè)0x1f(xx2x0

f(x1af(x。試確定常數(shù)af（DD

dxdy其中Dx2y211(x21(x2y21(x2y2f(x在1,yf(xx1,xt(t1x的平面圖形繞x軸旋轉(zhuǎn)一周所成的旋轉(zhuǎn)體體積為v(tt2f(tf(1)3

292高等數(shù)學(xué)標(biāo)準(zhǔn)模擬試卷（一）參考答案 12．a(chǎn)xcosaxsinax

23(sinxcosx)33214.2

1(1x2)ln(1x2)2

y

x2x1(axb)2=x2x2x1ax

x2x1axx2xx2x1axx2x1ax=lim(1a2)x2x2x1ax故1a2 a

f(x) 舍去aa1又12ab b2exexexex dxxd(ex 2 exexexex而ex1dx令tex2tex dt2(1 )dt2t2arctantt2 t2exexexexexexexexexex解：首先從所給的已知條件中求解出f(x),再求f(x)在區(qū)間0,2上的最方程兩端對x求導(dǎo)數(shù)，2xf(t)dt2xf(2x)2xf(2x)8x30x求導(dǎo)，2f(2x)24x2f(x3x22f(2)6,f13f(x ycos(t2)2t2sin(t2)1cos(t2)2t2t2sin(t2dy

d2

d(dy)d(t)

dx

2t(sint2p且平行于已知平面的平面13(x14yz40即3x4yz已知直線l的參數(shù)方程為：xt yt z代入1的方程，得t16知1與直線l的交點(diǎn)為Q(15,19,32)p、Qx1yz4 解x0f(xf(x1)af(x1x00x11f(x)1f(x1)1(x1)1(x1)21x(x1)(x f

0xf(xx0f(0)

1xf(0)limf(x)f(0)lim1(x1)(x2) x0 f(0)limf(x)f(0)lim(1x2) f(xx0處可導(dǎo)

f(0)f(0)，即a2f(0)11(x2y2)1(x2y2

2

1r2

rdr1(1r

)D

11r

0 r

01

201r1=2 1

arcsinr21 (1r4)212(2)依題意得v(t

tf

(x)dx