浙江省专升本历年真题卷.doc

》试卷年浙江省普通高校“专升本”联考《高等数学(一)2005一、填空题xsinxye??.函数。1的连续区间是2)x1(x?1?lim.。22??x?)?4x(x?xx轴在空间中的直线方程是。(1)。(2)过原点且与?1?12)1(x?1x?e,?2)1(x???f(x)____b?a?_____,x?11x?,f(x)?,函数,4??bx?1,x?1???处连续。2??2?,?3?2siny?r?dy??r是参数时,则)当。是常数,(1dxdy??r是常数,。是参数时,则(2)'0c)?f()f(b][a,x)fa,bc(x)?(y?时,)(在,且,,则当cx?处取得极大值。在''0)?(?0fxf(x)ca?x?b?c?x,,当时,时,(A)当''0)?(fx)?0f(xcx?a?bc??x,)当B,当时,时,(''0x)?f?f(x)0(ca?x?b?c?x,)当,当时,时,C(''0)?xf(x)?0(fcax??b?x?,,当(D)当时,xx?)f(x?,则20f(x?3h)?f(x?2h)00?lim)。(h0?h'''').()5fxf)4((x),D(3A()f(x),(B)f),(xC00002??x,xe?0?1???dxx?ff(xx?0)?0,。.设函数(,则积分)3??1?2x?,?ex?0?1(A)?1,(B)0(C),(D)???????b)(aba是()都发散,?n1?1?nn(A)发散(B)条件收敛(C)绝对收敛(D))1?x?y?(x的导数。?y?x?)中的极大值,极小值。在区间(-1,2nfd2xe)?xf(。的ndx10?。223xx??1?1?。5x2e?1线1??x2?dx2e?x?x。?yx?1的幂级数,并求出它的收敛区间。?12ydyd封x??y?。2dxdx22aa,?2,b?a3ba??22b?ab,a的是两个向量,且的值,。?12m?1?-?xsinsinxdxn,m是整数。.计算积分1,其中--220----32f(x)?4ax?3bx?2cx?d,----------a,b,c,da?b?c?d?0,满足其中常数------------f(x)在(0,1)内至少有一个根,1()证明函数------------2ac8b3?)xf()内只有一个根。1)当(2在(0时,证明函数,------------------------------------)卷A年高数(一)答案(2005----------------------------------)???(1,,0)?(0,1)(??.?y?zyxt??00,z?x?t,y?0x?)3(是参数),或者2),或者.(1(其中?0z?001?1???0,?.(25.(1)),。2)?1y?xln(x?xln分)3,.1解:令()x?1x(2'22x[y?xx?1)??ln(x?x?1)](则分)(721x?x?42')?4?3x?4x?x(3xy?xx?0,,驻点为分).解:2(2213''4?y?6x(法一),''0???4y(0)1?(0)y分)(,5(极大值),544''?)?4?0y()y(?((极小值),.7分)2733(法二)x-1)(-1,00)0,(434322),(432?2?'y正0负0正y-2递增1递减?527递增(5分)y?1x?y??0x?(极小值)当(时,时,7分)(极大值),:利用莱布尼兹公式nfd2x?[x?2nx?n(n?1)]e(7分)ndx0001111???[?]dx?dx?dx(3分):2(x?1)(x?2)x?2x?1x?3x?2?11?1?042x?lnln?(7分=)3?x11?2x2x1ee?1???dx?:5=分)(32x2xe?1e?112x)?1x?ln(?e?C(其中C是任意常数)(7分)??dxe(x?x?2)dx)e?1)e(2x???(xx?2(:3=分)00011xx?e2dxx?1)e(2)(3e1?=+=2-=2-003?3e?2e?2?1?e。=(7分)8:解:111?[]?y?)(2分1x?2?1x?121x?