15-16-1《数学分析(一)》期末考试试卷及参考答案.pdf

XXXX 学 院15 16 学年第 1 学期 数学分析（一）期末考试 试卷参考答案题号一二三四五六七总分核分人1 2 3 4 5 分值25 7 7 7 7 7 8 8 8 8 8 100 得分一、填空题（共5 个小题，每小题5 分，共 25 分）1、验证拉格朗日中值定理对函数lnfxx在区间 1,e 上的正确性：（填正确或不正确），若正确，则解 因为lnfxx在区间 1,e 上可导（且连续），所以拉格朗日中值定 理 对 函 数lnfxx 在 区 间 1,e 上 是 正 确 的。由 中 值 公 式 得()(1)()(1),f effe即11(1),e故1.e2、函数lnyxx的严格递增区间为解令ln10yx得1.xe故函数lnyxx的严 格递增区 间为1,).e3、曲线321yxx在其拐点处的切线方程为解 因232,yx6,yx所以(0)0,y易判断(0,(0)(0,1)y是曲线的拐点。因(0)2,y故所求切线方程为21.yx4、已知21()dfxxCx,则()f x解 在21()dfxxCx两边求导得1()2.fxx令1,tx得21,xt故22(),f tt即22().f xx5、31dxxx21l nl n12xxC.解原式222(1)d(1)xxxx x21dd1xxxxx2211d(1)d21xxxx21lnln12xxC二、基本题（共5 个小题，每小题7 分，共 35 分）10limx(1csc1xxe)解原式01 sinlimsin(1)xxxexx e201sinlimxxexx0coslim2xL Hxexx0sinlim2xL Hxex122、求曲线2(3)4(1)xyx的所有渐近线。解因为211(3)limlim,4(1)xxxyx所以直线1x是曲线的垂直渐近线。又因为2(3)1l i ml i m,4(1)4xxyxkxx x2(3)595lim()limlim.4(1)44(1)4xxxxxxbykxxx所以直线1544yx是曲线的斜渐近线。3、求函数4225yxx在区间 0,2 上的最大值和最小值。解 令3444(1)(1)0yxxx xx得函数在(0,2)内有一个驻点1.x因 为(0)5,(1)4,(2yyy所 以 函 数 在0,2上 的 最 大 值(2)1 3,My最小值(1)4.my4、求不定积分3d1xxx解 原式3(1)1d(1)xxx23(1)d(1)1d(1)xxxx121112xxC5、求不定积分arctan d.xx x解2222111arctan darctan d()arctand2221xxx xxxxxxx22211(1)1arctand221xxxxx2111arctanarctan222xxxxC211(1)arctan.22xxxC三、证明题（8 分）证明：当0 x时，2ln 1.2xxx证 令2()ln 1,2xf xxx则()f x在0,)上连续，在(0,)内可导，且21()10 (0),11xfxxxxx所以()f x在0,)上严格单调增。于是当0 x时，有()(0)0,f xf即2ln 1.2xxx四、应用题（8 分）如图,有三个生活小区(均可看成点)分别位于,A B C三点 处,ABAC,A到线 段BC的距离40AO,27ABO(参考数据:22tan373).今计划建一个生活垃圾中转站P,为方便运输,P准备建在线段AO(不含端点)上.设2(0,)7PBO,试将P到三个小区的距离之和y表示为的函数,并确定当取何值时,可使y最小?解tanABO20 3,BOAOcos20 3 cos,PCBPBOtan20 3tan,POBO2()yBPAOPO2sin4020 3,cos2(0,).7文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3易求得22sin120 3cosy,令0y,即1sin2,得2(0,)7内惟一驻点6.因为该问题的最小值必存在，故当6时，可使y最小.五、计算题（8 分）设()fx的一个原函数为ln xx,求(2)dxfxx.解 依题意可得2l n1l nl n()(),().xxxfxfxxCxxxd故111(2)dd(2)(2)(2)d222xfxxxfxxfxfxx11(2)(2)d(2)24xfxfxx211ln(2)1 ln(2)2(2)42xxxCxx12ln(2).8xCx六、综合题（本题共2 个小题，考生任选一题做，满分8分。若两题全做，则按得分最高的计分）1.讨论方程326930 xxx的实根个数，并指出每个根所在的区间。解 令32693fxxxx，则23129313fxxxxx，令0fx，得驻点1,3xx.列表讨论如下：x(,1)1(1,3)3(3,)fx00fx极大极小可见该函数在1x处取极大值11f，在3x处取得极小值33.f因为lim(),xf x110,f330,flim(),xf x所以由零点定理（也叫方程根的存在性定理）和函数的单调性知，方程有三个实根，分别在区间(,1)(1,3)(3,)、内。2.设1(ln),1fxx且(0)0,f求().fx解 令ln,xt则由1(ln)1fxx得1(),1tfte这样1(1)d(1)()ddln(1),111xxxxxxxeeef xxxxxeCeee由(0)0f可得ln 2,C故2()ln.1xf xxe七、证明题（本题共2 个小题，考生任选一题做，满分8分。若两题全做，则按得分最高的计分）1.设)(xf在,a b上连续，在(,)a b内可导，证明：至少存在一(,)a b，使得()()()().bf baf affba证 令()(),xxf x则由题设条件易知，()x在,a b上连续，在(,)a b内可导，这样由拉格朗日中值定理得，至少存在一(,)a b，使得文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 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ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3()()(),baba因为()()(),xf xxfx故上式即()()()().bf baf affba2.设)(xf在 2,1上连续，在)2,1(内可导，且,0)2()1(ff证明：至少存在一)2,1(，使得.0)()(3ff证 令3()(),xx f x则由题设条件易知，()x在2,1上连续，在)2,1(内 可导，且(1)(2)0,这 样 由 罗 尔 日 中 值 定 理 得，至 少存 在一(1,2)，使得()0,因为23()3()(),xx f xx fx 故上式即233()()0,ff又因1,所以.0)()(3ff文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 ZN6O7E1S4O3文档编码:CM8N1V6S6I3 HG1L3L5S1N6 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