江西师范大学高等代数 2013年考研专业课初试真题
2013年江西卷数学试题及答案(文)
2013·江西卷(文科数学)1． 复数z ＝i(－2－i)(i 为虚数单位)在复平面内所对应的点在( ) A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限1．D [解析] z ＝1－2i ，故选D.2． 若集合A ＝{x ∈|ax 2＋ax ＋1＝0}中只有一个元素，则a ＝( ) A ．4 B ．2 C ．0 D ．0或42．A [解析] 当a ＝0时，A ＝∅；当a ≠0时，Δ＝a 2－4a ＝0，则a ＝4，故选A. 3． 若sin α2＝33，则cos α＝( )A ．－23B ．－13C.13D.233．C [解析] cos α＝1－2sin 2 α2＝13，故选C.4． 集合A ＝{2，3}，B ＝{1，2，3}, 从A ，B 中各任意取一个数，则这两数之和等于4的概率是( )A.23B.12C.13D.164．C [解析] 从A ，B 中任取一个数，共有6种取法，其中两数之和为4的是(2，2)，(3，1)，故P ＝26＝13，故选C.5． 总体由编号为01，02，…，19，20的20个个体组成．利用下面的随机数表选取5个个体，选取方法是从随机数表第1行的第5列和第6列数字开始由左到右依次选取两个数字，则选出来的第5个个体的编号为( )7816 6572 0802 6314 0702 4369 9728 0198 3204 9234 4935 8200 3623 4869 6938 7481A.08 B ．07 C ．02 D ．015．D [解析] 选出来的5个个体编号依次为：08，02，14，07，01.故选D. 6． 下列选项中，使不等式x <1x <x 2成立的x 的取值范围是( )A ．(－∞，－1)B ．(－1，0)C ．(0，1)D ．(1，＋∞)6．A [解析] x －1x <0⇒x 2－1x <0⇒x <－1或0<x <1，x 2－1x >0⇒x <0或x >1，求交集得x <－1，故选A.7． 阅读如图1－1所示的程序框图，如果输出i ＝4，那么空白的判断框中应填入的条件是( )图1－1A ．S <8B ．S <9C ．S <10D ．S <117．B [解析] i ＝2，S ＝5，i ＝3，S ＝8，i ＝4，S ＝9，因输出i ＝4，故填S <9，故选B.8． 一几何体的三视图如图1－2所示，则该几何体的体积为( )图1－2A ．200＋9πB ．200＋18πC ．140＋9πD ．140＋18π8．A [解析] 该几何体上面是半圆柱，下面是长方体，半圆柱体积为12π·32·2＝9π，长方体体积为10×5×4＝200.故选A.9． 已知点A (2，0)，抛物线C ：x 2＝4y 的焦点为F ，射线F A 与抛物线C 相交于点M ，与其准线相交于点N ，则|FM |∶|MN |＝( )A ．2∶ 5B ．1∶2C ．1∶ 5D ．1∶39．C [解析] F A ：y ＝－12x ＋1，与x 2＝4y 联立，得x M ＝5－1，F A ：y ＝－12x ＋1，与y ＝－1联立，得N (4，－1)，由三角形相似知|FM ||MN |＝x M 4－x M ＝15，故选C.10． 如图1－3所示，已知l 1⊥l 2，圆心在l 1上、半径为1 m 的圆O 在t ＝0时与l 2相切于点A ，圆O 沿l 1以1 m/s 的速度匀速向上移动，圆被直线l 2所截上方圆弧长记为x ，令y ＝cos x ，则y 与时间t (0≤t ≤1，单位：s)的函数y ＝f (t )的图像大致为( )图1－3图1－410.B [解析]如图，设∠MOA ＝α，cos α＝1－t ，cos 2α＝2cos 2 α－1＝2t 2－4t ＋1，x ＝2α·1＝2α，y ＝cos x ＝cos 2α＝2t 2－4t ＋1，故选B.11． 若曲线y ＝x α＋1(α∈)在点(1，2)处的切线经过坐标原点，则α＝________．11．2 [解析] y ′＝αx α－1，y ′|x ＝1＝α，所以切线方程为y －2＝α(x －1)，该切线过原点，得α＝2.12． 某住宅小区计划植树不少于100棵，若第一天植2棵，以后每天植树的棵数是前一天的2倍，则需要的最少天数n (n ∈*)等于________．12．6 [解析] S n ＝2（1－2n ）1－2＝2n ＋1－2≥100，得n ≥6.13． 设f (x )＝3sin 3x ＋cos 3x ，若对任意实数x 都有|f (x )|≤a ，则实数a 的取值范围是________．13．a ≥2 [解析] |f (x )|max ＝2，则a ≥2. 14． 若圆C 经过坐标原点和点(4，0)，且与直线y ＝1相切，则圆C 的方程是________．14．(x －2)2＋⎝⎛⎭⎫y ＋322＝254 [解析] r 2＝4＋(r －1)2，得r ＝52，圆心为⎝⎛⎭⎫2，－32.故圆C的方程是(x －2)2＋⎝⎛⎭⎫y ＋322＝254. 15． 如图1－5所示，正方体的底面与正四面体的底面在同一平面α上，且AB ∥CD ，则直线EF 与正方体的六个面所在的平面相交的平面个数为________．图1－515．4 [解析] 直线EF 与正方体左右两个面平行，与其他四个面相交． 16． 正项数列{a n }满足：a 2n －(2n －1)a n －2n ＝0. (1)求数列{a n }的通项公式a n ；(2)令b n ＝1（n ＋1）a n ，求数列{b n }的前n 项和T n .16．解：(1)由a 2n －(2n －1)a n －2n ＝0，得(a n －2n )(a n ＋1)＝0. 由于{a n }是正项数列，所以a n ＝2n .(2)由a n ＝2n ，b n ＝1（n ＋1）a n ，则b n ＝12n （n ＋1）＝12⎝⎛⎭⎫1n －1n ＋1，T n ＝12⎝⎛⎭⎫1－12＋12－13＋…＋1n －1－1n ＋1n －1n ＋1 ＝12⎝⎛⎭⎫1－1n ＋1＝n 2（n ＋1）.17． 在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，已知sin A sin B ＋sin B sin C ＋cos 2B ＝1.(1)求证：a ，b ，c 成等差数列；(2)若C ＝2π3，求ab的值．17．解：(1)证明：由题意得sin A sin B ＋sin B sin C ＝2sin 2 B ，因为sin B ≠0，所以sin A ＋sin C ＝2sin B ，由正弦定理，有a ＋c ＝2b ，即a ，b ，c 成等差数列． (2)由C ＝2π3，c ＝2b －a 及余弦定理得(2b －a )2＝a 2＋b 2＋ab ，即有5ab －3b 2＝0，所以a b ＝35.18． 小波以游戏方式决定是去打球、唱歌还是去下棋．游戏规则为：以O 为起点，再从A 1，A 2，A 3，A 4，A 5，A 6(如图1－6)这6个点中任取两点分别为终点得到两个向量，记这两个向量的数量积为X ，若X >0就去打球，若X ＝0就去唱歌，若X <0就去下棋．(1)写出数量积X 的所有可能取值；(2)分别求小波去下棋的概率和不．去唱歌的概率．图1－6 18．解：(1)X 的所有可能取值为－2，－1，0，1. (2)数量积为－2的有OA 2→·OA 5→，共1种；数量积为－1的有OA 1→·OA 5→，OA 1→·OA 6→，OA 2→·OA 4→，OA 2→·OA 6→，OA 3→·OA 4→，OA 3→·OA 5→，共6种；数量积为0的有OA 1→·OA 3→，OA 1→·OA 4→，OA 3→·OA 6→，OA 4→·OA 6→，共4种； 数量积为1的有OA 1→·OA 2→，OA 2→·OA 3→，OA 4→·OA 5→，OA 5→·OA 6→，共4种． 故所有可能的情况共有15种． 所以小波去下棋的概率为P 1＝715；因为去唱歌的概率为P 2＝415，所以小波不去唱歌的概率P ＝1－P 2＝1－415＝1115.19．， 如图1－7所示，直四棱柱ABCD －A 1B 1C 1D 1中，AB ∥CD ，AD ⊥AB ，AB ＝2，AD ＝2，AA 1＝3，E 为CD 上一点，DE ＝1，EC ＝3.(1)证明：BE ⊥平面BB 1C 1C ； (2)求点B 1到平面EA 1C 1的距离．图1－719．解：(1)证明：过B 作CD 的垂线交CD 于F ，则BF ＝AD ＝2，EF ＝AB －DE ＝1，FC ＝2.在Rt △BEF 中，BE ＝ 3. 在Rt △CFB 中，BC ＝ 6.在△BEC 中，因为BE 2＋BC 2＝9＝EC 2，故BE ⊥BC . 由BB 1⊥平面ABCD 得BE ⊥BB 1. 所以BE ⊥平面BB 1C 1C .(2)三棱锥E －A 1B 1C 1的体积V ＝13·AA 1·S △A 1B 1C 1＝ 2.在Rt △A 1D 1C 1中，A 1C 1＝A 1D 21＋D 1C 21＝3 2.同理，EC 1＝EC 2＋CC 21＝3 2，A 1E ＝A 1A 2＋AD 2＋DE 2＝2 3. 故S △A 1C 1E ＝3 5.设点B 1到平面EA 1C 1的距离为d ，则三棱锥B 1－A 1C 1E 的体积V ＝13·d ·S △A 1C 1E ＝5d ，从而5d ＝2，d ＝105. 20．， 椭圆C ：x 2a 2＋y 2b 2＝1(a >b >0)的离心率e ＝32，a ＋b ＝3.(1)求椭圆C 的方程；(2)如图1－8所示，A ，B ，D 是椭圆C 的顶点，P 是椭圆C 上除顶点外的任意一点，直线DP 交x 轴于点N ，直线AD 交BP 于点M ，设BP 的斜率为k ，MN 的斜率为m .证明：2m －k 为定值．图1－820．解：(1)因为e ＝32＝c a， 所以a ＝23c ，b ＝13c ，代入a ＋b ＝3得，c ＝3，a ＝2，b ＝1， 故椭圆C 的方程为x 24＋y 2＝1.(2)方法一：因为B (2，0)，P 不为椭圆顶点，则直线BP 的方程为y ＝k (x －2)⎝⎛⎭⎫k ≠0，k ≠ ±12，①①代入x 24＋y 2＝1，解得P ⎝ ⎛⎭⎪⎫8k 2－24k 2＋1，－4k 4k 2＋1.直线AD 的方程为y ＝12x ＋1.②①与②联立解得M ⎝⎛⎭⎪⎫4k ＋22k －1，4k 2k －1.由D (0，1)，P ⎝ ⎛⎭⎪⎫8k 2－24k 2＋1，－4k 4k 2＋1，N (x ，0)三点共线知－4k4k 2＋1－18k 2－24k 2＋1－0＝0－1x －0，解得N ⎝ ⎛⎭⎪⎫4k －22k ＋1，0.所以MN 的斜率为m ＝4k2k －1－04k ＋22k －1－4k －22k ＋1＝4k （2k ＋1）2（2k ＋1）2－2（2k －1）2＝2k ＋14，则2m －k ＝2k ＋12－k ＝12(定值)．方法二：设P (x 0，y 0)(x 0≠0，±2)，则k ＝y 0x 0－2.直线AD 的方程为：y ＝12(x ＋2)，直线BP 的方程为：y ＝y 0x 0－2(x －2)，直线DP 的方程为：y －1＝y 0－1x 0x ，令y ＝0，由于y 0≠1可得N ⎝ ⎛⎭⎪⎫－x 0y 0－1，0，联立⎩⎨⎧y ＝12（x ＋2），y ＝y0x 0－2（x －2），解得M ⎝⎛⎭⎪⎫4y 0＋2x 0－42y 0－x 0＋2，4y 02y 0－x 0＋2，因此MN 的斜率为m ＝4y 02y 0－x 0＋24y 0＋2x 0－42y 0－x 0＋2＋x 0y 0－1＝4y 0（y 0－1）4y 20－8y 0＋4x 0y 0－x 20＋4＝4y 0（y 0－1）4y 20－8y 0＋4x 0y 0－（4－4y 20）＋4＝y 0－12y 0＋x 0－2. 所以2m －k ＝2（y 0－1）2y 0＋x 0－2－y 0x 0－2＝2（y 0－1）（x 0－2）－y 0（2y 0＋x 0－2）（2y 0＋x 0－2）（x 0－2）＝2（y 0－1）（x 0－2）－2y 20－y 0（x 0－2）（2y 0＋x 0－2）（x 0－2）＝2（y 0－1）（x 0－2）－12（4－x 20）－y 0（x 0－2）（2y 0＋x 0－2）（x 0－2）＝12(定值)．21．， 设函数f (x )＝⎩⎨⎧1ax ，0≤x ≤a ，11－a （1－x ），a <x ≤1.a 为常数且a ∈(0，1)．(1)当a ＝12时，求f ⎝⎛⎭⎫f ⎝⎛⎭⎫13； (2)若x 0满足f (f (x 0))＝x 0，但f (x 0)≠x 0，则称x 0为f (x )的二阶周期点．证明函数f (x )有且仅有两个二阶周期点，并求二阶周期点x 1，x 2；(3)对于(2)中的x 1，x 2，设A (x 1，f (f (x 1)))，B (x 2，f (f (x 2)))，C (a 2，0)，记△ABC 的面积为S (a )，求S (a )在区间⎣⎡⎦⎤13，12上的最大值和最小值．21．解：(1)当a ＝12时，f ⎝⎛⎭⎫13＝23， f ⎝⎛⎭⎫f ⎝⎛⎭⎫13＝f ⎝⎛⎭⎫23＝2⎝⎛⎭⎫1－23＝23. (2)f (f (x ))＝⎩⎪⎨⎪⎧1a2x ，0≤x ≤a 2，1a （1－a ）（a －x ），a 2<x ≤a ，1（1－a ）2（x －a ），a <x <a 2－a ＋1，1a （1－a ）（1－x ），a 2－a ＋1≤x ≤1.当0≤x ≤a 2时，由1a2x ＝x 解得x ＝0，因为f (0)＝0，故x ＝0不是f (x )的二阶周期点；当a 2<x ≤a 时，由1a （1－a ）(a －x )＝x 解得x ＝a－a 2＋a ＋1∈(a 2，a )， 因f ⎝⎛⎭⎫a －a 2＋a ＋1＝1a ·a －a 2＋a ＋1＝1－a 2＋a ＋1≠a －a 2＋a ＋1， 故x ＝a －a 2＋a ＋1为f (x )的二阶周期点；当a <x <a 2－a ＋1时，由1（1－a ）2(x －a )＝x解得x ＝12－a ∈(a ，a 2－a ＋1)，因f ⎝⎛⎭⎫12－a ＝11－a ·⎝⎛⎭⎫1－12－a ＝12－a ， 故x ＝12－a 不是f (x )的二阶周期点；当a 2－a ＋1≤x ≤1时， 由1a （1－a ）(1－x )＝x解得x ＝1－a 2＋a ＋1∈(a 2－a ＋1，1)， 因f ⎝⎛⎭⎫1－a 2＋a ＋1＝1（1－a ）·⎝⎛⎭⎫1－1－a 2＋a ＋1 ＝a －a 2＋a ＋1≠1－a 2＋a ＋1. 故x ＝1－a 2＋a ＋1为f (x )的二阶周期点． 因此，函数f (x )有且仅有两个二阶周期点，x 1＝a －a 2＋a ＋1，x 2＝1－a 2＋a ＋1. (3)由(2)得A ⎝⎛⎭⎫a －a 2＋a ＋1，a －a 2＋a ＋1，B ⎝⎛⎭⎫1－a 2＋a ＋1，1－a 2＋a ＋1， 则S (a )＝12·a 2（1－a ）－a 2＋a ＋1， S ′(a )＝12·a （a 3－2a 2－2a ＋2）（－a 2＋a ＋1）2， 因为a ∈⎣⎡⎦⎤13，12，有a 2＋a <1. 所以S ′(a )＝12·a （a 3－2a 2－2a ＋2）（－a 2＋a ＋1）2＝12·a [（a ＋1）（a －1）2＋（1－a 2－a ）]（－a 2＋a ＋1）2>0. (或令g (a )＝a 3－2a 2－2a ＋2，g ′(a )＝3a 2－4a －2＝3⎝ ⎛⎭⎪⎫a －2－103⎝ ⎛⎭⎪⎫a －2＋103， 因a ∈(0，1)，g ′(a )<0，则g (a )在区间⎣⎡⎦⎤13，12上的最小值为g ⎝⎛⎭⎫12＝58>0， 故对于任意a ∈⎣⎡⎦⎤13，12，g (a )＝a 3－2a 2－2a ＋2>0，S ′(a )＝12·a （a 3－2a 2－2a ＋2）（－a 2＋a ＋1）2>0) 则S (a )在区间⎣⎡⎦⎤13，12上单调递增，故S (a )在区间⎣⎡⎦⎤13，12上的最小值为S ⎝⎛⎭⎫13＝133，最大值为S ⎝⎛⎭⎫12＝120.。
2013年江西师范大学硕士研究生入学考试初试
2013年江西师范大学硕士研究生入学考试初试科目考 试 大 纲科目代码、名称: 722高等数学（决策学方向） 适用专业: 0701Z1决策学 一、考试形式与试卷结构（一）试卷满分 及 考试时间本试卷满分为 150 分，考试时间为180分钟。
（二）答题方式 答题方式为闭卷、笔试。
试卷由试题和答题纸组成；答案必须写在答题纸相应的位置上。
（三）试卷内容结构和题型结构（考试的内容比例及题型） 内容结构：微积分。
题型结构为计算题、求解题和证明题。
二、考查目标（复习要求）要求考生系统地理解高等数学的基本概念和基本理论，掌握高等数学的基本方法。
要求考生具有抽象思维能力、逻辑推理能力、数学运算能力和综合运用所学的知识分析问题和解决问题的能力。
三、考试内容与要求（一）函数、极限、连续考试内容函数的概念及表示法 函数的有界性．单调性．周期性和奇偶性 复合函数．反函数．分段函数和隐函数 基本初等函数的性质及其图形 初等函数 函数关系的建立数列极限与函数极限的定义及其性质 函数的左极限和右极限 无穷小量和无穷大量的概念及其关系 无穷小量的性质及无穷小量的比较 极限的四则运算 极限存在的两个准则：单调有界准则和夹逼准则 两个重要极限：0sin lim 1x x x →= 1lim 1xx e x →∞⎛⎫+= ⎪⎝⎭函数连续的概念 函数间断点的类型 初等函数的连续性 闭区间上连续函数的性质考试要求1．理解函数的概念，掌握函数的表示法，会建立应用问题的函数关系． 2．了解函数的有界性．单调性．周期性和奇偶性．3．理解复合函数及分段函数的概念，了解反函数及隐函数的概念． 4．掌握基本初等函数的性质及其图形，了解初等函数的概念． 5．了解数列极限和函数极限（包括左极限与右极限）的概念．6．了解极限的性质与极限存在的两个准则，掌握极限的四则运算法则，掌握利用两个重要极限求极限的方法． 7．理解无穷小的概念和基本性质．掌握无穷小量的比较方法．了解无穷大量的概念及其与无穷小量的关系． 8．理解函数连续性的概念（含左连续与右连续），会判别函数间断点的类型． 9．了解连续函数的性质和初等函数的连续性，理解闭区间上连续函数的性质（有界性、最大值和最小值定理．介值定理)，并会应用这些性质．（二）一元函数微分学 考试内容导数和微分的概念 导数的几何意义和经济意义 函数的可导性与连续性之间的关系 平面曲线的切线与法线 导数和微分的四则运算 基本初等函数的导数 复合函数．反函数和隐函数的微分法 高阶导数 一阶微分形式的不变性 微分中值定理 洛必达（L'Hospital ）法则 函数单调性的判别 函数的极值 函数图形的凹凸性．拐点及渐近线 函数图形的描绘 函数的最大值与最小值考试要求1．理解导数的概念及可导性与连续性之间的关系，了解导数的几何意义与经济意义（含边际与弹性的概念），会求平面曲线的切线方程和法线方程．2．掌握基本初等函数的导数公式．导数的四则运算法则及复合函数的求导法则，会求分段函数的导数 会求反函数与隐函数的导数．3．了解高阶导数的概念，会求简单函数的高阶导数．4．了解微分的概念，导数与微分之间的关系以及一阶微分形式的不变性，会求函数的微分．5．理解罗尔（Rolle ）定理．拉格朗日( Lagrange)中值定理．了解泰勒定理．柯西（Cauchy)中值定理，掌握这四个定理的简单应用．6．会用洛必达法则求极限．7．掌握函数单调性的判别方法，了解函数极值的概念，掌握函数极值、最大值和最小值的求法及其应用．8．会用导数判断函数图形的凹凸性（注：在区间(,)a b 内，设函数()f x 具有二阶导数．当()0f x ''>时，()f x 的图形是凹的；当()0f x ''<时，()f x 的图形是凸的），会求函数图形的拐点和渐近线．9．会描述简单函数的图形．（三）一元函数积分学 考试内容原函数和不定积分的概念 不定积分的基本性质 基本积分公式 定积分的概念和基本性质 定积分中值定理 积分上限的函数及其导数 牛顿一莱布尼茨（Newton- Leibniz ）公式 不定积分和定积分的换元积分法与分部积分法 反常（广义）积分 定积分的应用考试要求 1．理解原函数与不定积分的概念，掌握不定积分的基本性质和基本积分公式，掌握不定积分的换元积分法和分部积分法． 2．了解定积分的概念和基本性质，了解定积分中值定理，理解积分上限的函数并会求它的导数，掌握牛顿一莱布尼茨公式以及定积分的换元积分法和分部积分法．3．会利用定积分计算平面图形的面积．旋转体的体积和函数的平均值，会利用定积分求解简单的经济应用问题． 4．了解反常积分的概念，会计算反常积分． （四）多元函数微积分学 考试内容多元函数的概念 二元函数的几何意义 二元函数的极限与连续的概念 有界闭区域上二元连续函数的性质 多元函数偏导数的概念与计算 多元复合函数的求导法与隐函数求导法 二阶偏导数 全微分 多元函数的极值和条件极值．最大值和最小值 二重积分的概念．基本性质和计算 无界区域上简单的反常二重积分考试要求1．了解多元函数的概念，了解二元函数的几何意义．2．了解二元函数的极限与连续的概念，了解有界闭区域上二元连续函数的性质．3．了解多元函数偏导数与全微分的概念,会求多元复合函数一阶、二阶偏导数，会求全微分,会求多元隐函数的偏导数． 4．了解多元函数极值和条件极值的概念，掌握多元函数极值存在的必要条件，了解二元函数极值存在的充分条件，会求二元函数的极值，会用拉格朗日乘数法求条件极值，会求简单多元函数的最大值和最小值，并会解决简单的应用问题．5．了解二重积分的概念与基本性质，掌握二重积分的计算方法（直角坐标．极坐标）．了解无界区域上较简单的反常二重积分并会计算．（五）无穷级数 考试内容常数项级数收敛与发散的概念 收敛级数的和的概念 级数的基本性质与收敛的必要条件 几何级数与p 级数及其收敛性 正项级数收敛性的判别法 任意项级杰的绝对收敛与条件收敛 交错级数与莱布尼茨定理 幂级数及其收敛半径．收敛区间（指开区间）和收敛域 幂级数的和函数 幂级数在其收敛区间内的基本性质 简单幂级数的和函数的求法 初等函数的幂级数展开式考试要求1．了解级数的收敛与发散．收敛级数的和的概念．2．了解级数的基本性质和级数收敛的必要条件，掌握几何级数及p 级数的收敛与发散的条件，掌握正项级数收敛性的比较判别法和比值判别法．3．了解任意项级数绝对收敛与条件收敛的概念以及绝对收敛与收敛的关系，了解交错级数的莱布尼茨判别法． 4．会求幂级数的收敛半径、收敛区间及收敛域．5．了解幂级数在其收敛区间内的基本性质（和函数的连续性、逐项求导和逐项积分），会求简单幂级数在其收敛区间内的和函数．6．了解xe ．sin x ．cos x ．ln(1)x +及(1)x α+的麦克劳林（Maclaurin ）展开式． （六）常微分方程与差分方程 考试内容常微分方程的基本概念 变量可分离的微分方程 齐次微分方程 一阶线性微分方程 线性微分方程解的性质及解的结构定理 二阶常系数齐次线性微分方程及简单的非齐次线性微分方程 差分与差分方程的概念 差分方程的通解与特解 一阶常系数线性差分方程 微分方程的简单应用考试要求1．了解微分方程及其阶、解、通解、初始条件和特解等概念．2．掌握变量可分离的微分方程．齐次微分方程和一阶线性微分方程的求解方法． 3．会解二阶常系数齐次线性微分方程．4．了解线性微分方程解的性质及解的结构定理，会解自由项为多项式．指数函数．正弦函数．余弦函数的二阶常系数非齐次线性微分方程．5．了解差分与差分方程及其通解与特解等概念． 6．了解一阶常系数线性差分方程的求解方法． 7．会用微分方程求解简单的经济应用问题．参考教材《高等数学》（上、下册），同济大学数学教研室主编，高等教育出版社，2007年第六版。
2013考研数三真题及解析
2013年全国硕士研究生入学统一考试数学三试题一、选择题：1～8 小题，每小题4 分，共32 分，下列每小题给出的四个选项中，只有一项符合题目要求的，请将所选项前的字母填在答题纸．．．指定位置上.（1）当x 0时，用o(x) 表示比x高阶的无穷小，则下列式子中错误的是（）（A）x o x(2) o x( 3)（B）o x()o x(2) o x( 3)（C）o x(2) o x(2 ) o x( 2 )（D）o x() o x(2) o x( 2 )| x |x 1（2）函数f x( ) ）x x( 1)ln | x |（A）0（B）1（C）2（D）3（3）设D k 是圆域D {(x y, ) | x2 y2 1}位于第k 象限的部分，记I k (y x dxdy)k 1,2,3,4，D k则（）（A）I1 0（B）I2 0（C）I3 0（D）I4 0（4）设{a n}为正项数列，下列选项正确的是（）2（A ） 若a n a n 1,则(1)n 1a n 收敛n 1（B ）若(1)n 1a n 收敛，则a n a n 1n 1（C ）若a n 收敛，则存在常数 P 1,使 lim n a Pn 存在nn 1（D ） 若存在常数P 1，使 lim n a Pn 存在，则a n 收敛nn1（5）设矩阵 A,B,C 均为 n 阶矩阵，若 AB C ,则 可逆，则B（A ）矩阵 C 的行向量组与矩阵 A 的行向量组等价（B ）矩阵 C 的列向量组与矩阵 A 的列向量组等价（C ）矩阵 C 的行向量组与矩阵 B 的行向量组等价（D ）矩阵 C 的行向量组与矩阵 B 的列向量组等价1 a 12 0（6）矩阵a b a 与0 b 0相似的充分必要条件为1 a 100 0（A ） a 0,b2（B ） a 0,b 为任意常数 （C ） a 2, b（D ） a2,b 为任意常数（7）设 X 1， ，X 2X 3 是随机变量，且 X 1~N(0,1)，X2~N(0,2），X 23 ~ N (5,3 )2 ,P j P {2X j 2}( j 1,2,3), 则（ ）（A）P1 P2 P3（B）P2 P1 P3（C）P3 P1 P2（D）P1 P3 P2（8）设随机变量X 和Y 相互独立，则X 和Y 的概率分布分别为，则P X{Y 2} ( )（A）（B）（C）（D）二、填空题：914 小题，每小题 4 分，共24 分，请将答案写在答题纸．．．指定位置上.（9）设曲线y f x( ) 和y x2 x在点(0,1) 处有公共的切线，则lim nfn ________。
2010-2013年江西师范大学827真题
江西师范大学2010年硕士研究生入学考试试题（ A 卷）专业：学科教学（英语）科目：英语综合注：考生答题时，请写在考点下发的答题纸上，写在本试题纸或其他答题纸上的一律无效。
Ⅰ.Vocabulary: (20 points, 1 X 1)Directions: Choose the best word which has the closest meaning to the underli -ned word in the sentence.1.Hauchecome’s death was bought about by ________.A) remorse B) accusation C) rheumatism D) frustration2.Jack _______ crimson with embarrassment when his girl friend saw his dirty socks under the sheet.A) disputed B) fumbled C) flushed D) stopped3.The boy had a _______ expression because of silly mistakes he had made.A) rustic B) shamefaced C) incredulous D) desperate4.The poem handles the problem of instinct _______ intellect in man.A) versus B) positive C) implement D) academic5.The government _______ its policy of helping the unemployed.A) implemented B) enriched C) enrolled D) plagued6.Obviously, the Chairman’s marks at the conference were _______ and not planned.A) substantial B) spontaneous C) simultaneous D) synthetic7.Reporters and photographers alike took great _______ at the rude way the actor behaved during the interview.A) annoyance B) offence C) resentment D) irritation8.These continual _______ in temperature make it impossible to decide what to wear.A) transitions B) transformations C) exchanges D) fluctuations9.Susan has _______ the elbows of her son’s jacket with leather patches to make it more durable.A) reinforecd B) sustained C) steadied D) confirmed10.Although we tried to concentrate on the lecture, we were _______ by the noise from the next room.A) distracted B) displaced C) dispersed D) discarded11.The reason why so many children like to eat this new brand of biscuit is t-hat it is particularly sweet and _______.A) fragile B) feeble C) brisk D) crisp12.There is supposed to be a safety _______ which makes it impossible for trains to collide.A) appliance B) accessory C) machine D) mechanism13.For years now, the people of that faraway country have been cruelly _____ by a dictator.A) depressed B) immersed C) oppressed D) cursed14.Ever since the rise of industrialism, education has been _______ towards producing workers.A) harnessed B) hatched C) motivated D) geared15.He developed a _______ attitude after years of frustration in his career.A) sneaking B) disgusted C) drastic D) cynical16.They believed that this was not the _______ of their campaign for equality but merely the beginning.A) climax B) summit C) pitch D) maximum17.Several guests were waiting in the _______ for the front door to open.A) porch B) vent C) inlet D) entry18.As the mountains were covered with a _______ of cloud, we couldn’t see their tops.A) coating B) film C) veil D) shade19.We couldn’t really afford to buy a house so we got it on hire purchase and paid monthly _______.A) investments B) requirements C) arrangements D) installments20.The magician made us think he cut the girl into pieces but it was merely an _______.A) illusion B) impression C) image D) illumⅡ.Reading Comprehension: (40 points, 1 X 2)Directions: There are 4 passages in this part. Each passage is followed by so me questions or unfinished statements. For each of them there are 4 choices marked a, b, c, and d. You should decide on the best choice and write the an swer on the Answer Sheet.TEXT 1Pygmies are the earliest-known inhabitants of central Africa. They lived in t he Congo basin long before other groups migrated there--their presence confirm ed in ancient Egyptian records. They are gentle, peaceful people who conceal t he-mselves well in the rainforest. They have in recent times had to share with immigrant farmers. Pygmies do not farm, but trade meat, honey and other for -est products for knives, metal tools ,rice, corn and bananas. They are lighter skinned than some of their neighbors and different in stature. Adults are 1.2 to 1.5m tall.The Efe group of Pygmies of the Ituri forest in eastern Congo is one of the last to retain its original culture. The Efes, whose existence is threatened by l -ogging and farming in the forests in which they hunt, are among the Burundi group of Pygmies who live in north-eastern Congo.In Cameroon, there is a population of 35,000 Baka Pygmies,but this number is uncertain because of the group’s semi-nornadic lifestyle in wandering the ra -inforest in search of game and other foods. During the three-month rainy seas -on, when food is plentiful, the Baka leave their permanent villages to roam th e forest, rarely staying in one place more than a week.Men contract marriages during this crucial season; they prove their hunting a -bility by the game they bring home to the parents of a future wife.Men from farming tribes sometimes marry Pygmy women, although there is a ban again -st Pygmy men marrying women from farming tribes.Within the Baka culture, hunting elephant is one of the most important activ -ities,not only for food but for the symbolic meanings and prestige traditionally attached to it. Elephant hunting is linked to other cultural activities, including men’s initiation and women’s ritual songs.Hunting is performed with poisoned arrows, bows, crossbows, spears and traps. The Baka are interested in the out -side world while maintaining their identity and independence. And though they are attracted by much of what the outside world offers, they have always had access to the forest,a world that is completely their own.Their culture is robust enough to survive as long as the forest remains. Without it this culture will be meaningless.21.The fact that Pygmies are the earliest Central African inhabitants has been ______.A)established B)made C)discovered D)explored22.Pygmies are gentle and peaceful people who in stature is ________.A)unusually big C)normally smallB)normally big D)quite normal23.________fails to denote the name of a country.A)the Congo B)Ituri C)Burundi D)Cameroon24.Which of the following is NOT true according to the passage?A)A Baka Pygmy marriage takes place in the rainforestB)Of all Pygmies, the Efe group is the last group keeping its primitive cultureC)Pygmy men are forbidden to marry women from farming tribesD)Pygmy women are not forbidden to marry men from farming tribes25.Robust as Pygmy culture is, it can according to the text, by all means surv -ive ________.A)to the end of this centuryB)the next centuryC)the culture of farming tribes long in the outside worldD)as long as there is the rainforestTEXT 2The use of chemicals in almost all areas of life has become a commonplace phenmoenon. There is growing evidence ,however, that chemicals in the envir -onment, including pesticides, may contribute to some illnesses. While studies are still being conducted preliminary conclusions point to the verdict that such chemicals are indeed negatively impacting those humans to whom they are exp -osed.Children are especially vulnerable to toxic substancs. Pound for pound, they eat, drink and breathe more than adults,all of which expose them more heavily to those chemicals to which most individuals encounter on a daily basis. Fur-theremore, their bodies are still in developing stages, exacerbating the negative effects of those chemicals which negatively impact them.Of 50 types of pesticieds commonly used in American schools a study cond -ucted by the National Coalition Against the Misuse of Pesticides found thatmany caused negative reactions in laboratory animals. Such negative effects inc -luded kidney and liver damage, cancer, and neurological and reproductive prob -lems.These implications of these findings are far-reaching, Given the variety of ha -rmful effects attributed to those pesticides tested, one must pause and consider whether it is wise to continue their use in the nation’s schools,where children will be constantly exposed to them. Activists have lobbied for the elimination of such use with a degree of success, and recent findings, if supported by fur -ther analysis and confirmation, may help further the cause. A long term soluti on or alternative, however, remains elusive.In the short run, however, there may be some measures that can be taken to mitigate the harmful effects of dangerous pesticides. The American Medical Association’s Council on Scientific Affairs concluded in a 1997 report that giv-en the “particular uncertaint regarding the long-term health effects of low-dose pesticide exposures.”It is “prudent”for adults and children to limit their expos -ure and to “consider the use of the least toxic chemical pesticides or non-che-mical alternatives”.26.The citing of dangerous effects of pesticides on laboratory animals in paragr -aph 3 is used to suggest that _______.A)scientists are also concentrated with effects of pesticides on animalsB)cancer is the worst effect of the pesticidesC)most or all of the dangerous effects of pesticides are now knownD)the chemicals may also cause these effects in humans27.Children are more likely than adults to suffer the negative effects of pestici-des because ________.A)children absorb more pesticides than adults, proportionallyB)pesticides are used more in areas with childrenC)the pesticides used in schools are more dangerous than other pesticidesD)adults know more about pesticides than children28.The main point of this text is that _________.A)pesticides are dangerous and their uses need to be reevaluatedB)children suffer most from the effects of pesticidesC)schools are most responsible for pesticide-relate illnessesD)pesticides should be eliminates everywhere29.The author mentions the exaggerated effects of chemicals on children in par -agraph 2 to _____.A)prove that pesticides are dangerous to all humansB)show that children are more susceptible to many kinds of dangersC)suggest a solution to the problem of pesticidesD)identify one group that pesticides hurt in particular30.The author cites the American Medical Association’s advice in paragraph 4 in order to ________.A)present more factual details to clarify the issueB)present an alternative theory for considerationC)present a recommendation from a reputable sourceD)Present a solution drawn from all the evidence consideredTEXT 3It came as something of a surprise when Diana, Princess of Wales, made a trip to Angola in 1997,to support the Red Cross’s campaign for a total ban on all anti-personnel landmines. Within hours of arriving in Angola, television scre -ens around the world were filled with images of her comforting victims injure -d in explosions caused by landmines.”I knew the statistics”,she said,”But putti-ng a face to those figures brought the reality home to me; like when I met Sandra, a 13-year-old girl who had lost her leg, and people like her.”The Princess concluded with a simple message:”We must stop landmines”. And she used every opportunity during her visit to repeat this message. But, back in London, her views were not shared by some members of the British government,which refused to support a ban on these weapons. Angry politicians launched an attack on the princess in the press. They described her as”very ill -informed”and a “loose cannon”. The princess responed by brushing aside the Criticisms:”This is a distraction we do not need. All I’m trying to do is help”. Opposition parties, the media and public immediately voiced their support for Princess. To make matters worse for the government, it soon emerged that the Princess’s trip had been approved by the Foreign Office, and that she was in fact very well-informed about both the situation in Angola and the British go-vernment ‘s policy regarding landmines.The result was a severe embarrassment for the government. To try and limit the damage, the Foreign Secretary,Malcolm Rifkin, claimed taht the Princess’s views on landmines were not very different from government policy, and that it was “working towards”a worldwide ban. The Defense Secretary , Michael Portilo, claimed the matter was “a misinterpretation or misunderstanding.”For the Princess,the trip to this war-torn country was an excellent opportunity to use her popularity to show the world how much destruction and suffering lan -dmines can cause. She said that the experience had also given her the chance to get colser to people and their problems.31.Princess Diana paid a visit to Angola in 1997_______.A)to clarify the British government’s stand on landminesB)to establish her image as a friend of landmine victimsC)to investigate the sufferings of landmine victims thereD)to voice her support for a total ban of landmines32.What did Diana mean when she said “putting a face to those figures broug-ht the reality home to me”(Line 5, Para,1)?A)Meeting the landmine victims in person made her believe the statisticsB)She just couldn’t bear to meet the landmine victims face to faceC)The actual situation in Angola made her feel like going back homeD)Seeing the pain of the victims made she realized the seriousness of the situa -tion33.Some members of the British government criticized Diana because ______.A)she had not consulted the government before the visitB)she was ill-informed of the government’s policyC)they were actually opposed to banning landminesD)they believed that she had misinterpreted the situation in Angola34.How did Diana respond to the criticisms?A)She made more appearances on TVB)She paid no attention to themC)She rose to argue with her opponentsD)She met the 13-year-old girl as planned35.What did Princess Diana think of her visit to Angola?A)It had caused embarrassment to the British governmentB)It had greatly promoted her popularityC)It had brought her closer to the ordinary peopleD)It had affected her relations with the British governmentTEXT 4“History is written by the victors.”This famous phrase reverberates througho -ut the halls of history, constantly reminding us to take all that we learn with a grain of salt, knowing that the information provided for our dissemination was provided, shaped and influenced by this left to hold the pen that recorded it. In that respect, one of the worst crimes against history is the revision ofit, the altering of the record of the past so as to reflect the viewpoint of a biased group who stand to benefit from the altered version.By revising the lens by which history is judged, valuable information is lost, to the detriment of both students of the filed as well as the awareness that co-mes from experience. Without an accurately recorded account to serve as guidi -ng light, nations and societies are left to stumble their way about their affairs, ignorant of what has and hasn’t worked before, and unaware of what past ev -ents shaped and determined their present situation. Such dismal situations emer -ge from simple pride, as well as the desire of the revisionists to depict thems -elves in a better light to posterity or to cover up an embarrassing legacy, no matter the cost to the future.Recent attempts by nations involved in the second World War to minimize or erase altogether certain shameful incidents from their history textbook has been met with international outrage and protest, and rightly so. By allowing fu -ture generations to forget or never even learn about how their ancestors stumb -led on the path to progress, the experiences of those who suffered as a result of those mistakes are trivialized and made to be in its information. Both are heinous results for both nationals of that particular nation as well as those of the international community,whose stories intertwine to form the large picture.When a single string in the tapestry of world history is unraveled by revisio -n, the entire piece becomes a weaker one, subject to additional modification at the whim of those who would like to use history as a tool for their own purposes, even if it means fundamentally changing it. This outcome must be avoided at all costs, firstly by not allowing a precedent to be established that makes it acceptable, even in a single case, to commit the revision. Otherwise, humans as a race will fall prey to yet another oft-quoted phrase:”History, if fo -rgotten, is doomed to be repeated.”36.The first line of the text implies that _______.A)historical accounts are invariably colored by the views and stances of those who emerged victoriousB)those who have the power to do so will often influence recording of events to favor themselvesC)those who are defeated have little or no say in the documentation of their st ruggle, resulting in a biased account.D)the winners in a struggle have the moral obligation to accurately record eve-nts37.The author views the revision of history as _______.A)indisputably negative in all situationsB)generally harmful when done so to favor one side’s stanceC)always motivated by the desire to portray the reviser in a better lightD)Rendering the revised history useless for the purposeof analysis and learning38.In paragraph3, the author argues against historical revision with the assertion that ________.A)revision of World WarⅡevents has proven that such actions have a negative impactB)such revision results in an undeserved sense of national prideC)revising history has far-reaching effects beyond the borders of any one count -ryD)history is one of the primary concerns dealt with in the education system and should thus be pure39.In paragraph 4,”When a single string in a tapestry of world history is unra -veled by revision, the entire piece becomes a weaker one”means that ______.A).history is an intertwined series of events coming together to form a large pictureB)a loss of reliability in any single segment of history makes the entire histori -cal record suspectC)once one piece of history is revised, others soon followD)as soon as the integrity of the historical record is breached, it can never be fully recovered.40.The main point of text is that ________.A)revising history must be avoided in all situations at all costsB)the revision of history leads to a flawed perception resulting in loss of vital lessonsC)is revision of history goes on, the meaning behind the revised events will lo -se meaningD)historical revision is an international problem affecting all nations and people Ⅲ.Cloze (20 points, 1 X 2)When people __41__ to improve their breathing their initial thought is invar-iably to suck in the maximum possible draught of air __42__ you can’t pour wine into a full bottle. __43__you can’t fill the lungs with fresh air __44__ you’ve first drained them of every drop of stale air,__45__at the best of times only a sixth of the air in the lungs gets __46__with each fresh breath we take. If we breathe shallowly, or fail to clear the lunges of devitalized air, this poor rate of turnover declines __47__further.__48__always start your deep breathing exercises __49__collapsing the lungs as fully as possible. While you breatheout,imagine that you’re a hot air balloon collapsing slowly to the ground. This has a relaxing effect, particularly __50__ you quietly intone the world relax…relax…relax as you exhale.41.A)set forth B)set off C)set down D)set out42.A)But B)And C)Therefore D)Often43.A)At the same time B)In the same way C)More often than not D)Even if44.A)if B)unless C)after D)before45.A)Even B)Especially C)When D)Where46.A)changed B)changing C)to change D)change47.A)still B)more C)less D)farther48.A)Moreover B)So C)Yet D)Besides49.A)when B)in C)by D)on50.A)before B)even if C)if D)just beforeⅣ.Translation(40 points)Part A. Translate the following passage into English (20 points)依照中华人民共和国妇女权益保障法的规定，妇女在政治、文化、社会和家庭生活等方面享有与难自己平等的权利。
2013年考研数学一真题及答案解析
2013硕士研究生入学考试数学一真题及解析1. 已知极限0arctan lim kx x xc x→-=，其中k ，c 为常数，且0c ≠，则（） A. 12,2k c ==-B. 12,2k c == C. 13,3k c ==-D. 13,3k c ==答案（D ）解析：用洛必达法则222112111arctan 1111limlimlimlim(1)kk k k x x x x x xx x x cxkxkxx kx---→→→→--+-+====+因此112,k ck-==，即13,3kc ==2.曲面2cos()0x xy yz x +++=在点(0,1,1)-处的切平面方程为（ ） A. 2x y z -+=- B. 0x y z ++= C. 23x y z -+=- D. 0x y z --= 答案（A ）解析：法向量(0,1,1)(,,)(2sin()1,sin(),),|(1,1,1)x y z n F F F x y xy x xy z y n -==-+-+=-切平面的方程是：1(0)1(1)1(1)0x y z ---++=，即2x y z -+=-。
3.设1()2f x x =-，12()sin (1,2,)n b f x n xdx n π==⎰ ，令1()s i n nn S x b n x π∞==∑，则（ ）A .34B.14C. 14-D. 34-答案（C ）解析：根据题意，将函数在[1,1]-展开成傅里叶级数（只含有正弦，不含余弦），因此将函数进行奇延拓：1||,(0,1)2()1||,(1,0)2x x f x x x ⎧-∈⎪⎪=⎨⎪-+∈-⎪⎩，它的傅里叶级数为()s x ，它是以2为周期的，则当(1,1)x ∈-且()f x 在x 处连续时，()()s x f x =。
91111()()()()44444s s s f -=-=-=-=-。
4.设221:1L x y +=，222:2L x y +=，223:22L x y +=，224:22L x y +=为四条逆时针方向的平面曲线，记33()(2)(1,2,3,4)63ii L yxI y dx x dy i =++-=⎰ ，则{}1234ma x ,,,I I I I =A. 1IB. 2IC. 3I D 4I 答案（D ）解析：由格林公式，22(1)2iiD yI xdxdy=--⎰⎰ 14D D ⊂,在4D 内22102yx -->，因此14I I <24242222222\(1)(1)(1)222D D D D yyyI xdxdy xdxdy x dxdy=--=--+--⎰⎰⎰⎰⎰⎰在4D 外22102yx--<，所以24I I <32cos 2222223[0,1][0,2]2121/2/22323221(1)(12cos sin )22111122cos sin 224cos sin 24241!!111!!22442!!2422!!2x r y D r yI xdxdy r r rdrd d r dr d r dr d d θθθπππππθθθπθθθπθθθθπππ∈∈=--=--=--=-⋅⋅-⋅=-⋅⋅⋅⋅-⋅⋅⋅⋅⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰11124288ππππ=--=3cos 22sin 222224[0,1][0,2]2121/2/2232322(1)(1cos sin )2112cos sin 24cos sin 441!!11!!1324422!!242!!24442x r y r D r yI xdxdy r r rdrd d r dr d r dr d d θθθπππππθθθπθθθπθθθθπππππππ∈∈=----=--=-⋅-⋅=-⋅⋅⋅-⋅⋅⋅=--=⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰34I I <5.设A,B,C 均为n 阶矩阵，若AB=C ，且B 可逆，则（ ） A.矩阵C 的行向量组与矩阵A 的行向量组等价B 矩阵C 的列向量组与矩阵A 的列向量组等价 C 矩阵C 的行向量组与矩阵B 的行向量组等价D 矩阵C 的列向量组与矩阵B 的列向量组等价 6.矩阵1111a ab a a⎛⎫⎪ ⎪ ⎪⎝⎭与2000000b ⎛⎫⎪⎪ ⎪⎝⎭相似的充分必要条件为（ ） A. 0,2a b == B. 0,a b = 为任意常数 C. 2,0a b == D. 2,a b = 为任意常数7.设123,,X X X 是随机变量，且1(0,1)X N ，22(0,2)X N ，23(5,3)X N ，{}122(1,2,3)i P P X i =-≤≤=，则（ ）A. 123P P P >>B. 213P P P >>C. 322P P P >> D 132P P P >>8.设随机变量()X t n ,(1,)Y F n ，给定(00.5)a a <<，常数c 满足{}P X c a >=，则{}2P Y c>=( )（9）设函数y=f(x)由方程y-x=e x(1-y) 确定，则01lim [()1]n n f n→-＝ 。
2013年考研数学真题及参考答案(数学二)
π
2
, 则当 x → 0 时, α ( x ) 是
【 】 .
(A) 比 x 高阶的无穷小 (C) 与 x 同阶但不等价的无穷小 【答案】 答案】C.
(B) 比 x 低阶的无穷小 (D) 与 x 等价的无穷小
【考点】 考点】计算极限的方法:常用的等价无穷小.
【解析】 解析】 x sin α ( x) = cos x − 1 ~ −
(D) I 4 > 0
【解析】 解析】在第 II 象限除原点外被积函数 y − x > 0 ,因此 I 2 > 0 . 【评注】 评注】在第 IV 象限除原点外被积函数 y − x < 0 ,因此 I 4 < 0 ; 在第 I 象限和第 III 象限,根据轮换对称性得
I1 = I 3 = 0 .
(7)设 A, B, C 均为 n 阶矩阵,若 AB = C ,且 B 可逆,则 (A) 矩阵 C 的行向量组与矩阵 A 的行向量组等价 (C) 矩阵 C 的行向量组与矩阵 B 的行向量组等价 【答案】 答案】B. 【考点】 考点】向量组的线性表示方法. 【解析】 解析】将矩阵 A 和 C 按列分块,设 A = (α1 , α 2 ,⋯ , α n ) , B = (bij ) , C = (γ 1 , γ 2 ,⋯ , γ n ) . ①由 AB = C 组线性表示; 【 】 . (B) 矩阵 C 的列向量组与矩阵 A 的列向量组等价 (D) 矩阵 C 的列向量组与矩阵 B 的列向量组等价
π
6
≤θ ≤
π
6
),则 L 所围平面图形的面积为
．
【答案】 答案】
π
12
.
【考点】 考点】计算极坐标曲线所围图形的面积.
2013年考研数学真题及参考答案(数学一)
⑻ 设随机变量 X t ( n) ，Y F (1, n) ，给定 (0 0.5) ，常数 c 满足 P X c ， 则P Y c
2
（
）
（A） （B） 1 （C） 2 （D） 1 2 二、填空题：9～14 小题,每小题 4 分,共 24 分.请将答案写在答题纸 指定位置上. ．．． ⑼ 设函数 y f ( x) 由方程 y x e ⑽ 已知 y1 e
x3 x y )e 的极值. 3
z 0 ， z 2 所围成的立体为 . （Ⅰ）求曲面 的方程； （Ⅱ）求 的形心坐标.
（20） （本题满分 11 分） 设A
1 a 0 1 ，B ，当 a, b 为何值时，存在矩阵 C 使得 AC CA B ，并 1 0 1 b
ቤተ መጻሕፍቲ ባይዱ
(1 x 2
Di
y2 )dxdy . 2
2
1 2 1 y 0 x2 y 2 1 ， 所 以 被 积 函 数 在 2 2 1 1 D1 : x 2 y 2 1 内，恒有 f ( x, y ) 0 ；且 x 2 y 2 1 时，有 f ( x, y ) 0 2 2
(0,1, 1)
{1, 1,1} ,
于是切平面方程为 x ( y 1) ( z 1) 0 ，故应选（Ａ）. ⑶ 应选（C） ． 【分析】本题考查傅里叶级数的收敛定理.先将函数延拓成 ( 1,1) 上的奇函数 F ( x) .对
9 F ( x) 使用傅里叶级数的收敛定理(狄里赫雷定理)得到 S ( ) 的值. 4
（D） a 2, b 为任意常数
N (0,1) ， X 2
N (0, 22 ) , X 3
全国2013年1月高等教育自学考试线性代数试题02198答案
全国2013年1月高等教育自学考试线性代数试题02198答案一、单项选择题(本大题共5小题，每小题2分，共10分) 1．已知2阶行列式122121221232232a a a a a b b b b b -+=-=-+，则（ A ）A ． -6B ． -2C ．2D ． 6 2．若矩阵A 中有一个r 阶子式不为零，且所有1r +阶子式都不为零，则（ B ）A ．()r A r <B ．()r A r =C ．()1r A r >+D ． ()1r A r =+3．设向量组(1,0,0,),(0,1,0)T T αβ==，下列向量中可以表示,αβ线性组合的是（ C ）A ．(2,1,1)TB ．(2,0,1)TC ．(2,1,0)TD ． (0,1,1)T4．设线性方程组1231231232000x x x kx x x x x x ++=⎧⎪++=⎨⎪-+=⎩有非零解，则k 的值为（ D ）A ． -2B ． -1C ．1D ． 25．设12312001A x ⎛⎫⎪=- ⎪ ⎪⎝⎭，且A 的特征值为1，2，3，则x=（ D ） A ． -2 B ． 2 C ．3 D ．4二、填空题(本大题共10小题，每小题2分，共20分)6．行列式sin cos sin cos sin cos a a a a aa+-+= 17．设10110111111xa x a a --=+=-，则 -2 8．设A 为2阶矩阵，若将A 第二列的2倍加到第一列得到矩阵1234⎛⎫⎪⎝⎭，则A=3254-⎛⎫⎪-⎝⎭9．设A ，B 均为2阶可逆矩阵，则13A O O B -⎛⎫= ⎪⎝⎭11113A O O B ---⎛⎫ ⎪⎪ ⎪⎝⎭10．已知向量组123(1,1,0,),(3,0,),(1,2,3)T T T k ααα===线性相关，则k = -911．设12,a a 是非齐次线性方程组Ax b =的解，12,k k 是常数，若1122k a k a +也是Ax b =的一个解为，则12k k += 112．设线性方程组123231323x x x ax x b x x c -+=⎧⎪-=⎨⎪+=⎩有解，则数,,a b c 应满足2c a b =+ 13．设3阶矩阵A 的特征值为1，-2，3，则2A E += 100 14．若n 阶矩阵A 满足320E A A +=，则必有一个特征值为2315．二次型1,231223(,)f x x x x x x x =+的矩阵为1002110221002⎛⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪⎝⎭三、计算题（本大题共7小题，每小题9分，共63分）16. 计算行列式1231230100010001a a ab D b b ⎛⎫⎪⎪= ⎪⎪⎝⎭。
2013年考研数学一真题与解析完整版
2013硕士研究生入学考试数学一真题及解析来源：文都教育1. 已知极限0arctan lim k x x xc x→-=，其中k ，c 为常数，且0c ≠，则（） A. 12,2k c ==- B. 12,2k c == C. 13,3k c ==- D. 13,3k c ==答案（D ）解析：用洛必达法则2221121000011arctan 1111limlimlim lim (1)kk k k x x x x x xx x x c x kx kx x k x ---→→→→--+-+====+因此112,k c k-==，即13,3k c ==2.曲面2cos()0x xy yz x +++=在点(0,1,1)-处的切平面方程为（ ） A. 2x y z -+=- B. 0x y z ++= C. 23x y z -+=- D. 0x y z --= 答案（A ）解析：法向量(0,1,1)(,,)(2sin()1,sin(),),|(1,1,1)x y z n F F F x y xy x xy z y n -==-+-+=- 切平面的方程是：1(0)1(1)1(1)0x y z ---++=，即2x y z -+=-。
3.设1()2f x x =-，102()sin (1,2,)n b f x n xdx n π==⎰，令1()sin n n S x b n x π∞==∑，则（ ）A .34 B. 14 C. 14- D. 34- 答案（C ）解析：根据题意，将函数在[1,1]-展开成傅里叶级数（只含有正弦，不含余弦），因此将函数进行奇延拓：1||,(0,1)2()1||,(1,0)2x x f x x x ⎧-∈⎪⎪=⎨⎪-+∈-⎪⎩，它的傅里叶级数为()s x ，它是以2为周期的，则当(1,1)x ∈-且()f x 在x 处连续时，()()s x f x =。
91111()()()()44444s s s f -=-=-=-=-。
2013年考研数学一真题带答案
2013硕士研究生入学考试数学一真题及解析1. 已知极限0arctan lim k x x xc x →-=，其中k ，c 为常数，且0c ≠，则（）A. 12,2k c ==-B. 12,2k c ==C. 13,3k c ==-D. 13,3k c ==答案（D ）解析：用洛必达法则2221121000011arctan 1111lim lim lim lim (1)k k k k x x x x x x x x x c x kx kx x k x ---→→→→--+-+====+ 因此112,k c k -==，即13,3k c ==2.曲面2cos()0x xy yz x +++=在点(0,1,1)-处的切平面方程为（ ） A. 2x y z -+=- B. 0x y z ++= C. 23x y z -+=- D. 0x y z --= 答案（A ）解析：法向量(0,1,1)(,,)(2sin()1,sin(),),|(1,1,1)x y z n F F F x y xy x xy z y n -==-+-+=- 切平面的方程是：1(0)1(1)1(1)0x y z ---++=，即2x y z -+=-。
3.设1()2f x x =-，102()sin (1,2,)n b f x n xdx n π==⎰，令1()s i n n n S x b n x π∞==∑，则（ ） A .34 B. 14 C. 14- D. 34- 答案（C ）解析：根据题意，将函数在[1,1]-展开成傅里叶级数（只含有正弦，不含余弦），因此将函数进行奇延拓：1||,(0,1)2()1||,(1,0)2x x f x x x ⎧-∈⎪⎪=⎨⎪-+∈-⎪⎩，它的傅里叶级数为()s x ，它是以2为周期的，则当(1,1)x ∈-且()f x 在x 处连续时，()()s x f x =。
91111()()()()44444s s s f -=-=-=-=-。