2017-2018学年第一学期江苏省苏州市高三期中调研试卷数学(理)

注意事项：

1．本试卷共4页．满分160分，考试时间120分钟．

2．请将填空题的答案和解答题的解题过程写在答题卷上，在本试卷上答题无效． 3．答题前，务必将自己的姓名、学校、准考证号写在答题纸的密封线内．

一、填空题(本大题共14小题，每小题5分，共70分，请把答案直接填写在答卷纸相应的位置) ．．．1．已知集合U{1,2,3,4,5},A{1,3},B{2,3}，则AI(ðUB) ▲ ． 2．函数y 学 2017.11

1的定义域为 ▲ ．

ln(x1)3．设命题p:x4；命题q:x25x4≥0，那么p是q的 ▲ 条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分也不必要”）．

4．已知幂函数yx2mm(mN*)在(0,)是增函数，则实数m的值是 ▲ ． 5．已知曲线f(x)ax3lnx在(1,f(1))处的切线的斜率为2，则实数a的值是 ▲ ． 6．已知等比数列{an}中，a32，a4a616，则

2a7a9 ▲ ．

a3a5，则的值是 ▲ ． 12f(x)8．已知奇函数f(x)在(,0)上单调递减，且f(2)0，则不等式0的解集为 ▲ ．

x19．已知tan()2，则cos2的值是 ▲ ．

4x8,x≤2(a0且a1)的值域为[6,)，则实数a的取值范围是 ▲ ． 10．若函数f(x)logx5,x2a7．函数ysin(2x)(0)图象的一条对称轴是x211(nN*)，则b1b2Lb2017 ▲ ． 11．已知数列{an},{bn}满足a1,anbn1,bn12an112．设△ABC的内角A,B,C的对边分别是a,b,c，D为AB的中点，若bacosCcsinA且CD2，则△ABC面积的最大值是 ▲ ．

13．已知函数f(x)sin(x)，若对任意的实数[6f()f()0，则实数m的最小值是 ▲ ．

5,]，都存在唯一的实数[0,m]，使62高三数学 期中试卷 第 1 页 共 14 页

lnx,x014．已知函数f(x)，若直线yax与yf(x)交于三个不同的点A(m,f(m)),B(n,f(n)),

2x1,x≤0C(t,f(t))（其中mnt），则n12的取值范围是 ▲ ． m二、解答题(本大题共6个小题，共90分，请在答题卷区域内作答，解答时应写出文字说明、证明过程或

演算步骤) 15．(本题满分14分)

已知函数f(x)间的距离为

21sin(2ax)b(a0,b0)的图象与x轴相切，且图象上相邻两个最高点之242． 2（1）求a,b的值；

（2）求f(x)在[0,]上的最大值和最小值．

16．(本题满分14分)

在错误！未找到引用源。中，角A,B,C所对的边分别是a,b,c，已知sinBsinCmsinA(mR)，且

4a24bc0错误！未找到引用源。错误！未找到引用源。．

（1）当a2,m5时，求b,c的值； 4（2）若角A错误！未找到引用源。为锐角，求m的取值范围．

高三数学 期中试卷 第 2 页 共 14 页

17．(本题满分15分)

已知数列{an}的前n项和是Sn，且满足a11，Sn13Sn1(nN*)． （1）求数列{an}的通项公式； （2）在数列{bn}中，b13，bn1bnan1(nN*)，若不等式anbn≤n2对nN*有解，求实数an的取值范围．

18．(本题满分15分)

如图所示的自动通风设施．该设施的下部ABCD是等腰梯形，其中AB为2米，梯形的高为1米，CD是个半圆，为3米，上部CmD固定点E为CD的中点．MN是由电脑控制可以上下滑动的伸缩横杆（横

杆面积可忽略不计），且滑动过程中始终保持和CD平行．当MN位于CD下方和上方时，通风窗的形状均为矩形MNGH（阴影部分均不通风）． （1）设MN与AB之间的距离为x(0≤x关于x的函数yS(x)；

（2）当MN与AB之间的距离为多少米时，通风窗的通风面积S取得最大值？

5且x1)米，试将通风窗的通风面积S（平方米）表示成2

高三数学 期中试卷 第 3 页 共 14 页

19．(本题满分16分)

已知函数f(x)lnx,g(x)x2xm． （1）求过点P(0,1)的f(x)的切线方程；

（2）当m0时，求函数F(x)f(x)g(x)在(0,a]的最大值；

（3）证明：当m≥-3时，不等式f(x)g(x)x2(x2)ex对任意x[,1]均成立（其中e为自然对数的底数,e2.718...）．

20．(本题满分16分)

已知数列{an}各项均为正数，a11,a22，且anan3an1an2对任意nN*恒成立，记{an}的前n项和为Sn．

（1）若a33，求a5的值；

（2）证明：对任意正实数p，{a2npa2n1}成等比数列；

（3）是否存在正实数t，使得数列{Snt}为等比数列．若存在，求出此时an和Sn的表达式；若不存在，说明理由．

122017—2018学年第一学期高三期中调研试卷

数 学 (附加) 2017.11

注意事项：

1．本试卷共2页．满分40分，考试时间30分钟． 2．请在答题卡上的指定位置作答，在本试卷上作答无效．

3．答题前，请务必将自己的姓名、学校、考试证号填写在答题卡的规定位置．

高三数学 期中试卷 第 4 页 共 14 页

21．【选做题】本题包括A、B、C、D四小题，请选定其中两题，并在相应的答题区域内作答．若多做，．．．．．．．．．．．．．．．．．．．则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤． A．(几何证明选讲) （本小题满分10分）

如图，AB为圆O的直径，C在圆O上，CFAB于F，点D为线段CF上任意一点，延长AD交圆O于E，AEC300． （1）求证：AFFO；

（2）若CF3，求ADAE的值．

B．(矩阵与变换) （本小题满分10分）

CDAFOEBurr412u49已知矩阵A，2，求A的值． 21

C．(极坐标与参数方程) （本小题满分10分）

4

xt25

在平面直角坐标系中，直线l的参数方程为(t为参数)，以原点O为极点，x轴正半轴为

y2t5

极轴建立极坐标系，圆C的极坐标方程为2acos()(a0)．

4（1）求直线l和圆C的直角坐标方程；

（2）若圆C任意一条直径的两个端点到直线l的距离之和为5，求a的值．

D．(不等式选讲) （本小题满分10分）

设x,y均为正数，且xy，求证：2x1≥2y3．

x22xyy2

【必做题】第22、23题，每小题10分，共计20分．请在答题卡指定区域内作答，解答时应写出文字说．．．．．．．明、证明过程或演算步骤．

高三数学 期中试卷 第 5 页 共 14 页

22．(本小题满分10分)

在小明的婚礼上，为了活跃气氛，主持人邀请10位客人做一个游戏．第一轮游戏中，主持人将标有数字1,2,…,10的十张相同的卡片放入一个不透明箱子中，让客人依次去摸，摸到数字6,7,…,10的客人留下，其余的淘汰，第二轮放入1,2,…,5五张卡片，让留下的客人依次去摸，摸到数字3,4,5的客人留下，第三轮放入1,2,3三张卡片，让留下的客人依次去摸，摸到数字2,3的客人留下，同样第四轮淘汰一位，最后留下的客人获得小明准备的礼物．已知客人甲参加了该游戏． （1）求甲拿到礼物的概率；

（2）设表示甲参加游戏的轮数，求的概率分布和数学期望E()． ．．

23．(本小题满分10分)

（1）若不等式(x1)ln(x1)≥ax对任意x[0,)恒成立，求实数a的取值范围； （2）设nN*，试比较

111与ln(n1)的大小，并证明你的结论． L23n12017—2018学年第一学期高三期中调研试卷

数 学 参 考 答 案

一、填空题(本大题共14小题，每小题5分，共70分)

1．{1} 2．(1,2)U(2,) 3．充分不必要 4．1 5．

134 8．(2,0)U(1,2) 9． 10．(1,2]

531111． 12．21 13． 14．(1,e)

e201826．4 7．

二、解答题(本大题共6个小题，共90分) 15．(本题满分14分)

解：（1）∵f(x)图象上相邻两个最高点之间的距离为

， 2∴f(x)的周期为

2且a0，·，∴·····································································2分 2|a|22∴a2，··················································································································4分

高三数学 期中试卷 第 6 页 共 14 页

21sin(4x)b， 24212且b0，·又∵f(x)的图象与x轴相切，∴|b|······················································6分

2221；∴b··········································································································8分 2222sin(4x)（2）由（1）可得f(x)， 242∵x[0,]，∴4x[,]，

444421∴当4x，即x时，f(x)有最大值为；·················································11分

2444当4x，即x时，f(x)有最小值为0．························································14分

4216此时f(x)16．(本题满分14分)

解：由题意得bcma，a24bc0．···············································································2分

（1）当错误！未找到引用源。时，错误！未找到引用源。，

1b2b解得·······························································································6分 2；·1或c2c2a2(ma)a222222bca(bc)2bca22m23，·（2）cosA···························8分 2a2bc2bc22∵A错误！未找到引用源。为锐角，∴cosA2m23(0,1)，∴

3···················································11分 m22，·

2又由bcma可得m0，·························································································13分 ∴6m2．·····································································································14分 217．(本题满分15分)

解：（1）∵Sn13Sn1(nN*)，∴Sn3Sn11(nN*,n≥2)，

∴an13an(nN*,n≥2)，·························································································2分 又当n1时，由S23S11得a23符合a23a1，∴an13an(nN*)，······························3分 ∴数列{an}是以1为首项，3为公比的等比数列，通项公式为an3n1(nN*)；·····················5分 （2）∵bn1bnan13(nN*)，∴{bn}是以3为首项，3为公差的等差数列，····················7分 an高三数学 期中试卷 第 7 页 共 14 页

∴bn33(n1)3n(nN*)，·····················································································9分 ∴anbn≤n，即32n1n23n3n≤n，即≤n1对nN*有解，··································10分

32n23n(nN*)， 设f(n)n13(n1)23(n1)n23n2(n24n1)n1∵f(n1)f(n)，

3n33n∴当n≥4时，f(n1)f(n)，当n4时，f(n1)f(n)， ∴f(1)f(2)f(3)f(4)f(5)f(6)L， ∴[f(n)]maxf(4)∴≤4，···························································································14分 274．·············································································································15分 2718．(本题满分15分)

解：（1）当0≤x1时，过A作AKCD于K（如上图），

CDAB1，HM1x， 22AKMHHM1x由， 2，得DHDKDH22则AK1，DK∴HG32DH2x，

∴S(x)HMHG(1x)(2x)x2x2；·······························································4分 当1x5时，过E作ETMN于T，连结EN（如下图）， 22MN93则ETx1，TN(x1)2(x1)2，

242∴MN29(x1)2， 49(x1)2(x1)，······································································8分 4∴S(x)MNET2x2x2,0≤x1综上：S(x)································································9分 95；·2(x1),1x2(x1)42（2）当0≤x1时，S(x)x2x2(x)2129在[0,1)上递减， 4∴S(x)maxS(0)2；································································································11分

高三数学 期中试卷 第 8 页 共 14 页

952当1x时，S(x)2(x1)(x1)2≤242当且仅当(x1)(x1)29(x1)294，

2432591(1,)时取“”， (x1)2，即x424999，此时S(x)max2，∴S(x)的最大值为，············································14分 444321米时，通风窗的通风面积S取得最大值．·答：当MN与AB之间的距离为···················15分 4∴S(x)max19．(本题满分16分)

解：（1）设切点坐标为(x0,lnx0)，则切线方程为ylnx0将P(0,1)代入上式，得lnx00，x01，

∴切线方程为yx1；·······························································································2分 （2）当m0时，F(x)lnxx2x,x(0,)， ∴F(x)1(xx0)， x0(2x1)(x1)············································································3分 ,x(0,)，

x当0x1时，F(x)0，当x1时，F(x)0，

∴F(x)在(0,1)递增，在(1,)递减，·············································································5分 ∴当0a≤1时，F(x)的最大值为F(a)lnaa2a；

当a1时，F(x)的最大值为F(1)0；········································································7分 （3）f(x)g(x)x2(x2)ex可化为m(x2)exlnxx，

设h(x)(x2)exlnxx,x[,1]，要证m≥-3时mh(x)对任意x[,1]均成立， 只要证h(x)max3，下证此结论成立．

12121······················································8分 x1时，x10，·

2111设u(x)ex，则u(x)ex20，∴u(x)在(,1)递增，

xx211又∵u(x)在区间[,1]上的图象是一条不间断的曲线，且u()e20，u(1)e10，

22∵h(x)(x1)(ex)，∴当

1x∴x0(,1)使得u(x0)0，即ex0121，lnx0x0，····················································11分 x0当x(,x0)时，u(x)0，h(x)0；当x(x0,1)时，u(x)0，h(x)0；

12高三数学 期中试卷 第 9 页 共 14 页

∴函数h(x)在[,x0]递增，在[x0,1]递减， ∴h(x)maxh(x0)(x02)ex0lnx0x0(x02)1212····························14分 2x012x0，

x0x0∵y12212x在x(,1)递增，∴h(x0)12x01223，即h(x)max3，

x0x2∴当m≥-3时，不等式f(x)g(x)x2(x2)ex对任意x[,1]均成立．··························16分 20．(本题满分16分)

解：（1）∵a1a4a2a3，∴a46，又∵a2a5a3a4，∴a5分

123······································2a49；·

2aaan1an2（2）由nn3，两式相乘得anan1an3an4an1an22an3，

an1an4an2an3∵an0，∴anan4an22(nN*)，

从而{an}的奇数项和偶数项均构成等比数列，···································································4分 设公比分别为q1,q2，则a2na2q2n12q2n1，a2n1a1q1n1q1n1，······································5分 又∵

an3an1aa2q，∴4222，即q1q2，···························································6分 =an2ana3a1q1设q1q2q，则a2npa2n1q(a2n2pa2n3)，且a2npa2n10恒成立，

数列{a2npa2n1}是首项为2p，公比为q的等比数列，问题得证；····································8分 （3）法一：在（2）中令p1，则数列{a2na2n1}是首项为3，公比为q的等比数列，

3k ,q1∴S2k(a2ka2k1)(a2k2a2k3)(a2a1)3(1qk)，

1q,q13k2qk1 ,q1，·····································································10分 S2k1S2ka2k3(1qk)k11q2q,q1且S11,S23,S33q,S433q，

2(S2t)(S1t)(S3t),∵数列{Snt}为等比数列，∴ 2(S3t)(S2t)(S4t),高三数学 期中试卷 第 10 页 共 14 页

22t6q(1t),(3t)(1t)(3qt),即，即 2tq3,(3qt)(3t)(33qt),t1解得（t3舍去），·························································································13

q4分

∴S2k4k122k1，S2k122k11， 从而对任意nN*有Sn2n1， 此时Snt2n，

Snt2为常数，满足{Snt}成等比数列，

Sn1tn1当n≥2时，anSnSn12n22n1，又a11，∴an2n1(nN*)，

综上，存在t1使数列{Snt}为等比数列，此时an2n1,Sn2n1(nN*)．······················16分 法二：由（2）知，则a2n2qn1，a2n1qn1，且S11,S23,S33q,S433q，

2(S2t)(S1t)(S3t),∵数列{Snt}为等比数列，∴ 2(S3t)(S2t)(S4t),22t6q(1t),(3t)(1t)(3qt),即，即 2tq3,(3qt)(3t)(33qt),t1解得（t3舍去），·······················································································11

q4分

∴a2n2qn122n1，a2n122n2，从而对任意nN*有an2n1，····································13分 ∴Sn2222此时Snt2n，

012n112n2n1， 12Snt2为常数，满足{Snt}成等比数列，

Sn1t综上，存在t1使数列{Snt}为等比数列，此时an2n1,Sn2n1(nN*)．······················16分 21．【选做题】本题包括A、B、C、D四小题，请选定其中两题，并在相应的答题区域内作答．若多做，．．．．．．．．．．．．．．．．．．．则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤． A．(几何证明选讲，本小题满分10分)

解：（1）证明 ：连接OC,AC，∵AEC300，∴AOC2AEC600，

CDE高三数学 期中试卷 第 11 页 共 14 页

AFOB又OAOC，∴AOC为等边三角形，

∵CFAB，∴CF为AOC中AO边上的中线，

∴AFFO；······································································5分 （2）解：连接BE，

∵CF3，AOC是等边三角形， ∴可求得AF1，AB4，

∵AB为圆O的直径，∴AEB90o，∴AEBAFD， 又∵BAEDFA，∴AEB∽AFD，∴

ADAF， ABAE即ADAEABAF414．··················································································10分 B．(矩阵与变换，本小题满分10分) 解：矩阵A的特征多项式为f()122223， 1令f()0，解得矩阵A的特征值11,23，····························································2分

uur11当11时特征向量为1，当23时特征向量为2，·····································6分

11ur4uuruur又∵132，······························································································8分

2uur3501∴A132250．···········································································10分

314949149uruuruurC．(极坐标与参数方程，本小题满分10分)

解：（1）直线l的普通方程为x2y20；··········································································3分

a2a2a2圆C的直角坐标方程为(x)(y)；·······························································6分

222（2）∵圆C任意一条直径的两个端点到直线l的距离之和为5，

aa2|55∴圆心C到直线l的距离为，即2，·······················································8分 225|解得a3或a．·······························································································10分 D．(不等式选讲，本小题满分10分) 证：∵x0,y0,xy0，

∴2x13112y2(xy) 222x2xyy(xy)高三数学 期中试卷 第 12 页 共 14 页

(xy)(xy)1123(xy)≥33，

(xy)2(xy)2∴2x1····················································································10分 ≥2y3．

x22xyy222．(本题满分10分)

解：（1）甲拿到礼物的事件为A，

在每一轮游戏中，甲留下的概率和他摸卡片的顺序无关，

13211

，

2532101······················································································3分 答：甲拿到礼物的概率为；·10····································································4分 （2）随机变量的所有可能取值是1,2,3,4．·

则P(A)1P1，

2121P2，

2551311P3，

253101321P4，

2535随机变量的概率分布列为：

 P 1 2 3 4 1 21 51 101 5·············································8分 所以E()12323．(本题满分10分)

121511····································································10分 42．105解：（1）原问题等价于ln(x1)ax≥0对任意x[0,)恒成立， x1x1aax令g(x)ln(x1)，则g'(x)，

(x1)2x1当a≤1时，g'(x)x1a≥0恒成立，即g(x)在[0,)上单调递增，

(x1)2∴g(x)≥g(0)0恒成立；

当a1时，令g'(x)0，则xa10，

∴g(x)在(0,a1)上单调递减，在(a1,)上单调递增， ∴g(a1)g(0)0，即存在x0使得g(x)0，不合题意；

综上所述，a的取值范围是(,1]．················································································4分 （2）法一：在（1）中取a1，得ln(x1)x(x(0,))， x1高三数学 期中试卷 第 13 页 共 14 页

令x(nN*)，上式即为ln(即ln(n1)lnn1nn11， )nn11，·····························································································7分 n11ln2ln1,2ln3ln21,∴ 3LLln(n1)lnn1,n1111····················································10分 Lln(n1)(nN*)．

23n1111法二：注意到ln2，ln3，„„，

223111故猜想L···································································5分 ln(n1)(nN*)，·

23n1上述各式相加可得

下面用数学归纳法证明该猜想成立．

证明：①当n1时，ln2，成立；·············································································6分

12111Lln(k1)， 23k1x在（1）中取a1，得ln(x1)(x(0,))，

x111k2令x······································································8分 (kN*)，有ln()，·

k2k1k1②假设当nk时结论成立，即那么，当nk1时，

11111k2Lln(k1)ln(k1)ln()ln(k2)，也成立； 23k1k2k2k1111由①②可知，L····································································10分 ln(n1)．·

23n1

高三数学 期中试卷 第 14 页 共 14 页