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西城2016高三数学期末

西城2016高三数学期末
西城2016高三数学期末

2015-2016学年北京市西城区高三(上)期末数学试卷(理科)

一、选择题:本大题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项.

1.设集合A={x|x>1},集合B={a+2},若A∩B=?,则实数a的取值范围是()A.(﹣∞,﹣1]B.(﹣∞,1]C.[﹣1,+∞)D.[1,+∞)

2.下列函数中,值域为R的偶函数是()

A.y=x2+1 B.y=e x﹣e﹣x C.y=lg|x| D.

3.设命题p:“若,则”,命题q:“若a>b,则”,则()

A.“p∧q”为真命题B.“p∨q”为假命题

C.“¬q”为假命题D.以上都不对

4.“”是“数列{a n}为等比数列”的()

A.充分不必要条件B.必要不充分条件

C.充要条件 D.既不充分也不必要条件

5.一个几何体的三视图如图所示,那么这个几何体的表面积是()

A.B.C.D.

6.设x,y满足约束条件,若z=x+3y的最大值与最小值的差为7,则实数m=()A.B. C.D.

7.某市乘坐出租车的收费办法如下:

不超过4千米的里程收费12元;超过4千米的里程按每千米2元收费(对于其中不足千米的部分,若其小于0.5千米则不收费,若其大于或等于0.5千米则按1千米收费);当车程超过4千米时,另收燃油附加费1元.

相应系统收费的程序框图如图所示,其中x(单位:千米)为行驶里程,y(单位:元)为所收费用,用[x]表示不大于x的最大整数,则图中①处应填()

A.B.C. D.

8.如图,正方形ABCD的边长为6,点E,F分别在边AD,BC上,且DE=2AE,CF=2BF.如果对于常数λ,在正方形ABCD的四条边上,有且只有6个不同的点P使得成立,那么λ的取值范围是()

A.(0,7)B.(4,7)C.(0,4)D.(﹣5,16)

二、填空题:本大题共6小题,每小题5分,共30分.

9.已知复数z满足z(1+i)=2﹣4i,那么z=.

10.在△ABC中,角A,B,C所对的边分别为a,b,c.若A=B,a=3,c=2,则cosC=.

11.双曲线C:的渐近线方程为;设F1,F2为双曲线C的左、右

焦点,P为C上一点,且|PF1|=4,则|PF2|=.

12.如图,在△ABC中,∠ABC=90°,AB=3,BC=4,点O为BC的中点,以BC为直径的半圆与AC,AO分别相交于点M,N,则AN=;=.

13.现有5名教师要带3个兴趣小组外出学习考察,要求每个兴趣小组的带队教师至多2人,但其中甲教师和乙教师均不能单独带队,则不同的带队方案有种.(用数字作答)

14.某食品的保鲜时间t(单位:小时)与储藏温度x(单位:℃)满足函数关系

且该食品在4℃的保鲜时间是16小时.

已知甲在某日上午10时购买了该食品,并将其遗放在室外,且此日的室外温度随时间变化如图所示.给出以下四个结论:

①该食品在6℃的保鲜时间是8小时;

②当x∈[﹣6,6]时,该食品的保鲜时间t随着x增大而逐渐减少;

③到了此日13时,甲所购买的食品还在保鲜时间内;

④到了此日14时,甲所购买的食品已然过了保鲜时间.

其中,所有正确结论的序号是.

三、解答题:本大题共6小题,共80分.解答应写出必要的文字说明、证明过程或演算步骤.

15.已知函数,x∈R.

(Ⅰ)求f(x)的最小正周期和单调递增区间;

(Ⅱ)设α>0,若函数g(x)=f(x+α)为奇函数,求α的最小值.

16.甲、乙两人进行射击比赛,各射击4局,每局射击10次,射击命中目标得1分,未命

(Ⅰ)若从甲的4局比赛中,随机选取2局,求这2局的得分恰好相等的概率;

(Ⅱ)如果x=y=7,从甲、乙两人的4局比赛中随机各选取1局,记这2局的得分和为X,求X的分布列和数学期望;

(Ⅲ)在4局比赛中,若甲、乙两人的平均得分相同,且乙的发挥更稳定,写出x的所有可能取值.(结论不要求证明)

17.如图,在四棱锥P﹣ABCD中,底面ABCD是平行四边形,∠BCD=135°,侧面PAB⊥底面ABCD,∠BAP=90°,AB=AC=PA=2,E,F分别为BC,AD的中点,点M在线段PD 上.

(Ⅰ)求证:EF⊥平面PAC;

(Ⅱ)若M为PD的中点,求证:ME∥平面PAB;

(Ⅲ)如果直线ME与平面PBC所成的角和直线ME与平面ABCD所成的角相等,求的值.

18.已知函数f(x)=x2﹣1,函数g(x)=2tlnx,其中t≤1.

(Ⅰ)如果函数f(x)与g(x)在x=1处的切线均为l,求切线l的方程及t的值;

(Ⅱ)如果曲线y=f(x)与y=g(x)有且仅有一个公共点,求t的取值范围.

19.已知椭圆C:的离心率为,点在椭圆C上.

(Ⅰ)求椭圆C的方程;

(Ⅱ)设动直线l与椭圆C有且仅有一个公共点,判断是否存在以原点O为圆心的圆,满足此圆与l相交两点P1,P2(两点均不在坐标轴上),且使得直线OP1,OP2的斜率之积为定值?若存在,求此圆的方程;若不存在,说明理由.

20.在数字1,2,…,n(n≥2)的任意一个排列A:a1,a2,…,a n中,如果对于i,j∈N*,i<j,有a i>a j,那么就称(a i,a j)为一个逆序对.记排列A中逆序对的个数为S(A).

如n=4时,在排列B:3,2,4,1中,逆序对有(3,2),(3,1),(2,1),(4,1),则S (B)=4.

(Ⅰ)设排列C:3,5,6,4,1,2,写出S(C)的值;

(Ⅱ)对于数字1,2,…,n的一切排列A,求所有S(A)的算术平均值;

(Ⅲ)如果把排列A:a1,a2,…,a n中两个数字a i,a j(i<j)交换位置,而其余数字的位置保持不变,那么就得到一个新的排列A':b1,b2,…,b n,求证:S(A)+S(A')为奇数.

2015-2016学年北京市西城区高三(上)期末数学试卷(理

科)

参考答案与试题解析

一、选择题:本大题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项.

1.设集合A={x|x>1},集合B={a+2},若A∩B=?,则实数a的取值范围是()A.(﹣∞,﹣1]B.(﹣∞,1]C.[﹣1,+∞)D.[1,+∞)

【考点】交集及其运算.

【专题】计算题;集合思想;集合.

【分析】由A与B,以及两集合的交集为空集,确定出a的范围即可.

【解答】解:∵A={x|x>1},集合B={a+2},若A∩B=?,

∴a+2≤1,即a≤﹣1,

则实数a的范围为(﹣∞,﹣1],

故选:A.

【点评】此题考查了交集及其运算,熟练掌握交集的定义是解本题的关键.

2.下列函数中,值域为R的偶函数是()

A.y=x2+1 B.y=e x﹣e﹣x C.y=lg|x| D.

【考点】函数奇偶性的判断.

【专题】计算题;规律型;转化思想;函数的性质及应用.

【分析】判断函数的奇偶性然后求解值域,推出结果即可.

【解答】解:y=x2+1是偶函数,值域为:[1,+∞).

y=e x﹣e﹣x是奇函数.

y=lg|x|是偶函数,值域为:R.

的值域:[0,+∞).

故选:C

【点评】本题考查函数的奇偶性的判断以及函数的值域,是基础题.

3.设命题p:“若,则”,命题q:“若a>b,则”,则()

A.“p∧q”为真命题B.“p∨q”为假命题

C.“¬q”为假命题D.以上都不对

【考点】复合命题的真假.

【专题】对应思想;综合法;简易逻辑.

【分析】分别判断出p,q的真假,从而判断出复合命题的真假即可.

【解答】解:命题p:“若,则”是假命题,

命题q:“若a>b,则”如:a=1,b=﹣1,

故命题q是假命题,

故p∨q是假命题,

故选:B.

【点评】本题考察了复合命题的判断,是一道基础题.

4.“”是“数列{a n}为等比数列”的()

A.充分不必要条件B.必要不充分条件

C.充要条件 D.既不充分也不必要条件

【考点】必要条件、充分条件与充要条件的判断.

【专题】等差数列与等比数列.

【分析】根据等比数列的性质,对于数列{a n},“数列{a n}为等比数列”可以推出

““”,对于反面,我们可以利用特殊值法进行判断;

【解答】解:若数列{a n}是等比数列,根据等比数列的性质得:

反之,若“”,当a n=0,此式也成立,但数列{a n}不是等比数列,

∴“”是“数列{a n}为等比数列”的必要不充分条件,

故选B.

【点评】此题主要考查等比数列的性质及必要条件、充分条件和充要条件的定义,是一道基础题.

5.一个几何体的三视图如图所示,那么这个几何体的表面积是()

A.B.C.D.

【考点】由三视图求面积、体积.

【专题】计算题;空间位置关系与距离;立体几何.

【分析】由已知中的三视图可得:该几何体是一个以主视图为底面的四棱柱,结合柱体表面积公式,可得答案.

【解答】解:由已知中的三视图可得:该几何体是一个以主视图为底面的四棱柱,

其底面面积为:×(1+2)×2=3,

底面周长为:2+2+1+=5+,

高为:2,

故四棱柱的表面积S=2×3+(5+)×2=,

故选:B

【点评】本题考查的知识点是由三视图,求体积和表面积,根据已知的三视图,判断几何体的形状是解答的关键.

6.设x,y满足约束条件,若z=x+3y的最大值与最小值的差为7,则实数m=()

A.B. C.D.

【考点】简单线性规划.

【专题】计算题;对应思想;数形结合法;不等式的解法及应用.

【分析】由约束条件画出可行域,化目标函数为直线方程的斜截式,数形结合得到最优解,联立方程组求得最优解的坐标,进一步求出最值,结合最大值与最小值的差为7求得实数m 的值.

【解答】解:由约束条件作出可行域如图,

联立,解得A(1,2),

联立,解得B(m﹣1,m),

化z=x+3y,得.

由图可知,当直线过A时,z有最大值为7,

当直线过B时,z有最大值为4m﹣1,

由题意,7﹣(4m﹣1)=7,解得:m=.

故选:C.

【点评】本题考查简单的线性规划,考查了数形结合的解题思想方法,是中档题.

7.某市乘坐出租车的收费办法如下:

不超过4千米的里程收费12元;超过4千米的里程按每千米2元收费(对于其中不足千米的部分,若其小于0.5千米则不收费,若其大于或等于0.5千米则按1千米收费);当车程超过4千米时,另收燃油附加费1元.

相应系统收费的程序框图如图所示,其中x(单位:千米)为行驶里程,y(单位:元)为所收费用,用[x]表示不大于x的最大整数,则图中①处应填()

A.B.C. D.

【考点】程序框图;分段函数的应用;函数模型的选择与应用.

【专题】应用题;函数的性质及应用;算法和程序框图.

【分析】根据已知中的收费标准,求当x>4时,所收费用y的表达式,化简可得答案.【解答】解:由已知中,超过4千米的里程按每千米2元收费(对于其中不足千米的部分,若其小于0.5千米则不收费,若其大于或等于0.5千米则按1千米收费);

当车程超过4千米时,另收燃油附加费1元.

可得:当x>4时,所收费用y=12+[x﹣4+]×2+1=,

故选:D

【点评】本题考查的知识点是分段函数的应用,函数模型的选择与应用,难度中档.

8.如图,正方形ABCD的边长为6,点E,F分别在边AD,BC上,且DE=2AE,CF=2BF.如

果对于常数λ,在正方形ABCD的四条边上,有且只有6个不同的点P使得成立,那么λ的取值范围是()

A.(0,7)B.(4,7)C.(0,4)D.(﹣5,16)

【考点】平面向量数量积的运算.

【专题】函数思想;数形结合法;平面向量及应用.

【分析】建立坐标系,逐段分析的取值范围及对应的解,

【解答】解:以DC为x轴,以DA为y轴建立平面直角坐标系,如图,则E(0,4),F(6,4).

(1)若P在CD上,设P(x,0),0≤x≤6.∴=(﹣x,4),=(6﹣x,4).

∴=x2﹣6x+16,∵x∈[0,6],∴7≤≤16.

∴当λ=7时有一解,当7<λ≤16时有两解.

(2)若P在AD上,设P(0,y),0≤y≤6.∴=(0,4﹣y),=(6,4﹣y).

∴=(4﹣y)2=y2﹣8y+16,∵0≤y≤6,∴0≤≤16.

∴当λ=0或4<λ≤16,有一解,当0<λ≤4时有两解.

(3)若P在AB上,设P(x,6),0≤x≤6.=(﹣x,﹣2),=(6﹣x,﹣2).

∴=x2﹣6x+4,∵0≤x≤6.∴﹣7≤≤4.

∴当λ=﹣7时有一解,当﹣7<λ≤2时有两解.

(4)若P在BC上,设P(6,y),0≤y≤6,∴=(﹣6,4﹣y),=(0,4﹣y).

∴=(4﹣y)2=y2﹣8y+16,∵0≤y≤6,∴0≤≤16.

∴当λ=0或4<λ≤16,有一解,当0<λ≤4时有两解.

综上,∴0<λ<4.

故选:C.

【点评】本题考查了平面向量的数量积计算,二次函数的根的个数判断.属于中档题.

二、填空题:本大题共6小题,每小题5分,共30分.

9.已知复数z满足z(1+i)=2﹣4i,那么z=﹣1﹣3i.

【考点】复数代数形式的乘除运算.

【专题】计算题;方程思想;数学模型法;数系的扩充和复数.

【分析】把已知的等式变形,然后利用复数代数形式的乘除运算化简得答案.

【解答】解:由z(1+i)=2﹣4i,得

故答案为:﹣1﹣3i.

【点评】本题考查复数代数形式的乘除运算,是基础的计算题.

10.在△ABC中,角A,B,C所对的边分别为a,b,c.若A=B,a=3,c=2,则cosC=.

【考点】余弦定理.

【专题】计算题;转化思想;分析法;解三角形.

【分析】由已知可求b的值,利用余弦定理即可求值得解.

【解答】解:∵A=B,a=3,c=2,可得:b=3,

∴cosC===.

故答案为:.

【点评】本题主要考查了等腰三角形的性质,考查了余弦定理的应用,属于基础题.

11.双曲线C:的渐近线方程为;设F1,F2为双曲线C的左、右

焦点,P为C上一点,且|PF1|=4,则|PF2|=12.

【考点】双曲线的简单性质.

【专题】计算题;方程思想;综合法;圆锥曲线的定义、性质与方程.

【分析】双曲线C:中a=4,b=2,可得渐近线方程为,由题意P在双曲线的左支上,则|PF2|﹣|PF1|=2a=8,即可得出结论.

【解答】解:双曲线C:中a=4,b=2,则渐近线方程为,

由题意P在双曲线的左支上,则|PF2|﹣|PF1|=2a=8,

∴|PF2|=12

故答案为:,12.

【点评】本题考查双曲线的方程与性质,考查双曲线的定义,比较基础.

12.如图,在△ABC中,∠ABC=90°,AB=3,BC=4,点O为BC的中点,以BC为直径的

半圆与AC,AO分别相交于点M,N,则AN=;=.

【考点】与圆有关的比例线段.

【专题】选作题;方程思想;综合法;推理和证明.

【分析】利用勾股定理、切割线定理,即可得出结论.

【解答】解:由题意,AO==,

由切割线定理可得9=AN?(+2),

∴AN=.

AC==5,

由切割线定理可得9=AM?5,∴AM=,

∴MC=,

∴=.

故答案为:,.

【点评】本题考查勾股定理、切割线定理,考查学生的计算能力,属于中档题.

13.现有5名教师要带3个兴趣小组外出学习考察,要求每个兴趣小组的带队教师至多2人,但其中甲教师和乙教师均不能单独带队,则不同的带队方案有54种.(用数字作答)【考点】排列、组合的实际应用.

【专题】计算题;整体思想;数学模型法;排列组合.

【分析】第一类,把甲乙看做一个复合元素,和另外的3人分配到3个小组中,第二类,先把另外的3人分配到3个小组,再把甲乙分配到其中2个小组,根据分类计数原理可得

【解答】解:第一类,把甲乙看做一个复合元素,和另外的3人分配到3个小组中(2,1,1),C42A33=36种,

第二类,先把另外的3人分配到3个小组,再把甲乙分配到其中2个小组,A33C32=18种,根据分类计数原理可得,共有36+18=54种,

故答案为:54.

【点评】本题考查了分类计数原理,关键是分类,特殊元素特殊处理,属于中档题.14.某食品的保鲜时间t(单位:小时)与储藏温度x(单位:℃)满足函数关系

且该食品在4℃的保鲜时间是16小时.

已知甲在某日上午10时购买了该食品,并将其遗放在室外,且此日的室外温度随时间变化如图所示.给出以下四个结论:

①该食品在6℃的保鲜时间是8小时;

②当x∈[﹣6,6]时,该食品的保鲜时间t随着x增大而逐渐减少;

③到了此日13时,甲所购买的食品还在保鲜时间内;

④到了此日14时,甲所购买的食品已然过了保鲜时间.

其中,所有正确结论的序号是①④.

【考点】命题的真假判断与应用.

【专题】数形结合;数形结合法;函数的性质及应用;简易逻辑.

【分析】根据食品在4℃的保鲜时间是16小时.求出k值,进而逐一分析四个结论的真假,可得答案.

【解答】解:∵食品的保鲜时间t(单位:小时)与储藏温度x(单位:℃)满足函数关系

且该食品在4℃的保鲜时间是16小时.

∴24k+6=16,即4k+6=4,解得:k=﹣,

∴,

当x=6时,t=8,故①该食品在6℃的保鲜时间是8小时,正确;

②当x∈[﹣6,0]时,保鲜时间恒为64小时,当x∈(0,6]时,该食品的保鲜时间t随看x 增大而逐渐减少,故错误;

③到了此日10时,温度超过8度,此时保鲜时间不超过4小时,故到13时,甲所购买的食品不在保鲜时间内,故错误;

④到了此日14时,甲所购买的食品已然过了保鲜时间,故正确,

故正确的结论的序号为:①④,

故答案为:①④.

【点评】本题以命题的真假判断为载体,考查了函数在实际生活中的应用,难度中档.

三、解答题:本大题共6小题,共80分.解答应写出必要的文字说明、证明过程或演算步骤.

15.已知函数,x∈R.

(Ⅰ)求f(x)的最小正周期和单调递增区间;

(Ⅱ)设α>0,若函数g(x)=f(x+α)为奇函数,求α的最小值.

【考点】三角函数的周期性及其求法;正弦函数的单调性.

【专题】转化思想;综合法;三角函数的图像与性质.

【分析】(Ⅰ)由条件利用三角恒等变换化简函数的解析式,再利用正弦函数的单调性求得函数f(x)的单调递增区间.

(Ⅱ)由题意可得g(0)=0,即,由此求得α的最小正值.

【解答】(Ⅰ)解:

==

=,

所以函数f(x)的最小正周期.

由,k∈Z,

得,

所以函数f(x)的单调递增区间为,k∈Z.

(Ⅱ)解:由题意,得,

因为函数g(x)为奇函数,且x∈R,所以g(0)=0,即,

所以,k∈Z,解得,k∈Z,验证知其符合题意.

又因为α>0,所以α的最小值为.

【点评】本题主要考查三角恒等变换,正弦函数的单调性,正弦函数的奇偶性,属于基础题.

16.甲、乙两人进行射击比赛,各射击4局,每局射击10次,射击命中目标得1分,未命中目标得0分.两人4局的得分情况如下:

(Ⅰ)若从甲的4局比赛中,随机选取2局,求这2局的得分恰好相等的概率;

(Ⅱ)如果x=y=7,从甲、乙两人的4局比赛中随机各选取1局,记这2局的得分和为X,求X的分布列和数学期望;

(Ⅲ)在4局比赛中,若甲、乙两人的平均得分相同,且乙的发挥更稳定,写出x的所有可能取值.(结论不要求证明)

【考点】离散型随机变量及其分布列;离散型随机变量的期望与方差.

【专题】计算题;转化思想;综合法;概率与统计.

【分析】(Ⅰ)从甲的4局比赛中,随机选取2局,基本事件总数n=,这2局的得分恰

好相等基本数件个数m=2,由此能求出从甲的4局比赛中,随机选取2局,且这2局得分恰好相等的概率.

(Ⅱ)X的所有可能取值为13,15,16,18,分别求出相应的概率,由此能求出X的分布列和数学期望.

(Ⅲ)由已知条件能写出x的可能取值为6,7,8.

【解答】(本小题满分13分)

(Ⅰ)解:记“从甲的4局比赛中,随机选取2局,且这2局的得分恰好相等”为事件A,…由题意,得,

所以从甲的4局比赛中,随机选取2局,且这2局得分恰好相等的概率为.…

(Ⅱ)解:由题意,X的所有可能取值为13,15,16,18,…

且,,,,…

所以.…

(Ⅲ)解:x的可能取值为6,7,8.…

【点评】本题考查概率的求法,考查离散型随机变量的分布列和数学期望的求法,是中档题,在历年高考中都是必考题型之一.

17.如图,在四棱锥P﹣ABCD中,底面ABCD是平行四边形,∠BCD=135°,侧面PAB⊥底面ABCD,∠BAP=90°,AB=AC=PA=2,E,F分别为BC,AD的中点,点M在线段PD 上.

(Ⅰ)求证:EF⊥平面PAC;

(Ⅱ)若M为PD的中点,求证:ME∥平面PAB;

(Ⅲ)如果直线ME与平面PBC所成的角和直线ME与平面ABCD所成的角相等,求的值.

【考点】直线与平面所成的角;直线与平面平行的判定;直线与平面垂直的判定.

【专题】空间位置关系与距离;空间角.

【分析】(Ⅰ)证明AB⊥AC.EF⊥AC.推出PA⊥底面ABCD,即可说明PA⊥EF,然后证明EF⊥平面PAC.

(Ⅱ)证明MF∥PA,然后证明MF∥平面PAB,EF∥平面PAB.即可阿门平面MEF∥平面PAB,从而证明ME∥平面PAB.

(Ⅲ)以AB,AC,AP分别为x轴、y轴和z轴,如上图建立空间直角坐标系,求出相关点的坐标,平面ABCD的法向量,平面PBC的法向量,利用直线ME与平面PBC所成的角和此直线与平面ABCD所成的角相等,列出方程求解即可

【解答】(本小题满分14分)

(Ⅰ)证明:在平行四边形ABCD中,因为AB=AC,∠BCD=135°,∠ABC=45°.

所以AB⊥AC.

由E,F分别为BC,AD的中点,得EF∥AB,

所以EF⊥AC.…

因为侧面PAB⊥底面ABCD,且∠BAP=90°,

所以PA⊥底面ABCD.…

又因为EF?底面ABCD,

所以PA⊥EF.…

又因为PA∩AC=A,PA?平面PAC,AC?平面PAC,

所以EF⊥平面PAC.…

(Ⅱ)证明:因为M为PD的中点,F分别为AD的中点,

所以MF∥PA,

又因为MF?平面PAB,PA?平面PAB,

所以MF∥平面PAB.…

同理,得EF∥平面PAB.

又因为MF∩EF=F,MF?平面MEF,EF?平面MEF,

所以平面MEF∥平面PAB.…

又因为ME?平面MEF,

所以ME∥平面PAB.…

(Ⅲ)解:因为PA⊥底面ABCD,AB⊥AC,所以AP,AB,AC两两垂直,故以AB,AC,AP

分别为x轴、y轴和z轴,如上图建立空间直角坐标系,

则A(0,0,0),B(2,0,0),C(0,2,0),P(0,0,2),D(﹣2,2,0),E(1,1,0),

所以,,,…

设,则,

所以M(﹣2λ,2λ,2﹣2λ),,

易得平面ABCD的法向量=(0,0,1).…

设平面PBC的法向量为=(x,y,z),

由,,得

令x=1,得=(1,1,1).…

因为直线ME与平面PBC所成的角和此直线与平面ABCD所成的角相等,

所以,即,…

所以,

解得,或(舍).…

【点评】本题考查直线与平面所成角的求法,直线与平面平行的判定定理以及性质定理的应用,平面与平面平行的判定定理的应用,考查转化思想以及空间想象能力逻辑推理能力的应用.

18.已知函数f(x)=x2﹣1,函数g(x)=2tlnx,其中t≤1.

(Ⅰ)如果函数f(x)与g(x)在x=1处的切线均为l,求切线l的方程及t的值;(Ⅱ)如果曲线y=f(x)与y=g(x)有且仅有一个公共点,求t的取值范围.

【考点】利用导数研究曲线上某点切线方程.

【专题】综合题;分类讨论;分析法;导数的概念及应用.

【分析】(Ⅰ)分别求得f(x),g(x)的导数,求得切线的斜率,解方程可得t=1,即可得到切线的斜率和切点坐标,可得切线的方程;

(Ⅱ)设函数h(x)=f(x)﹣g(x),“曲线y=f(x)与y=g(x)有且仅有一个公共点”等价于“函数y=h(x)有且仅有一个零点”.对h(x)求导,讨论①当t≤0时,②当t=1时,③当0<t<1时,求出单调区间,即可得到零点和所求范围.

【解答】解:(Ⅰ)求导,得f′(x)=2x,,(x>0).

由题意,得切线l的斜率k=f′(1)=g′(1),

即k=2t=2,解得t=1.

又切点坐标为(1,0),

所以切线l的方程为2x﹣y﹣2=0;

(Ⅱ)设函数h(x)=f(x)﹣g(x)=x2﹣1﹣2tlnx,x∈(0,+∞).

“曲线y=f(x)与y=g(x)有且仅有一个公共点”等价于

“函数y=h(x)有且仅有一个零点”.

求导,得.

①当t≤0时,由x∈(0,+∞),得h'(x)>0,

所以h(x)在(0,+∞)单调递增.

又因为h(1)=0,所以y=h(x)有且仅有一个零点1,符合题意.

所以h(x)在(0,1)上单调递减,在(1,+∞)上单调递增,

所以当x=1时,h(x)min=h(1)=0,

故y=h(x)有且仅有一个零点1,符合题意.

③当0<t<1时,令h'(x)=0,解得.

↘↗

所以h(x)在上单调递减,在上单调递增,

所以当时,.

因为h(1)=0,,且h(x)在上单调递增,

所以.

又因为存在,,

所以存在x0∈(0,1)使得h(x0)=0,

所以函数y=h(x)存在两个零点x0,1,与题意不符.

综上,曲线y=f(x)与y=g(x)有且仅有一个公共点时,t的范围是{t|t≤0,或t=1}.

【点评】本题考查导数的运用:求切线的方程和单调区间、极值和最值,考查函数的零点问题的解法,注意运用构造法,通过导数求得单调性,同时考查分类讨论的思想方法,属于中档题.

19.已知椭圆C:的离心率为,点在椭圆C上.

(Ⅰ)求椭圆C的方程;

(Ⅱ)设动直线l与椭圆C有且仅有一个公共点,判断是否存在以原点O为圆心的圆,满足此圆与l相交两点P1,P2(两点均不在坐标轴上),且使得直线OP1,OP2的斜率之积为定值?若存在,求此圆的方程;若不存在,说明理由.

【考点】圆锥曲线的定值问题;椭圆的标准方程.

【专题】圆锥曲线的定义、性质与方程;圆锥曲线中的最值与范围问题.

【分析】(Ⅰ)利用离心率列出方程,通过点在椭圆上列出方程,求出a,b然后求出椭圆的方程.

(Ⅱ)当直线l的斜率不存在时,验证直线OP1,OP2的斜率之积.

当直线l的斜率存在时,设l的方程为y=kx+m与椭圆联立,利用直线l与椭圆C有且只有一个公共点,推出m2=4k2+1,通过直线与圆的方程的方程组,设P1(x1,y1),P2(x2,y2),结合韦达定理,求解直线的斜率乘积,推出k1?k2为定值即可.

【解答】(本小题满分14分)

(Ⅰ)解:由题意,得,a2=b2+c2,…

又因为点在椭圆C上,

所以,…

解得a=2,b=1,,

所以椭圆C的方程为.…

(Ⅱ)结论:存在符合条件的圆,且此圆的方程为x2+y2=5.…

证明如下:

假设存在符合条件的圆,并设此圆的方程为x2+y2=r2(r>0).

当直线l的斜率存在时,设l的方程为y=kx+m.…

由方程组得(4k2+1)x2+8kmx+4m2﹣4=0,…

因为直线l与椭圆C有且仅有一个公共点,

所以,即m2=4k2+1.…

由方程组得(k2+1)x2+2kmx+m2﹣r2=0,…

则.

设P1(x1,y1),P2(x2,y2),则,,…

设直线OP1,OP2的斜率分别为k1,k2,

所以

=,…

将m2=4k2+1代入上式,得.

要使得k1k2为定值,则,即r2=5,验证符合题意.

所以当圆的方程为x2+y2=5时,圆与l的交点P1,P2满足k1k2为定值.…

当直线l的斜率不存在时,由题意知l的方程为x=±2,

此时,圆x2+y2=5与l的交点P1,P2也满足.

综上,当圆的方程为x2+y2=5时,圆与l的交点P1,P2满足斜率之积k1k2为定值.…

【点评】本题考查椭圆的标准方程的求法,直线与椭圆的位置关系的综合应用,考查转化思想以及计算能力.

20.在数字1,2,…,n(n≥2)的任意一个排列A:a1,a2,…,a n中,如果对于i,j∈N*,i<j,有a i>a j,那么就称(a i,a j)为一个逆序对.记排列A中逆序对的个数为S(A).

如n=4时,在排列B:3,2,4,1中,逆序对有(3,2),(3,1),(2,1),(4,1),则S (B)=4.

(Ⅰ)设排列C:3,5,6,4,1,2,写出S(C)的值;

(Ⅱ)对于数字1,2,…,n的一切排列A,求所有S(A)的算术平均值;

(Ⅲ)如果把排列A:a1,a2,…,a n中两个数字a i,a j(i<j)交换位置,而其余数字的位置保持不变,那么就得到一个新的排列A':b1,b2,…,b n,求证:S(A)+S(A')为奇数.【考点】数列与函数的综合.

【专题】新定义;分类讨论;分析法;排列组合.

【分析】(Ⅰ)由逆序对的定义,列举即可得到所求值为10;

(Ⅱ)考察排列D:d1,d2,…,d n﹣1,d n,运用组合数可得排列D中数对(d i,d j)共有个,即可得到所有S(A)的算术平均值;

(Ⅲ)讨论(1)当j=i+1,即a i,a j相邻时,(2)当j≠i+1,即a i,a j不相邻时,由新定义,运用调整法,可得S(A)+S(A')为奇数.

【解答】解:(Ⅰ)逆序对有(3,1),(3,2),(5,4),(5,1),(5,2),(4,1),(4,2),(6,4),(6,1),(6,2)则S(C)=10;

(Ⅱ)考察排列D:d1,d2,…,d n﹣1,d n与排列D1:d n,d n﹣1,…,d2,d1,

(完整版)2020年北京市海淀区高三期末英语试卷(含答案)_202001102004361

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2018年北京市海淀区高三(上)期末英语试题(附答案)

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2020北京海淀高三(上)期末英语试卷和答案 第一部分:听力理解(共三节,30 分) 第一节(共5 小题;每小题1.5 分,共7.5 分) 听下面 5 段对话,每段对话有一道小题,从每题所给的A、B、C 三个选项中选出最佳选项,听完每段对话后,你将有10 秒钟的时间来回答有关小题和阅读下一小题。每段对话你将听一遍。 1.How will the speakers go to London? A. By air. B. By ship. C. By coach. 2.What is the woman’s brother? A. A project manager. B. A wildlife photographer. C. A government official. 3.Where will the man go? A. To the butcher’s. B. To the baker’s. C. To the grocer’s. 4. What does the woman suggest the man do? A. Go to bed. B. Watch a match. C. See the dentist. 5. How much will the man pay? A.$10. B.$18. C.$20. 第二节(共10 小题;每小题 1.5 分,共15 分) 听下面4段对话或独白。每段对话或独白后有几道小题,从每题所给的A、B、C三个选项中选出最佳选项。听每段对话或独白前,你将有5秒钟的时间阅读每小题。听完后,每小题将给出5秒钟的作答时间。每段对话或独白你将听两遍。听第6段材料,回答第6至7题。 6. Why does the woman call? A. To make a reservation. B. To fill in a position. C. To talk to her friend. 7. What is the man doing? A. Ask for help. B. Arguing with a friend. C. Talking about an employee. 听第7段材料,回答第8至9题。 8. Who is the man talking? A. Tourists. B. Cleaners. C. V olunteers. 9. Where will the signs be set up? A. Around the lake. B. In the picnic areas. C. Along the jogging paths. 听第8 段材料,回答第10 至12 题。 10. Why was the man stopped by police? A. He followed a vehicle too close. B. He cut in between two vehicles. C. He caused a really bad accident. 11. What does a ghost car mean? A. A car secretly following others. B. A police car that is unmarked. C. A car whose owner gets mad. 12. What did the man get in the end? A. A warning. B. A ticket. C. A fine. 听第9段材料,回答第13至15题。 13. What are the speakers talking about? A. Intelligence changes. B. Climate differences. C. Creativity analysis. 14. In which season are people probably the smartest?

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