湖南省湖南师大附中、长沙市一中等六校2014届高三下学期4月联考试题数学(理) 扫描版含答案

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2014届高三年级师大附中联考试卷分析(第四次)

13 推理与证明
14 （理）函数，不等式、（文）立体几何
填空题（共20分）
15
（理）圆锥曲线，椭圆的离心率、（文）三角函数
16 （理）函数、（文）导数
17
（理）等差数列，数列求和、（文）数列，数列求和
（理）直线与平面垂直、二面角的余弦值、
18 直线与平面平行、（文）棱锥：平面与平面
垂直、棱锥的体积
教学中可以从以下几个方面来改进： 1、立足于教材。 2、教学中凸现学生的主观能动性，教会学生思考生的计算能力。
选择题（共60分）
8
（理）平面向量，数列、（文）平面向量，等差数列的性质
9
（理）立体几何，空间直角坐标系、（文）不等式，指数函数和对数函数的性质
10
（理）函数的周期性，三角函数的周期、（文）三角函数的性质
11
（理）圆锥曲线：直线与双曲线的位置关系、（文）立体几何，空间坐标系
12
（理）立体几何，三视图、（文）导数，线性规划
数形结合
计算能力弱
不会数形结合、计算能力弱
不会数形结合公式记忆不住不会分析，不会数形结合计算能力弱公式记忆不住，计算容易出错计算能力及公式的记忆
计算能力弱
忘记知识点，空间向量坐标系的建立出错
分析解决问题能力弱
知识点记忆不牢靠知识点记忆不牢靠
文科班
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题型的设置及计算量上，能够看出本次考试与前两次型设置上与高考相比仍有一定差距。存在的问题主要抽象能力差（如第12小题）。数学概念学习是数学学运用概念解决数学问题。在这方面教师在教学中已经不理想。（2）运算能力弱。（如第2小题）运算能力运算不过关，因而不能有效地对试题作答，这样的情正确理解造成失误。（如第1小题）（4）数学理性思思维能力考查的要求越来越高，特别是试题对理性思维方法考查为重点。答卷中暴露出学生在这一方面存在

14年第四次月考.doc

长沙市六中2014届高三第四次月考数学试题文科数学总分150分时量120分钟一、选择题（本大题共9小题，每小题5分，共45分.在每小题给出的四个选项中，只有一项是符合题目要求的）.1.复数1iz i=+(i 为虚数单位)在复平面上对应的点位于( ) A.第一象限 B.第二象限 C.第三象限 D.第四象限2.已知集合{}21|()1,|6802xA xB x x x ⎧⎫=≤=-+≤⎨⎬⎩⎭,则“x A ∈”是“x B ∈”的（）A 充分而不必要条件B 必要而不充分条件C 充要条件D 既不充分也不必要条件3. 已知某几何体的三视图（单位：dm ）如图所示，则该几何体的体积（dm 3）是（）A ．3464π+ B π+1664 C ．π+464 D π+4324.曲线sin x xy e=在0x =处的切线斜率为( ) A 0 B 1 C12D 1- 5．要得到函数cos(2)3y x π=-的图像,只需将函数sin 2y x =的图像( )A 向左平移12π个单位长度 B 向右平移12π个单位长度 C 向左平移6π个单位长度 D 向右平移6π个单位长度（第3题图）侧视图俯视图正视图6．已知点(,)P x y 是不等式组02323x x y x y ≥⎧⎪+≥⎨⎪+≤⎩表示的平面区域内一动点，且(1,1)A -,O 为坐标原点,则1t OA OP =∙+的最小值是（）A ．2-B ．3-C ．12-D ．1 7. 在ABC ∆中,角 A B C 、、所对的边长分别为a b c 、、，若sin cos a B C +1sin cos 2c B A b =,且a b >,则B ∠=( ) A6π B 3πC 23πD 56π8. 数列{}n a 中,1(1)(43)n n a n +=--,其前n 项和为n S ,则2211S S -=( )A 85-B 85C 65-D 659．已知函数1)(2--=bx ax x f ，其中] 2 , 0 (∈a ，] 2 , 0 (∈b ，在其取值范围内任取实数a 、b ，则函数)(x f 在区间) , 1 [∞+上为增函数的概率为（） A ．21 B ．31 C ．32 D ．43二、填空题（本大题共6小题，每小题5分，共30分，把答案填在题中横线上）10．设全集U=R ,A={x ∈N |110x ≤≤},B={x ∈R |260x x +-=},则右图中阴影部分表示的集合为11．已知等差数列{}n a 的前n 项和为n S ，若32,4,a 成等比数列，则5S =12．已知向量(2,4)a =,(1,1)b =,若向量()b a b λ⊥+,则实数λ=12.如右图若某算法框图如图所示，则输出的结果为；14．若正数,x y 满足3x y xy +=，则34x y +的最小值为15. 已知函数()|2|f x x x m =--仅有3个零点分别为123,,x x x ,则(1) m 的取值范围是 ; (2)123x x x ++ 的取值范围是。

湖南省长沙市湖南师范大学附属中学2024-2025学年高三上学期月考(一)数学试题及答案

大联考湖南师大附中2025届高三月考试卷（一）数学命题人：高三数学备课组审题人：高三数学备课组时量：120分钟满分：150分一､选选选：本选共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的，1. 已知{}()260,{lg 10}Axx x B x x =+−≤=−<∣∣，则A B = （）A. {}32xx −≤≤∣ B. {32}x x −≤<∣ C. {12}x x <≤∣D. {12}x x <<∣2. 若复数z 满足()1i 3i z +=−+（i 是虚数单位），则z 等于（）A.B.54C.D.3. 已知平面向量()()5,0,2,1ab ==−，则向量a b +在向量b上投影向量为（）A. ()6,3−B. ()4,2−C. ()2,1−D. ()5,04. 记n S 为等差数列{}n a 的前n 项和，若396714,63a a a a +==，则7S =（） A. 21B. 19C. 12D. 425. 某校高二年级下学期期末考试数学试卷满分为150分，90分以上（含90分）为及格．阅卷结果显示，全年级1200名学生的数学成绩近似服从正态分布，试卷的难度系数（难度系数=平均分/满分）为0.49，标准差为22，则该次数学考试及格的人数约为（）附：若()2,X Nµσ∼，记()()p k P k X k µσµσ=−≤≤+，则()()0.750.547,10.683p p ≈≈．A 136人 B. 272人C. 328人D. 820人6. 已知()π5,0,,cos ,tan tan 426αβαβαβ∈−=⋅=，则αβ+=（） A.π6 B.π4C.π3D.2π37. 已知12,F F 是双曲线22221(0)x y a b a b−=>>的左､右焦点，以2F 为圆心，a 为半径的圆与双曲线的一条的.渐近线交于,A B 两点，若123AB F F >，则双曲线的离心率的取值范围是（）A.B.C. (D. (8. 已知函数()220log 0x a x f x x x ⋅≤= > ，，，，若关于x 的方程()()0f f x =有且仅有两个实数根，则实数a 的取值范围是（） A. ()0,1B. ()(),00,1−∞∪C. [)1,+∞D. ()()0,11,+∞二､多选题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分9. 如图，在正方体111ABCD A B C D −中，E F M N ，，，分别为棱111AA A D AB DC ，，，的中点，点P 是面1B C 的中心，则下列结论正确的是（）A. E F M P ，，，四点共面B. 平面PEF 被正方体截得的截面是等腰梯形C. //EF 平面PMND. 平面MEF ⊥平面PMN10. 已知函数()5π24f x x=+，则（）A. ()f x 的一个对称中心为3π,08B. ()f x 的图象向右平移3π8个单位长度后得到的是奇函数的图象 C. ()f x 在区间5π7π,88上单调递增 D. 若()y f x =在区间()0,m 上与1y =有且只有6个交点，则5π13π,24m∈11. 已知定义在R 上的偶函数()f x 和奇函数()g x 满足()()21f x g x ++−=，则（）A. ()f x 的图象关于点()2,1对称B. ()f x 是以8为周期的周期函数C. ()20240g =D.20241(42)2025k f k =−=∑ 三､填空题：本题共3小题，每小题5分，共15分.12. 6(31)x y +−的展开式中2x y 的系数为______．13. 已知函数()f x 是定义域为R 的奇函数，当0x >时，()()2f x f x ′−>，且()10f =，则不等式()0f x >的解集为__________.14. 已知点C 为扇形AOB 弧AB 上任意一点，且60AOB ∠=，若(),R OC OA OB λµλµ=+∈，则λµ+的取值范围是__________.四､解答题：本题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤.15. ABC V 的内角,,A B C 的对边分别为,,a b c ，已知22cos a b c B +=．（1）求角C ；（2）若角C 的平分线CD 交AB于点,D AD DB =，求CD 的长．16. 已知1ex =为函数()ln af x x x =的极值点．（1）求a 的值；（2）设函数()ex kxg x =，若对()120,,x x ∀∈+∞∃∈R ，使得()()120f x g x −≥，求k 的取值范围． 17. 已知四棱锥P ABCD −中，平面PAB ⊥底面,ABCD AD∥,,,2,BC AB BC PA PB AB AB BC AD E ⊥==为AB 的中点，F 为棱PC 上异于,P C 的点．的（1）证明：BD EF ⊥；（2）试确定点F 的位置，使EF 与平面PCD18. 在平面直角坐标系xOy 中，抛物线21:2(0)C ypx p =>的焦点到准线的距离等于椭圆222:161C x y +=的短轴长，点P 在抛物线1C 上，圆222:(2)E x y r −+=（其中01r <<）.（1）若1,2r Q =为圆E 上的动点，求线段PQ 长度的最小值；（2）设()1,D t 是抛物线1C 上位于第一象限的一点，过D 作圆E 的两条切线，分别交抛物线1C 于点,M N .证明：直线MN 经过定点.19. 龙泉游泳馆为给顾客更好的体验，推出了A 和B 两个套餐服务，顾客可选择A 和B 两个套餐之一，并在App 平台上推出了优惠券活动，下表是该游泳馆在App 平台10天销售优惠券情况．日期t 12345678910销售量千张 1.9 1.98 2.2 2.36 2.43 2.59 2.68 2.76 2.7 04经计算可得：10101021111 2.2,118.73,38510i i i i i i i y y t y t ======∑∑∑ （1）因为优惠券购买火爆，App 平台在第10天时系统出现异常，导致当天顾客购买优惠券数量大幅减少，已知销售量y 和日期t 呈线性关系，现剔除第10天数据，求y 关于t 的经验回归方程结果中的数值用分数表示；（2）若购买优惠券的顾客选择A 套餐的概率为14，选择B 套餐的概率为34，并且A 套餐可以用一张优惠券，B 套餐可以用两张优惠券，记App 平台累计销售优惠券为n 张的概率为n P ，求n P ；（3）记（2）中所得概率n P 的值构成数列{}()N n P n ∗∈.①求n P 的最值；②数列收敛的定义：已知数列{}n a ，若对于任意给定的正数ε，总存在正整数0N ，使得当0n N >时，n a a ε−<，（a 是一个确定的实数），则称数列{}n a 收敛于a .根据数列收敛的定义证明数列{}n P 收敛．..参考公式： ()()()1122211ˆˆ,n ni ii ii i n n i i i i x x y y x y nx yay bx x xx nx====−−−==−−−∑∑∑∑.大联考湖南师大附中2025届高三月考试卷（一）数学命题人：高三数学备课组审题人：高三数学备课组时量：120分钟满分：150分一､选选选：本选共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的，1. 已知{}()260,{lg 10}Axx x B x x =+−≤=−<∣∣，则A B = （）A. {}32xx −≤≤∣ B. {32}x x −≤<∣ C. {12}x x <≤∣ D. {12}x x <<∣【答案】D 【解析】【分析】通过解一元二次不等式和对数函数的定义域，求出集合,A B ，再求交集．【详解】集合{}()32,{lg 10}{12}A x x B x x x x =−≤≤=−<=<<∣∣∣，则{12}A B xx ∩=<<∣，故选：D ．2. 若复数z 满足()1i 3i z +=−+（i 是虚数单位），则z 等于（）A.B.54C.D.【答案】C 【解析】【分析】由复数的除法运算计算可得12i z =−+，再由模长公式即可得出结果. 【详解】依题意()1i 3i z +=−+可得()()()()3i 1i 3i 24i12i 1i 1i 1i 2z −+−−+−+====−+++−，所以z =. 故选：C3. 已知平面向量()()5,0,2,1a b ==−，则向量a b +在向量b上的投影向量为（）A. ()6,3−B. ()4,2−C. ()2,1−D. ()5,0【答案】A 【解析】【分析】根据投影向量的计算公式即可求解.【详解】()()7,1,15,a b a b b b +=−+⋅==所以向量a b +在向量b 上的投影向量为()()236,3||a b b b bb +⋅==− .故选：A4. 记n S 为等差数列{}n a 的前n 项和，若396714,63a a a a +==，则7S =（） A. 21 B. 19C. 12D. 42【答案】A 【解析】【分析】根据等差数列的性质，即可求解公差和首项，进而由求和公式求解.【详解】{}n a 是等差数列，396214a a a ∴+==，即67a =，所以67769,a a a a == 故公差76162,53d a a a a d =−=∴=−=−，()767732212S ×∴=×−+×=，故选：A5. 某校高二年级下学期期末考试数学试卷满分为150分，90分以上（含90分）为及格．阅卷结果显示，全年级1200名学生的数学成绩近似服从正态分布，试卷的难度系数（难度系数=平均分/满分）为0.49，标准差为22，则该次数学考试及格的人数约为（）附：若()2,X Nµσ∼，记()()p k P k X k µσµσ=−≤≤+，则()()0.750.547,10.683p p ≈≈．A. 136人B. 272人C. 328人D. 820人【答案】B 【解析】【分析】首先求出平均数，即可得到学生的数学成绩2~(73.5,22)X N ，再根据所给条件求出(5790)P X ≤≤，即可求出(90)P X ≥，即可估计人数．【详解】由题得0.4915073.5,22µσ=×==，()()(),0.750.547p k P k X k p µσµσ=−≤≤+≈ ，()5790P X ∴≤≤ ()0.750.547p ≈，()()900.510.5470.2265P X ≥×−，∴该校及格人数为0.22651200272×≈（人），故选：B ． 6. 已知()π5,0,,cos ,tan tan 426αβαβαβ∈−=⋅=，则αβ+=（） A.π6 B.π4C.π3D.2π3【答案】D 【解析】【分析】利用两角差的余弦定理和同角三角函数的基本关系建立等式求解，再由两角和的余弦公式求解即可．【详解】由已知可得5cos cos sin sin 6sin sin 4cos cos αβαβαβαβ⋅+⋅=⋅ =⋅ ，解得1cos cos 62sin sin 3αβαβ⋅=⋅=，，()1cos cos cos sin sin 2αβαβαβ∴+=⋅−⋅=−，π,0,2αβ∈，()0,παβ∴+∈， 2π,3αβ∴+=，故选：D ．7. 已知12,F F 是双曲线22221(0)x y a b a b−=>>的左､右焦点，以2F 为圆心，a 为半径的圆与双曲线的一条渐近线交于,A B 两点，若123AB F F >，则双曲线的离心率的取值范围是（）A.B.C. (D. (【答案】B 【解析】【分析】根据双曲线以及圆的方程可求得弦长AB =，再根据不等式123AB F F >整理可得2259c a <，即可求得双曲线的离心率的取值范围.【详解】设以()2,0F c 为圆心，a 为半径的圆与双曲线的一条渐近线0bx ay −=交于,A B 两点，则2F 到渐近线0bx ay −=的距离d b，所以AB =，因为123AB F F >，所以32c ×>，可得2222299a b c a b −>=+，即22224555a b c a >=−，可得2259c a <，所以2295c a <，所以e <，又1e >，所以双曲线的离心率的取值范围是 .故选：B8. 已知函数()220log 0x a x f x x x ⋅≤= > ，，，，若关于x 的方程()()0f f x =有且仅有两个实数根，则实数a 的取值范围是（） A. ()0,1 B. ()(),00,1−∞∪C. [)1,+∞D. ()()0,11,+∞【答案】C 【解析】【分析】利用换元法设()u f x =，则方程等价为()0f u =，根据指数函数和对数函数图象和性质求出1u =，利用数形结合进行求解即可．【详解】令()u f x =，则()0f u =．�当0a =时，若()0,0u f u ≤=；若0u >，由()2log 0f u u==，得1u =．所以由()()0ff x =可得()0f x ≤或()1f x =．如图所示，满足()0f x ≤的x 有无数个，方程()1f x =只有一个解，不满足题意；�当0a ≠时，若0≤u ，则()20uf u a =⋅≠；若0u >，由()2log 0f u u==，得1u =．所以由()()0ff x =可得()1f x =，当0x >时，由()2log 1f x x==，可得2x =，因为关于x 的方程()()0f f x =有且仅有两个实数根，则方程()1f x =在(,0∞−]上有且仅有一个实数根，若0a >且()(]0,20,xx f x a a ≤=⋅∈，故1a ≥；若0a <且()0,20xx f x a ≤=⋅<，不满足题意．综上所述，实数a 的取值范围是[)1,+∞，故选：C ．二､多选题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分9. 如图，在正方体111ABCD A B C D −中，E F M N ，，，分别为棱111AA A D AB DC ，，，的中点，点P 是面1B C 的中心，则下列结论正确的是（）A. E F M P ，，，四点共面B. 平面PEF 被正方体截得的截面是等腰梯形C. //EF 平面PMND. 平面MEF ⊥平面PMN【答案】BD 【解析】【分析】可得过,,E F M 三点的平面为一个正六边形，判断A ；分别连接,E F 和1,B C ，截面1C BEF 是等腰梯形，判断B ；分别取11,BB CC 的中点,G Q ，易证EF 显然不平行平面QGMN ，可判断C ；EM ⊥平面PMN ，可判断D.【详解】对于A ：如图经过,,E F M 三点的平面为一个正六边形EFMHQK ，点P 在平面外，,,,E F M P ∴四点不共面，∴选项A 错误；对于B ：分别连接,E F 和1,B C ，则平面PEF 即平面1C BEF ，截面1C BEF 是等腰梯形，∴选项B 正确；对于C ：分别取11,BB CC 的中点,G Q ，则平面PMN 即为平面QGMN ，由正六边形EFMHQK ，可知HQ EF ，所以MQ 不平行于EF ，又,EF MQ ⊂平面EFMHQK ，所以EF MQ W = ，所以EF I 平面QGMN W =，所以EF 不平行于平面PMN ，故选项C 错误；对于D ：因为,AEM BMG 是等腰三角形，45AME BMG ∴∠=∠=°， 90EMG ∴∠=°，EMMG ∴⊥，,M N 是,AB CD 的中点，易证MN AD ∥，由正方体可得AD ⊥平面11ABB A ，MN ∴⊥平面11ABB A ，又ME ⊂平面11ABB A ，EM MN ∴⊥，,MG MN ⊂ 平面PMN ，EM ∴⊥平面GMN ，EM ⊂ 平面MEF ，∴平面MEF ⊥平面,PMN 故选项D 正确．��BD ．10. 已知函数()5π24f x x=+，则（）A. ()f x 的一个对称中心为3π,08B. ()f x 的图象向右平移3π8个单位长度后得到的是奇函数的图象 C. ()f x 在区间5π7π,88上单调递增 D. 若()y f x =在区间()0,m 上与1y =有且只有6个交点，则5π13π,24m∈【答案】BD 【解析】【分析】代入即可验证A ，根据平移可得函数图象，即可由正弦型函数的奇偶性求解B ，利用整体法即可判断C ，由5πcos 24x+求解所以根，即可求解D.【详解】对于A ，由35π3π2π0848f =+×=≠，故A 错误；对于B ，()f x 的图象向右平移3π8个单位长度后得： 3π3π5ππ228842y f x x x x=−−++，为奇函数，故B 正确；对于C ，当5π7π,88x∈时，则5π5π2,3π42x +∈ ，由余弦函数单调性知，()f x 在区间5π7π,88 上单调递减，故C 错误；对于D ，由()1f x =，得5πcos 24x+ππ4x k =+或ππ,2k k +∈Z ， ()y f x =在区间()0,m 上与1y =有且只有6个交点，其横坐标从小到大依次为：ππ5π3π9π5π,,,,,424242，而第7个交点的横坐标为13π4， 5π13π24m ∴<≤，故D 正确. 故选：BD11. 已知定义在R 上的偶函数()f x 和奇函数()g x 满足()()21f x g x ++−=，则（）A. ()f x 的图象关于点()2,1对称B. ()f x 是以8为周期的周期函数C. ()20240g =D.20241(42)2025k f k =−=∑ 【答案】ABC 【解析】【分析】根据函数奇偶性以及所满足的表达式构造方程组可得()()222f x f x ++−=，即可判断A 正确；利用对称中心表达式进行化简计算可得B 正确，可判断()g x 也是以8为周期的周期函数，即C 正确；根据周期性以及()()42f x f x ++=计算可得20241(42)2024k f k =−=∑，可得D 错误. 【详解】由题意()()()(),f x f x g x g x −=−=−，且()()()00,21g f x g x =++−=，即()()21f x g x +−=①，用x −替换()()21f x g x ++−=中的x ，得()()21f x g x −+=②，由①+②得()()222f x f x ++−=，所以()f x 的图象关于点(2,1)对称，且()21f =，故A 正确；由()()222f x f x ++−=，可得()()()()()42,422f x f x f x f x f x ++−=+=−−=−，所以()()()()82422f x f x f x f x +=−+=−−= ，所以()f x 是以8为周期的周期函数，故B 正确；由①知()()21g x f x =+−，则()()()()882121g x f x f x g x +=++−=+−=，故()()8g x g x +=，因此()g x 也是以8为周期的周期函数，所以()()202400g g ==，C 正确；又因为()()42f x f x ++−=，所以()()42f x f x ++=，令2x =，则有()()262f f +=，令10x =，则有()()10142,f f +=…，令8090x =，则有()()809080942f f +=，所以1012(2)(6)(10)(14)(8090)(8094)2222024f f f f f f ++++++=+++=个所以20241(42)(2)(6)(10)(14)(8090)(8094)2024k f k f f f f f f =−=++++++=∑ ，故D 错误.故选：ABC【点睛】方法点睛：求解函数奇偶性、对称性、周期性等函数性质综合问题时，经常利用其中两个性质推得第三个性质特征，再进行相关计算.三､填空题：本题共3小题，每小题5分，共15分.12. 6(31)x y +−的展开式中2x y 的系数为______．【答案】180− 【解析】【分析】根据题意，由条件可得展开式中2x y 的系数为213643C C (1)⋅−，化简即可得到结果．【详解】在6(31)x y +−的展开式中，由()2213264C C 3(1)180x y x y ⋅⋅−=−，得2x y 的系数为180−. 故答案为：180−．13. 已知函数()f x 是定义域为R 的奇函数，当0x >时，()()2f x f x ′−>，且()10f =，则不等式()0f x >的解集为__________.【答案】()()1,01,−∪+∞ 【解析】【分析】根据函数奇偶性并求导可得()()f x f x ′′−=，因此可得()()2f x f x ′>，可构造函数()()2xf x h x =e并求得其单调性即可得()f x 在()1,+∞上大于零，在()0,1上小于零，即可得出结论. 【详解】因为()f x 为奇函数，定义域为R ，所以()()f x f x −=−，两边同时求导可得()()f x f x ′′−−=−，即()()f x f x ′′−=且()00f =，又因为当0x >时，()()2f x f x ′−>，所以()()2f x f x ′>. 构造函数()()2xf x h x =e，则()()()22x f x f x h x ′−′=e ，所以当0x >时，()()0,h x h x ′>在()0,∞+上单调递增，又因为()10f =，所以()()10,h h x =在()1,+∞上大于零，在()0,1上小于零，又因为2e 0x >，所以()f x 在()1,+∞上大于零，在()0,1上小于零，因为()f x 为奇函数，所以()f x 在(),1∞−−上小于零，在()1,0−上大于零，综上所述，()0f x >的解集为()()1,01,−∪+∞. 故答案为：()()1,01,−∪+∞14. 已知点C 为扇形AOB 的弧AB 上任意一点，且60AOB ∠=，若(),R OC OA OB λµλµ=+∈，则λµ+的取值范围是__________.【答案】【解析】【分析】建系设点的坐标，再结合向量关系表示λµ+，最后应用三角恒等变换及三角函数值域求范围即可. 【详解】方法一：设圆O 的半径为1，由已知可设OB 为x 轴的正半轴，O 为坐标原点，过O 点作x 轴垂线为y 轴建立直角坐标系，其中()()1,1,0,cos ,sin 2A B C θθ ，其中π,0,3BOC θθ ∠=∈ ，由(),R OC OA OB λµλµ=+∈，即()()1cos ,sin 1,02θθλµ =+，整理得1cos sin 2λµθθ+=，解得cos λµθ=，则ππcos cos ,0,33λµθθθθθ+=++=+∈，ππ2ππ,,sin 3333θθ+∈+∈所以λµ +∈ . 方法二：设k λµ+=，如图，当C 位于点A 或点B 时，,,A B C 三点共线，所以1k λµ=+=；当点C 运动到AB的中点时，k λµ=+，所以λµ +∈故答案为：四､解答题：本题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤.15. ABC V 的内角,,A B C 的对边分别为,,a b c ，已知22cos a b c B +=．（1）求角C ；（2）若角C 的平分线CD 交AB于点,D AD DB =，求CD 的长．【答案】（1）2π3C = （2）3CD = 【解析】【分析】（1）利用正弦定理及两角和的正弦定理整理得到()2cos 1sin 0C B +=，再利用三角形的内角及正弦函数的性质即可求解；（2）利用正弦定理得出3b a =，再由余弦定理求出4a =，12b =，再根据三角形的面积建立等式求解．【小问1详解】由22cos a b c B +=，根据正弦定理可得2sin sin 2sin cos A B C B +=，则()2sin sin 2sin cos B C B C B ++=，所以2sin cos 2cos sin sin 2sin cos B C B C B C B ++=，整理得()2cos 1sin 0C B +=，因为,B C 均为三角形内角，所以(),0,π,sin 0B C B ∈≠，因此1cos 2C =−，所以2π3C =．【小问2详解】因为CD 是角C的平分线，AD DB=所以在ACD 和BCD △中，由正弦定理可得，,ππsin sin sin sin 33AD CD BD CDA B ==，因此sin 3sin BADA BD==，即sin 3sin B A =，所以3b a =，又由余弦定理可得2222cos c a b ab C =+−，即222293a a a =++，解得4a =，所以12b =．又ABCACD BCD S S S =+△△△，即111sin sin sin 222ab ACB b CD ACD a CD BCD ∠∠∠=⋅⋅+⋅⋅，即4816CD =，所以3CD =． 16. 已知1ex =为函数()ln af x x x =的极值点．（1）求a 的值；（2）设函数()ex kxg x =，若对()120,,x x ∀∈+∞∃∈R ，使得()()120f x g x −≥，求k 的取值范围．【答案】（1）1a = （2）(]()10,−∞−+∞ , 【解析】【分析】（1）直接根据极值点求出a 的值；（2）先由（1）求出()f x 的最小值，由题意可得是求()g x 的最小值，小于等于()f x 的最小值，对()g x 求导，判断由最小值时的k 的范围，再求出最小值与()f x 最小值的关系式，进而求出k 的范围．【小问1详解】()()111ln ln 1a a f x ax x x x a x xα−−==′+⋅+，由1111ln 10e e e a f a −=+=′，得1a =，当1a =时，()ln 1f x x =′+，函数()f x 在10,e上单调递减，在1,e∞ +上单调递增，所以1ex =为函数()ln af x x x =的极小值点，所以1a =．【小问2详解】由（1）知min 11()e ef x f ==−．函数()g x 的导函数()()1e xg x k x −=−′ �若0k >，对()1210,,x x k ∞∀∈+∃=−，使得()()12111e 1e k g x g f x k=−=−<−<−≤，即()()120f x g x −≥，符合题意． �若()0,0kg x =，取11ex =，对2x ∀∈R ，有()()120f x g x −<，不符合题意．�若0k <，当1x <时，()()0,g x g x ′<在(),1∞−上单调递减；当1x >时，()()0,g x g x ′>在(1,+∞)上单调递增，所以()min ()1ekg x g ==，若对()120,,x x ∞∀∈+∃∈R ，使得()()120f x g x −≥，只需min min ()()g x f x ≤，即1e ek ≤−，解得1k ≤−．综上所述，k 的取值范围为(](),10,∞∞−−∪+．17. 已知四棱锥P ABCD −中，平面PAB ⊥底面,ABCD AD ∥,,,2,BC AB BC PA PB AB AB BC AD E ⊥==为AB 的中点，F 为棱PC 上异于,P C 的点．（1）证明：BD EF ⊥；（2）试确定点F 的位置，使EF 与平面PCD【答案】（1）证明见解析（2）F 位于棱PC 靠近P 的三等分点【解析】【分析】（1）连接,,PE EC EC 交BD 于点G ，利用面面垂直的性质定理和三角形全等，即可得证；（2）取DC 的中点H ，以E 为坐标原点，分别以,,EB EH EP 所在直线为,,x y z 轴建立，利用线面角公式代入即可求解．小问1详解】如图，连接,,PE EC EC 交BD 于点G ．因为E 为AB 的中点，PA PB =，所以PE AB ⊥．因为平面PAB ⊥平面ABCD ，平面PAB ∩平面,ABCD AB PE =⊂平面PAB ，所以PE ⊥平面ABCD ，因为BD ⊂平面ABCD ，所以PE BD ⊥．因为ABD BCE ≅ ，所以CEB BDA ∠∠=，所以90CEB ABD ∠∠+= ，所以BD EC ⊥，因为,,PE EC E PE EC ∩=⊂平面PEC ，所以BD ⊥平面PEC ．因为EF ⊂平面PEC ，所以BD EF ⊥．【小问2详解】如图，取DC 的中点H ，以E 为坐标原点，分别以,,EB EH EP 所在直线为,,x y z 轴建立空间直角坐标系，【设2AB =，则2,1,BC AD PA PB ====则()()()()0,0,1,1,2,0,1,1,0,0,0,0P C D E −，设(),,,(01)F x y z PF PC λλ=<<，所以()(),,11,2,1x y z λ−=−，所以,2,1x y z λλλ===−，即(),2,1F λλλ−．则()()()2,1,0,1,2,1,,2,1DC PC EF λλλ==−=−，设平面PCD 的法向量为(),,m a b c =，则00DC m PC m ⋅=⋅=，，即2020a b a b c += +−= ，，取()1,2,3m =−− ，设EF 与平面PCD 所成的角为θ，由cos θ=sin θ=．所以sin cos ,m EF m EF m EF θ⋅===整理得2620λλ−=，因为01λ<<，所以13λ=，即13PF PC = ，故当F 位于棱PC 靠近P 的三等分点时，EF 与平面PCD18. 在平面直角坐标系xOy 中，抛物线21:2(0)C ypx p =>的焦点到准线的距离等于椭圆222:161C x y +=的短轴长，点P 在抛物线1C 上，圆222:(2)E x y r −+=（其中01r <<）.（1）若1,2r Q =为圆E 上的动点，求线段PQ 长度的最小值；（2）设()1,D t 是抛物线1C 上位于第一象限的一点，过D 作圆E 的两条切线，分别交抛物线1C 于点,M N .证明：直线MN 经过定点.【答案】（1（2）证明见解析【解析】【分析】（1）根据椭圆的短轴可得抛物线方程2y x =，进而根据两点斜率公式，结合三角形的三边关系，即可由二次函数的性质求解，（2）根据两点坐标可得直线,MN DM 的直线方程，由直线与圆相切可得,a b 是方程()()()2222124240r x r x r −+−+−=的两个解，即可利用韦达定理代入化简求解定点. 【小问1详解】由题意得椭圆的方程：221116y x +=，所以短半轴14b = 所以112242p b ==×=，所以抛物线1C 的方程是2y x =. 设点()2,P t t ，则111222PQ PE ≥−=−=≥，所以当232ι=时，线段PQ . 【小问2详解】()1,D t 是抛物线1C 上位于第一象限的点，21t ∴=，且()0,1,1t D >∴设()()22,,,M a a N b b ，则：直线()222:b a MN y a x a b a −−=−−，即()21y a x a a b −=−+，即()0x a b y ab −++=. 直线()21:111a DM y x a −−=−−，即()10x a y a −++=. 由直线DMr =，即()()()2222124240r a r a r −+−+−=..同理，由直线DN 与圆相切得()()()2222124240r b r b r −+−+−=. 所以,a b 是方程()()()2222124240r x r x r −+−+−=的两个解， 22224224,11r r a b ab r r −−∴+==−− 代入方程()0x a b y ab −++=得()()222440x y r x y +++−−−=， 220,440,x y x y ++= ∴ ++= 解得0,1.x y = =− ∴直线MN 恒过定点()0,1−.【点睛】圆锥曲线中定点问题的两种解法（1）引进参数法：先引进动点的坐标或动线中系数为参数表示变化量，再研究变化的量与参数何时没有关系，找到定点.（2）特殊到一般法：先根据动点或动线的特殊情况探索出定点，再证明该定点与变量无关.技巧：若直线方程为()00y y k x x −=−,则直线过定点()00,x y ;若直线方程为y kx b =+ (b 为定值),则直线过定点()0,.b 19. 龙泉游泳馆为给顾客更好的体验，推出了A 和B 两个套餐服务，顾客可选择A 和B 两个套餐之一，并在App 平台上推出了优惠券活动，下表是该游泳馆在App 平台10天销售优惠券情况．日期t 1 2 3 4 5 6 7 8 9 10 销售量千张 1.9 1.98 2.2 2.36 2.43 259 2.68 2.76 2.7 0.4经计算可得：10101021111 2.2,118.73,38510i i i i i i i y y t y t ======∑∑∑. （1）因为优惠券购买火爆，App 平台在第10天时系统出现异常，导致当天顾客购买优惠券数量大幅减少，已知销售量y 和日期t 呈线性关系，现剔除第10天数据，求y 关于t 的经验回归方程结果中的数值用分数表示；..（2）若购买优惠券的顾客选择A 套餐的概率为14，选择B 套餐的概率为34，并且A 套餐可以用一张优惠券，B 套餐可以用两张优惠券，记App 平台累计销售优惠券为n 张的概率为n P ，求n P ；（3）记（2）中所得概率n P 的值构成数列{}()Nn P n ∗∈. ①求n P 的最值；②数列收敛的定义：已知数列{}n a ，若对于任意给定的正数ε，总存在正整数0N ，使得当0n N >时，n a a ε−<，（a 是一个确定的实数），则称数列{}n a 收敛于a .根据数列收敛的定义证明数列{}n P 收敛．参考公式： ()()()1122211ˆˆ,n ni ii i i i n n ii i i x x y y x y nx y ay bx x x x nx ====−−−==−−−∑∑∑∑. 【答案】（1）673220710001200y t + （2）433774n n P =+⋅−（3）①最大值为1316，最小值为14；②证明见解析【解析】【分析】（1）计算出新数据的相关数值，代入公式求出 ,ab 的值，进而得到y 关于t 的回归方程；（2）由题意可知1213,(3)44n n n P P P n −−=+≥，其中12113,416P P ==，构造等比数列，再利用等比数列的通项公式求解；（3）①分n 为偶数和n 为奇数两种情况讨论，结合指数函数的单调性求解；②利用数列收敛的定义，准确推理、运算，即可得证．【小问1详解】解：剔除第10天的数据，可得2.2100.4 2.49y ×−==新， 12345678959t ++++++++=新，则9922111119.73100.4114,73,38510285i i i i t y t = =−×==−= ∑∑新新，所以912922119114,7395 2.4673ˆ2859560009i i i i t y t y b t t == − −×× ==−× − ∑∑新新新新新，可得6732207ˆ 2.4560001200a =−×=，所以6732207ˆ60001200y t +. 【小问2详解】解：由题意知1213,(3)44n n n P P P n −−=+≥，其中12111313,444416P P ==×+=，所以11233,(3)44n n n n P P P P n −−−+=+≥，又由2131331141644P P ++×，所以134n n P P − +是首项为1的常数列，所以131,(2)4n n P P n −+=≥ 所以1434(),(2)747n n P P n −−=−−≥，又因为1414974728P −=−=−，所以数列47n P − 是首项为928−，公比为34−的等比数列，故1493()7284n n P −−=−−，所以1934433()()2847774n n n P −=−−+=+−. 【小问3详解】解：①当n 为偶数时，19344334()()28477747n n n P −=−−+=+⋅>单调递减，最大值为21316P =；当n 为奇数时，19344334()()28477747n n n P −=−−+=−⋅<单调递增，最小值为114P =，综上可得，数列{}n P 的最大值为1316，最小值为14. ②证明：对任意0ε>总存在正整数0347[log ()]13N ε=+，其中 []x 表示取整函数，当 347[log ()]13n ε>+时，347log ()34333333()()()7747474n n n P εε−=⋅−=⋅<⋅=，所以数列{}n P 收敛.【点睛】知识方法点拨：与新定义有关的问题的求解策略：1、通过给出一个新的定义，或约定一种新的运算，或给出几个新模型来创设新问题的情景，要求在阅读理解的基础上，依据题目提供的信息，联系所学的知识和方法，实心信息的迁移，达到灵活解题的目的；2、遇到新定义问题，应耐心读题，分析新定义的特点，弄清新定义的性质，按新定义的要求，“照章办事”，逐条分析、运算、验证，使得问题得以解决.方法点拨：与数列有关的问题的求解策略：3、若新定义与数列有关，可得利用数列的递推关系式，结合数列的相关知识进行求解，多通过构造的分法转化为等差、等比数列问题求解，求解过程灵活运用数列的性质，准确应用相关的数列知识.。

【原创·纯word版解析系列】数学理卷·2014届湖南省六校高三下学期4月联考试题(2014.04)

湖南省六校2014届高三下学期4月联考数学（理）试题本试题卷包括选择题、填空题和解答题三部分。

时量120分钟，满分150分。

【试卷综析】本试题是一份高三测试的好题，涉及范围广，包括集合、正态分布、复数、函数、解三角形、、排列组合、导数、方程、定积分、线性规划、充要条件、三视图、程序框图、直线、倾斜角、数列、平面向量、双曲线、离心率、三角函数、概率、几何证明、不等式选讲、参数方程与极坐标等高考核心考点，又涉及了概率统计、三角向量、立体几何、解析几何、导数应用等必考解答题型。

本题难易程度涉及合理，梯度分明；既有考查基础知识的经典题目，又有考查能力的创新题目；从9,10,16等题能看到命题者在创新方面的努力，从17,18，19三题看出考基础，考规范；从20题可以看出考融合，考传统；从16,21两题可以看出，考拓展，考创新。

一、选择题（本大题共10小题，每小题5分，共50分．在每小题给出的四个选项中，只有一项是符合题目要求的） 1．复数z 满足()()211i z i -+=+，其中i 为虚数单位，则在复平面上复数z 对应的点所在的象限为A ．第一象限B ．第二象限C ．第三象限D ．第四象限【知识点】复数运算及其几何意义【答案解析】D 由()()()2211111i i z i z i i +-+=+⇒==-+- 故选D【思路点拨】复数除法运算最好考出来，一定要掌握。

2．已知随机变量X 服从正态分布N （3，1），且P （l≤X≤5）=0．682 6，则P （X>5）= A ．0．158 8 B ．0．158 7 C ．0．158 6 D ．0．158 5【知识点】正态分布，概率。

【答案解析】B ()10.682650.15872P x ->== 【思路点拨】注意公式及对称性。

3．如图所示，程序框图（即算法流程图）运算的结果是 A ．5 B ．6 C ．7 D ．8 【知识点】程序框图，求和及周期性【答案解析】C 由12320S n =+++⋅⋅⋅+> 17n += 【思路点拨】一步步推导，弄清要的那个n 。

湖南师大附中2014届高三月考试卷

湖南师大附中2014届高三月考试卷(二)数学（理科）命题：湖南师大附中高三数学备课组一、选择题：本大题共10小题，每小题5分，共50分，在每个小题给出的四个选项中，只有一项是符合题目要求的． 1．在复平面内,复数21iz i=+(i 为虚数单位)的共轭复数对应的点位于（） A ．第一象限 B ．第二象限C ．第三象限D ．第四象限【解析】∵22(1)112i i i z i i -===++，其共轭复数是1i -，故选D ． 2．若角α的终边落在直线x ＋y ＝0上，则ααααcos cos 1sin 1sin 22-+-的值等于（） A ．2- B ．2C ．2-或2D ．0【解析】原式＝ααααcos |sin ||cos |sin +，由题意知角α的终边在第二、四象限，αsin 与αcos 的符号相反，所以原式等于0，故选D ．3．如图是一个几何体的三视图，则该几何体的体积是（） A ．54 B ．27 C ．18 D ． 9 【解析】由三视图可知，该几何体是一个四棱锥，其体积为1633183.⨯⨯⨯=故选C4.我校高三年级共有24个班，学校为了解同学们的心理状况，将每个班编号，依次为1到24，现用系统抽样方法，抽取4个班进行调查，若抽到的编号之和为48，则抽到的最小编号为（）A ．2 B.3 C ．4 D ．5 【解析】设最小编号为x ，则462636483,x x ++⨯+⨯==.故选B5．若等边△ABC 的边长为23，平面内一点M 满足CM ＝16CB ＋23,CA 则MA MB ⋅＝（）A ．1-BC ．2-D ．【解析】建立如图所示的直角坐标系，根据题设条件可知A (0,3)，B (－3，0)，M (0,2)，∴(0,1)MA =,(2)MB =-．∴2MA MB ⋅=-．故选C ．6．设函数)(x f ＝22,0,,0x x bx c x ->⎧⎨++≤⎩，若0)2(),0()4(=-=-f f f ，则关于x 的不等式)(x f ≤1的解集为（）A ．(,3][1,)-∞--+∞B ．[3,1]--C ．[3,1](0,)--+∞D ．[3,)-+∞【解析】当x ≤0时，f (x )＝x 2＋bx ＋c 且f (－4)＝f (0)，故其对称轴为x ＝－b2＝－2，∴b ＝4.又f (－2)＝4－8＋c ＝0，∴c ＝4，当x ≤0时，令x 2＋4x ＋4≤1有－3≤x ≤－1；当x ＞0时，)(x f ＝－2≤1显然成立，故不等式的解集为[3,1](0,)--+∞．故选C ．7．已知x ，y 取值如下表：从所得的散点图分析可知：y 与x 线性相关，且0.95y x a =+，则13x =时，y ＝( ) A ．1.45B ．13.8C ．13D ．12.8 【解析】依题意得，x ＝16×(0＋1＋4＋5＋6＋8)＝4，y ＝16×(1.3＋1.8＋5.6＋6.1＋7.4＋9.3)＝5.25.又直线y ^＝0.95x ＋a 必过样本中心点(x ，y )，即点(4,5.25)，于是有5.25＝0.95×4＋a ，由此解得a ＝1.45，从而当13x =时，计算得13.8y =，故选B ． 8．已知函数2()()f x x ax b a b =++∈R ，的值域为[0)+∞，，若关于x 的不等式()f x c <的解集为(6)m m +，，则实数c 的值为（） A ．0B ．3C ．6D ．9【解析】由值域为[0)+∞，，当2=0x ax b ++时有240a b =-=V ，即24a b =， ∴2222()42a a f x x ax b x ax x ⎛⎫=++=++=+ ⎪⎝⎭．∴2()2a f x x c ⎛⎫=+< ⎪⎝⎭解得2a x <+<22a a x <<．∵不等式()f x c <的解集为(6)m m +，，∴)()622aa --==，解得9c =．故选D ．9．已知双曲线)0,0(12222>>=-b a b y a x 的左、右焦点分别为21,F F ，若在双曲线的右支上存在一点P ，使得213PF PF =，则双曲线的离心率e 的取值范围为（）A ．B ．2]C ．(1,2]D ．(1,)+∞【解析】由213PF PF =及双曲线定义得12||||2PFPF a -=，从而1||3PF a =，2||PF a =，又12||2F F c =．因为点P 在右支上运动，所以1212||||||PF PF F F +≥，得42a c ≥，即2ca≤，又1e >．故填12e <≤．故选C ．10．已知函数()f x 满足：()114f =，()()()()()4,f x f y f x y f x y x y R =++-∈，则(2015)f =（）A ．12B ．14C ．14-D ．0【解析】取1x =，0y =得21)0(=f ．法一：通过计算)........4(),3(),2(f f f ，得周期为6，从而1(2015)4f =．法二：取x n =，1y =,有()(1)(1)f n f n f n =++-,同理(1)(2)()f n f n f n +=++，联立得(2)(1)f n f n +=--，所以T =6，故1(2015)(5)4f f ==．故选B ．二、填空题：本大题共5小题，每小题5分，共25分． 11．已知曲线12y x -=在点(1,1)处的切线为直线l ，则l 与两坐标轴围成的三角形的面积为．【解析】94．11．从区间[]5,5-内随机取出一个数x ，从区间[]3,3-内随机取出一个数y ，则使得4x y +≤的概率是．【解析】1213．将6位志愿者分成4组，其中两个组各2人，另两个组各1人，分赴四个不同学校支教，不同的分配方案有种（用数字作答）．【解析】先分组，考虑到有2个是平均分组，得221164212222C C C C A A 两个两人组两个一人组，再全排列得：221146421422221080C C C C A A A ⋅⋅=，故填1080． 14．已知函数()sin(2)f x x ϕ=+，其中ϕ为实数，若()()6f x f π≤对x R ∈恒成立，且sin 0ϕ<，则()f x 的单调递增区间是．【解析】若()()6f x f π≤对x R ∈恒成立，则()s i n ()163f ππϕ=+=，所以,32k k Z ππϕπ+=+∈，,6k k Z πϕπ=+∈.又sin 0ϕ<，所以(21),6k k Z πϕπ=++∈，代入()sin(2)f x x ϕ=+，得()sin(2)6f x x π=-+，由3222262k x k πππππ+++剟，得263k x k ππππ++剟，故填2,()63k k k Z ππππ⎡⎤++∈⎢⎥⎣⎦． 15．将自然数按如下图排列，其中处于从左到右第m 列从下到上第n 行的数记为(,)A m n ，如(3,1)4A =，(4,2)12A =，则(1,)A n =________，(10,10)A =________．【解析】填(1)2n n +，181 由(1,2)(1,1)2(1,3)(1,2)3(1,4)(1,3)4(1,)(1,1)A A A A A A A n A n n-=⎧⎪-=⎪⎪-=⎨⎪⎪--=⎪⎩，相加得(1)(2)(1,)(1,1)2n n A n A -+-=，从而(1)(1,)2n n A n +=．类似可求出(1)(,1)12m m A m -=+．进而(,2)(,1)1(,3)(,2)2(,4)(,3)3(,)(,1)1A m A m m A m A m m A m A m m A m n A m n m n -=+⎧⎪-=+⎪⎪-=+⎨⎪⎪--=+-⎪⎩，得(1)(1)(2)(,)122m m n m n A m n --+=++．10(101)(101)(2010)(10,10)118122A --+∴=++=三、解答题：本大题共6个小题，共75分．解答应写出文字说明、证明过程或演算步骤．16．（本题满分12分）在平面直角坐标系xOy 中，点21(,cos )2θP 在角α的终边上，点2(sin ,1)θ-Q 在角β的终边上，且12⋅=-OP OQ ．（1）求θ2cos 的值；（2）求)sin(βα+的值．解：（1）因为12⋅=-OP OQ ，所以2211sin cos 22θθ-=-，即2211(1cos )cos 22θθ--=-，所以22cos 3θ=，所以21cos 22cos 13θθ=-=． ………………5分（2）因为 22cos 3θ=，所以21sin 3θ=，所以)32,21(P 点，)1,31(-Q 点，又点12(,)23P 在角α的终边上，所以54sin =α，53cos =α ． …………………8分同理 10103sin -=β，1010cos =β， …………………10分所以sin()sin cos cos sin αβαβαβ+=+43(55=+⨯=12分 17．（本题满分12分）坛子中有6个阄，其中有3个标记是“中奖”，另外3个标记是“谢谢参与”，甲、乙、丙三人分两轮按甲、乙、丙、甲、乙、丙顺序依次抽取，当有人摸到“中奖”阄时，摸奖随即结束．（1）若按有放回抽取，甲、乙、丙中奖的概率分别是多少？（2）若按不放回抽取，甲、乙、丙中奖的概率分别是多少？（3）按不放回抽取，第一轮摸奖时有人中奖则可获得奖金10000元，第二轮摸奖时才中奖可获得奖金6000元，求甲、乙、丙三人所获奖金总额ξ的分布列与数学期望．B解：（1）若按有放回抽取，设甲、乙、丙中奖分别为事件A 1、B 1、C 1，则甲中奖的概率为)(1A P 16921)21(213=⋅+=乙中奖的概率为)(1B P 32921)21(21214=⋅+⋅=丙中奖的概率为)(1C P 64921)21(2121215=⋅+⋅⋅= ．．．．．．．．．．．．．．．．．．．．．．．．．4分（2）若按不放回抽取，设甲、乙、丙中奖分别为事件A 2、B 2、C 2，则甲中奖的概率为)(2A P 201141526363=⋅⋅+=乙中奖的概率为)(2B P 1035363=⋅=丙中奖的概率为20310320111)()(1)(222=--=--=B P A P C P ．．．．．．．．．．．．8分（3）设三人获得的奖金总额为ξ，ξ的可能取值有10000和6000元两种情况．法一：201415263)6000(=⋅⋅==ξP 20192011)6000(1)10000(=-==-==ξξP P 法二：按不放回抽取，第一轮摸奖时甲、乙、丙中奖分别为事件A 3、B 3、C 3，则==)10000(ξP )()()()(333333C P B P A P C B A P ++=++2019435263536363=⋅⋅+⋅+=． 20120191)10000(1)6000(=-==-==ξξP P ． ∴奖金总额ξ的分布列为故奖金总额的数学期望98002060002010000=⨯+⨯=ξE 元．．．．．．．．．12分 18．（本题满分12分）如图,四面体A-BCD 中， AD BCD ⊥面,BC CD ⊥，2,AD BD ==M 是AD 的中点,P 是BMD ∆的外心,点Q 在线段AC 上,且 4AC QC =.(1)证明://PQ 平面BCD ;(2)若二面角D BM C --的大小为060,求四面体A-BCD 的体积.【解析】证明(Ⅰ)方法一:如图,取MD 的中点F ,且M 是AD 中点,所以3AF FD =.因为P 是BM 中点,所以//PF BD ;又因为(Ⅰ)4AC QC =且3AF FD =,所以//QF BD ,所以面//PQF 面BDC ,且PQ ⊂面BDC ,所以//PQ 面BDC ;方法二:如图所示,取BD 中点O ,且P 是BM 中点,所以1//2PO MD ;取CD 的四等分点H ,使3DH CH =, 且4AC QC =,所以11////42QH AD MD ,所以////PO QH PQ OH ∴,且OH BCD ⊂面,所以//PQ 面BDC ; ………………6分 (Ⅱ)如图8所示,由已知得到面ADB ⊥面BDC ,过C 作CG BD ⊥于G ,所以CG BMD ⊥,过G 作GH BM ⊥于H ,连接CH ,所以CHG ∠就是C BM D --的二面角;由已知得到3BM ==,设BDC α∠=,所以cos ,sin ,sin ,,CD CG CBCD CG BC BD CD BDαααααα===⇒=== 在RT BCG ∆中,2sin BGBCG BG BCααα∠=∴=∴=,所以在Rt BHG ∆中,13HG =∴=,所以在Rt CHG ∆中tan tan 603CG CHGHG ∠==== tan (0,90)6060BDCααα∴=∈∴=∴∠=BCD A BCDCD BC S V ∆-∴==∴== (12)注：用向量法解答酌情计分19．（本题满分13分）甲、乙两超市同时开业，第一年的年销售额都为a 万元. 甲超市前n (n ∈N*)年的总销售额为)2(22+-n n a 万元；从第二年起，乙超市第n 年的销售额比前一年的销售额多a n 1)32(-万元.（Ⅰ）设甲、乙两超市第n 年的年销售额分别为a n ，b n 万元，求a n ，b n 的表达式；（Ⅱ）若在同一年中，某一超市的年销售额不足另一超市的年销售额的50%，则该超市将于当年底被另一超市收购. 若今年（2014年）为第一年，问：在今后若干年内，乙超市能否被甲超市收购？若能，请推算出在哪一年底被收购；若不能，请说明理由. 【解】（Ⅰ）设甲超市前n 年的总销售额为n S 万元，则2(2)2n a S n n =-+. 当2≥n 时，221(2)[(1)(1)2](1)22n n n a aa S S n n n n n a -=-=-+----+=-. 所以(1)(1)(2)n an a n a n =⎧=⎨-≥⎩ ………………3分由题设，当2n ≥时，a b b n n n 11)32(--=-，且1b a =，则)()()(123121--++-+-+=n n n b b b b b b b b2121()22223()()3[1()]2333313nn n a a a a a a --=++++==--L . 显然1b a =满足上式，故23[1()]3nn b a =-. ………………7分（Ⅱ）若乙超市能被甲超市收购，则1(2)2n n b a n <≥，即213[1()](1)32n a n a -<-.因为a ＞0，则216[1()]3n n ->-，即26()70(2)3nn n +⋅->≥.设2()6()7(2)3nf n n n =+⋅-≥，则1222(1)()16[()()]12()0333n n n f n f n ++-=+-=-⋅>，即f (n ＋1)＞f (n )，所以f (n )是增函数. ………………11分因为76522128(6)6()111033148f =⋅-=-=-<，72(7)6()03f =⋅>，则当n ≤6时，f (n )＜0，当n ≥7时，f (n )＞0.故到第7年底，即2020年底乙超市将被甲超市收购. ……………13分20．（本题满分13分）已知椭圆22221(0)x y a b a b+=>>的左、右焦点分别为1F 、2F ，若以2F 为圆心，b c -为半径作圆2F ，过椭圆上一点P 作此圆的切线，切点为T ，且PT 的最小值不小于)2a c -. （1）证明：椭圆上的点到2F 的最短距离为a c -；（2）求椭圆的离心率e 的取值范围；（3）设椭圆的短半轴长为1，圆2F 与x 轴的右交点为Q ，过点Q 作斜率为(0)k k >的直线l 与椭圆相交于A 、B 两点，若OA ⊥OB ，求直线l 被圆2F 截得的弦长L 的最大值。

湖南省2014届高三四校联考

湖南省2014届高三四校联考历史试题时量：90分钟满分：100分第I卷选择题（共50分）一、选择题（本大题共25小题，每小题2分，共计50分。

每小题列出的四个选项中，只有一项符合题意。

）1．西周确立的宗法制在春秋战国时期走向式微，但其君主世袭的继承惯例得到了部分延续。

下列史实中符合上述惯例的是A．李世民通过玄武门之变夺得帝位B．兄长宋太祖传位宋太宗以确保政权平稳过渡C．明太祖因太子亡而指定嫡长孙继位D．清康熙帝死后皇四子继位延续百年定局2．李贽驳斥轻视和贬低商人的传统观念，指出：“且商贾亦何可鄙之有？挟数万之赀，经风涛之险，受辱于关吏，忍诟于市易，辛勤万状，所挟者重，所得者末。

”更有甚者，商人“必交结于卿大夫之门，然后可以收其利而远其害”。

这反映出当时A．官商勾结，贿赂成风B．商业资本力量雄厚，商品经济发达C．封建制度严重阻碍商品经济发展D．商业发展严重依赖于封建官府3．一般观点认为：秦朝确立了同心圆的体制架构，而汉代以后确立了同心圆的文化思想架构。

右图是《小中华思想下的华夷秩序》，在这一秩序下的朝鲜、日本、越南等国都曾自称为“小中华”，意即“文化认同、文明继承者”等。

这一思想秩序的出现最有可能是得益于A．西周实行的分封制B．秦朝开创的中央集权制度C．汉朝开通丝绸之路D．汉朝实行独尊儒术的思想政策4．明朝的何良俊在《四友斋丛说》中有这样一段记载，嘉靖年间的一位宦官说：“我辈在顺门上久，见时事几复矣。

昔日张先生进朝，我们多要打个躬。

后至夏先生，我们只平着眼看望。

今严先生与我们拱拱手，方始进去。

”这则材料说明的实质问题是A．宦官的权力随着年龄增长而增长B．内阁首辅在朝臣中的权威逐渐下降C．明朝的礼仪制度日益规范D．皇帝的威权被宦官分割5．1087年，王岩叟在一个奏章中讲到地主与佃户的关系：“富民如客为佃户，每岁未收获间，借贷周给，无所不至，一失抚存，明年必去而之他。

”促使地主与佃户关系发生变化的主要原因是A．佃户人身依附关系有所削弱B．商品经济的发展．资本主义萌芽的产生C．南宋为缓和阶级矛盾，保护佃户的自由．D．生产力水平的发展和广大农民的长期斗争6．1920年梁启超在《清代学术概论》中指出：“此等论调，由今日观之，固甚普通甚至肤浅，然在二百六七十年前，则真极大胆之创举也。

湖南省长沙市第一中学2023-2024学年高三下学期月考(七)数学试题试卷(七)

长沙市一中2024 届高三月考试卷(七)数学试卷一、单项选择题: 本题共8 小题, 每小题5 分, 共40 分. 在每小题给出的四个选项中, 只有一项是符合题目要求的.1. 样本数据15 、13 、12 、31 、29 、23 、43 、19 、17 、38 的中位数为( )(A) 19 (B) 23 (C) 21 (D) 182. 已知集合A = {x''' e x2 −2x ≤ 1}, B = {−1, 0, 1}, 则集合A ∩ B 的非空子集个数为( )(A) 4 (B) 3 (C) 8 (D) 73. 已知实部为3 的复数z 满足z · (1 −2i) 为纯虚数, 则|z| = ( )(D) √54. 已知数列{a n } 满足a n = 3n −b (n ∈ N* , b ∈ R), 则“b < 3”是“{|a n |} 是递增数列”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. 已知tan θ= 2, 则 sin 2θ= ( )(A) (B) 2 (C) 1 (D)6. 过抛物线E: y2 = 2px (p > 0) 的焦点F 的直线交E 于点A, B , 交E 的准线l 于点C , AD ⊥ l , 点D 为垂足.若F 是AC 的中点, 且|AF | = 3, 则|AB| = ( )(A) 4 (B) 2√3 (C) 3√2 (D) 37. 已知双曲线C: kx2 −y2 = 1 的左焦点为F , P (3m, −4m) (m > 0) 为C 上一点, 且P 与F 关于C 的一条渐近线对称, 则C 的离心率为( )(A) (B) √3 (C) 2 (D)√58. 已知函数f(x) 的定义域为R, 且满足f(x) + f(3 −x) = 4, f(x) 的导函数为g(x), 函数y = g(x −1) 的图象关于点(2, 1) 中心对称, 则f + g(2024) = ( )(A) 3 (B) −3 (C) 1 (D) −1二、多项选择题: 本题共3 小题, 每小题6 分, 共18 分. 在每小题给出的选项中, 有多项符合题目要求, 全部选对的得6 分, 部分选对的得部分分, 有选错的得0 分.9. 已知函数cos 2x + sin 2x, 则( )(A) 函数f (x −关于原点对称(B) 曲线y = f(x) 的对称轴为x = + , k ∈ Z2 cos2 θ + 4 sin2 θ(C) f (x) 在区间单调递减(D) 曲线y = f (x) 在点(0, f (0)) 处的切线方程为2x −2y + 1 = 010. 已知二面角A −CD −B 的大小为, AC ⊥ CD , BD ⊥ CD , 且CD = 1, AC + BD = 2, 则( )(A) △ABD 是钝角三角形(B) 异面直线AD 与BC 可能垂直(C) 线段AB 长度的取值范围是[2, √5) (D) 四面体A −BCD 体积的最大值为11. 甲、乙两同学参加普法知识对抗赛, 规则是每人每次从题库中随机抽取一题回答. 若回答正确, 得 1 分, 答题继续; 若回答错误, 得0 分, 同时换成对方进行下一轮答题. 据经验统计, 甲、乙每次答题正确的概率分别是和 , 且第1 题的顺序由抛掷硬币决定. 设第i 次答题者是甲的概率为P i , 第i 次回答问题结束后中甲的得分是K i , 则( )(A) P2 =(C) P i+1= P i+ P i+ K i−1三、填空题: 本题共3 小题, 每小题5 分, 共15 分.12. (x + 3y)(x −y)8 的展开式中x3 y6 的系数为.13. 已知动点P 在圆M : (x −m + 1)2 + (y −m)2 = 1 上, 动点Q 在曲线y = ln x 上. 若对任意的m ∈ R, |PQ| ≥ n恒成立, 则n 的最大值是.14. 已知正六棱锥的高是底面边长的2√3 倍, 侧棱长为√13, 正六棱柱内接于正六棱锥, 即正六棱柱的所有顶点均在正六棱锥的侧棱或底面上, 则该正六棱柱的外接球表面积的最小值为.四、解答题: 本题共5 小题, 共77 分. 解答应写出文字说明、证明过程或演算步骤.15. 盒中有形状、大小均相同的卡片6 张, 卡片依次标记数字1, 2, 2, 3, 3, 3.(1) 若随机一次取出两张卡片, 求这两张卡片标记数字之差为1 的概率;(2) 若每次随机取出两张卡片后不放回, 直到将所有标记数字为2 的卡片全部取出, 记此时盒中剩余的卡片数量X , 求X 的分布列和E(X).16. 如图三棱锥P −ABC 中, PA = BC , AB = PC , AC ⊥ PB.(1) 证明: AB = BC;(2) 若平面PAC ⊥ 平面ABC , AC = √2AB , 求二面角A −PB −C 的余弦值.PA CB17. 已知定义在 (0, π) 上的函数 f (x) = cos 2 x + sin x.(1) 求 f (x) 的极大值点;(2) 证明: 对任意x 4 − x 2 + 1. 18. 已知椭圆的上、下顶点分别为 A(0, 1), B(0, −1), 其右焦点为 F , 且 F #---A -→ · B #---A -→ = F #---A -→ · F #---B -→ .(1) 求椭圆 C 的方程;(2) 若点 P (2, −1), 在直线 BP 上存在两个不同的点 P 1 , P 2 满足 P #---P ---1→ · P #---P ---2→ = P #---B -→2 . 若直线 AP 1 与直线 AP 2 分别交 C 于点 M , N (异于点 A), 证明: P , M , N 三点共线.19. 定义 △ABC 三边长分别为 a, b, c, 则称三元无序数组 (a,b, c) 为三角形数. 记 D 为三角形数的全集, 即 (a,b, c) ∈D.(1) 证明:“ (a,b, c) ∈ D ”是“(√a, √b, √c) ∈ D ”的充分不必要条件;(2) 若锐角 △ABC 内接于圆 O , 且 x O #---A -→ + y O #---B -→ + z O #---C -→ = 0, 设 I = (x,y, z) (x, y, z > 0).① 若 I = (3, 4, 5), 求 S △AOB : S △AOC ;② 证明: I ∈ D.。

湖南省长沙市湖南师大附中2024-2025学年高三上学期第三次月考数学试题(含解析)

湖南师大附中2025届高三月考试卷（三）数学时量：120分钟满分：150分得分：________________一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.集合的真子集个数是（）A.7B.8C.15D.162.“”是“”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3.已知角的终边上有一点的坐标是，其中，则（）A.B.C.D.4.设向量，满足，等于（）A. B.2C.5D.85.若无论为何值，直线与双曲线总有公共点，则的取值范围是（）A. B.C.，且 D.，且6.已知函数的图象关于原点对称，且满足，且当时，，若，则等于（）A.B.C. D.7.已知正三棱台所有顶点均在半径为5的半球球面上，且棱台的高为（）A.1B.4C.7D.1或78.北宋数学家沈括博学多才、善于观察.据说有一天，他走进一家酒馆，看见一层层垒起的酒坛，不禁想到：{}0,1,2,311x -<240x x -<αP ()3,4a a 0a ≠sin2α=4372524252425-a b a b += a b -=a b ⋅ θsin cos 10y x θθ⋅+⋅+=2215x y m -=m 1m ≥01m <≤05m <<1m ≠1m ≥5m ≠()2f x ()()130f x f x ++-=()2,4x ∈()()12log 2f x x m =--+()()2025112f f -=-m 132323-13-111ABC A B C -AB =11A B =“怎么求这些酒坛的总数呢？”经过反复尝试，沈括提出对于上底有个，下底有个，共层的堆积物（如图所示），可以用公式求出物体的总数，这就是所谓的“隙积术”，相当于求数列，的和.若由小球堆成的上述垛积共7层，小球总个数为238，则该垛积最上层的小球个数为（）A.2B.6C.12D.20二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9.若，则下列正确的是（）A. B.C. D.10.对于函数和，下列说法中正确的有（）A.与有相同的零点B.与有相同的最大值点C.与有相同的最小正周期D.与的图象有相同的对称轴11.过点的直线与抛物线交于，两点，抛物线在点处的切线与直线交于点，作交于点，则（）A.B.直线恒过定点C.点的轨迹方程是D.的最小值为选择题答题卡题号1234567891011得分ab cd n()()()2266n nS b d a b d c c a⎡⎤=++++-⎣⎦ab()()()()()()11,22,,11a b a b a n b n cd+++⋅++-+-=2024220240122024(12)x a a x a x a x+=++++2024a=20240120243a a a+++=012320241a a a a a-+-++=12320242320242024a a a a-+--=-()sin cosf x x x=+()sin cos22g x x xππ⎛⎫⎛⎫=---⎪ ⎪⎝⎭⎝⎭()f x()g x()f x()g x()f x()g x()f x()g x()0,2P2:4C x y=()11,A x y()22,B x yC A2y=-N NM AP⊥AB M5OA OB⋅=-MNM()22(1)10y x y-+=≠ABMN答案三、填空题：本题共3小题，每小题5分，共15分.12.已知复数，的模长为1，且，则________.13.在中，角，，所对的边分别为，，已知，，，则________.14.若正实数是函数的一个零点，是函数的一个大于e 的零点，则的值为________.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.（本小题满分13分）现有某企业计划用10年的时间进行技术革新，有两种方案：贷款利润A 方案一次性向银行贷款10万元第1年利润1万元，以后每年比前一年增加的利润B 方案每年初向银行贷款1万元第1年利润1万元，以后每年比前一年增加利润3000元两方案使用期都是10年，贷款10年后一次性还本付息（年末结息），若银行贷款利息均按的复利计算.（1）计算10年后，A 方案到期一次性需要付银行多少本息？（2）试比较A 、B 两方案的优劣.（结果精确到万元，参考数据：，）16.（本小题满分15分）如图，四棱锥中，底面为等腰梯形，.点在底面的射影点在线段上.（1）在图中过作平面的垂线段，为垂足，并给出严谨的作图过程；（2）若.求平面与平面所成锐二面角的余弦值.17.（本小题满分15分）1z 2z 21111z z +=12z z +=ABC ∆A B C a b c 5a =4b =()31cos 32A B -=sin B =1x ()2e e xf x x x =--2x ()()()3e ln 1e g x x x =---()122e ex x -25%10%101.12.594≈101.259.313≈P ABCD -ABCD 222AD AB BC ===P Q AC A PCD H 2PA PD ==PAB PCD已知函数，为的导数.（1）证明：当时，；（2）设，证明：有且仅有2个零点.18.（本小题满分17分）在平面直角坐标系中，已知椭圆的两个焦点为、，为椭圆上一动点，设，当时，.（1）求椭圆的标准方程.（2）过点的直线与椭圆交于不同的两点、（在，之间），若为椭圆上一点，且，①求的取值范围；②求四边形的面积.19.（本小题满分17分）飞行棋是大家熟悉的棋类游戏，玩家通过投掷骰子来决定飞机起飞与飞行的步数.当且仅当玩家投郑出6点时，飞机才能起飞.并且掷得6点的游戏者可以连续投掷骰子，直至显示点数不是6点.飞机起飞后，飞行步数即骰子向上的点数.（1）求甲玩家第一轮投掷中，投郑次数的均值）（2）对于两个离散型随机变量，，我们将其可能出现的结果作为一个有序数对，类似于离散型随机变量的分布列，我们可以用如下表格来表示这个有序数对的概率分布：（记，）()e sin cos x f x x x =+-()f x '()f x 0x ≥()2f x '≥()()21g x f x x =--()g x xOy 2222:1(0)x y C a b a b+=>>1F 2F P C 12F PF θ∠=23πθ=12F PF ∆C ()0,2B l M N M B N Q C OQ OM ON =+ OBMOBNS S OMQN X 11()()lim ()n n k k E X kP k kP k ∞→∞==⎛⎫== ⎪⎝⎭∑∑ξη()()()11,m i i ijj p x p x p x y ξ====∑()()()21,njjiji p y p y p x y η====∑ξη1x 2x ⋯nx 1y ()11,p x y ()21,p x y ⋯()1,n p x y ()21p y 2y ()12,p x y ()22,p x y()2,n p x y ()22p y1若已知，则事件的条件概率为.可以发现依然是一个随机变量，可以对其求期望.（ⅰ）上述期望依旧是一个随机变量（取值不同时，期望也不同），不妨记为，求；（ⅱ）若修改游戏规则，需连续掷出两次6点飞机才能起飞，记表示“甲第一次未能掷出6点”表示“甲第一次掷出6点且第二次未能掷出6点”，表示“甲第一次第二次均掷出6点”，为甲首次使得飞机起飞时抛掷骰子的次数，求.⋯⋯⋯⋯⋯⋯my ()1,m p x y ()2,m p x y ⋯(),n m p x y ()2m p y ()11p x ()12p x()1n p x i x ξ={}j y η={}{}{}()()1,,j i i j jii i P y x p x y Py x P x p x ηξηξξ=======∣i x ηξ=∣{}{}1mi j j i j E x y P y x ηξηξ===⋅==∑∣∣()()111,mj i j i i y p x y p x ==⋅∑ξ{}E ηξ∣{}E E ηξ⎡⎤⎣⎦∣0ξ=1ξ=2ξ=ηE η湖南师大附中2025届高三月考试卷（三）数学参考答案题号1234567891011答案CACBBDABBCACDBC一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.C 【解析】集合共有（个）真子集.故选C.2.A 【解析】解不等式，得，解不等式，得，所以“”是“”的充分不必要条件.3.C 【解析】根据三角函数的概念，，，故选C.4.B 【解析】.5.B 【解析】易得原点到直线的距离，故直线为单位圆的切线，由于直线与双曲线总有公共点，所以点必在双曲线内或双曲线上，则.6.D 【解析】依题意函数的图象关于原点对称，所以为奇函数，因为，故函数的周期为4，则，而，所以由可得，而，所以，解得.7.A 【解析】上下底面所在外接圆的半径分别为，，过点，，，的截面如图：{}0,1,2,342115-=240x x -<04x <<11x -<02x <<11x -<240x x -<44tan 33y a x a α===22sin cos 2tan 24sin211tan 25ααααα===+()2211()()1911244a b a b a b ⎡⎤⋅=+--=⨯-=⎣⎦ 1d ==2215x y m -=()1,0±01m <≤()f x ()f x ()()()133f x f x f x +=--=-()f x ()()20251f f =()()11f f -=-()()2025112f f -=-()113f =()()13f f =-()121log 323m --=13m =-13r =24r =A 1A 1O 2O，，，故选A.8.B 【解析】由题意，得，，则由得，整理得，所以.因为，为正整数，所以或6.因此有或而无整数解，因此.故选B.二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9.BC 【解析】对于A ：令，则，故A 错误；对于B ：令，则，故B 正确；对于C ：令，则，故C 正确；对于D ，由，两边同时求导得，令，则，故D 错误.故选BC.10.ACD 【解析】，.令，则，；令，则，，两个函数的零点是相同的，故选项A 正确.的最大值点是，，的最大值点是，，两个函数的最大值虽然是相同的，但最大值点是不同的，故选项B 不正确.由正弦型函数的最小正周期为可知与有相同的最小正周期，故选项C 正确.曲线的对称轴为，，曲线的对称轴为，，两个函数的图象有相同的对称轴，故选项D 正确.故选ACD.11.BC 【解析】作图如下：24OO ==13OO ==211h OO OO ∴=-=6c a =+6d b =+()()()772223866b d a b dc c a ⎡⎤++++-=⎣⎦()()()()77262126623866b b a b b a a a ⎡⎤++++++++-=⎣⎦()321ab a b ++=773aba b +=-<a b 3ab =6,3a b ab +=⎧⎨=⎩5,6.a b ab +=⎧⎨=⎩63a b ab +=⎧⎨=⎩6ab =0x =01a =1x =20240120243a a a +++= 1x =-012320241a a a a a -+-++= 2024220240122024(12)x a a x a x a x +=++++ 202322023123202420242(12)232024x a a x a x a x ⨯⨯+=++++ 1x =-12320242320244048a a a a -++-=- ()4f x x π⎛⎫=+ ⎪⎝⎭()3244g x x x πππ⎛⎫⎛⎫=--=-⎪ ⎪⎝⎭⎝⎭()0f x =4x k ππ=-+k ∈Z ()0g x =34x k ππ=+k ∈Z ()f x 24k ππ+k ∈Z ()g x 324k ππ-+k ∈Z 2πω()f x ()g x 2π()y f x =4x k ππ=+k ∈Z ()y g x =54x k ππ=+k ∈Z设直线的方程为（斜率显然存在），，，联立消去整理可得，由韦达定理得，，A.，，故A 错误；B.抛物线在点处的切线为，当时，，即，直线的方程为，整理得，直线恒过定点，故B 正确；C.由选项B 可得点在以线段为直径的圆上，点除外，故点的轨迹方程是，故C 正确；D.，则，，，则，设，，当单调递增，所以，故D 错误.故选BC.三、填空题：本题共3小题，每小题5分，共15分.AB 2y tx =+211,4x A x ⎛⎫ ⎪⎝⎭222,4x B x ⎛⎫ ⎪⎝⎭22,4,y tx x y =+⎧⎨=⎩x 2480x tx --=124x x t +=128x x =-221212444x x y y =⋅=1212844OA OB x x y y ⋅=+=-+=- C A 21124x x x y ⎛⎫=+ ⎪⎝⎭2y =-11121244282222x x x x x t x x =-=-=+=-()2,2N t -MN ()122y x t t +=--xy t=-MN ()0,0M OP O M ()22(1)10y x y -+=≠2MN AB ===22ABMN ===m =m ≥12ABm MN m ⎛⎫=- ⎪⎝⎭()1f m m m =-m ≥()2110f m m=+>'m ≥()f m min ()f m f==12.1【解析】设，，因为，所以.因为，，所以，所以，所以，，所以.【解析】在中，因为，所以.又，可知为锐角且.由正弦定理，，于是.将及的值代入可得，平方得，故.14.e 【解析】依题意得，，即，，，即，，，，，又，，同构函数：，则，又，，，，又，，单调递增，，.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.【解析】（1）A 方案到期时银行贷款本息为（万元）.……（3分）()1i ,z a b a b =+∈R ()2i ,z c d cd =+∈R 21111z z +=1222111z z z z z z +=111z z =221z z =121z z +=()()i i i 1a b c d a c b d -+-=+-+=1a c +=0b d +=()()12i 1z z a c b d +=+++=ABC ∆a b >A B >()31cos 32A B -=A B -()sin A B -=sin 5sin 4A aB b ==()()()5sin sin sin sin cos cos sin 4B A A B B A B B A B B ⎡⎤==-+=-+-⎣⎦()cos A B -()sin A B -3sin B B =2229sin 7cos 77sin B B B ==-sin B =1211e e 0xx x --=1211e e xx x -=10x >()()322e ln 1e 0x x ---=()()322e ln 1e x x --=2e x >()()()131122e e e e ln 1x x x x x ∴-==--()()()11122e e ln 1e x x x x +∴-=--()()()21ln 11112e e ln 1e e x x x x -++⎡⎤∴-=--⎣⎦2ln 1x > 2ln 10x ->∴()()1e e ,0x F x x x +=->()()312ln 1e F x F x =-=()()111e e e e e 1e x x x x F x x x +++=-+'=-+0x > 0e e 1x ∴>=e 10x ∴->1e 0x x +>()0F x ∴'>()F x 12ln 1x x ∴=-()()()31222222e ln 1e e e eeex x x x ---∴===()1010110%26⨯+≈（2）A 方案10年共获利：（万元），……（5分）到期时银行贷款本息为（万元），所以A 方案净收益为：（万元），……（7分）B 方案10年共获利：（万元），……（9分）到期时银行贷款本息为（万元），……（11分）所以B 方案净收益为：（万元），……（12分）由比较知A 方案比B 方案更优.……（13分）16.【解析】（1）连接，有平面，所以.在中，.同理，在中，有.又因为，所以，，所以，，故，即.又因为，，平面，所以平面.平面，所以平面平面.……（5分）过作垂直于点，因为平面平面，平面平面，且平面，有平面.……（7分）（2）依题意，.故为，的交点，且.所以过作直线的平行线，则，，，两两垂直，以为原点建立如图所示空间直角坐标系，()1091.2511125%(125%)33.31.251-+++++=≈- 1010(110%)25.9⨯+≈33.325.97-≈()()101010.31 1.3190.310123.52⨯-⨯++++⨯=⨯+= ()()10109 1.11.11(110%)(110%)110%17.51.11-++++++=≈- 23.517.56-≈PQ PQ ⊥ABCD PQ CD ⊥ACD ∆2222cos 54cos AC AD CD AD CD ADC ADC =+-⋅⋅∠=-∠ABC ∆222cos AC ABC =-∠180ABC ADC ∠+∠= 1cos 2ADC ∠=()0,180ADC ∠∈ 60ADC ∠=AC =222AC CD AD +=AC CD ⊥PQ AC Q = PQ AC ⊂PAC CD ⊥PAC CD ⊂PCD PCD ⊥PAC A AH PC H PCD ⊥PAC PCD PAC PC =AH ⊂PAC AH ⊥PCD AQ DQ ==Q AC BD 2AQ ADCQ BC==23AQ AC ==PQ ==C PQ l l AC CD C则：，，，，所以，，，.设平面的法向量为，则取.同理，平面的法向量，，……（14分）故所求锐二面角余弦值为.……（15分）17.【解析】（1）由，设，则，当时，设，，，，和在上单调递增，，，当时，，，则，函数在上单调递增，，即当时，.()1,0,0D P ⎛ ⎝()A 12B ⎛⎫- ⎪ ⎪⎝⎭()1,0,0CD = CP ⎛= ⎝ 0,AP ⎛= ⎝ 1,2BP ⎛= ⎝ PCD (),,m x y z =)0,0,m CD x m CP y ⎧⋅==⎪⎨⋅=+=⎪⎩()0,m =- PAB )1n =-1cos ,3m n m n m n ⋅==13()e cos sin xf x x x =+'+()e cos sin xh x x x =++()e sin cos xh x x x =+'-0x ≥()e 1x p x x =--()sin q x x x =-()e 10x p x ='-≥ ()1cos 0q x x ='-≥()p x ∴()q x [)0,+∞()()00p x p ∴≥=()()00q x q ≥=∴0x ≥e 1x x ≥+sin x x ≥()()()e sin cos 1sin cos sin 1cos 0xh x x x x x x x x x =-+≥+-+=-++≥'∴()e cos sin x h x x x =++[)0,+∞()()02h x h ∴≥=0x ≥()2f x '≥（2）由已知得.①当时，，在上单调递增，又，，由零点存在定理可知，在上仅有一个零点.……（10分）②当时，设，则，在上单调递减，，，，在上单调递减，又，，由零点存在定理可知在上仅有一个零点，综上所述，有且仅有2个零点.……（15分）18.【解析】（1）设，为椭圆的焦半距，，，当时，最大，此时或，不妨设，当时，得，所以，又因为，所以，.从，而椭圆的标准方程为.……（3分）（2）由题意，直线的斜率显然存在.设，.……（4分），同理，..……（6分）联立，……（8分）()e sin cos 21xg x x x x =+---0x ≥()()e cos sin 220x g x x x f x =+='+--'≥ ()g x ∴[)0,+∞()010g =-< ()e 20g πππ=->∴()g x [)0,+∞0x <()2sin cos (0)e x x xm x x --=<()()2sin 10exx m x -=≤'()m x ∴(),0-∞()()01m x m ∴>=e cos sin 20x x x ∴++-<()e cos sin 20x g x x x ∴=++-<'()g x ∴(),0-∞()010g =-< ()e 20g πππ--=+>∴()g x (),0-∞()g x ()00,P x y c C 12122F PF p S c y ∆=⋅⋅00y b <≤ 0y b =12F PF S ∆()0,P b ()0,P b -()0,P b 23πθ=213OPF OPF π∠=∠=c =12F PF S bc ∆==1b =c =2a =∴C 2214x y +=l ()11: 2.,l y kx M x y =+()22,N x y 1112OBM S OB x x ∆∴=⋅=2OBN S x ∆=12OBM OBN S xS x ∆∆∴=()22222,141612044y kx k x kx x y =+⎧⇒+++=⎨+=⎩，.……（9分）又，，，同号..，，.令，则，解得，.……（12分）（3），.且四边形为平行四边形.由（2）知，，.而在椭圆上，.化简得.……（14分）线段，……（15分）到直线的距离……（16分）.……（17分）()()222Δ(16)4121416430k k k∴=-⨯⨯+=->234k ∴>1221614k x x k -+=+ 12212014x x k=>+1x ∴2x ()()2222122121212216641421231414k x x x x k k x x x x k k -⎛⎫ ⎪++⎝⎭∴===++++234k > ()2226464164,1331434k k k ⎛⎫∴=∈ ⎪⎛⎫+⎝⎭+ ⎪⎝⎭211216423x x x x ∴<++<()120x x λλ=≠116423λλ<++<()1,11,33λ⎛⎫∈ ⎪⎝⎭()1,11,33OBM OBN S S ∆∆⎛⎫∴∈ ⎪⎝⎭ OQ OM ON =+()1212,Q x x y y ∴++OMQN 1221614k x x k -+=+()121224414y y k x x k ∴+=++=+22164,1414k Q k k -⎛⎫∴ ⎪++⎝⎭Q C 2222164441414k k k -⎛⎫⎛⎫∴+⨯= ⎪ ⎪++⎝⎭⎝⎭2154k =∴MN ====O MN d ==OMQN S MN d ∴=⋅==四边形19.【解析】（1），，2，3，…，所以，，2，3，…，记，则.作差得：，所以，.故.……（6分）（2）（ⅰ）所有可能的取值为：，.且对应的概率，.所以，又，所以.……（12分）（ⅱ），；，；，，，故.……（17分）()11566k P X k -⎛⎫==⨯ ⎪⎝⎭1k =()56k k k P X k ⋅==1k =()21111512666nn k kP k n =⎛⎫=⨯+⨯++⨯ ⎪⎝⎭∑ 211112666n n S n =⨯+⨯++⨯ 2311111126666n n S n +=⨯+⨯++⨯ 1211111511111111661666666556616n n n n n n n S n n ++⎛⎫- ⎪⎛⎫⎛⎫⎝⎭=+++-⨯=-⨯=-+ ⎪⎪⎝⎭⎝⎭- 611155566n n n S ⎡⎤⎛⎫⎛⎫=⋅-+⎢⎥ ⎪⎪⎝⎭⎝⎭⎢⎥⎣⎦()16615556n nn k kP k S n =⎛⎫⎛⎫==-+ ⎪⎪⎝⎭⎝⎭∑116616()()lim ()lim 5565nn n n k k E X kP k kP k n ∞→∞→∞==⎡⎤⎛⎫⎛⎫⎛⎫===-+=⎢⎥ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦∑∑{}E ηξ∣{}i E x ηξ=∣1,2,,i n = {}{}()()()1ii i p E E x p x p x ηξηξξ=====∣∣1,2,,i n = {}()()()()()111111111[{}],,nnm n m i i j i j i j i j i i j i j i E E E x p x y p x y p x y p x y p x ηξηξ=====⎛⎫==⋅=⋅= ⎪ ⎪⎝⎭∑∑∑∑∑∣∣()()()()21111111,,,n m m n mn mj i j j i j j i j j j i j j i j i j y p x y y p x y y p x y y p y E η=======⎛⎫⋅=⋅==⋅= ⎪⎝⎭∑∑∑∑∑∑∑{}E E E ηξη⎡⎤=⎣⎦∣{}01E E ηξη==+∣156p ={}12E E ηξη==+∣2536p ={}22E η==3136p ={}()()5513542122636363636E E E E E E ηηηηηξ⎡⎤==++++⨯=+⎣⎦∣42E η=。

湖南省长沙市湖南师范大学附属中学2024届高三上学期月考卷(四)数学

湖南师大附中2024届高三月考试卷（四）数学时量：120分钟满分：150分一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知复数12i z =+，其中i 为虚数单位，则复数2z 在复平面内对应的点的坐标为（）A.(4,5)- B.(4,3)C.(3,4)- D.(5,4)）2.若随机事件A ，B 满足1()3P A =，1()2P B =，3()4P A B = ，则(|)P A B =（）A.29B.23C.14D.168.设{}n a 是公比不为1的无穷等比数列，则“{}n a 为递减数列”是“存在正整数0N ，当0n N >时，1n a <”的（）A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件4.设0,2πα⎛⎫∈ ⎪⎝⎭，0,2πβ⎛⎫∈ ⎪⎝⎭，且1tan tan cos αβα+=，则（）A.22παβ+=B.22παβ-=C.22πβα-=D.22πβα+=5.若52345012345(12)(1)(1)(1)(1)(1)x a a x a x a x a x a x -=+-+-+-+-+-，则下列结论中正确的是（）A.01a = B.480a =C.50123453a a a a a a +++++= D.()()10024135134a a a a a a -++++=6.函数1()2cos[(2023)]|1|f x x x π=++-在区间[3,5]-上所有零点的和等于（）A.2B.4C.6D.87.点M 是椭圆22221x y a b+=（0a b >>）上的点，以M 为圆心的圆与x 轴相切于椭圆的焦点F ，圆M 与y 轴相交于P ，Q ，若PQM △是钝角三角形，则椭圆离心率的取值范围是（）A.(0,2B.⎛ ⎝C.⎫⎪⎪⎭D.(2-8.已知函数22,0,()4|1|4,0,x x f x x x ⎧=⎨-++<⎩…若存在唯一的整数x ，使得()10f x x a -<-成立，则所有满足条件的整数a 的取值集合为（）A.{2,1,0,1}-- B.{2,1,0}-- C.{1,0,1}- D.{2,1}-二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分、9.已.知双曲线C过点且渐近线为y x =，则下列结论正确的是（）A.C 的方程为2213x y -= B.CC.曲线2e1x y -=-经过C 的一个焦点D.直线10x --=与C 有两个公共点10.已知向量a ，b满足|2|||a b a += ，20a b a ⋅+= 且||2a = ，则（）A.||8b = B.0a b += C.|2|6a b -=D.4a b ⋅= 11.如图、正方体1111ABCD A B C D -的棱长为2，点M 是其侧面11ADD A 上的一个动点（含边界），点P 是线段1CC 上的动点，则下列结论正确的是（）A.存在点P ，M ，使得二面角M DC P --大小为23πB.存在点P ，M ，使得平面11B D M 与平面PBD 平行C.当P 为棱1CC的中点且PM =时，则点M 的轨迹长度为23πD.当M 为1A D 中点时，四棱锥M ABCD -12.若存在实常数k 和b ，使得函数()F x 和()G x 对其公共定义域上的任意实数x 都满足：()F x kx b +…和()G x kx b +…恒成立，则称此直线y kx b =+为()F x 和()G x 的“隔离直线”.已知函数2()f x x =（x ∈R ），1()g x x=（0x <），()2eln h x x =（e 2.718≈），则下列选项正确的是（）A.()()()m x f x g x =-在x ⎛⎫∈ ⎪⎝⎭时单调递增B.()f x 和()g x 之间存在“隔离直线”，且b 的最小值为–4C.()f x 和()g x 之间存在“隔离直线”，且k 的取值范围是[4,1]-D.()f x 和()h x之间存在唯一的“隔离直线”ey =-三、填空题：本题共4小题，每小题5分，共20分.13.已知函数()y f x =的图象在点(1,(1))M f 处的切线方程是122y x =+，则(1)(1)f f +'=___________.14.如图，由3个全等的钝角三角形与中间一个小等边三角形DEF 拼成的一个较大的等边三角形ABC ，若3AF =，sin ACF ∠=，则DEF △的面积为___________.15.已知数列{}n a 的首项132a =，且满足1323n n n a a a +=+.若123111181n a a a a ++++< ，则n 的最大值为___________.16.在棱长为3的正方体1111ABCD A B C D -中，点E 满足112A E EB =，点F 在平面1BC D 内，则|1||A F EF +的最小值为___________.四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.17.（10分）已知函数2()2cos 2xf x x m ωω=++（0ω>）的最小值为–2.（1）求函数()f x 的最大值；（2）把函数()y f x =的图象向右平移6πω个单位长度，可得函数()y g x =的图象，且函数()y g x =在0,8π⎡⎤⎢⎥⎣⎦上单调递增，求ω的最大值.18.（12分）为了丰富在校学生的课余生活，某校举办了一次趣味运动会活动，学校设置项目A “毛毛虫旱地龙舟”和项目B “袋鼠接力跳”.甲、乙两班每班分成两组，每组参加一个项目，进行班级对抗赛.第一个比赛项目A 采取五局三胜制（即有一方先胜3局即获胜，比赛结束）；第二个比赛项目B 采取领先3局者获胜。

湖南省湖南师大附中、长沙市一中等六校2014届高三下学期4月联考试题化学 Word版含答案.pdf

湖南省2014届高三六校联考化学能力试题 11．化学兴趣小组利用如图装置进行电化学实验，下列预期现象正确的是 A．X和Y与电流表连接，将Zn换成Fe，测得电流更大 B．X和Y与电流表连接，将盐桥换成铜线，电流表指针偏转 C．X和Y分别与电源“+”、“一”极相连，Cu极质量增大 D．X和Y分别与电源“一”、“+”极相连，盐桥中阴离子向Zn极移动 12．—C4 H9和—C2 H5O取代苯环上的氢原子，形成的有机物中能与金属钠反应的有 A．8种 B．18种 C．24种 D．36种 13．常温下，用0．100 0 mol·L－1 NaOH溶液分别滴定20．00 mL 0．100 0 mol·L－1 HA和HB溶液，得到2条滴定曲线，如下图所示（图1表示HA，图2表示HB），下列说洼正确的是 A．HA是弱酸，HB是强酸 B．达到E点时，对应混合溶液中c（Na+）=c（B－）+c（HB） C．达到B、D点时，两溶液pH均为7，反应消耗的n（HA）=n（HB） D．当达到E点后，继续滴加NaOH溶液，对应混合溶液中各离子浓度由大到小的顺序均为 c（Na+）> c（B－）>c（OH－）>c（H+） 26．（14分，每空2分）用软锰矿（主要成分为MnO2，还有少量含铁、铜、汞的化合物）制备MnSO4电解液继而提取锰的工业流程如下（1）已知锰粉的主要成分为MnO，反应①的化学方程式为。

（2）实验室可用软锰矿与浓盐酸反应制Cl2，比较MnO2、C12、Fe3+的氧化性强弱：。

（3）溶液I中加入一定质量的MnO2以测定Fe2+的浓度，加入MnO2有关反应的离子方程式为。

（4）已知90℃时水的离子积常数Kw=3．6×10－13，Fe(OH)3的Ksp=6．4×10－24，要使滤渣A沉淀完全，需用氨水调pH至少为（已知lg 3≈0．5）。

（5）滤渣B的主要成分为。

湖南省师大附中、长沙市一中等六校2014届高三下学期4月联考语文试题(word版)

湖南省师大附中、长沙市一中等六校2014届高三下学期4月联考语文试题本试卷共7道大题，21道小题（2道选做题任选1小题）。

时量150分钟，满分150分。

一、语言文字运用（12分，每小题3分）1．下列词语中加点的字，读音全都正确的一组是2．下列词语中，没有错别字的一纽是A．分辨夙愿金榜提名因材施教B．契约嬉戏暇不掩瑜完璧归赵C．贷款笃信贻笑大方游刃有余D．蓊郁鹊跃孤苦伶仃不胫而走3．下列各句中，没有语病的一句是A．成为亚洲第一位单打大满贯冠军之后，李娜又进一步打破了此前由伊达公子保持的亚洲球员单打最高排名，从而取代了这位仍在巡回赛中征战的老将，成为亚洲网球的代言入。

B．时至今日，改革进入攻坚期，新旧问题交织、利益主体多样；改革越深入，就越要触及深层问题、体制弊端，各个领域的改革越是相互推动、相互制约、相互关联。

C．秀兰·邓波儿用自己的精湛演技和无限活力激励了观众，更重要的是，她还是一个有社会责任感、关心他人的好公民，她一定会被非常多的人所怀念。

D．我们反对铺张浪费，不仅是因为我们的国力还远未达到富裕、我们的发展还在艰难爬坡，而是因为勤俭节约是我们建设国家的根本方针，是任何时候都不可丢弃的传家宝4．依次填入下面一段文字画横线处的语句，衔接最恰当的一组是至于所有的花，。

所有的蕊，。

所有的树，。

而风，去记忆、垂询。

①交给风去纵宠②已交给蝴蝶去点数③交给檐前的老风铃④交给蜜蜂去编册A．④②①③B．②①④③C．④①②③D．②④①③二、文言文阅读（22分。

其中，选择题12分，每小题3分；翻译题10分）阅读下面的文言文，完成5～9题。

甘罗者，甘茂孙也。

茂既死后，甘罗年十二，事秦相文信侯吕不韦。

秦始皇帝使刚成君蔡泽于燕．三年而燕王喜使太子丹入质于秦。

秦堡堡鸶芰垒塑盏，欲与燕共伐赵，以广河旦圭超张唐谓文信侯日：‚臣尝为秦昭王伐赵〃赵怨臣，曰：‘得唐者与百里之地。

，今之燕必经赵，臣不可以行。

‛文信侯不快，未有以强也。

湖南省湖南师大附中、长沙市一中等六校高三英语下学期

湖南省2014届高三六校联考英语试题总分:150分时量:120分钟Part I Listening Comprehension （30 marks）Section A（ 22．5 marks）Directions: In this section, you' II hear six conversations between two speakers．For each conversation, there are several questions and each question is followed by three choices marked A, B and C．Listen carefully and then choose the best answer for each question．You will hear each conversation TWICE．Conversation 1l．Where does the man play tennis?A．At the university．B．At the community center．C．At the club．2．How often does the woman swim?A．Once a week．B．Five times a week．C．Three times a week．Conversation 23．Whose photo are the speakers talking about?A．The woman's friends'．B．The woman's colleagues'．C．The woman's family's．4．What does Mary do?A．She is an accountant．B．She is a business woman．C．She is a lawyer．Conversation 35．What is the man doing?A．Preparing for a trip．B．Writing an article．C．Selling his articles．6．Where did the man go last month?A．Amazon．B．Asia．C．America．Conversation 47．Where is Rick working now?A．In a travel company．B．In a hotel．C．In a law office．8．How does Rick feel about his trip?A．Excited． B．Curious．C．Nervous．9．When will Rick probably leave?A．In three days．B．In a week．C．In a month．Conversation 510．What is the man worried about?A．His children's friends．B．His children's behaviors．C．His children's health．1l．What do we know about the man's children?A．They do their schoolwork till midnight．B．They are interested in computer games．C．They feel bored about pop music．12．How many children does the woman have?A. 2．B．3． c．4．Conversation 613．Which of the following is TRUE about the boy?A．He got an A in maths．B．He hasn't studied hard enough at history．C．He likes history．14．What can we learn about the boy's study of government?A．He has a new teacher．B．He has got a B in it．C．He finds it confusing．15．What is the boy's mother's attitude to his report card?A．She doesn't take it too hard．B．She doesn't excuse him．C．She thinks it very bad．Section B （7．5 marks）Directions: In．this section, you will hear a short passage, Listen carefully and then fill in the numbered blanks with the information you have heard． Fill in each blank with NO MORE THAN THREF, WORDS．You'll hear the short passage TWICE．Part Ⅱ Language Knowledge （45 marks）Section A （15 marks）Directions．．For each of the following unfinished sentences there are four choices marked A,B,C and D．Choose the one that best completes the sentence．Example:The wild flowers looked like a soft orange blanket the desert A．covering B．covered C．cover D．to coverThe answer is A．21．We the computer problem, and got back online to continue our research．A．have solved B．solve C．would solve D．solved22． Making a silly mistake at a job interview is sometimes very embarrassing, butif ， it may become a driving force．A．properly handled B．properly handlingC．you properly handled D．you properly handling23．The little girl showed no anxiety before the competition．She seemed for it pretty wellA．to prepare B．to have preparedC．having prepared D．preparing24． I first met Mr．Li three years ago．He had just returned from abroad and English in Senior One then．A．taught B．would teach C．was teaching D．had taught25． Only when at least 50% of our energy is provided by grain a balancedand healthy food consumption．A．we had B．did we have C．we have D．do we have26．The Beijing transport authority released the news buyers of green carscan now receive a maximum amount of 5700 yuan from the central government．A．when B．that C．if, D．which27．—How can you catch a taxi at home?—Don-t worry．I have opened an account on DiDi Dache．I for a taxidriver to pick me up here．A．was waiting B．have waited C．am waiting D．had waited28． Mr．Jenkins wanted to buy a computer as soon as possible, he decidedto shop around to see which shop had the best deal．A．While B．As C．When D．Once29．Gertrude was upset when one of her cats overturned a glass of wine, adot on her favorite carpet．A．leave B．left C．to leave D．leaving30．—What a heavy rain ?—Yes．I to take an umbrella with me, but I didn't ．A．had planned B．would plan C．have planned D．will plan31．—Mary doesn't mind lending you her e-dictionary．—She I have already borrowed Jane's．A．can't B．mustn't C．needn't D．shouldn't32．____ your support, I could have missed the first prize．A．Should it not be for B．Had it not been forC．If it had been for D． If it should be for33．Today is my grandfather's l00th birthday, an occasion we are going tocelebrate with all the family and a few friends in for the evening．A．who B．which C．where D．when34．It is very important to recognize what kind of person you are and which specialqualities make you different from in your club．A．anyone else B．the other C．someone else D．others35． He appreciated a chance to make a presentation in the annual seminar on Comparative Literature．A．giving B ．to give C．being given D．to be givenSection B (18 marks)Directions: For each blank in the following passages there are four words or phrases marked A, B, C and D．Fill in each blank with the word or phrase that best fits the context．Annie Dillard, the author of TEACHING A STONE TO TALK, tells us a sad story about a British Arctic expedition（远征）that took ship in 1845 to explore the Northwest Passage around the Canadian Arctic to the Pacific Ocean．Neither of the two ships and none of the 138 men aboard 36 ．In her book Captain Sir John Franklin prepared as if they were 37 in a pleasure travel rather than a challenging journey through one of earth's 38 ．environments． He packed a 1,200 volume library, china place settings and glass wine cups． Years later, some of them were found near some 39 bodies．The voyage was hopeless when the ships sailed into 40 cold waters and became trapped in ice．First 41 coated the decks, the poles and the rope．Then water froze around the rudders （舵） , and the ships became 42 in the frozen sea desperately．Sailors set out to search for help, but soon 43 to severe Arctic weather and died of exposure to its harsh winds and low temperatures．For the next twenty years, 44 of the expedition were found all over the frozen landscape．Dillard reports that on a voyage that was to last 2-3 years the crew did not prepare for the chance of the ships being ice-locked and the captain 45 just a 12-day supply of coal for the spare steam engines．Historians may doubt the wisdom of such a journey．But I have a question: Are we, too, prepared for the important voyage we've started-the journey we- call " 46 ”?I want to be as ready as possible for what may lie ahead．As long as we are alive, our journey is not over．And in large part, the success of our journey will be 47 by our regular and systematic preparation．36．A．continued B．returned C．existed D．survived 37．A．resulting B．consisting C．engaging D．ending 38．A．toughest B．excellent C．commonest D．superb 39．A．burned B．rotten C．swelled D．frozen40．A．slightly B．lightly C．ordinarily D．extremely41．A．ice B．wave C．steam D．water 42．A．lost B．moved C．locked D．opened43．A．put up B．gave in C．brought in D．broke up 44．A．library B．place settings C．wine cups D．remains 45．A．donated B．carried C．expected D．answered46．A．trouble B．happiness C．life D．death47. A. determined B. damaged C. defeated D. decoratedSection C（12 marks）Directions: Complete the following passage by filling in each black with one word that best fits the context．Nearly all of us have dreamed about winning the big prize in a lottery（彩票） We dream about 48 we would do with the money, but we 49 stop to think about how the money would change us．For most of us, our way of life is closely linked 50 our economic circumstances．The different parts of our lives fit together like a jigsaw（拼板）- work, home, friends, hobbies and sports make up our world． 51 sudden fortune would change it all and break the jigsaw．For example , some people like the idea of 52 having to work, but winners have found that without work there is no reason to get up in the morning．In seems great to move to a bigger house in a wealthy area, 53 if you do that, you leave old friends behind．Usually, though winners are advised not to publicize _ 54 _ addresses and phone numbers, begging letters still arrive． 55 they are not careful, most of their money will be spent on the protection of their possessions．Art Ⅲ Reading Comprehension（30 marks）Directions: Read the following three passages． Each passage is followed by several questions or unfinished statements． For each of them there are four choices marked A, B, C, and D．Choose the one that fits according to the information given in passage．AMy son drew a smiley face underneath the day of the family calendar．A1l I could manage was a weak smile and an acceptance of the unavoidable event．While he was counting down the days until he left home, I still felt sad and unwi11ing to accept the situation．For 18 years, we had prepared for this moment．But there's still some anxiety in ev ery parent“ was he really ready ?” The boy—man was on his way to becoming a man—boy．A few days later,1 walked in his room and felt it was an empty area of oxygen．Then I looked around and laughed．My missing scissors Just lay on the ground．Sadly, a single sock hung over the chair like a flag．An old and torn Michael Jordan poster hung on the wall．When looking through the possessions he left behind, I found Fred, his teddy bear ,which made the tonsillectomy （扁桃体切除术）seem a little less painful for him．He had marched into the hospital, holding the teddy bear tightly with great determination．And then there were loads of books, like the star wars series, which had fueled his young mind and actually helped motivate his ambitions．Now, he was off to serve his nation．I also remember him sticking his arms out like wings on my shoulders, making sounds like a pilot．He was five then．Who knew he would be a new soldier of the Air National Guard?For a while l wonder if this is a lifetime transition（转变）．when l was a girl baby, I needed my parents for everything．Then I turned into foolish independence andlearned the meaning of “no " and“don't put that in your mouth”．Each year I grew a little more independent，finding myself on my own．But in my eyes my son is forever a kid．I never made much money, and I never really cared if my boy did either．I hope that he will face the challenges in his life bravely, tolerate the frustration and never forget that this is his home．56．What impressed the author most when she entered her son's room?A．The unpleasant Smell．B．The complete emptiness．C．The superstar's poster．D．The lost scissors．57．Which of the following about Fred, the teddy bear is right?A．It was the boy's favorite toy．B．It motivated the boy's ambition．C．It showed that the boy didn’t lack sense of security．D．It gave the boy encouragement before the operation．58．The author wrote the fifth paragraph____．A．to indicate her son's dream came trueB．to show her son had various hobbiesC．to explain things always went as plannedD．to stress strong desire could lead to success59．What can be inferred from the underlined sentences?A．The author was trying her best to be a good mother．B．The author was improving her self-protection awareness．C．The author was struggling to depend less on her parents．D．The author was always saying no to her parents．60．What-s the passage mainly about?A．Children-s growing up making parents painful．B．The author's delight and unwillingness of her son-s growing up．C．Children's unwillingness to accept their growing up．D．The author-s ignorance of her son's growing up．BCollege visits are important．Before you commit years of your life and money to a school, be sure you're choosing a place," that is a good match for your personality and interests．You can-t get the "feel' of a school from any guidebook, so be sure to visit the campus．Below are a few tips for getting the most out of your college visit．1．Explore on Your ownOf course you should take the official campus tour, but be sure to allow time to hang around on your own． The trained tour guides will show you a school's selling points．But the o1dest and prettiest buildings don't give you the entire plctureof a college．Try to walk the extra mile and get the complete picture of the campus．2．Read the Bulletin BoardsWhen you visit the student center, academic buildings and residence halls．take a few minutes to read the bulletin boards．They provide an easy way to see what's happening on campus．The ads for lectures, clubs and plays can tell you what's going on outside the classrooms．3．Eat in the Dining HallYou can get a good feel for student life by eating in the dining hall．Try to sit with students if you can．Do the students seem happy or stressed? Also, is the food good? Are there adequate healthy options? Many admission offices will give prospective students coupons（优惠券）for free meals in the dining halls．4． Visit a Class in Your MajorIf you know what you want to study, a class visit makes a lot of sense．You'll get to observe other students in your field and see how involved they are in classroom discussion．Try to stay after class for a few minutes and chat with the students to get their impressions of their professors and major．Be sure to call in advance to schedule a classroom visit—most colleges don't allow visitors to drop in on class unannounced．5．Talk to Lots of StudentsYour campus tour guide has been trained to market the sch001．Try to talk to students who aren't getting paid to impress you．These important conversations can often provide you with information about c01lege life that isn't part of the admission script．Few university officials wi11 tell you if their students spend all weekend drinking of studying, but a group of random students might.For more tips，please go to collegeapps．about．com．61. The best way to know about a college in an official campus t伽r is____．A．to walk around the campus aloneB．to take pictures of the campusC．to Visit the oldest and prettiest buildingsD．to hire a tour guide62．Reading the bulletin boards can help youA．decide the major you want to studyB．find out the after-class activities on campusC．get free coupons from the admission officesD．find out the options of healthy food63．If you plan to attend a class on campus,A．drop in on a class in your majorB．have talks with the students in the classC． make an appointment in advanceD．ask how the students like their teachers and majors64． To get a real idea of the college during a visit, it's best to talk to A．famous professors B．campus tour guidesC．admission officers D．random students65．Where does the passage probably come from?A．A news report．B．A tourist poster．C．An educational journal．D．A geographical magazine．CWikipedia is an online encyclopedia（百科全书）．This name is a combination of“wiki" and “encyclopedia"． Wiki is the Hawaiian word for quick． Wiki websites are designed to enable users to make addition or edit any page Of the site．Wikipedia has been accused of exhibiting systemic prejudice and mistakes．Critics argue that Wikipedia's open nature and a lack of proper sources for much of the information makes it unreliable,Mention Wikipedia within the walls of academy and you'll find no shortage of opinions．That-s no surprise．Can an online encyclopedia that is edited by anyone be trusted as a credible information source? Should students be encouraged to use this tool? And is it even possible to discourage its users?To find out, we performed a simple experiment．We selected 100 terms from the U．S．history textbooks, We chose a mix of items that students might be asked to research for a test or paper and we entered each item into Google to find out which websites the search engine suggests as the most useful links． The results are shocking． Google listed Wikipedia as the number one: 87 times out of 100．Several conclusions can be drawn from this finding． First, people searching for information about these historical terms are finding the entries from Wikipedia helpful．Second, 6'banning" the use of Wikipedia appears hopelessly simple-minded．Jimmy Wales, one of Wikipedia's founders, told the New York Times, "They might as well suggest not listening to rock & roll either"．To our untrained eyes, the information from Wikipedia appeared just as reliable．The reason why the content is mostly reliable is probably that these terms are rather mainstream．The high-school level content is less likely to be wrong than the subjects studied in graduate schools．So when primary and secondary students are researching history, Wikipedia is still a proper place to start．66．Which of the following can best describe Wikipedia?A．Wikipedia is from the Hawaiian language with the meaning of prejudice and mistakes．B．Wikepedia is an online encyclopedia edited freely．C．Wikipedia is an encyclopedia under public attack．D． Wikipedia isa t60l for enjoying rock & roll．67．What does "it" in the second paragraph refer to?A．Wikipedia's open nature．B．Wikipedia's lack of proper sources．C．WikipediaD．Information on the Internet．68．From the underlined sentence we can conclude that ．A．Wikipedia serves only primary and secondary studentsB．people are always ready to find faults with othersC．Wikipedia has a range of topics and is rich in knowledgeD．people hold different opinions about Wikipedia69． What does the writer imply in the last paragraph?` A． Wikipedia can be useful in certain aspects．B．Students should be discouraged from using Wikipedia．C．The information from Wikipedia is totally reliable．D．Wikipedia cannot be trusted at all．70．What could be the best title for the passage?A．Wikipedia-a perfect encyclopediaB．Wikipedia-a misleading encyclopediaC．Wikipedia-a paper encyclopediaD．Wikipedia-a helpful online encyclopediaPartⅣ Writing （45 marks）Section A（10 marks）Directions: Read the following passage．Fill in the numbered blanks by using the information from the passage．Write NO MORE THAN THREE WORDS for each answer．In recent years, Smart Grid Technology is being promoted by many countries．It means the application of a digital system to the electricity network．Smart Grid Technology aims to provide the electricity industry with a better understanding of power generation and demand, and use the information to create a more efficient power network．The digital system can be used for companies to collect information about 'supply arid demand and improve engineers' ability to manage the system．With better information about the electricity demand, the network will be able to increase the amount of electricity delivered per unit generated, leading to potential decreases in fuel needs and carbon emission（排放）．Moreover, the digital system will assist in reducing maintenance costs．Smart Grid Technology offers benefits to consumers too．Through Smart meter, the equipment used by consumers to record power use in real time, the technology makes it possible to charge differently the appliances（电器）used at different times．Therefore, consumers may use their appliances, for example, washing machines at night．Smart meters can also be connected to the Internet or the telephone system, making it more convenient for consumers to switch appliances on or off when they are out．With these changes comes a range of challenges．The first involves managing the supply and demand．Sources of renewable energy, such as wind, wave and solar, are unpredictable．With alternative sources, this is far more difficult, and may lead to the failure of the system．A second problem is the fact that many renewable power generation sources are located in remote areas, such as windy uplands and coastal regions, which lack electrical infrastructure．Therefore, it is necessary to build new infrastructure there．Although Smart Grid Technology is still in its early years, people believe thatby the year 2050, changes in the energy supply will have transformed our homes, our roads and our behavior．Isn’t it a bright future?Section B（10 marks）Directions: Read the following passage．Aaswer the questions according to the information given in the passage．Several years ago, I had to replace a receptionist for my unit． Before I advertised for the position,1 was approached by another manager, my wife's friend, and asked if l would consider a transfer from his staff．It appeared from his description of this employee, Maria, that she was having a lot of problems performing her duties, had difficulties with her boss, and did not appear to respond to any attempts her boss had made to settle the issues．Now, after listening to his explanation, I really had severe doubts． What crossed my mind, after this discussion, was that he wanted me to take a problem off his hands．I told him l would think about it and would let him know the next day．After thinking more about this potential performance issue, I decided that l would agree to speak with Maria privately before I made any decision．And, what I found out during this interview left me deep in thought．As it turned out, the performance issue appeared to result from working for a boss who never encouraged his employees, and was constantly changing the workload and routines．The end result was a totally ruined employee who was unconfident and hated to see a new workday start．I decided to accept her transfer to my unit． What started out as a shy, withdrawnand frightened employee turned out to be one of the most productive and loyal employees that I had ever worked with．Maria couldn't do enough and was constantly asking for more responsibility．I was more than happy to give her what she needed-praise for work well done, respect as a person, and encouragement for doing well in anything she dealt with．Sometimes, we are guilty of holding others back from accomplishing or doing or developing as they should or could． It may be a wise idea that we all look in the mirror, from time to time, to see what we are really like．81．According to the first paragraph, why did the author plan to advertise?（No more than 8 words）（2 marks）To replace a receptionist for his unit．/He wanted a new receptionist．82．How did the author get to know about Maria after hearing her boss's description of her? （No more than 11 words）（2 marks）By speaking with her privately before he made any decision．／He spoke with her privately before he made any decision．83．In Maria's eyes, how did her former boss treat his employees? （No more than 14 words）（3 marks）He never encouraged his employees and was constantly changing the workload and routines．84．What reaction did the author have towards Maria's good performances in her new unit? （No more than 8 words）（3 marks）He praised, respected and encouraged her．/ He gave her praise, respect and encouragement．Section C (25 marks )Directions : Write an English composition according to the instructions given below in Chinese.请以“My Most Unforgettable Experience in High School”为题，描述高中三年中让你最难忘的一段经历以及该经历对你的影响或启发。

湖南省湖南师大附中长沙市一中等六校高三英语下学期4月联考试题牛津译林版1

湖南省2014届高三六校联考英语试题总分:150分时量:120分钟Part I Listening Comprehension （30 marks）Section A（ 22．5 marks）Directions: In this section, you' II hear six conversations between two speakers．For each conversation, there are several questions and each question is followed by three choices marked A, B and C．Listen carefully and then choose the best answer for each question．You will hear each conversation TWICE．Conversation 1l．Where does the man play tennis?A．At the university．B．At the community center．C．At the club．2．How often does the woman swim?A．Once a week． B．Five times a week． C．Three times a week．Conversation 23．Whose photo are the speakers talking about?A．The woman's friends'．B．The woman's colleagues'．C．The woman's family's．4．What does Mary do?A．She is an accountant．B．She is a business woman．C．She is a lawyer．Conversation 35．What is the man doing?A．Preparing for a trip． B．Writing an article． C．Selling hisarticles．6．Where did the man go last month?A．Amazon． B．Asia． C．America．Conversation 47．Where is Rick working now?A．In a travel company． B．In a hotel． C．In a law office．8．How does Rick feel about his trip?A．Excited． B．Curious． C．Nervous．9．When will Rick probably leave?A．In three days． B．In a week． C．In a month．Conversation 5110．What is the man worried about?A．His children's friends．B．His children's behaviors．C．His children's health．1l．What do we know about the man's children?A．They do their schoolwork till midnight．B．They are interested in computer games．C．They feel bored about pop music．12．How many children does the woman have?A. 2． B．3． c．4．Conversation 613．Which of the following is TRUE about the boy?A．He got an A in maths．B．He hasn't studied hard enough at history．C．He likes history．14．What can we learn about the boy's study of government?A．He has a new teacher．B．He has got a B in it．C．He finds it confusing．15．What is the boy's mother's attitude to his report card?A．She doesn't take it too hard．B．She doesn't excuse him．C．She thinks it very bad．Section B （7．5 marks）Directions: In．this section, you will hear a short passage, Listen carefully and then fill in the numbered blanks with the information you have heard． Fill in each blank with NO MORE THAN THREF, WORDS．You'll hear the short passage TWICE．） Language Knowledge （45 marksPart Ⅱ）（15 marksSection Achoices four are unfinished sentences there following For Directions．．each of the．marked A,B,C and D．Choose the one that best completes the sentenceExample:The wild flowers looked like a soft orange blanket the desertto cover．cover C．． Acovering Bcovered ．D2The answer is A．21．We the computer problem, and got back online to continue our research．A．have solved B．solve C．would solve D．solved22． Making a silly mistake at a job interview is sometimes very embarrassing, butif ， it may become a driving force．A．properly handled B．properly handlingC．you properly handled D．you properly handling23． The little girl showed no anxiety before the competition． She seemed for it pretty wellA．to prepare B．to have preparedC．having prepared D．preparing24． I first met Mr．Li three years ago．He had just returned from abroad and English in Senior One then．A．taught B．would teach C．was teaching D．had taught25． Only when at least 50% of our energy is provided by grain a balanced and healthy food consumption．A．we had B．did we have C．we have D．do we have26．The Beijing transport authority released the news buyers of green cars can now receive a maximum amount of 5700 yuan from the central government．A．when B．that C．if, D．which27．—How can you catch a taxi at home?—Don-t worry．I have opened an account on DiDi Dache．I for a taxi driver to pick me up here．A．was waiting B．have waited C．am waiting D．had waited28． Mr．Jenkins wanted to buy a computer as soon as possible, he decidedto shop around to see which shop had the best deal．A．While B．As C．When D．Once29．Gertrude was upset when one of her cats overturned a glass of wine, adot on her favorite carpet．A．leave B．left C．to leave D．leaving30．—What a heavy rain ?—Yes．I to take an umbrella with me, but I didn't ．A．had planned B．would plan C．have planned D．will plan31．—Mary doesn't mind lending you her e-dictionary．—She I have already borrowed Jane's．A．can't B．mustn't C．needn't D．shouldn't32．____ your support, I could have missed the first prize．A．Should it not be for B．Had it not been forC．If it had been for D． If it should be for33．Today is my grandfather's l00th birthday, an occasion we are going to celebrate with all the family and a few friends in for the evening．A．who B．which C．where D．when34．It is very important to recognize what kind of person you are and which special qualities make you different from in your club．3A．anyone else B．the other C．someone else D．others35． He appreciated a chance to make a presentation in the annual seminar on Comparative Literature．A．giving B ．to give C．being given D．to be givenSection B (18 marks)Directions: For each blank in the following passages there are four words or phrases marked A, B, C and D．Fill in each blank with the word or phrase that best fitsthe context．Annie Dillard, the author of TEACHING A STONE TO TALK, tells us a sad story abouta British Arctic expedition（远征）that took ship in 1845 to explore the Northwest Passage around the Canadian Arctic to the Pacific Ocean．Neither of the two ships and none of the 138 men aboard 36 ．In her book Captain Sir John Franklin prepared as if they were 37 in a pleasure travel rather than a challenging journey through one of earth's38 ．environments． He packed a 1,200 volume library, china place settings and glass wine cups． Years later, some of them were found near some 39 bodies．The voyage was hopeless when the ships sailed into 40 cold waters and became trapped in ice．First 41 coated the decks, the poles and the rope．Then water froze around the rudders （舵） , and the ships became 42 in the frozen sea desperately．Sailors set out to search for help, but soon 43 to severe Arctic weather and died of exposure to its harsh winds and low temperatures．For the next twenty years, 44 of the expedition were found all over the frozen landscape．Dillard reports that on a voyage that was to last 2-3 years the crew did not prepare for the chance of the ships being ice-locked and the captain 45 just a 12-day supply of coal for the spare steam engines．Historians may doubt the wisdom of such a journey．But I have a question: Are we, too, prepared for the important voyage we've started-the journey we- call 46 ”?I want to be as ready as possible for what may lie ahead．As long as we are alive, our journey is not over．And in large part, the success of our journey will be 47 by our regular and systematic preparation．36．A．continued B．returned C．existed D．survived 37．A．resulting B．consisting C．engaging D．ending38．A．toughest B．excellent C．commonest D．superb39．A．burned B．rotten C．swelled D．frozen40．A．slightly B．lightly C．ordinarily D．extremely41．A．ice B．wave C．steam D．water42．A．lost B．moved C．locked D．opened43．A．put up B．gave in C．brought in D．broke up44．A．library B．place settings C．wine cups D．remains 45．A．donated B．carried C．expected D．answered46．A．trouble B．happiness C．life D．death447. A. determined B. damaged C. defeated D. decoratedSection C（12 marks）Directions: Complete the following passage by filling in each black with one word that best fits the context．Nearly all of us have dreamed about winning the big prize in a lottery（彩票） We dream about 48 we would do with the money, but we 49 stop to think about how the money would change us．For most of us, our way of life is closely linked 50 our economic circumstances．The different parts of our lives fit together like a jigsaw（拼板）- work, home, friends, hobbies and sports make up our world． 51 sudden fortune would change it all and break the jigsaw．For example , some people like the idea of 52 having to work, but winners have found that without work there is no reason to get up in the morning．In seems great to move to a bigger house in a wealthy area, 53 if you do that, you leave old friends behind．Usually, though winners are advised not to publicize _ 54 _ addresses and phone numbers, begging letters still arrive． 55 they are not careful, most of their money will be spent on the protection of their possessions．Art Ⅲ Reading Comprehension（30 marks）Directions: Read the following three passages． Each passage is followed by several questions or unfinished statements． For each of them there are four choices marked A, B, C, and D．Choose the one that fits according to the information given in passage．AMy son drew a smiley face underneath the day of the family calendar．A1l I could manage was a weak smile and an acceptance of the unavoidable event．While he was counting down the days until he left home, I still felt sad and unwi11ing to accept the situation．For 18 years, we had prepared for this moment．But there's still some anxiety in every parent“ was he really ready ?” The boy—man was on his way to becoming a man—boy．A few days later,1 walked in his room and felt it was an empty area of oxygen．Then I looked around and laughed．My missing scissors Just lay on the ground．Sadly, a single sock hung over the chair like a flag．An old and torn Michael Jordan poster hung on the wall．When looking through the possessions he left behind, I found Fred, his teddybear ,which made the tonsillectomy （扁桃体切除术）seem a little less painful forhim．He had marched into the hospital, holding the teddy bear tightly with great determination．And then there were loads of books, like the star wars series, which had fueled his young mind and actually helped motivate his ambitions．Now, he was off to serve his nation．I also remember him sticking his arms out like wings on my shoulders, making sounds like a pilot．He was five then．Who knew he would be a new soldier of theAir National Guard?For a while l wonder if this is a lifetime transition（转变）．when l was a girl baby, I needed my parents for everything．Then I turned into foolish independence and5learned the meaning of “no and“don't put that in your mouth”．Each year I grewa little more independent，finding myself on my own．But in my eyes my son is forever a kid．I never made much money, and I never really cared if my boy did either．I hope that he will face the challenges in his life bravely, tolerate the frustration and never forget that this is his home．56．What impressed the author most when she entered her son's room?A．The unpleasant Smell．B．The complete emptiness．C．The superstar's poster．D．The lost scissors．57．Which of the following about Fred, the teddy bear is right?A．It was the boy's favorite toy．B．It motivated the boy's ambition．C．It showed that the boy didn't lack sense of security．D．It gave the boy encouragement before the operation．58．The author wrote the fifth paragraph____．A．to indicate her son's dream came trueB．to show her son had various hobbiesC．to explain things always went as plannedD．to stress strong desire could lead to success59．What can be inferred from the underlined sentences?A．The author was trying her best to be a good mother．B．The author was improving her self-protection awareness．C．The author was struggling to depend less on her parents．D．The author was always saying no to her parents．60．What-s the passage mainly about?A．Children-s growing up making parents painful．B．The author's delight and unwillingness of her son-s growing up．C．Children's unwillingness to accept their growing up．D．The author-s ignorance of her son's growing up．BCollege visits are important．Before you commit years of your life and money to a school, be sure you're choosing a place, that is a good match for your personality and interests．You can-t get theeel' of a school from any guidebook, so be sureto visit the campus．Below are a few tips for getting the most out of your college visit．1．Explore on Your ownOf course you should take the official campus tour, but be sure to allow time to hang around on your own． The trained tour guides will show you a school's selling points．But the o1dest and prettiest buildings don't give you the entire plcture6of a college．Try to walk the extra mile and get the complete picture of the campus．2．Read the Bulletin BoardsWhen you visit the student center, academic buildings and residence halls．take a few minutes to read the bulletin boards．They provide an easy way to see what's happening on campus．The ads for lectures, clubs and plays can tell you what's going on outside the classrooms．3．Eat in the Dining HallYou can get a good feel for student life by eating in the dining hall．Try to sit with students if you can．Do the students seem happy or stressed? Also, is the food good? Are there adequate healthy options? Many admission offices will give prospective students coupons（优惠券）for free meals in the dining halls．4． Visit a Class in Your MajorIf you know what you want to study, a class visit makes a lot of sense．You'll get to observe other students in your field and see how involved they are in classroom discussion．Try to stay after class for a few minutes and chat with the studentsto get their impressions of their professors and major．Be sure to call in advance to schedule a classroom visit—most colleges don't allow visitors to drop in on class unannounced．5．Talk to Lots of StudentsYour campus tour guide has been trained to market the sch001．Try to talk to students who aren't getting paid to impress you．These important conversations can often provide you with information about c01lege life that isn't part of the admission script．Few university officials wi11 tell you if their students spend all weekend drinking of studying, but a group of random students might.For more tips，please go to collegeapps．about．com．61. The best way to know about a college in an official campus t伽r is____．A．to walk around the campus aloneB．to take pictures of the campusC．to Visit the oldest and prettiest buildingsD．to hire a tour guide62．Reading the bulletin boards can help youA．decide the major you want to studyB．find out the after-class activities on campusC．get free coupons from the admission officesD．find out the options of healthy food63．If you plan to attend a class on campus,A．drop in on a class in your majorB．have talks with the students in the classC． make an appointment in advanceD．ask how the students like their teachers and majors64． To get a real idea of the college during a visit, it's best to talk toA．famous professors B．campus tour guidesC．admission officers D．random students65．Where does the passage probably come from?7A．A news report．B．A tourist poster．C．An educational journal．D．A geographical magazine．CWikipedia is an online encyclopedia（百科全书）．This name is a combination of“wiki and “湥祣汣灯摥慩． Wiki is the Hawaiian word for quick． Wiki websites are designed to enable users to make addition or edit any page Of the site．Wikipedia has been accused of exhibiting systemic prejudice and mistakes．Critics argue that Wikipedia's open nature and a lack of proper sources for much of the information makes it unreliable,Mention Wikipedia within the walls of academy and you'll find no shortage of opinions．That-s no surprise．Can an online encyclopedia that is edited by anyone be trusted as a credible information source? Should students be encouraged to use this tool? And is it even possible to discourage its users?To find out, we performed a simple experiment．We selected 100 terms from the U．S．history textbooks, We chose a mix of items that students might be asked to research for a test or paper and we entered each item into Google to find out which websites the search engine suggests as the most useful links． The results are shocking． Google listed Wikipedia as the number one: 87 times out of 100．Several conclusions can be drawn from this finding． First, people searching for information about these historical terms are finding the entries from Wikipedia helpful．Second, 6'banning the use of Wikipedia appears hopelesslysimple-minded．Jimmy Wales, one of Wikipedia's founders, told the New York Times, 呜敨?業桧?獡眠汥?畳杧獥?潮?楬瑳湥湩?潴爠捯?…潲汬攠瑩敨屲．To our untrained eyes, the information from Wikipedia appeared just as reliable．The reason why the content is mostly reliable is probably that these terms are rather mainstream．The high-school level content is less likely to be wrong than the subjects studied in graduate schools． So when primary and secondary students are researching history, Wikipedia is still a proper place to start．66．Which of the following can best describe Wikipedia?A．Wikipedia is from the Hawaiian language with the meaning of prejudice andmistakes．B．Wikepedia is an online encyclopedia edited freely．C．Wikipedia is an encyclopedia under public attack．D． Wikipedia isa t60l for enjoying rock & roll．67．What does it in the second paragraph refer to?A．Wikipedia's open nature．B．Wikipedia's lack of proper sources．C．WikipediaD．Information on the Internet．68．From the underlined sentence we can conclude that ．A．Wikipedia serves only primary and secondary students8B．people are always ready to find faults with othersC．Wikipedia has a range of topics and is rich in knowledgeD．people hold different opinions about Wikipedia69． What does the writer imply in the last paragraph?` A． Wikipedia can be useful in certain aspects．B．Students should be discouraged from using Wikipedia．C．The information from Wikipedia is totally reliable．D．Wikipedia cannot be trusted at all．70．What could be the best title for the passage?A．Wikipedia-a perfect encyclopediaB．Wikipedia-a misleading encyclopediaC．Wikipedia-a paper encyclopediaD．Wikipedia-a helpful online encyclopediaPartⅣ Writing （45 marks）Section A（10 marks）Directions: Read the following passage．Fill in the numbered blanks by using the information from the passage．Write NO MORE THAN THREE WORDS for each answer．In recent years, Smart Grid Technology is being promoted by many countries．It means the application of a digital system to the electricity network．Smart Grid Technology aims to provide the electricity industry with a better understanding of power generation and demand, and use the information to create a more efficient power network．The digital system can be used for companies to collect information about 'supply arid demand and improve engineers' ability to manage the system．With better information about the electricity demand, the network will be able to increase the amount of electricity delivered per unit generated, leading to potential decreases in fuel needs and carbon emission（排放）．Moreover, the digital system will assist in reducing maintenance costs．Smart Grid Technology offers benefits to consumers too．Through Smart meter, the equipment used by consumers to record power use in real time, the technology makesit possible to charge differently the appliances（电器）used at different times．Therefore, consumers may use their appliances, for example, washing machines at night．Smart meters can also be connected to the Internet or the telephone system, making it more convenient for consumers to switch appliances on or off when they are out．With these changes comes a range of challenges．The first involves managing the supply and demand．Sources of renewable energy, such as wind, wave and solar, are unpredictable．With alternative sources, this is far more difficult, and may lead to the failure of the system．A second problem is the fact that many renewable power generation sources are located in remote areas, such as windy uplands and coastal regions, which lack electrical infrastructure．Therefore, it is necessary to build new infrastructure there．Although Smart Grid Technology is still in its early years, people believe that9by the year 2050, changes in the energy supply will have transformed our homes, our roads and our behavior．Isn't it a bright future?Section B（10 marks）Directions: Read the following passage．Aaswer the questions according to the information given in the passage．Several years ago, I had to replace a receptionist for my unit． Before I advertised for the position,1 was approached by another manager, my wife's friend, and askedif l would consider a transfer from his staff．It appeared from his description of this employee, Maria, that she was having a lot of problems performing her duties, had difficulties with her boss, and did not appear to respond to any attempts her boss had made to settle the issues．Now, after listening to his explanation, I really had severe doubts． What crossed my mind, after this discussion, was that he wanted me to take a problem off his hands．I told him l would think about it and would let him know the next day．After thinking more about this potential performance issue, I decided that l would agree to speak with Maria privately before I made any decision．And, what I found out during this interview left me deep in thought．As it turned out, the performance issue appeared to result from working for a boss who never encouraged his employees, and was constantly changing the workload and routines．The end result was a totally ruined employee who was unconfident and hated to see a new workday start．I decided to accept her transfer to my unit． What started out as a shy, withdrawn10and frightened employee turned out to be one of the most productive and loyal employees that I had ever worked with．Maria couldn't do enough and was constantlyasking for more responsibility．I was more than happy to give her what sheneeded-praise for work well done, respect as a person, and encouragement for doing well in anything she dealt with．Sometimes, we are guilty of holding others back from accomplishing or doing or developing as they should or could． It may be a wise idea that we all look in the mirror, from time to time, to see what we are really like．81．According to the first paragraph, why did the author plan to advertise?（No more than 8 words）（2 marks）To replace a receptionist for his unit．/He wanted a new receptionist．82．How did the author get to know about Maria after hearing her boss's description of her? （No more than 11 words）（2 marks）By speaking with her privately before he made any decision．／He spoke with her privately before he made any decision．83．In Maria's eyes, how did her former boss treat his employees? （No more than14 words）（3 marks）He never encouraged his employees and was constantly changing the workload and routines．84．What reaction did the author have towards Maria's good performances in her new unit? （No more than 8 words）（3 marks）He praised, respected and encouraged her．/ He gave her praise, respect and encouragement．Section C (25 marks )Directions : Write an English composition according to the instructions given below in Chinese.请以“My Most Unforgettable Experience in High School”为题，描述高中三年中让你最难忘的一段经历以及该经历对你的影响或启发。

湖南省湖南师大附中、长沙市一中等六校高三化学下学期

湖南省2014届高三六校联考化学能力试题时量：150分钟满分：300分本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第33—40题为选考题，其他题为必考题。

考生作答时，将答案答在答题卡上，在本试卷上答题无效。

可能用到的相对原子质量：H～l C～12 O～16 Cu～64第I卷选择题（共126分）一、选择题（本题共13小题，每小题6分，共78分。

在每小题列出的四个选项中，只有一项是符合题目要求的）7．下列说法正确的是A．PM2．5表示每立方米空气中直径小于或等于2．5毫米的颗粒物的含量B．将浸泡过高锰酸钾溶液的硅藻土放入新鲜的水果箱内是为了催熟水果“夺命快递”事件源于氟乙酸甲酯泄漏，其结构简式为CH2 FCOOCH3 C．2013年11月29日，D．绿色化学的核心是利用化学原理治理污染8．设NA代表阿伏加德罗常数的值，下列叙述正确的是A．18 g D2O含电子数目为10NAB．1 L 0．1 mol／L NH4NO3溶液中NHt数目为0．1NAC．1 mol CaC2固体含离子数目为3NAD．ag C2 H4和C3H6混合气体含原子数目为3aNA／149．短周期元素A、B、C，D、E原子序数依次增大，其中A、C同主族，B、E同主族，A的最外层电子数与次外层电子数相等，E是太阳能转化为电能的常用材料，则下列说法正确的是A．最价氧化物对应水化物碱性强弱：A>CB．C、D的简单阳离子均可促进水的电离C．B、E的最高价氧化物具有相似的物理性质和化学性质D．含D的盐溶于水一定显酸性10．下列实验操作正确且能达到预期目的的是11．化学兴趣小组利用如图装置进行电化学实验，下列预期现象正确的是A．X和Y与电流表连接，将Zn换成Fe，测得电流更大B．X和Y与电流表连接，将盐桥换成铜线，电流表指针偏转C．X和Y分别与电源“+”、“一”极相连，Cu极质量增大D．X和Y分别与电源“一”、“+”极相连，盐桥中阴离子向Zn极移动12．—C4 H9和—C2 H5O取代苯环上的氢原子，形成的有机物中能与金属钠反应的有A．8种B．18种C．24种D．36种13．常温下，用0．100 0 mol·L－1 NaOH溶液分别滴定20．00 mL 0．100 0 mol·L－1 HA 和HB溶液，得到2条滴定曲线，如下图所示（图1表示HA，图2表示HB），下列说洼正确的是A．HA是弱酸，HB是强酸B．达到E点时，对应混合溶液中c（Na+）=c（B－）+c（HB）C．达到B、D点时，两溶液pH均为7，反应消耗的n（HA）=n（HB）D．当达到E点后，继续滴加NaOH溶液，对应混合溶液中各离子浓度由大到小的顺序均为c（Na+）> c（B－）>c（OH－）>c（H+）26．（14分，每空2分）用软锰矿（主要成分为MnO2，还有少量含铁、铜、汞的化合物）制备MnSO4电解液继而提取锰的工业流程如下（1）已知锰粉的主要成分为MnO，反应①的化学方程式为。

湖南省湖南师大附中、长沙市一中等六校高三历史下学期4月联考试题新人教版

湖南省2014届高三六校联考试题历试题本试题卷分选择题和非选择题两部分，时量150分钟，满分300分。

第I卷选择题（共140分）一、选择题（本大题共35小题，每小题4分，共计140分。

每小题列出的四个选项中，只有一项符合题意。

）24．《礼记》中说：“山川神祗，有不举祭者为不敬，不敬者，君削以地；宗庙有不顺者为不孝，不孝者，看绌以爵。

变礼易乐者为不从，不从者君流。

革制度衣服者为畔，畔者君讨。

”这表明西周的礼乐制度（B）A．为君主大权独揽的意识服务 B．有利于形成天下一家的文化心理C．具有威慑性与持久性的特点 D．实现了国家对土地和人口的控制25．有学者认为，“新的哲学流派促进了对孑L子学说和其他古代思想家理解的争鸣。

他们力求证明本土思想体系如儒家和道家思想与外来思想如佛教相比的优越性。

”这里的“哲学流派”是指（B）A．董仲舒的新儒学B．程朱理学C．启蒙思想D．乾嘉学派26．明代文人的作品逐步具有了商品属性，郑板桥公开开出自己的书画价格（时称“笔榜”）。

袁枚给某盐商题跋22个字，得了2 000两银子，成为最善经营文化产业的文人。

这一现象反映了（D）A．商人开始重视文化修莠 B．驵清时期文化产业兴盛C．社会拜金逐利风气盛行’D．经济发展冲击传统观念27．1912年2月12日，溥仪颁布《宣统帝退位诏书》：“予亦何忍因一姓之尊荣，拂兆民之好恶。

是用外观大势，内审舆情，特率皇帝将统治权公诸全国，定为共和立宪国体。

”据此，有学者认为这是中国版的“光荣革命”。

这一认识的依据是两者都（C）A．体面地保留了封建王室及其待遇 B．演绎了一场没有流血的宫廷政变C．是资产阶级以和平手段夺取政权 D．以法律手段有效限制了君主权力28．1944年美日莱特湾大海战，日本运用“神风特攻战术”，共出动“神风”飞机55架以自杀式攻击，重创美国军舰。

日本南九州市“知览特攻和平会馆”中，满目都是军旗、军刀、血书。

一众遗书中，“玉碎一‘报君（天皇）。

湖南省湖南师大附中长沙市一中等六校高三下学期4月联考试题语文含答案

湖南省2014届高三六校联考试题语文试题本试卷共7道大题，21道小题（2道选做题任选1小题）。