2015届考前针对训练题(一)

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安徽省淮北市2015届高三第一次模拟考试数学试题 (理科)

安徽省淮北市2015届高三第一次模拟考试数学试题（理科）2015.1.24考生注意事项：1．答题前，考生务必在试题卷、答题卡规定的地方填写自己的座位号、姓名。

考生要认真核对答题卡上粘贴的条形码的“考场座位号、姓名”与考生本人考场座位号、姓名是否一致。

2. 本试卷满分150分，考试时间120分钟。

3．考生务必在答题卷上答题，考试结束后交回答题卷。

第I 卷 (选择题共50分)一．选择题（本大题共10小题，每小题只有一个正确答案，每小题5分）1.已知,,x y R i ∈为虚数单位，且(2)1x i y i --=+，则(1)x y i ++的值为（）。

A ．4B ． 4-C ． 44i +D ．2i 2.已知n X m log =，则1>mn 是1>X 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. 已知棱长为1的正方体的俯视图是边长为1正方形，则其主视图的面积不可能是（）A.2 B.212- C. 1 D. 433 4. 等差数列{}n a 有两项m a 和()k a m k ≠，满足11,m k a a k m==，则该数列前mk 项之和为 ( ) A. 12mk - B 2mk C 12mk + D 12mk +5.下列命题正确的是（）A.函数)32sin(π+=x y 在区间)6,3(ππ-内单调递增B.函数x x y 44sin cos -=的最小正周期为π2C.函数)3cos(π+=x y 的图像是关于点)0,6(π成中心对称的图形 D.函数)3tan(π+=x y 的图像是关于直线6π=x 成轴对称的图形6.已知实数x ，y 满足200,0x y x y y k +≥⎧⎪-≤⎨⎪≤≤⎩设y x m +=，若m 的最大值为6，则m 的最小值为（）A ．—3B ．—2C ．—1D ．07. 某项实验，要先后实施6个程序，其中程序A 只能出现在第一或最后一步，程序B 和C 在实施时必须相邻，问实验顺序的编排方法共有（） A ．34种B ．48种C ．96种D ．144种8. 若函数)(x f 的导函数是34)(2+-='x x x f ，则函数)()(x a f x g = (0<a<1)的单调递减区间是( )A 、 []0,3log a ，[)+∞,1B 、(]),0[,3log ,+∞∞-aC 、[]a a ,3 D 、[]1,3log a9. 若对任意[]5,0∈x ，不等式x n xx m 514241+≤+≤+恒成立，则一定有（） A ． 31,21-≥≤n m B ．31,21-≥-≤n m C ．31,21≥-≤n m D ．31,21->-<n m10.已知ABC ∆的外接圆的圆心为O ,满足：CB n CA m CO +=,234=+n m ，346,则=∙( )A. 36B. 24C. 243D. 312 二、填空题（每小题5分，共25分）11. 执行如图所示的程序框图，若输入A 的值为2，则输出的P 值为12. 在52512⎪⎭⎫ ⎝⎛-x x 的二项展开式中，x 的系数为13．已知),0(,,,,+∞∈≠∈+y x n m R n m ，则有yx n m y n x m ++≥+222)(，且当yn x m =时等号成立，利用此结论，可求函数x x x f -+=1334)(，)1,0(∈x 的最小值为14. 已知正方体ABCD-A 1B 1C 1D 1的棱长为2，M 、N 分别为AD 、CC 1的中点，O 为上底面A 1B 1C 1D 1的中心，则三棱锥O-MNB 的体积是。

2015届高考数学一轮总复习阶段性测试题1(集合与常用逻辑用语)

阶段性测试题一(集合与常用逻辑用语)本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

满分150分。

考试时间120分钟。

第Ⅰ卷(选择题共60分)一、选择题(本大题共12个小题，每小题5分，共60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．(文)(2014·甘肃临夏中学、金昌市二中期中)设集合A＝{x|x>1}，B＝{x|x(x－2)<0}，则A∩B 等于()A．{x|x>2}B．{x|0<x<2}C．{x|1<x<2} D．{x|0<x<1}[答案] C[解析]∵B＝{x|x(x－2)<0}＝{x|0<x<2}，∴A∩B＝{x|1<x<2}．(理)(2014·福建省闽侯二中、永泰二中、连江侨中、长乐二中联考)已知全集U＝R，集合M＝{x|x2－x＝0}，N＝{x|x＝2n＋1，n∈Z}，则M∩N为()A．{0} B．{1}C．{0,1} D．∅[答案] B[解析]∵M＝{x|x2－x＝0}＝{0,1}，N＝{x|x＝2n＋1，n∈Z}中的元素是奇数，∴M∩N＝{1}，选B.2．(2014·威海期中)已知集合A＝{－1,1}，B＝{m|m＝x＋y，x∈A，y∈A}，则集合B等于() A．{－2,2} B．{－2,0,2}C．{－2,0} D．{0}[答案] B[解析]∵x∈A，y∈A，A＝{－1,1}，m＝x＋y，∴m的取值为－2,0,2，即B＝{－2,0,2}，故选B.3．(2014·山西曲沃中学期中)集合A＝{x|(x－1)(x＋2)≤0}，B＝{x|x<0}，则A∪B＝()A．(－∞，0] B．(－∞，1]C．[1,2] D．[1，＋∞)[答案] B[解析]∵A＝{x|－2≤x≤1}，B＝{x|x<0}，∴A∪B＝{x|x≤1}，故选B.4．(文)(2014·山东省德州市期中)若U＝{1,2,3,4,5,6}，M＝{1,2,4}，N＝{2,3,6}，则∁U(M∪N)＝()A．{1,2,3} B．{5}C．{1,3,4} D．{2}[答案] B[解析] ∵U ＝{1,2,3,4,5,6}，M ∪N ＝{1,2,3,4,6}， ∴∁U (M ∩N )＝{5}．(理)(2014·文登市期中)已知集合A ＝{x |log 4x <1}，B ＝{x |x ≥2}，则A ∩(∁R B )＝( ) A ．(－∞，2) B ．(0,2) C ．(－∞，2] D ．[2,4)[答案] B[解析] ∵A ＝{x |log 4x <1}＝{x |0<x <4}，B ＝{x |x ≥2}，∴∁R B ＝{x |x <2}，所以A ∩∁R B ＝(0,2)，故选B.5．(文)(2014·福州市八县联考)命题“有些实数的绝对值是正数”的否定是( ) A ．∀x ∈R ，|x |>0 B ．∃x 0∈R ，|x 0|>0 C ．∀x ∈R ，|x |≤0 D ．∃x 0∈R ，|x 0|≤0[答案] C[解析] 由词语“有些”知原命题为特称命题，故其否定为全称命题，因为命题的否定只否定结论，所以选C.(理)(2014·甘肃临夏中学期中)命题“存在x ∈Z ，使x 2＋2x ＋m ≤0成立”的否定是( ) A ．存在x ∈Z ，使x 2＋2x ＋m >0 B ．不存在x ∈Z ，使x 2＋2x ＋m >0 C ．对于任意x ∈Z ，都有x 2＋2x ＋m ≤0 D ．对于任意x ∈Z ，都有x 2＋2x ＋m >0 [答案] D[解析] 特称命题的否定是全称命题．6．(文)(2014·河北冀州中学期中)下列命题中的真命题是( ) A ．∃x ∈R ，使得sin x ＋cos x ＝32B ．∀x ∈(0，＋∞)，e x >x ＋1C ．∃x ∈(－∞，0)，2x <3xD ．∀x ∈(0，π)，sin x >cos x [答案] B[解析] ∵sin x ＋cos x ＝2sin(x ＋π4)∈[－2，2]，32>2，∴不存在x ∈R ，使sin x ＋cos x ＝32成立，故A 错；令f (x )＝e x －x －1(x ≥0)，则f ′(x )＝e x －1，当x >0时，f ′(x )>0，∴f (x )在[0，＋∞)上单调递增，又f (0)＝0，∴x >0时，f (x )>0恒成立，即e x >x ＋1对∀x ∈(0，＋∞)都成立，故B 正确；在同一坐标系内作出y ＝2x 与y ＝3x 的图象知，C 错误；当x ＝π4时，sin x ＝22＝cos x ，∴D 错误，故选B.(理)(2014·山东省德州市期中)下面命题中，假命题是( ) A ．∀x ∈R,3x >0B ．∃α，β∈R ，使sin(α＋β)＝sin α＋sin βC ．∃m ∈R ，使f (x )＝mxm 2＋2m 是幂函数，且在(0，＋∞)上单调递增D ．命题“∃x ∈R ，x 2＋1>3x ”的否定是“∀x ∈R ，x 2＋1>3x ” [答案] D[解析] 由指数函数性质知，对任意x ∈R ，都有3x >0，故A 真；当α＝π3，β＝2π时，sin(α＋β)＝sin α＋sin β成立；故B 真；要使f (x )＝mxm 2＋2m 为幂函数，应有m ＝1，∴f (x )＝x 3，显然此函数在(0，＋∞)上单调递增，故C 真；D 为假命题，“>”的否定应为“≤”．7．(文)(2014·甘肃省金昌市二中期中)a 、b 为非零向量，“a ⊥b ”是“函数f (x )＝(x a ＋b )·(x b －a )为一次函数”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[答案] B[解析] ∵f (x )＝(x a ＋b )·(x b －a )＝x 2a ·b ＋x (|b |2－|a |2)－a ·b ，当f (x )为一次函数时，a ·b ＝0且|b |2－|a |2≠0，∴a ⊥b ，当a ⊥b 时，f (x )未必是一次函数，因为此时可能有|a |＝|b |，故选B.(理)(2014·江西临川十中期中)已知平面向量a ，b 满足|a |＝1，|b |＝2，a 与b 的夹角为60°，则“m ＝1”是“(a －m b )⊥a ”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 [答案] C[解析] ∵|a |＝1，|b |＝2，〈a ，b 〉＝60°，∴a ·b ＝1×2×cos60°＝1，(a －m b )⊥a ⇔(a －m b )·a ＝0⇔|a |2－m a ·b ＝0⇔m ＝1，故选C.8．(2014·江西都昌一中月考)已知全集U ＝{1,2,3,4,5,6}，集合A ＝{2,3,4}，集合B ＝{2,4,5}，则右图中的阴影部分表示( )A ．{2,4}B ．{1,3}C ．{5}D ．{2,3,4,5} [答案] C[解析] 阴影部分在集合B 中，不在集合A 中，故阴影部分为B ∩(∁U A )＝{2,4,5}∩{1,5,6}＝{5}，故选C.9．(2014·华安、连城、永安、漳平一中，龙海二中，泉港一中六校联考)已知m ，n 是两条不同的直线，α，β，γ是三个不同的平面，下列命题正确的是( )A ．若m ∥α，n ∥α，则m ∥nB ．若α⊥β，α⊥γ，则β∥γC ．若m ∥α，m ∥β，则α∥βD ．若m ⊥α，m ⊥β，则α∥β [答案] D[解析] m ∥α，n ∥α时，m 与n 可平行，也可相交或异面，故A 错误；由正方体相邻三个面可知，α⊥β，α⊥γ时，β与γ可能相交，故B 错；当α∩β＝l ，m ⊄α，m ⊄β，m ∥l 时，m ∥α，m ∥β，故C 错，故选D.10．(2014甘肃临夏中学期中)已知函数f (x )＝x ＋b cos x ，其中b 为常数．那么“b ＝0”是“f (x )为奇函数”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[答案] C[解析] 当b ＝0时，f (x )＝x 为奇函数，故满足充分性；当f (x )为奇函数时，f (－x )＝－f (x )，∴－x ＋b cos x ＝－x －b cos x ，从而2b cos x ＝0，∵此式对任意x ∈R 都成立，∴b ＝0，故满足必要性，选C.11．(2014·海南省文昌市检测)下列命题中是假命题．．．的是( ) A ．∃m ∈R ，使f (x )＝(m －1)·xm 2－4m ＋3是幂函数，且在(0，＋∞)上单调递减B ．∀a >0，函数f (x )＝ln 2x ＋ln x －a 有零点C ．∃α，β∈R ，使cos(α＋β)＝cos α＋sin βD ．∀φ∈R ，函数f (x )＝sin(2x ＋φ)都不是偶函数 [答案] D[解析] ∵f (x )为幂函数，∴m －1＝1，∴m ＝2，f (x )＝x －1，∴f (x )在(0，＋∞)上递减，故A 真；∵y ＝ln 2x ＋ln x 的值域为[－14，＋∞)，∴对∀a >0，方程ln 2x ＋ln x －a ＝0有解，即f (x )有零点，故B真；当α＝π6，β＝2π时，cos(α＋β)＝cos α＋sin β成立，故C 真；当φ＝π2时，f (x )＝sin(2x ＋φ)＝cos2x为偶函数，故D 为假命题．12．(2014·黄冈中学检测)已知集合M ＝{(x ，y )|y ＝f (x )}，若对于任意(x 1，y 1)∈M ，存在(x 2，y 2)∈M ，使得x 1x 2＋y 1y 2＝0成立，则称集合M 是“理想集合”，则下列集合是“理想集合”的是( )A ．M ＝{(x ，y )|y ＝1x }B ．M ＝{(x ，y )|y ＝cos x }C ．M ＝{(x ，y )|y ＝x 2－2x ＋2}D ．M ＝{(x ，y )|y ＝log 2(x －1)} [答案] B[解析] 设A (x 1，y 1)，B (x 2，y 2)，则由x 1x 2＋y 1y 2＝0知OA ⊥OB ，由理想集合的定义知，对函数y ＝f (x )图象上任一点A ，在图象上存在点B ，使OA ⊥OB ，对于函数y ＝1x ，图象上点A (1,1)，图象上不存在点B ，使OA ⊥OB ；对于函数y ＝x 2－2x ＋2图象上的点A (1,1)，在其图象上也不存在点B ，使OA ⊥OB ；对于函数y ＝log 2(x －1)图象上的点A (2,0)，在其图象上不存在点B ，使OA ⊥OB ；而对于函数y ＝cos x ，无论在其图象上何处取点A ，总能在其位于区间[－π2，π2]的图象上找到点B ，使OA ⊥OB ，故选B.第Ⅱ卷(非选择题共90分)二、填空题(本大题共4个小题，每小题4分，共16分，把正确答案填在题中横线上．) 13．(文)(2014·高州四中质量检测)已知函数f (x )＝x 2＋mx ＋1，若命题“∃x 0>0，f (x 0)<0”为真，则m 的取值范围是________．[答案] (－∞，－2)[解析] 由条件知⎩⎪⎨⎪⎧－m 2>0，m 2－4>0，∴m <－2.(理)(2014·福州市八县联考)已知命题p ：m ∈R ，且m ＋1≤0，命题q ：∀x ∈R ，x 2＋mx ＋1>0恒成立，若p ∧q 为假命题且p ∨q 为真命题，则m 的取值范围是________．[答案] m ≤－2或－1<m <2[解析] p ：m ≤－1，q ：－2<m <2，∵p ∧q 为假命题且p ∨q 为真命题，∴p 与q 一真一假，当p 假q 真时，－1<m <2，当p 真q 假时，m ≤－2，∴m 的取值范围是m ≤－2或－1<m <2.14．(文)(2014·安徽程集中学期中)以下四个命题：①在△ABC 中，内角A ，B ，C 的对边分别为a ，b ，c ，且b sin A ＝a cos B ，则B ＝π4；②设a ，b 是两个非零向量且|a ·b |＝|a ||b |，则存在实数λ，使得b ＝λa ；③方程sin x －x ＝0在实数范围内的解有且仅有一个；④a ，b ∈R 且a 3－3b >b 3－3a ，则a >b ；其中正确的是________．[答案] ①②③④[解析] ∵b sin A ＝a cos B ，∴sin B sin A ＝sin A cos B ，∵sin A ≠0，∴sin B ＝cos B ，∵B ∈(0，π)，∴B ＝π4，故①正确； ∵|a ·b |＝||a |·|b |·cos 〈a ，b 〉|＝|a |·|b |，∴|cos 〈a ，b 〉|＝1，∴a 与b 同向或反向，∴存在实数λ，使b ＝λa ，故②正确；由于函数y ＝sin x 的图象与直线y ＝x 有且仅有一个交点，故③正确；∵(a 3－3b )－(b 3－3a )＝(a 3－b 3)＋3(a －b )＝(a －b )(a 2＋ab ＋b 2＋3)>0，∵a 2＋ab ＋b 2＋3>0，∴a －b >0，∴a >b ，故④正确．(理)(2014·屯溪一中期中)下列几个结论：①“x <－1”是“x <－2”的充分不必要条件； ②⎠⎛01(e x ＋sin x )d x ＝e －cos1；③已知a >0，b >0，a ＋b ＝2，则y ＝1a ＋4b 的最小值为92；④若点(a,9)在函数y ＝3x 的图象上，则tan a π3的值为－3；⑤函数f (x )＝2sin(2x －π3)－1的对称中心为(k π2＋π6，0)(k ∈Z )其中正确的是________．(写出所有正确命题的序号) [答案] ②③④[解析] x <－1⇒/ x <－2，x <－2⇒x <－1，故①错误；⎠⎛01(e x ＋sin x )d x ＝(e x －cos x )|10＝e －cos1，故②正确；∵a >0，b >0，a ＋b ＝2，∴y ＝1a ＋4b ＝12(a ＋b )(1a ＋4b )＝12(5＋b a ＋4a b )≥12(5＋2b a ·4a b )＝92，等号在⎩⎪⎨⎪⎧b a ＝4a b ，a ＋b ＝2，即a ＝23，b ＝43时成立，故③正确；∵(a,9)在函数y ＝3x 的图象上，∴3a ＝9，∴a＝2，∴tan 2π3＝－tan π3＝－3，故④正确；f (x )＝2sin(2x －π3)－1的对称中心不落在x 轴上，故⑤错．正确答案为②③④.15．(2013·福建文，16)设S ，T 是R 的两个非空子集，如果存在一个从S 到T 的函数y ＝f (x )满足：(1)T ＝{f (x )|x ∈S }；(2)对任意x 1，x 2∈S ，当x 1<x 2时，恒有f (x 1)<f (x 2)，那么称这两个集合“保序同构”．现给出以下3对集合： ①A ＝N ，B ＝N *；②A ＝{x |－1≤x ≤3}，B ＝{x |－8≤x ≤10}； ③A ＝{x |0<x <1}，B ＝R .其中，“保序同构”的集合对的序号是________．(写出所有“保序同构”的集合对的序号) [答案] ①②③[解析] 由(1)知T 是定义域为S 的函数y ＝f (x )的值域；由(2)知f (x )为增函数，因此对于集合A 、B ，只要能够找到一个增函数y ＝f (x )，其定义域为A ，值域为B 即可．对于①，A ＝N ，B ＝N *，可取f (x )＝x ＋1，(x ∈A )；对于②，A ＝{x |－1≤x ≤3}，B ＝{x |－8≤x ≤10}，可取f (x )＝92x －72(x ∈A )；对于③，A ＝{x |0<x <1}，B ＝R ，可取f (x )＝tan(x －12)π(x ∈A )．16．(文)(2014·合肥八中联考)给出下列四个命题： ①∃α，β∈R ，α>β，使得tan α<tan β；②若f (x )是定义在[－1,1]上的偶函数，且在[－1,0]上是增函数，θ∈(π4，π2)，则f (sin θ)>f (cos θ)；③在△ABC 中，“A >π6”是“sin A >12”的充要条件；④若函数y ＝f (x )的图象在点M (1，f (1))处的切线方程是y ＝12x ＋2，则f (1)＋f ′(1)＝3，其中所有正确命题的序号是________．[答案] ①④[解析] ①当α＝3π4，β＝π3时，tan α<0<tan β，∴①为真命题；∵f (x )是[－1,1]上的偶函数，在[－1,0]上单调递增，∴在[0,1]上单调递减，又θ∈(π4，π2)，∴1>sin θ>cos θ>22，从而f (sin θ)<f (cos θ)，∴②为假命题；③当A ＝5π6时，A >π6成立，但sin A ＝12，∴③为假命题；④由条件知f ′(1)＝12，f (1)＝12×1＋2＝52，∴f (1)＋f ′(1)＝3，∴④为真命题．(理)(2014·银川九中一模)给出下列命题： ①已知a ，b 都是正数，且a ＋1b ＋1>ab，则a <b ；②已知f ′(x )是f (x )的导函数，若∀x ∈R ，f ′(x )≥0，则f (1)<f (2)一定成立； ③命题“∃x ∈R ，使得x 2－2x ＋1<0”的否定是真命题； ④“x ≤1且y ≤1”是“x ＋y ≤2”的充要条件．其中正确命题的序号是________．(把你认为正确命题的序号都填上) [答案] ①②③[解析] ①∵a ，b 是正数，∴a ＋1>0，b ＋1>0，∵a ＋1b ＋1>ab ，∴b (a ＋1)>a (b ＋1)，∴b >a ，即a <b ，∴①正确；②∵对任意x ∈R ，f ′(x )≥0，∴f (x )在R 上为增函数， ∴f (1)<f (2)，∴②正确；③“∃x ∈R ，使得x 2－2x ＋1<0”的否定为“∀x ∈R ，x 2－2x ＋1≥0”，∵x ∈R 时，x 2－2x ＋1＝(x －1)2≥0成立，∴③正确；④当x ≤1且y ≤1时，x ＋y ≤2成立；当x ＝3，y ＝－2时，满足x ＋y ≤2，∴由“x ＋y ≤2”推不出“x ≤1且y ≤1”，∴④错误．三、解答题(本大题共6个小题，共74分，解答应写出文字说明，证明过程或演算步骤．) 17．(本小题满分12分)(文)(2014·福州市八县联考)A ＝{x |x 2－2x －8<0}，B ＝{x |x 2＋2x －3>0}，C ＝{x |x 2－3ax ＋2a 2<0}，(1)求A ∩B ；(2)试求实数a 的取值范围，使C ⊆(A ∩B )．[解析] (1)依题意得：A ＝{x |－2<x <4}，B ＝{x |x >1或x <－3}， ∴A ∩B ＝{x |1<x <4}．(2)①当a ＝0时，C ＝∅，符合C ⊆(A ∩B )； ②当a >0时，C ＝{x |a <x <2a }，要使C ⊆(A ∩B )，则⎩⎪⎨⎪⎧a ≥12a ≤4，解得1≤a ≤2；③当a <0时，C ＝{x |2a <x <a }，∵a <0，C ⊆(A ∩B )不可能成立，∴a <0不符合题设． ∴综上所述得：1≤a ≤2或a ＝0.(理)(2014·甘肃临夏中学期中)记函数f (x )＝lg(x 2－x －2)的定义域为集合A ，函数g (x )＝3－|x |的定义域为集合B .(1)求A ∩B ；(2)若C ＝{x |x 2＋4x ＋4－p 2<0，p >0}，且C ⊆(A ∩B )，求实数p 的取值范围．[解析] (1)由条件知，x 2－x －2>0，∴A ＝{x |x <－1，或x >2}，由g (x )有意义得3－|x |≥0，所以B ＝{x |－3≤x ≤3}，∴A ∩B ＝{x |－3≤x <－1，或2<x ≤3}；(2)∵C ＝{x |x 2＋4x ＋4－p 2<0}(p >0)，∴C ＝{x |－2－p <x <－2＋p }， ∵C ⊆(A ∩B )，∴－2－p ≥－3，且－2＋p ≤－1， ∴0<p ≤1，∴实数p 的取值范围是{p |0<p ≤1}．18．(本小题满分12分)(2014·山东省菏泽市期中)已知命题p ：关于x 的不等式|x －1|>m －1的解集为R ，命题q ：函数f (x )＝(5－2m )x 是R 上的增函数，若p 或q 为真命题，p 且q 为假命题，求实数m 的取值范围．[解析] 不等式|x －1|>m －1的解集为R ，须m －1<0，即p 是真命题时，m <1；函数f (x )＝(5－2m )x 是R 上的增函数，须5－2m >1，即q 是真命题时，m <2. ∵p 或q 为真命题，p 且q 为假命题， ∴p 、q 中一个为真命题，另一个为假命题． (1)当p 真，q 假时，m <1且m ≥2，此时无解； (2)当p 假，q 真时，m ≥1且m <2，此时1≤m <2，因此1≤m <2.19．(本小题满分12分)(文)(2014·灵宝实验高中月考)设命题p ：实数x 满足x 2－4ax ＋3a 2<0，其中a <0；命题q ：实数x 满足x 2＋2x －8>0且綈p 是綈q 的必要不充分条件，求实数a 的取值范围．[解析] 由x 2－4ax ＋3a 2<0及a <0得，3a <x <a ， ∴p ：3a <x <a ；由x 2＋2x －8>0得，x <－4或x >2，∴q ：x <－4或x >2.∵綈p 是綈q 的必要不充分条件， ∴p 是q 的充分不必要条件，∴a ≤－4.(理)(2014·福建省闽侯二中、永泰二中、连江侨中、长乐二中联考)设命题p ：实数x 满足(x －a )(x －3a )<0，其中a >0，命题q ：实数x 满足x －3x －2≤0.(1)若a ＝1，且p ∧q 为真，求实数x 的取值范围；(2)若綈p 是綈q 的充分不必要条件，求实数a 的取值范围． [解析] (1)∵a ＝1，∴不等式化为(x －1)(x －3)<0，∴1<x <3；由x －3x －2≤0得，2<x ≤3，∵p ∧q 为真，∴2<x <3. (2)∵綈p 是綈q 的充分不必要条件， ∴q 是p 的充分不必要条件，又q ：2<x ≤3，p ：a <x <3a ，∴⎩⎪⎨⎪⎧a ≤2，3a >3，∴1<a ≤2.20．(本小题满分12分)(2014·马鞍山二中期中)设命题p ：f (x )＝2x －m 在区间(1，＋∞)上是减函数；命题q ：x 1，x 2是方程x 2－ax －2＝0的两个实根，且不等式m 2＋5m －3≥|x 1－x 2|对任意的实数a ∈[－1,1]恒成立，若(綈p )∧q 为真，试求实数m 的取值范围．[解析] 对命题p ：x －m ≠0，又x ∈(1，＋∞)，故m ≤1，对命题q ：|x 1－x 2|＝(x 1＋x 2)2－4x 1x 2＝a 2＋8对a ∈[－1,1]有a 2＋8≤3， ∴m 2＋5m －3≥3⇒m ≥1或m ≤－6. 若(綈p )∧q 为真，则p 假q 真，∴⎩⎪⎨⎪⎧m >1，m ≥1或m ≤－6，∴m >1. 21．(本小题满分12分)(2014·河北冀州中学期中)设集合A 为函数y ＝ln(－x 2－2x ＋8)的定义域，集合B 为函数y ＝x ＋1x ＋1的值域，集合C 为不等式(ax －1a )(x ＋4)≤0的解集．(1)求A ∩B ；(2)若C ⊆∁R A ，求a 的取值范围．[解析] (1)由于－x 2－2x ＋8>0，解得A ＝(－4,2)，又y ＝x ＋1x ＋1＝(x ＋1)＋1x ＋1－1，当x ＋1>0时，y ≥2(x ＋1)·1x ＋1－1＝1；当x ＋1<0时，y ≤－2(x ＋1)·1x ＋1－1＝－3.∴B ＝(－∞，－3]∪[1，＋∞)， ∴A ∩B ＝(－4，－3]∪[1,2)． (2)∵∁R A ＝(－∞，－4]∪[2，＋∞)，由(ax －1a)(x ＋4)≤0，知a ≠0，当a >0时，由(ax －1a )(x ＋4)≤0，得C ＝[－4，1a 2]，不满足C ⊆∁R A ；当a <0时，由(ax －1a )(x ＋4)≤0，得C ＝(－∞，－4]∪[1a 2，＋∞)，欲使C ⊆∁R A ，则1a 2≥2，解得：－22≤a <0或0<a ≤22，又a <0，所以－22≤a <0，综上所述，所求a 的取值范围是[－22，0)． 22．(本小题满分14分)(2014·九江市七校第一次联考)“城中观海”是近年来国内很多大中型城市内涝所致的现象，究其原因，除天气因素、城市规划等原因外，城市垃圾杂物也是造成内涝的一个重要原因．暴雨会冲刷城市的垃圾杂物一起进入下水道，据统计，在不考虑其他因素的条件下，某段下水道的排水量V (单位：立方米/小时)是杂物垃圾密度x (单位：千克/立方米)的函数．当下水道的垃圾杂物密度达到2千克/立方米时，会造成堵塞，此时排水量为0；当垃圾杂物密度不超过0.2千克/立方米时，排水量是90立方米/小时；研究表明，0.2≤x ≤2时，排水量V 是垃圾杂物密度x 的一次函数．(1)当0≤x ≤2时，求函数V (x )的表达式；(2)当垃圾杂物密度x 为多大时，垃圾杂物量(单位时间内通过某段下水道的垃圾杂物量，单位：千克/小时)f (x )＝x ·V (x )可以达到最大，求出这个最大值．[解析] 当0.2≤x ≤2时，排水量V 是垃圾杂物密度x 的一次函数，设为V (x )＝mx ＋n ，将(0.2,90)，(2,0)代入得V (x )＝－50x ＋100，V (x )＝⎩⎪⎨⎪⎧90(0≤x ≤0.2)，－50x ＋100(0.2<x ≤2).(2)f (x )＝x ·V (x )＝⎩⎪⎨⎪⎧90x (0≤x ≤0.2)，－50x (x －2)(0.2<x ≤2).当0≤x ≤0.2时，f (x )＝90x ，最大值为1.8千克/小时；当0.2≤x ≤2时，f (x )＝50x (2－x )≤50，当x ＝1时，f (x )取到最大值50，所以，当杂物垃圾密度x ＝1千克/立方米，f (x )取得最大值50千克/小时．。

2015届高考语文考前专题提升训练：名言名句word版含答案2

2015届高考语文考前专题提升训练：名言名句1.【山东省济宁市2015届高三第一次模拟】补写出下列名篇名句中的空缺部分。

（6分）（1）子曰：“▲，小人同而不和。

” (《论语》) 昔我往矣，▲；今我来思，雨雪霏霏。

（《诗经·采薇》）（2）老当益壮，宁移白首之心?穷且益坚，▲。

(王勃《滕王阁序》) ▲，唯见江心秋月白。

（白居易《琵琶行》）（3）居庙堂之高则忧其民，▲。

（范仲淹《岳阳楼记》）▲，归去，也无风雨也无晴。

（苏轼《定风波》）【答案】（1）君子和而不同杨柳依依（2）不坠青云之志东船西舫悄无言（3）处江湖之远则忧其君回首向来萧瑟处（6分。

每句1分，有错别字该句不得分）2.【山东省临沂市第一中学2015届高三下学期二轮阶段性检测】补写下列名篇名句中空缺部分。

（6分）（1）但以刘日薄西山，气息奄奄，，朝不虑夕。

《陈情表》（2），切问而近思，仁在其中矣。

《论语》（3），百年多病独登台。

《登高》（4），一夫当关，万夫莫开。

《蜀道难》（5）此情可待成追忆，________________________。

《锦瑟》（6）且夫水之积也不厚，______________________。

《逍遥游》【答案】（1）人命危浅（2）博学而笃志（3）万里悲秋常作客（4）剑阁峥嵘而崔嵬，（5）只是当时已惘然（6）则其负大舟也无力3.【山东省枣庄市第九中学2015届高三上学期期末考试】补写出下列名句名篇的空缺部分。

（6分）（1）。

有三秋桂子，十里荷花。

（柳永《望海潮》）（2）？邻之厚，君之薄也。

（《烛之武退秦师》）（3）江间波浪兼天涌，。

（杜甫《秋兴八首》）（4）既自以心为行役，？（陶渊明《归去来兮辞》）（5）背绳墨以追曲兮，。

（屈原《离骚》）（6）闾阎扑地，；舸舰弥津，青雀黄龙之舳。

（王勃《滕王阁序》）【答案】（1）重湖叠巘清嘉（2）焉用亡郑以陪邻（3）塞上风云接地阴（4）奚惆怅而独悲（5）竞周容以为度（6）钟鸣鼎食之家4.补写出下列名篇名句的空缺部分。

2015山西中考适应性历史试题+答案

一、选择题（在每小题的四个选项中，只有一项最符合题意，请选出并在答题卡上将该项涂黑。

每小题2分，共30分）1.山西翼城县大河口西周墓地是我国2010年度“十大考古新发现”之一。

下列物品最有可能出土于该墓葬的是A.青铜器 B.铁农具C.唐三彩D.石佛像2.2000多年前，东起中国长安、西达意大利罗马的古丝绸之路曾是连接中国与亚欧各国的贸易通道，作为经济全球化的早期版本，这条贸易通道被誉为全球最重要的商贸大动脉。

21世纪的今天，中国倡导建设“丝绸之路经济带”，受到国内外各界的高度关注。

古今丝绸之路的共同作用有①都造福于沿途各国人民②都有利于东西方经济、文化交流③都促进欧亚各国商业贸易的发展④都对中国的兴盛起到了积极作用A.②③④B.①②③④C.①②④D.①③④3.如何发现德才兼备的优秀人才，让其参与国家管理，是历代朝廷的头等大事。

隋朝创立的这项制度不仅给古代官员和公务员录用方式带来重大影响，而且还影响了此后中国人读书的行为。

该项制度是A.分封制B.科举制C.郡县制D.行省制4.下表为西晋至宋代南北方户数统计表（单位：万户）。

表中户数变化最能说明这一时期A.经济重心南移趋势B.政府统治比较开明C.城市经济趋于活跃D.民族融合逐渐加强5.元朝和清朝都是少数民族政权入主中原，它们的共同点有①都建立起了全国性统一政权②都加强对西藏地区的管辖③都是幅员辽阔的多民族国家④对外都实行闭关锁国政策A.①③④B.②③④C.①②④D.①②③6.“民生”问题事关各国经济发展、统治稳定，2015年全国人大会议上李克强总理提出“立国之道，惟在富民”。

下面史实能说明这一观点的有①唐太宗轻徭薄赋②人民公社化运动③罗斯福新政④列宁新经济政策A.①②④B.①③④C.②③④D.①②③7.某班同学收集了有关世界反法西斯战争的一组图片，下列是他们对图片所反映史实的表述，其中不正确的是图一山西平型关大捷纪念馆图二山东台儿庄大战纪念馆图三山西狮脑山百团大战纪念碑图四法国诺曼底登陆纪念碑A.图一反映的战役是抗战以来的第一次大捷B.图二反映的战役是国共合作的重大胜利成果C.图三反映的是抗战以来中国军队主动出击日军的最大规模战役D.图四反映的战役迫使德国陷入两线夹击，加速了德国法西斯的灭亡山西省2015年中考考前适应性训练试题文科综合注意事项：1.文科综合由历史和思想品德两部分组成，满分150分，考试时间150分钟。

河北省石家庄市2015届高三毕业班教学质量检测(一)数学(理)试题 Word版含答案

石家庄市2015届高三复习教学质量检测（一）高三数学（理科）（时间120分钟，满分150分）一、选择题（每小题5分，共60分） 1．复数21i i =- A ．1i + B ．1i - C ．1i - D ．12i -2．已知集合2{|230}A x x x =--≤，{0,1,2,3,4}B =，则AB =A ．{1,2,3}B ．{0,1,2,3}C ．{1,0,1,2,3}-D ．{0,1,2}3．已知向量(2,6)=--a ，||=b ，10⋅=-a b ，则向量a 与b 的夹角为 A ．150︒ B ．30-︒ C ．120︒ D ．60-︒4．已知双曲线2221()4x y a R a -=∈的右焦点与抛物线212y x =的焦点重合，则该双曲线的离心率为A ．35 B C D 5．设()f x 是定义在R 上的周期为3的函数，当[2,1)x ∈-时，242,20(),,01x x f x x x ⎧--≤≤=⎨<<⎩，则5()2f =A ．1-B ．1C ．12D ．0 6．设a 、b 表示不同的直线，α、β、γ表示不同的平面，则下列命题中正确的是 A ．若a α⊥且a b ⊥，则//b α B ．若γα⊥且γβ⊥，则//αβ C ．若//a α且//a β，则//αβ D ．若//γα且//γβ，则//αβ7．已知函数3()sin34(,)f x a x bx a R b R =++∈∈，'()f x 为()f x 的导函数，则(2014)(2014)'(2015)'(2015)f f f f +-+--=A ．8B ．2014C ．2015D ．08．为了得到函数3cos 2y x =的图象，只需把函数3sin(2)6y x π=+的图象上所有的点A ．向右平行移动3π个单位长度 B ．向右平行移动6π个单位长度 C ．向左平行移动3π个单位长度 D ．向左平行移动6π个单位长度9．阅读如下程序框图，运行相应的程序，则程序运行后输出的结果为A ．7B ．9C ．10D ．11 10．二项式71(2)x x+的展开式中31x的系数是A ．42B ．168C ．84D ．2111．某几何体的三视图如右图，若该几何体的所有顶点都在一个球面上，则该球面的表面积为A ．4πB ．283π C ．443π D ．20π侧视图俯视图正视图12．设函数()2(,x f x e x a a R e =+-∈为自然对数的底数)，若曲线sin y x =上存在点00(,)x y ，使得00(())f f y y =，则a 的取值范围是A ．1[1,1]e e --++B ．[1,1]e +C ．[,1]e e +D ．[1,]e二、填空题（每题5分，共20分）13．曲线23(x y e e =+为自然对数的底数)在0x =处的切线方程为_____.14．实数,x y 满足402200,0x y x y x y +-≤⎧⎪-+≥⎨⎪≥≥⎩，则x y -的最小值为_____.15．已知圆22:1C x y +=，过第一象限内一点(,)P a b 作圆C 的两条切线，切点分别为A B 、，若60APB ∠=︒，则a b +的最大值为_____.16．观察右图的三角形数阵，依此规律，则第61行的第2个数是_____.... ... ... ...11　27　40　40　27　119　18　22　18　97　11　11　75　6　53　31三、解答题（本大题共6个小题，共７０分，解答应写出文字说明、证明过程或演算步骤）．17.（本小题满分10分）在ABC ∆中，角A 、B 、C 的对边长分别为a 、b 、c ，且3a =，2b =，2A B =，求co s B 和c 的值．18．（本小题满分12分）已知{}n a 为公差不为0的等差数列，13a =，且1a 、4a 、13a 成等比数列．（I ）求数列{}n a 的通项公式；（II ）若2n n n b a =，求数列{}n b 的前n 项和．19.（本小题满分12分）某学校为了解学生身体发育情况，随机从高一年级中抽取40人作样本，测量出他们的身高（单位：cm ），身高分组区间及人数见下表：2148ba[175,180][170,175)[165,170)[160,165)[155,160)人数分组（I ）求a 、b 的值并根据题目补全频率分布直方图；（II ）在所抽取的40人中任意选取两人，设Y 为身高不低于170cm 的人数，求Y 的分布列及期望．20.（本小题满分12分）如图所示，在四棱锥P ABCD -中，底面ABCD 为正方形，侧棱PA ⊥底面ABCD ，1PA AD ==，E 、F 分别为PD 、AC 的中点．（I ）求证：//EF 平面PAB ；（II ）求直线EF 与平面ABE 所成角的大小．21．（本小题满分12分）定长为3的线段AB 的两个端点A 、B 分别在x 轴、y 轴上滑动，动点P 满足2BP PA =．（I ）求点P 的的轨迹曲线C 的的方程；（II ）若过点(1,0)的直线与曲线C 交于M 、N 两点，求OM ON ⋅的最大值．22．(本小题满分12分)已知函数2()ln ,f x x x ax a R =+-∈．（I ）若3a =，求()f x 的单调区间；（II ）若()f x 有两个极值点1x 、2x ，记过点11(,())A x f x ，22(,())B x f x 的直线的斜率为k ，问是否存在a ，使22ak a =-？若存在，求出a 的值；若不存在，请说明理由．石家庄市2015届高三第一次质量检测数学理科答案一、选择题：1-5CBCDA 6-10DADBC 11-12BA 二、填空题：13．24y x =+ 14．1- 15． 16.3602 三、解答题 17.因为c=2,不合题意舍去，所以52c =.....................................10分 18．解(1)设{}n a 的公差为d ，由题意得2(33)3(312)d d +=+，得2d =或0d =(舍)，……………………2分所以{}n a 的通项公式为3(1)221n a n n =+-=+……………………4分（2）2(21)2nnn n b a n ==+123325272(21)2n n S n =+++++………………①…………②……………………6分222222,............2sin sin sin 3cos .............62sin 2494cos 2629100 (85)2c= (92)==∴===+-+-==-+==a bA B A BA aB B b a c b c B ac cc c c 解：分sinA=sin2B=2sinBcosB.........4分分分解得或分23412325272(21)2n n S n +=+++++①－②得123132222222(21)2n n n S n +-=++++-+…………………8分1+12(12)22(21)2122(21)2n n n n n +-=+-+-=---……………………10分 ∴1(21)22n n S n +=-+……………………12分19. 解：（1）解：a=6 b=10……………………………2分……….5分（2）P （Y=0）=13063240228=C CP （Y=1）=6528240112128=C C C P （Y=2）=13011240212=C C35E （P ）=.…………………………12分20(1)分别取PA 和AB 中点M 、N ，连接MN 、ME 、NF ，则=NF ∥12AD ,=ME ∥12AD ,所以=NF ∥ME ,∴四边形M E F N为平行四边形.-------------2∴EF MN ∥,又,EF PAB ⊄平面,MN PAB ⊂平面∴EF ∥PAB 平面.- ------------4(2) 由已知得，底面ABCD 为正方形，侧棱PA ⊥底面ABCD ,所以AP AB AD ，，两两垂直．如图所示，以A 为坐标原点,分别以,,为轴轴，轴，z y x 的正方向，建立空间直角坐标系x y z A -，所以(001),(000),B (1P A C D ，，，，，，，，，,1111(0),(0)2222E F ，，，，,所以,11(0)22EF =-，，, 11(0),(100)22AE AB ==，，，，,- ------------6设平面ABE 法向量(,,)n a b c =,0,0,n AE n AB ==所以11022b c a ⎧+=⎪⎨⎪=⎩令1,0,1b a c ===-则所以(0,1,1)n =-为平面ABE 的一个法向量 -------------8 设直线EF 与平面ABE 所成角为α, 于是1sin cos ,2EF n EF n EF nα=<>==.-------------10所以直线EF 与平面ABE 所成角为6π. -------------12 解法2:在平面PAD 内作EH ∥PA H 于, 因为侧棱PA ⊥底面ABCD ,所以EH ⊥底面ABCD . -------------6E 为PD 的中点,12EH =,1111224ABFS =⨯⨯= 11111334224E ABF ABF V S EH -==⨯⨯= -------------8设点F 到平面ABE 的距离为h,E ABF F ABE V V --=1112224ABES AB AE =⨯⨯=⨯⨯=1133ABFABES EH Sh =,h =-------------10设直线EF 与平面ABE 所成角为α,1sin 2h EF α==,所以直线EF 与平面ABE 所成角为6π. -------------1221.解：（1）设A （0x ，0），B （0，0y ），P （,x y ），由2BP PA =得，00(,)2(,)x y y x x y -=--，即000032()223x x x x x y y y y y⎧=-=⎧⎪⇒⎨⎨-=-⎩⎪=⎩，————————————————————2分又因为22009x y +=，所以223()(3)92x y +=，化简得：2214x y +=，这就是点P 的轨迹方程。

2015届高考语文一轮专题训练：专题08 名句默写1

2015届高考语文一轮专题训练：专题08 名句默写11、（14届安徽合肥高三第三次质检）默写。

（任选6空，超过6空，按前6空评分）（6分）(1)民生各有所乐兮，。

（屈原《离骚》）(2)土地平旷，。

（陶渊明《桃花源记》）(3)虽无丝竹管弦之盛，，亦足以畅叙幽情。

（王羲之《兰亭集序》）(4) ，归雁入胡天。

（王维《使至塞上》）(5)鼎铛玉石，金块珠砾，，秦人视之，亦不甚惜。

（杜牧《阿房宫赋》）(6) ，微冷，山头斜照却相迎。

（苏轼《定风波》）(7)满地黄花堆积，憔悴损，? （李清照《声声慢》）(8) ，梦入芙蓉浦。

（周邦彦《苏幕遮》）2、（14届湖北黄冈高三4月统考）补写出下列名篇名句中的空缺部分。

（限选其中的5小题作答，如答题超过5个，按所答的前5个小题计分）（5分）（1），如闻仙乐耳暂明。

（白居易《琵琶行》）（2）心非木石岂无感？。

（鲍照《拟行路难》）（3），忧伤以终老。

（《涉江采芙蓉》）（4）蟹六跪而二螯，，用心躁也。

（荀子《劝学》）（5）怀良辰以孤往，。

（陶渊明《归去来兮辞》）（6），小桥流水人家。

（马致远《天净沙·秋思》）（7）吴楚东南坼，。

（杜甫《登岳阳楼》）（8），出则无敌国外患者，国恒亡。

（《孟子·生于忧患死于安乐》）【答案】⑴今夜闻君琵琶语⑵吞声踯躅不敢言⑶同心而离居⑷非蛇鳝之穴无可寄托者⑸或植杖而耘籽⑹枯藤老树昏鸦⑺乾坤日夜浮⑻入则无法家拂士【解析】3．（14届吉林长春高三第三次调研）补写出下列名篇名句中的空缺部分。

（6分）（1）余则缊袍敝衣处其间，略无慕艳意，，。

（宋濂《送东阳马生序》）（2），百年多病独登台。

，潦倒新停浊酒杯。

（杜甫《登高》）（3）桂棹兮兰桨，。

，望美人兮天一方。

（苏轼《赤壁赋》）4、（14届陕西西安中学高三第三次质检）补写出下列名篇名句中的空缺部分。

（6分）【小题1】蟹六跪而二螯，，用心躁也。

（荀子《劝学》）【小题2】背负青天，,而后乃今将图南。

【恒心】2015届湖南省株洲市高三教学质量统一检测(一)英语试题及参考答案【首发word版】

绝密★启用前株洲市2015届高三年级教学质量统一检测（一）英语试题本试卷分为四个部分，包括听力、语言知识运用、阅读理解和书面表达。

满分150分，考试时间120分钟。

Part I Listening comprehension (30 marks)Section A (22.5 marks)Directions: In this section, you will 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 11. How many nights will they stay in the hotel?A. Three.B. Four.C. Five.2. Who needs a room with a shower?A. The man.B. The daughter.C. The son.Conversation 23. Who wants to buy a new dress?A. Jenny.B. Ann.C. Lucy.4. Where will the speakers go first?A. A clothes store.B. The City Library.C. A bookstore. Conversation 35. Where are the speakers?A. At a post office.B. At a clothing store.C. At the airport.6. How much will the woman pay?A. $ 16.60.B. $ 16.20.C. $ 12.50.Conversation 47. Which museum will they go to?A. An art museum.B. A children’s museum.C. A history museum.8. Who is Henry?A. Jim’s brother.B. The woman’s husband.C. The woman’s father.9. Why do they have to get there earlier?A. It’s too far.B. Jim wants to have more fun.C. The parking lot there is small. Conversation 510. What kind of computer does Tom prefer?A. A desktop computer.B. A notebook computer.C. An iPad.11. Why does Tom like this kind of computer?A. It is much cheaper.B. It is light and easy to carry.C. It has a big keyboard.12. What is the advantage of Rose’s computer?A. It’s less expensive.B. It has a battery.C. It takes up little space. Conversation 613. What is the man busy doing recently?A. Running for Student President.B. Designing posters.C. Giving free notebooks.14. What’s Emily studying?A. Science.B. Literature.C. Public relations.15. When will they meet for the ideas of Mark’s campaign?A. Tomorrow morning.B. Tomorrow afternoon.C. Next Monday. 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 three words.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.21. Humor, if ________ properly, will help us break the ice and gain affection of others in socialcommunication.A. usingB. usedC. to useD. to be used22. If you________ the failed experiences, you would not have made such a mistake in yourhomework.A. refer toB. have referred toC. referred toD. had referred to23. Taking a gap year is a good chance for students to learn skills and gain life experience, ______them an edge in the job market.A. givenB. to giveC. givingD. having given24. The mixture of his improved self-confidence and never-ending efforts is ________ lies behindhis rapid progress in his study.A. whatB. thatC. whereD. which25. Reading for fun means much to us and we _______ never tell where its influence stops.A. mustB. needC. canD. should26. The Romany prefer to move and stay in small groups________ they can protect and preservetheir culture and freedom.A. even thoughB. as ifC. as soon asD. so that27. Happiness will be within our reach if positive thinking ________ into daily routine and becomesan important part of our world.A. adoptsB. is adoptedC. will adoptD. will be adopted28. Ann_________ to the company for the position of senior clerk and she is waiting anxiously forthe result.A. will applyB. will be applyingC. appliedD. had applied29. It is generally acknowledged that only by having acceptance of the cultures of different countries_________ good relationships with each other.A. can we developB. we can developC. we developedD. did we develop30. Wherever you study after graduating from high school, _________ regular contacts with yourfamily.A. keepB. to keepC. keepingD. kept31. Zhuzhou offers visitors outstanding natural beauty coupled with excellent customer service,_________ makes for an unforgettable experience.A. whatB. whichC. thatD. whom32. What makes tomorrow’s successful citizens_________ the ability to cope with a changing worldas well as the responsibility toward a community.A. isB. areC. hasD. have33. It was the deep love and encouragement of my teammates_________ helped me gain the greateststrength and meet the challenges in life.A. whatB. whichC. whoD. that34. Lyn gave up the chance of going abroad, but he feels he _________ a right decision.A. makesB. has madeC. had madeD. is making35. There is no doubt that _________ a goal, one needs not only knowledge but also goodpersonalities.A. achieveB. achievingC. to achieveD. achieved Section B (18 marks)Directions: For each blank in the following passage 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.It was a terribly cold Christmas Eve with heavy snow outside. I stayed in bed, feeling upset because there wasn’t enough money to buy me the36 that I’d wanted that year.Mum came back with some snow in her hair. “Come on, Matt!” she said. “Dress37 ! I t’sfreezing cold out tonight.”38 , I went out in the cold with Mum and climbed up the sled beside her. She pulled it around the house and then stopped in front of the woodshed(柴房). She went in and 39 with an armload of wood.“I rode by Jensen’s today,” said Mum, “Little Jakey was out digging around in the snow. They’re out of wood, Matt.”Jensen lived about two miles down the road. Her husband died 3 months ago, 40 three children, Jakey being only 8 years old.We loaded the sled high with 41 . Then we went to the store and my mother took down some meat, a sack of flour and a smaller sack of shoes.We rode the two miles to Jensen’s in 42 . All the way I wondered why Mum bought them some shoes and candy as we didn’t have much money. Jensen had closer neighbors than us. It shouldn’t have been our 43 .We got to Jensen’s house and unloaded the wood as quietly as possible. Then we took the meat, flour and shoes to the door.We 44 . Jensen opened the door and let us in. “We brought you a few things, Jensen,” Mum said, 45 the sack of flour and the meat on the table. Then Mum handed her the sack of shoes. Jensen bit her lower lip to keep it from trembling, tears welling out and down her cheeks.“We also46 a load of wood, Jensen,” Mum added. She turned to me and said: “Matt, go and bring some in.”I wasn’t the same person when I went back out to get the wood. Just then the scarf didn’t seem47 . The look on Jensen’s face and the smiles of her three children was the best Christmas gift of my life.36. A. scarf B. overcoat C. shoes D. handbag37. A. fashionably B. casually C. comfortably D. warmly38. A. Unsteadily B. Unhurriedly C. Unwillingly D. Uncomplainingly39. A. stepped away B. came out C. broke down D. fell over40. A. adopting B. taking C. overlooking D. leaving41. A. wood B. meat C. flour D. grass42. A. surprise B. danger C. silence D. harmony43. A. custom B. concern C. comment D. courage44. A. signed B. chased C. knocked D. hesitated45. A. taking away B. getting back C. giving out D. putting down46. A. prepared B. bought C. borrowed D. required47. A. expensive B. important C. ordinary D. availableSection C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Leaving for college can be a big challenging learning experience. Maybe this is the first time for you to live with people 48. ________ aren’t your family members. Because of the differences in values and personalities, unavoidably, conflicts may easily arise between roommates.So how can you develop a successful roommate relationship 49.________ fewer conflicts? First, lose your shyness 50.________ open up to your roommates by talking about the common interest, which can help you learn each other’s different cultural or social backgrounds. 51.________, be flexible with your roommates and adjust your thinking to new situations. Don’t get stuck with 52.________thinking patterns and habits that you bring from home. At last, 53.________ your roommate does something that bothers you, d on’t let 54.________ turn into a bad situation. Try to work together to reach a compromise that 55.________ can both live with.In a word, a harmonious roommate relationship lies in the friendly communication, mutual respect and acceptance of others’ differences.Part ⅢReading 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 best according to the information given in the passage.AKarl Fleming joined the military because he needed a change in his life. He had a successful career with a shipping company but he wanted to do something more. He found that something in the U.S. army. Fleming began his service in 2009 and never looked back.A few years later, Fleming volunteered to go to Afghanistan. There, he worked as a bodyguard for the U.S. Army Corps of Engineers. He enjoyed it, except for the almost-nightly rocket attacks. Karl was never hit di rectly by a rocket, but he didn’t need to be to feel its effects. The rockets caused severe shaking, shaking so bad that Fleming was left with many injuries. He was also diagnosed with concussions(脑震荡) and Post-Traumatic Stress Disorder(PTSD).Fleming said he was down but not out. Once he returned from Afghanistan, Fleming underwent one test after another. At first, Fleming said he thought he could recover or be able to return to duty and realize his dream of becoming an officer. But then came the news he had never imagined: Fleming would never be an officer because he was too injured to continue.Fleming said he was depressed after learning his military career was over. Add that to the memory loss, extreme anxiety and the many painful medical procedures he was already experiencing. He rarely ventured outside on the weekend. Instead, he preferred to sleep in. All that changed, however, with Fleming’s service dog, Kuchar.Fleming said he had heard dogs could help people suffering from PTSD, so he started doing research. Karl eventually selected K9s for Warriors, which is where he met Kuchar, a yellow lab. Fleming and Kuchar trained together for weeks, before returning to Fort Benning. K9s for Warriors provided Fleming with Kuchar and the training for free.Life with Kuchar has been life-changing. Fleming doesn’t sleep in any more because Kuchar won’t let him. Instead, they venture out into a world Karl was once afraid of — a world for Fleming that now seems impossible to imagine without Kuchar by his side.56. Karl Fleming joined the army because ___________.A. he had a successful careerB. he loved to be a soldierC. he wanted to have a changeD. he expected to work in a shipping company57. From Paragraph 2, we can learn Karl Fleming was __________.A. forced to go to AfghanistanB. struck directly by a rocket one dayC. satisfied with everything in the armyD. injured because of the shaking from the rockets58. Fleming suffered from the following illnesses except ___________.A. concussionsB. PTSDC. memory lossD. a heart attack59. Karl Fleming recovered with the help of ___________.A. an experienced bodyguardB. a well-trained dogC. a laboratory engineerD. a military officer60. The passage is mainly about Fleming’s__________.A. change of lifeB. service in the militaryC. medical proceduresD. experiences in AfghanistanBThe snow cap of Mount Kilimanjaro, famous in literature and beloved by tourists, initially formed some 11,000 years ago, but will be gone in two decades, according to researchers who say the ice fields on Africa’s highest mountain shrank by 80 percent in the past century.Lonnie G. Thompson, one professor from Ohio State University, said measurements using modern navigation satellites show that the oldest ice layers on the famous mountain were deposited during an extremely wet period starting about 11,700 years ago. The mountain appears in literature,most notably Ernest Hemingway’s “The Snows of Kilimanjaro” and some ancient beliefs in Africa hold the mountain to be a sacred place.But a temperature rise in recent years is destroying the 150-foot-high blocks of ice that gave Kilimanjaro its unique white cap. “The ice will be gone by about 2030,” said Thompson. The disappearing ice already has reduced the amount of water in some Tanzanian rivers and the government fears that when Kilimanjaro is bald of snow the tourists will stop coming.“Kilimanjaro is the number one foreign currency earner for the government of Tanzania,” said Thompson. “It has its own international airport and some 20,000 tourists every year. The question is how many will come if there ar e no ice fields on the mountain.”Africa was not alone in the global drought. Thompson said other records show that civilizations during this period collapsed in India, the Middle East and South America.Researchers put markers on the ice field blocks in 1962 and Thompson said measurements using satellites show the summit of the ice has been lowered by about 56 feet in 40 years. The margin of the ice also has moved back more than six feet in the past two years, much smaller than before.“That’s more than tw o meter’s worth of ice lost from a wall 164 feet (50 meters) high,” said Thompson. “That’s an enormous amount of ice.”61. The snow cap of Mount Kilimanjaro ________.A. will disappear in two centuriesB. first developed some 11,000 years agoC. resulted in a temperature rise in recent yearsD. has decreased to 80 percent over the past century62. According to Thompson, the disappearing of the snow may mainly influence________.A. the local water supplyB. the local tourismC. the weather in the surrounding areasD. the government foreign currency exchange63. The underlined word“margin” in Paragraph 6 means__________.A. centerB. topC. edgeD. back64. The writer’s tone in this article is___________.A. concernedB. casualC. doubtfulD. angry65. Which of the following can be the best title for this passage?A. Risks of Visiting Mount KilimanjaroB. A Sacred Place— KilimanjaroC. Africa’s Highest Mountai nD. Kilimanjaro Snow Cap May Melt SoonCHow many times a day do you check your email? When you wake up? Before bed? A dozen times in between? The technology that was supposed to simplify our lives has become the vital time-suck: the average teen spends more than seven hours a day using technological devices, plus an additional hour just text-messaging friends.The advantage of technological devices is connectedness: email lets us respond on the go, and we are in touch with more people during more hours of the day than at any other time in history. But is it possible we’re more lonely than ever, too? That’s what MIT professor Sherry Turkle observes in her new book, Alone Together, a fascinating portrait of our changing relationship with technology. Turkle details the ways technology has redefined our comprehension of closeness and loneliness—and warns us of the danger of accepting such virtual(虚拟的) relationships in place of lasting emotional connections.For Turkle, the biggest worry is the effect all these shallow connections have on our development. Is technology offering us the lives we want to live? “We’re texting people at a distance,” says Turkle, “We’re using lifeless objects to convince ourselves that even when we’re alone, we feel together. And then when we’re with each other, we put ourselves in situations where we are alone—constantly on our mobile devices. It’s what I call a perfect storm of confusion about what’s important in our human connections.”What can’t be denied is that technology, whatever its faults, makes life a whole lot easier. It allows us to communicate with more people in less time and makes conversation simple. But it can also be seductive(具有诱惑性的), providing more stimulation than our natural lives. There are usually some unhealthy videos online which remove teenagers’ attention from their schoolwork. Besides, some online activities make people addicted, which occupied their daily life and affected their ability to form real-world relationships. “Technology can be more immediately satisfying than the labor of building an intima te relationship,” said one high school student, “Every time I text, I start to have some hap py feelings.”But are any of those feelings equal to the kind we feel when engaged in real, face-to-face communication? Online, you can neglect others’ feelings. In a text message, you can avoid eye contact. A number of studies have found that this generation of teens is less sympathetic than ever. That doesn’t spell disaster, says Turkle—but it does mean we might want to start thinking about the way we want to live.66. From the first paragraph we can infer that_________.A. email checking helps people wake up earlyB. technological device production has been simplifiedC. using technological devices costs teens much timeD. people communicate mainly by text-messaging now67. Turkle’s new book mainly discussed________.A. ways to draw a fascinating portraitB. how technology influences human relationshipsC. the dangers of accepting emotional connectionsD. the advantages of technology68. What worries Turkle most is that more and more people are_________.A. starting to accept emotional connections in place of virtual connectionsB. convincing themselves by using fewer lifeless objects in connectionsC. dropping the use of technological devices for connection with each otherD. being affected by the shallow connections through the mobile devices69. Which of the following is True according to the passage?A. Others’ feeling s can be ignored in online communication.B. No stimulation is provided in natural life connections.C. People always send text messages to avoid eye contact.D. It may be a disaster that teens are less sympathetic than ever.70. What is the main purpose of the passage?A. To instruct people how to do with emails.B. To stress the importance of technology.C. To promote a wider use of technological devices.D. To lead us to consider what’s important in human connection s.Part IV 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.A consumer complaint or customer complaint is an expression of dissatisfaction on a consumer’s behalf to a responsible party. It can also be described in a positive sense as a report from a consumer providing documentation about problems with a product or service.So what are the common reasons for customer complaints? The most common complaints about retail(零售) stores fall into several aspects. First, they have to circle the filled parking lot endlessly, which is a waste of time and a test for their patience. They also can’t stand cluttered shelves, over-loaded racks, out-of-stock items and long check-out lines. Worst of all, sometimes some salespeople are rude, turning their mood into a bad one.In fact, some modern business consultants urge businesses to view customer complaints as a gift but not a trouble. Some retailers, however,ignore complaints or deal with them dishonestly,which can cause a chain of events like bad reputation, leave their business with fewer and fewer customers. The resulting “snowball effect” can be disastrous to retailers. In the most severe cases, it can even cause companies to shut down.Increasing competition is forcing companies to take more effective measures to satisfy customers and better their customer service. During peak shopping hours, some moonlighting(业余兼职的) local police have been employed as parking attendants by some retailers to solve the parking problems. Some hire flag wavers to direct customers to parking spaces that are empty. This guidance can avoid confrontation between those eyeing the same parking space. Retailers can relieve the headache by redesigning store layouts, pre-stocking sales items, hiring the cashiers with much experience, and having sales representatives on hand to answer questions. Most importantly, salespeople should be trained to deal with angry customers with politeness.Try their best to resolve the problem if they can.Quickly and properly solving customer complaints can help retailers smooth over issues and their business can grow and prosper.Title: 71.______Definition72. _______73. _______●an expression of consumers’ dissatisfaction●74. __________ on a problem with a product or serviceFilled parking lots, cluttered shelves, over-loaded racks,out-of-stock items, long check-out lines and 75. ________●77. ________ as parking attendants●Hiring flag wavers to direct customers to 78. ________●Redesigning store layouts and pre-stocking sales items●Hiring 79. ____________●Having sales representatives answer questions●Training salespeople to deal with angrycustomers80._____________Effects bad reputation →76.________→companies shutting downSection B (10 marks)Directions: Read the following passage. Answer the questions according to the information given in the passage.My grandmother was a master gardener that could make anything bloom. Even me.She spent most of her life living on a farm in the mountains of North Caroline, where she got married, raised four children, and watched the changing of the seasons. When I was 12 years old, my dad gave up working downtown and moved back to the farm to turn to gardening. I visited on weekends to keep them company.Every time my farm chores were done, I was free to climb the mountain, singing songs and gathering flowers. Sometimes the plants scratched me. My grandmother would say: “Beauty has a price. I hope it was worth it.” I would say, “Yes, ma’am. It was.” Then I’d scratch some more. At dusk, we’d sort the flowers and make bouquets(花束): One for the living room, one for the kitchen, and three for the bedrooms.Even then as a child, I knew that what I desired most from my grandmother was not her flowers but her time. She has been gone for decades, but sometimes when I reach down to pick a flower or pull a weed, I see her hand, not mine. I thought I’d grow up to be a gardener as well. I informed myself, someday, when my children had children，I would be a gardening grandma. Then the grandbabies started showing up, and I discovered I would much rather chase after them than go digging.The truth is, I’m no gardener. I’m a picker, not a planter. I don’t need to plant a garden. My children are my flowers. They delight me and complete me with a beauty that is worth any price.My grandmother and I differ in lots of ways, but from her, I do learn what a grandmother means.I also learn that I need attend to my grandbabies with time and water them with love. I hope that, one day, when they hold their first grandchild, they might see my hand.81. Why did the author go to the farm on weekends?(No more than 8 words) (2 marks) ____________________________________________________________________________ 82. How did the author and her grandmother deal with the collected flowers?(No more than 12 words) (3 marks) ____________________________________________________________________________ 83. What did the author want most from her grandmother when she was a kid?(No more than 4 words) (2 marks) ____________________________________________________________________________ 84. What does the author learn from her grandmother?(No more than 15 words) (3 marks) ____________________________________________________________________________ Section C (25 marks)Directions: Write an English composition according to the instructions given below.每个人都行走在自己的路上, 有着不同的经历和收获。

【成套特辑】2015届高三高考备考必备专题精选训练：专项阅读理解(含新题详解)

阅读理解4题1（2014届河北邯郸高三摸底考试）Recently I fully understood that a little favor could really make a big difference. My daughter and I were, accidentally, in time to catch a falling heart just before it hit the ground. It all started when Charlene, one of my co-volunteers at the library, asked if I could do her duty on Tuesday since her doctor appointment had been unexpectedly changed. I agreed, which meant I was home on Friday instead of volunteering as previously planned.My daughter, Mary, managers a book store in town. Mid-morning, she called to ask me for help. Would I buy a fifty dollar gift card, birthday card, and cake for one of Mary’s assistant managers, Cindy?Mary explained she had to call Cindy in to take the place of another assistant manager who was sick, but felt terrible about it when she discovered it was Cindy’s birthday. Cindy insisted that it was just fine. But that was not what Mary thought, so she gathered enough money from other employees to throw a surprise party for Cindy. Since neither Mary, nor any of the other employees, could leave to pick up the goodies, they were turning to me for help. Days later, Mary told me the wonderful rest of the story, “Cindy cried and cried when we surprised her. After the party, Cindy told us that her boyfriend had chosen that morning to break up with her. To top that off, she only had 26 cents left in her bank account.”We were surprised at the series of unexpected events that took place in order to circle Cindy with love and catch her falling heart just before it hit the ground.36. Why did Charlene ask a favor of the writer?A. She had to see the doctor on Friday.B. She wanted to celebrate her daughter’s birthday.C. She had an unexpected appointment on Tuesday.D. She needed to prepare for her daughter’s birthday.37. What can we learn about Mary?A. She was a learned career woman.B. She had uneasy relationships with her assistants.C. She knew Cindy’s love story before the surprise party.D. She played the most important role in this circle of love.38. Which of the following couldn’t be the reason why Cindy cried?A. She had to work on her birthday.B. She had little money in her bank account.C. She was greatly moved by Mary’s care and kindness.D. She and her boyfriend parted that very morning.39. T he underlined word “goodies” refers to_________.A. Gifts for CindyB. Good storiesC. Close friendsD. Goods on sale【文章大意】本文是一篇记叙文。

济南市高三5月针对性训练理综物理试题.doc

高中物理学习材料（灿若寒星**整理制作）济南市2015届高三5月针对性训练理综物理试题物理答案题号 14 15 16 17 18 19 20 答案 AC A C AD B CD BD21.（1）0.15 （2分）（2）33.0；（2分） 800 （2分）22. （1）00011ER r R R ER U ++= （2分）（2）作图（为一条直线）；（2分）（3）0.0044（0.0043——0.0046）；（2分） 0.70（或0.69——0.71）（2分）（4）1.53（1.45——1.55）；（2分） 8.00（4.50——16.5）（2分）23. （1）由牛顿第二定律 ma f mg F =-- …………2分2m/s 6=a上升高度 221at h = ……………………2分解得 m 75=h ……………………1分（2）下落过程中 1ma f mg =- ……………………2分21m/s 8=a落地时速度 H a v 122= ……………………2分解得 m/s 40=v ……………………1分（3）恢复升力后向下减速运动过程 2ma f mg F =+- ………2分22m/s 10=a 设恢复升力时的速度为v m ，则有 H a v a v m m =+221222 ………2分m/s 3540=m v 由 11t a v m = ……………………2分解得 s 3551=t ……………………2分 24. （1）粒子的轨迹如图所示，已知qamv B 42= 由 rv m q v B 2= ……………………2分得 a r 22=下磁场区域中弦长 a r l 445sin 2=︒=所以第一次击中点的坐标a a a a x =--=24 ……………2分（2）开孔位置a na x +=6 ……………4分或 a na x 56+= （n =0，1，2，3，……）……4分（答出 a x = 给2分；答出a x 5= 给2分；答出 a x = 和a x 5= 给4分）（3）若开孔位置在a x =，所用时间为va a v r a v a t 2324 2322221ππ+=++= ……………2分所以在a na x +=6处开孔，粒子运动的时间表达式为va n v a v r v r a v a n v a a t n 22)54(2)34() 21 2322322( 23241πππππ+⨯++=++⨯+++= (n=0，1，2，……) ……………2分若开孔在位置a x 5=，所用时间为va v r v r a v a t 22)54(21 23223222πππ+=⨯++⨯+= ……………2分所以在a na x 56+=处开孔，粒子运动的时间表达式为 va n t n t n 22)54()1122π++=+=（）（（n=0，1，2，……） …………2分（或写成 va n nt t n 22)54(22π+== （n=1，2，3，……））37.（1）BC （4分）（2）对理想气体Ⅰ，由玻意尔定律S L P S L P ＇1110= …………2分S gm P P A +=01 …………1分代入数据得 m L 1.01=＇对理想气体Ⅱ，由玻意尔定律S L P S L P ＇2220= …………2分S gm P P B +=12 …………1分代入数据得 m L 05.02=＇故活塞A 移动的距离m L L L L d 05.0)()(2121=--+=＇＇…………2分 38．（1）BD （4分）（2）负（2分） 40m/s （2分）（3）光在棱镜中的光路如图所示（2分）由图可知，α=60º i =180º－30º－120º=30º 由i rn sin sin = 即得 22s i n s i n ==i n r故r =45º （2分）39．（1）14 （1分） 13 （1分） α 30° O C B A i r（2）①碰撞时由动量守恒定律可得： 100)(2)2(v m m mv v m m ++=+ …………2分解得： 1012v v = …………1分由于物体P 与P 1、P 2之间的力为内力，三者整体由动量守恒定律可得：02(2)(2)m m v m m m v +=++ …………2分解得：2034v v = …………1分②整个过程能量守恒定律可得： 222102111()2(2)22222m m v mv m m m v mg xμ++=+++⋅⋅ …………3分解得：232v x g μ= …………1分。

泸溪一中2015届高三数学①综合训练试题

泸溪一中2015届高三数学综合训练①试题姓名得分一、选择题：（本大题共10道小题，每小题5分，共50分。

每小题只有一个正确答案）1. 若{}{}21,20A x x B x x x =<=+>，则A B = ( )A.()0,1B.(),2-∞-C.()2,0-D.()(),20,1-∞-2. 若复数z 的实部为1，且z =2，则复数z 的虚部是 ( ) A.3-B.3±C.3i ±D.3i3. 已知,,a b c R ∈，命题“若3a b c ++=，则2223a b c ++≥”的否命题是 ( ) A. 若22233a b c a b c ++≠++<，则 B. 若22233a b c a b c ++=++<，则 C. 若22233a b c a b c ++≠++≥，则 D. 若22233a b c a b c ++≥++=，则 4. 执行如右图所示的程序框图，若输入的x 的值为2，则输出的x 的值为 ( )A.3B.126C.127D.128 5.在ABC ∆中，内角A ，B ，C 所对的边长分别为,,,a b c1sin cos sin cos 2a B C c B Ab a b +=>∠，且，则B= ( )A.6πB.3πC.23πD.56π 6. 函数()sin ln f x x x =⋅的部分图象为 ( )7. 设0,1a b >>，若3121a b a b +=+-，则的最小值为 ( ) A.2.3 B.8 C.43D.423+8. 下列说法正确．．的是 ( ) A.样本10，6，8，5，6的标准差是3.3. B.“p q ∨为真”是“p q ∧为真”的充分不必要条件； C.已知点()2,1A -在抛物线()220y px p =>的准线上，记其焦点为F ，则直线AF 的斜率等于4-D.设有一个回归直线方程为ˆ2 1.5yx =-，则变量x 每增加一个单位，ˆy 平均减少1.5个单位； 9. 将函数()()sin 222f x x ππθθ⎛⎫=+-<< ⎪⎝⎭的图象向右平移()0ϕϕ>个单位长度后得到函数()g x 的图象，若()(),f x g x 的图象都经过点30,2P ⎛⎫⎪ ⎪⎝⎭，则ϕ的值可以是 ( )A.53πB.56πC.2πD. 6π10. 双曲线221x y m-=的离心率2e =，则以双曲线的两条渐近线与抛物线2y mx =的交点为顶点的三角形的面积为 ( ) A.3B.93C.273D.363二、填空题(本大题共5道小题，每小题5分，共25分.)11. 在区间[]2,3-上随机选取一个数X ，则1X ≥的概率等于__________.12. 若实数,x y 满足24010,1x y x y x y x +-≤⎧⎪--≤+⎨⎪≥⎩则的取值范围为 .13. 某三棱锥的正视图与俯视图如右图所示，则其侧视图的面积为_________.14. 已知圆O 过椭圆22162x y +=的两焦点且关于直线10x y -+=对称，则圆O 的方程为 .15. 定义在R 上的奇函数()()()[]()402f x f x f x f x +==满足，且在，上 ()1,01294146sin ,12x x x f f x x π⎧-≤≤⎪⎛⎫⎛⎫+=⎨⎪ ⎪<≤⎝⎭⎝⎭⎪⎩，则 . 三、解答题：（本大题共6小题，共75分. 解答应写出文字说明、证明过程或演算步骤.）16.（本小题满分12分）已知函数()()4cos sin 04f x x x πωωω⎛⎫=⋅+> ⎪⎝⎭的最小正周期为π.（1）求ω的值；（2）讨论()f x 在区间0,2π⎡⎤⎢⎥⎣⎦上的单调性.17.（本小题满分12分）参加市数学调研抽测的某高三学生成绩分析的茎叶图和频率分布直方图均受到不同程度的破坏，但可见部分信息如下，据此解答如下问题：（1）求参加数学抽测的人数n ，抽测成绩的中位数及分数分别在[)80,90，[]90,100内的人数；（2）若从分数在[]80,100内的学生中任选两人进行调研谈话，(文科做）求恰好有一人分数在[]90,100内的概率.(理科做）求分数在[)80,90内人数§的分布列和期望.18.（本小题满分12分）如图几何体中，四边形ABCD 为矩形，36,2,AB BC BF CF DE EF ======4//EF AB ,G 为FC 的中点，M 为线段CD 上的一点，且2CM =. （1）证明：AF //面BDG ；（2）证明：面BGM ⊥面BFC ；（3）求三棱锥F BMC -的体积V .19.（本小题满分13分）已知等差数列{}n a 的前n 项和为n S ，且248,40a S ==. 数列{}n b 的前n 项和为*230n n n T T b n N -+=∈且，.（1）求数列{}{},n n a b 的通项公式；（2）设,,n n n a n c b n ⎧=⎨⎩为奇数为偶数，求数列{}n c 的前21n +项和21+n M .20.（本小题满分13分）已知函数()1ln 1.a f x x ax x+=++- （1）当1a =时，求曲线()y f x =在点()()2,2f 处切线方程；（2）当102a -≤≤时，讨论()f x 单调性.21.（本小题满分13分）已知椭圆()2222:10x y C a b a b +=>>的离心率为12，右焦点2F 到直线1:340l x y +=的距离为35.（1）求椭圆C 的方程；（2）过椭圆右焦点2F 斜率为()0k k ≠的直线l 与椭圆C 相交于F E ,两点，A 为椭圆的右顶点，直线AF AE ,分别交直线3x =于点N M ,，线段MN 的中点为P ，记直线2PF 的斜率为k '，求证：k k '⋅为定值.泸溪一中2015届高三数学(文）综合训练①参考答案及评分标准一、选择题（每小题5分，共50分） DBACA ADDBC二、填空题（每小题5分，共25分）11．25 12．[1,3] 13．2 14．22(1)5x y +-= 15．516 三、解答题：16．（本小题满分12分）解:(Ⅰ) 2()4cos sin()22sin cos 22cos 4f x x x x x x πωωωωω=⋅+=⋅+2(sin 2cos 2)2x x ωω=++2sin(2)24x πω=++，…………3分因为()f x 的最小正周期为π，且0ω>，从而有22ππω=，故1ω=. ………………………6分 (Ⅱ) 由(Ⅰ)知，()2sin(2)24f x x π=++，，时，当]45,4[)42(]2,0[ππππ∈+∈x x ………………………8分当2442x πππ≤+≤，即08x π≤≤时，()f x 单调递增；当52244x πππ≤+≤，即82x ππ≤≤时，()f x 单调递减. ……………11分综上可知，上单调递减，上单调递增；在在]28[]8,0[)(πππx f .………………12分17．（本小题满分12分）解：（Ⅰ）分数在[)50,60内的频数为2，由频率分布直方图可以看出，分数在[]90,100内同样有2人． ……………………………………………2分，由2100.008n=⨯，得25n = ， ……………………………………………3分茎叶图可知抽测成绩的中位数为73 ． …………………………………4分∴分数在[)80,90之间的人数为()25271024-+++= ……………………5分参加数学竞赛人数25n =，中位数为73，分数在[)80,90、[]90,100内的人数分别为4 人、2 人． ………………………………………6分（Ⅱ）设“在[]80,100内的学生中任选两人，恰好有一人分数在[]90,100内”为事件M ，将[)80,90内的4人编号为a b c d ,,, ；[]90,100内的2人编号为A B ,，在[]80,100内的任取两人的基本事件为：,,ab ac ad aA aB ,,,bc bd ,,,bA bB ,cd cA cB dA dB AB ,,,,,共15个，…………………………………………9分其中，恰好有一人分数在[]90,100内的基本事件有,aA aB ,,bA bB ,,cA cB dA ,,dB ,共8个，故所求的概率得()8=15P M ， …………………11分答：恰好有一人分数在[]90,100内的概率为815． ………………………12 18．（本小题满分12分）解：（Ⅰ）连接AC 交BD 于O 点，则O 为AC 的中点，连接OG ,因为点G 为CF 中点，所以OG 为AFC ∆的中位线,所以//OG AF ，……2分AF ⊄面BDG ， OG ⊂面BDG ，∴//AF 面BDG ……………………………………5分（Ⅱ）连接FM ,2BF CF BC ===，G 为CF 的中点,BG CF ∴⊥,2CM =，4DM ∴=,//EF AB ，ABCD 为矩形, ………………7分//EF DM ∴，又4EF =，EFMD ∴为平行四边形, ………………8分2FM ED ∴==，FCM ∴∆为正三角形 MG CF ∴⊥，MG BG G =CF ∴⊥面BGM ,CF ⊂面BFC ,∴面BGM ⊥面BFC ． …………………………10分（Ⅲ）11233F BMC F BMG C BMG BMG BMG V V V S FC S ---=+=⨯⨯=⨯⨯, 因为3GM BG ==，22BM =,所以122122BMG S =⨯⨯=,所以22233F BMC BMC V S -=⨯=．…………………………12分19．（本小题满分13分）解：（Ⅰ）由题意，1184640a d a d +=⎧⎨+=⎩，得14,44n a a n d =⎧∴=⎨=⎩． ………3分230n n T b -+=，113n b ∴==当时，，…………4分112230n n n T b --≥-+=当时，，两式相减，得12,(2)n n b b n -=≥数列{}n b 为等比数列，132n n b -∴=⋅． ………7分（Ⅱ）14 32n n nn c n -⎧=⎨⋅⎩为奇数为偶数， 211321242()()n n n P a a a b b b ++=+++++++ …………9分[44(21)]6(14)(1)214n n n ++-=⋅++-……………11分 2122482n n n +=+++…………13分20．（本小题满分13分）解：（Ⅰ）当1a =时，2()ln +1f x x x x =+-，此时'212()+1f x x x=-， …………2分 '12(2)+1124f =-=，又2(2)ln 2+21ln 2+22f =+-=，所以切线方程为：(ln 2+2)2y x -=-，整理得：ln 20x y -+=； …………………………5分CABDE FGM O（Ⅱ）2'222111(1)(1)()a ax x a ax a x f x a x x x x ++--++-=+-==， …6分当0a =时，'21()x f x x-=，此时，在(0,1)上'()0f x <,()f x 单调递减，在(1,)+∞上'()0f x >,()f x 单调递增； …………………… 8分当102a -≤<时，'21()(1)()a a x x a f x x ++-=，当11a a +-=,即12a =-时2'2(1)()02x f x x -=-≤在(0,)+∞恒成立，所以()f x 在(0,)+∞单调递减； ………………………10分当102a -<<时，11aa +->，此时在1(0,1),(,)a a+-+∞上'()0f x <,()f x 单调递减，()f x 在1(1,)aa+-上'()0f x >单调递增； ……………………12分综上所述：当0a =时，()f x 在(0,1)单调递减，()f x 在(1,)+∞单调递增；当102a -<<时， ()f x 在1(0,1),(,)a a +-+∞单调递减，()f x 在1(1,)a a+-单调递增；当12a =-时()f x 在(0,)+∞单调递减. ……………………………13分21．（本小题满分14分）解：（Ⅰ）由题意得21==a c e ，2233534c =+，……………………………2分所以1c =，2=a ，所求椭圆方程为13422=+y x ． …………………… 4分（Ⅱ）设过点()21,0F 的直线l 方程为：)1(-=x k y ，设点),(11y x E ，点),(22y x F ， …………………………………5分将直线l 方程)1(-=x k y 代入椭圆134:22=+y x C ，整理得：01248)34(2222=-+-+k x k x k ………………………………… 6分因为点2F 在椭圆内，所以直线l 和椭圆都相交，0∆>恒成立，且3482221+=+k k x x 341242221+-=⋅k k x x …………………………8分直线AE 的方程为：)2(211--=x x y y ，直线AF 的方程为：)2(222--=x x y y 令3=x ，得点11(3,)2y M x -，22(3,)2yN x -，所以点P 的坐标12121(3,())222y y x x +--， ………………………………… 9分直线2PF 的斜率为)22(41130)22(21'22112211-+-=---+-=x y x yx y x y k4)(24)(32414)(2)(241212121212121211212++-++-⋅=++-+-+=x x x x k x x k x kx x x x x y y y x x y ，……… 11分将34124,34822212221+-=+=+k k x x k k x x 代入上式得： 222222224128234134343'412844244343k k k k k k k k k k kk k -⋅-⋅+++=⋅=---+++，所以'k k ⋅为定值43-． ………………………………… 13分。

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高三A年级物理考前针对训练题（一）
选择题（1-5题为单选，6-8题为多选）
1、电磁感应现象在生活及生产中的应用非常普遍，下列不属于电磁感应现象及其应用的
是（）

2、物体以初速度V0从A点出发，沿光滑水平轨道向前滑行，途中经过一小段粗糙程度恒定的轨道
并滑离，关于该物体在通过轨道的粗糙部分的前后，下列判读中正确的是：（）
A、初速度V0越大，物体动能的减少量越大
B、初速度V0越大，物体动能的减少量越小
C、初速度V0越大，物体速度的减少量越大
D、初速度V0越大，物体速度的减少量越小
3、如图所示，同一竖直面内有上下两条用相同材料做成的水平轨道MN、PQ，两个完全相同的物
块A、B放置在两轨道上，A在B物块正上方，A、B之间用一细线相连．在细线的中点O施加
拉力，使A、B一起向右做匀速直线运动，则F的方向是（图中②表示水平方向）：（）
A、沿①方向
B、沿②方向
C、沿③方向
D、沿①②③方向都可以

4．某月球探测卫星先贴近地球表面绕地球做匀速圆周运动，此时其动能为Ek1，周期为T1；再控制
它进行一系列变轨，最终进入贴近月球表面的圆轨道做匀速圆周运动，此时其动能为Ek2，周期

为T2，若地球的质量为M1，月球的质量为M2，则21MM为（）

A．1221KKEETT B．31221KKEETT C．32121KKEETT D．2121KKEETT
5．下表列出了某种型号轿车的部分数据,表格右侧图为轿车中用于改变车速的档位.手推变速杆到
达不同档位可获得不同的运行速度,从“1～5”逐档速度增大，R是倒车档。试问若轿车在额定

功率下,要以最大动力上坡,变速杆应推至哪一档？当轿车以最高速度运行时,轿车的牵引力约
为多大（）
A．“5”档、8000N B．“5”档、2000N
C．“1”档、4000N D．“1”档、2000N

6
. 如图所示，有两个质量均为m，带等量正、负电荷的小球P、Q，Q被固定在倾角为a的光滑直
角斜面的竖直面上，当P小球被放置在与Q小球同一竖直面内且等高位置的斜面上时，P恰好
静止，此时它们之间的距离为L，则下列说法正确的是（）
A、将Q球下移一小段距离，P球仍可能静止
B、由题设条件能算出P、Q小球所带的电荷量
C、将Q球沿水平线向左移一小段距离，P球将沿斜面向上移动
D、先设法使P球静止，将Q球置于斜面顶端后，再释放P球，P球将沿斜面下滑

7．如图所示，空间有一垂直纸面向外的磁感应强度为0.5 T的匀强磁场，一质量为0.2 kg且足够长
的绝缘木板静止在光滑水平面上，在木板左端放置一质量为m=0.1 kg、带正电q = 0.2C的滑块，
滑块与绝缘木板之间动摩擦因数为0.5，滑块受到的最大静摩擦力可认为等于滑动摩擦力．现对
木板施加方向水平向左，大小为F=0.6 N的恒力，g取10 m/s2．则滑块（）
A．开始做匀加速运动，然后做加速度减小的加速运动，最后做匀速直线运动
B．一直做加速度为2 m/s2的匀加速运动,直到滑块飞离木板为止
C．速度为6m/s时，滑块开始减速
D．最终做速度为10m/s的匀速运动

8、研究表明，蜜蜂是依靠蜂房、采蜜地点和太阳三个点来定位的，蜜蜂飞行时就是根据这三个位
置关系呈8字型运动来告诉同伴蜜源的方位。某兴趣小组用带电粒子在如图所示的电场和磁场中
模拟蜜蜂的8字形运动，即在y＞0的空间中和y＜0的空间内同时存在着大小相等，方向相反的
匀强电场，上、下电场以x轴为分界线，在y轴左侧和图中竖直虚线MN右侧均无电场，但有方
向垂直纸面向里、和向外的匀强磁场，MN与y轴的距离为2d．一重力不计的负电荷从y轴上的
P（0，d）点以沿x轴正方向的初速度v
0
开始运动，经过一段时间后，电子又以相同的速度回到

P
点，则下列说法正确的是（）

题2图
F
B
A．电场与磁场的比值为0
B．电场与磁场的比值为02

C．带电粒子运动一个周期的时间为0022dd

D．带电粒子运动一个周期的时间为
00

42dd



