上海市交大附中2018-2019学年高三上9月开学摸底考试数学试题
2018届上海市交大附中下学期高三开学考试数学试题
上海交大附中2017-2018学年第二学期高三数学摸底考试卷一、填空题（第1题至第6题，每题4分；第7题至12题，每题5分，共54分）1. 已知a 是实数，i 是虚数单位，若21(1)z a a i =-++是纯虚数，则a =____________．2. 已知二元一次方程组111222a x b y c a x b y c ì+=ïïíï+=ïî的增广矩阵是111113骣-÷ç÷ç÷ç÷桫，则此方程组的解是____________．3. 为了普及环保知识，增强环保意识，某大学随机抽取30名学生参加环保知识测试，得分（10分制）的频数分布统计图如图所示，如果得分值的中位数为a ，众数为b ，平均数为c ，则a 、b 、c 中的最大者是____________．4. 命题：“若a 、b 、c 成等比数列，则2b ac =”及其逆命题、否命题、逆否命题（这四个命题）中正确的个数是____________．5. 已知正数a 、b 满足430a b +=，使得14a b+取最小值的实数对(,)a b 是____________． 6. 不等式10x x ->的解集为____________．7. 已知2(1)(1)(1)f x x x +=-?，则1(1)f x -+=____________．8. 在平面直角坐标系xOy 中，以Ox 正半轴为始边的钝角a 的终边与圆22:4O x y +=交于点11(,)P x y ，点P 沿圆顺时针移动23p个单位弧长后到达点22(,)Q x y ，则12y y +的取值范围是____________．9. 已知三棱锥V ABC -，底面是边长为2的正三角形，VA ^底面ABC V ，2VA =，D 是VB 中点，则异面直线VC 、AD 所成角的大小为____________．（用反三角函数表示） 10. 如表给出一个“等差数阵”：其中每行、每列都是等差数列，ij a 表示位于第i 行第j 列的数，则112在这“等差数阵”中出现的次数为____________．11. 在平面直角坐标系xOy 中，设(1,1)A -，,B C 是函数1(0)y x x=>图像上的两点，且ABC V 为正三角形，则ABC V 的高为____________．12. 如图，平面上两点(0,1),(3,6)P Q ，在直线y x =上取两点,M N 使MN =，且使PM MN NQ++的值取最小，则N 的坐标为____________．二、选择题（每题5分，共20分）13. 某电商设计了一种红色，打开每个红包都会获得三种福卡（“和谐”、“爱国”、“敬业”）中的一种，若集齐三种卡片可获得奖励，小明现在有4个此类红包，则它获奖的概率为（ ）A.38B.58C.49D.7914. 设,,l m n 表示三条直线，,,a b g 表示三个平面，给出下列四个命题： ①若,l m a a ^^，则//l m ；②若m b Ì，n 是l 在b 内的射影，m l ^，则m n ^； ③若,//m m n a Ì，则//n a ；④若,a g b g ^^，则//a b ，其中真命题为（ ） A. ①②B. ①②③C. ②③④D. ①③④15. 在平面内，定点,,,A B C O 满足2OA OB OC ===uu r uu u r uu u r ，AC AB BC BA OB AC AB BC BA骣骣鼢珑鼢珑鼢珑-=?鼢珑鼢珑鼢鼢珑桫桫uuu r uu u r uu u r uu ruu u r uuu r uu u r uu u r uu r 0=，动点P 满足1,AP PM MC ==uu u r uuu r uuu r ，则2BM uuu r 的最大值是（ ）A.434B.494C.374D.37216. 设等差数列{}n a 满足：22223535317cos cos sin sin cos2sin()a a a a a a a --=+，4,2k a k Z p 刮且公差(1,0)d ?，若当且仅当8n =时，数列{}n a 的前n 项和n S 取得最大值，则首项1a 的取值范围是（ ）A. 3,22p p 轾犏犏臌B. 3,22pp 骣÷ç÷ç÷ç桫C. 7,24pp 轾犏犏臌D. 7,24pp 骣÷ç÷ç÷ç桫三、解答题17. （本题满分14分）已知函数221()cos sin ,(0,)2f x x x x p =-+?． （1）求()f x 的单调递增区间；（2）设ABC V 为钝角三角形，角A所对边a =角B 所对边5b =，若()0f A =，求ABC V 的面积．18. （本题满分14分，第1小题6分，第2 小题8分）如图1，在高为2的梯形ABCD 中，//,2,5AB CD AB CD ==，过A 、B 分别作,A E C D B F C D^^，垂足分别为E 、F ，已知1DE =，将梯形ABCD 沿AE 、BF 同侧折起，使得,//AF BD DE CF ^，得空间几何体ADE BCF -，如图2． （1）证明：//BE 面ACD ； （2）求三棱锥B ACD -的体积．19. （本题满分14分）某小区有一块三角形空地，如图ABC V ，其中180AC =米，90BC =米，90C?o ，开发商计划在这片空地上进行绿化和修建运动场所，在ABC V 内的点P 处有一服务站（其大小可忽略不计），开发商打算在AC 边上选一点D ，，然后过点P 和点D 画一分界线与边AB 相交于点E ，在ADE V 区域内绿化，在四边形BCDE 区域内修建运动场所，现已知点P 处的服务站与AC 距离为10米，与BC 距离为100米，设DC d =米，试问d 取何值时，运动场所面积最大？20. （本题满分16分，第1小题4分，第2小题6分，第3小题6分）已知O 为坐标原点，圆22:(1)16M x y ++=，定点(1,0)F ，点N 是圆M 上一动点，线段NF 的垂直平分线交圆M 的半径MN 于点O ，点Q 的轨迹为E ． （1）求曲线E 的方程；（2）已知点P 是曲线E 上但不在坐标轴上的任意一点，曲线E 与y 轴的交点分别为1B 、2B ，某直线1B P 和2B P 分别与x 轴交于C 、D 两点，请问线段之积OC OD ×是否为定值？如果是请求出定值，如果不是请说明理由；（3）在（2）的条件下，若点C 坐标为(1,0)-，过点C 的直线l 与曲线E 相交于A 、B 两点，求ABD V 面积的最大值．21. （本题满分18分，第1小题4分，第2小题6分，第3小题8分）设()f x 是定义在[,]a b 上的函数，若存在(,)x a b Î，使得()f x 在[,]a x 单调递增，在[,]x b 上单调递减，则称()f x 为[,]a b 上的单峰函数，x 为峰点，包含峰点的区间称为含峰区间，其含峰区间的长度为：b a -．（1）判断下列函数中，哪些是“[0,1]上的单峰函数”？若是，指出峰点；若不是，说出原因；2123241()2,()121,()log ,()sin 42f x x x f x x f x x f x x 骣÷ç=-=--=+=÷ç÷ç桫； （2）若函数3()(0)f x ax x a =+<是[1,2]上的单峰函数，求实数a 的取值范围； （3）若函数()f x 是区间[0,1]上的单峰函数，证明：对于任意的1212,(0,1),x x x x ?，若12()()f x f x ³，则2(0,)x 为含峰区间；若12()()f x f x £，则1(,1)x 为含峰区间；试问当12,x x 满足何种条件时，所确定的含峰区间的长度不大于0.6．参考答案一、填空题 1. 12. 21x y ì=ïïíï=ïî 3.c4. 2个5. 15,154骣÷ç÷ç÷ç桫 6. (0,1)(1,)+?U7. )2[0,)x -??8.9. 1arccos410. 711. 212. 99,44骣÷ç÷ç÷ç桫二、选择题 13. C 14. A 15. B 16. D三、解答题17. （1）,2pp 轹÷ê÷÷êøë； （218. （1）证明略； （2）2319. 6020. （1）22143x y +=； （2）定值为4； （3）92 21. （1）①21()2f x x x =-是[0,1]上的单峰函数，峰点为14； ②2()121f x x =--不是[0,1]上的单峰函数； ③321()log 2f x x 骣÷ç=+÷ç÷ç桫不是[0,1]上的单峰函数； ④4()sin 4f x x =是[0,1]上的单峰函数，峰点为8p（2）11,312a 骣÷ç?-÷ç÷ç桫 （3）证明略；120.40.6x x ì=ïïíï=ïî。
2018年上海市交大附中高考数学模拟试卷含参考答案
2018年上海市交大附中高考数学模拟试卷一、填空题（本大题共12小题, 1-6题每题4分, 7～12题每题5分, 满分54分）1．（4分）函数f（x）＝的定义域为．2．（4分）双曲线3x2﹣y2＝12的两渐近线的夹角大小为．3．（4分）用行列式解线性方程组, 则D y的值为．4．（4分）湖面上浮着一个球, 湖水结冰后将球取出, 冰上留下一个直径为24厘米, 深为8厘米的空穴, 则这个球的半径为厘米．5．（4分）直线2x+y﹣4＝0经过抛物线y2＝2px的焦点, 则抛物线的准线方程是．6．（4分）已知函数y＝sin（ωx+φ）（ω＞0, 0＜φ≤）的部分图象如图所示, 则点P （ω, φ）的坐标为．7．（5分）设函数f（x）＝的反函数为f﹣1（x）, 若, 则f（a+4）＝．8．（5分）二项展开式（2x+3）7中, 在所有的项的系数、所有的二项式系数中随机选取一个, 恰好为奇数的概率为．9．（5分）在平面直角坐标系xOy内, 曲线|x+1|+|x﹣3|+|y|＝7所围成的区域的面积为．10．（5分）已知梯形ABCD中, AD＝DC＝CB＝AB, P是BC边上一点, 且＝x+y, 当P在BC边上运动时, x+y的最大值是．11．（5分）求方程2sin x﹣sec x+tan x﹣1＝0在x∈[0, 2π]的解集．12．（5分）已知底面为正方形且各侧棱长均相等的四棱锥V﹣ABCD可绕着AB任意旋转, AB⊂平面α, M是CD的中点, AB＝2, VA＝, 点V在平面α上的射影点为O, 则|OM|的最大值为．二、选择题（本大题共4道小题, 每小题只有一个正确选项, 答对得5分, 满分20分）13．（5分）下列以t为参数的方程所表示的曲线中, 与曲线xy＝1完全一致的是（）A．B．C．D．14．（5分）已知无穷数列{a n}是公比为q的等比数列, S n为其前n项和, 则“0＜|q|＜1”是“存在M＞0, 使得|S n|＜M对一切n∈N*恒成立”的（）条件A．充分不必要B．必要不充分C．充要D．既不充分也不必要15．（5分）已知z均为复数, 则下列命题不正确的是（）A．若z＝, 则z为实数B．若z2＜0, 则z为纯虚数C．若|z+1|＝|z﹣1|, 则z为纯虚数D．若z3＝1, 则＝z216．（5分）直线l在平面α内, 直线m平行于平面α, 且与直线l异面, 动点P在平面α上, 且到直线l、m距离相等, 则点P的轨迹为（）A．直线B．椭圆C．抛物线D．双曲线三、解答题（本大题共5道小题, 每一问均需写出必要步骤, 满分共76分）17．（12分）如图, 在正三棱柱ABC﹣A1B1C1中, AA1＝A1B1＝4, D、E分别为AA1与A1B1的中点．（1）求异面直线C1D与BE所成角的大小；（2）求四面体BDEC1的体积．18．（14分）某工厂在制造产品时需要用到长度为698m的A型和长度为518m的B型两种钢管．工厂利用长度为4000m的钢管原材料, 裁剪成若干A型和B型钢管．假设裁剪时损耗忽略不计, 裁剪后所剩废料与原材料的百分比称为废料率．（1）有两种裁剪方案的废料率小于4.5%, 请说明这两种方案并计算它们的废料率；（2）工厂现有100根原材料钢管, 一根A型和一根B型钢管为一套毛胚, 按（1）中的方案裁剪, 最多可裁剪多少套毛胚？最终的废料率为多少？19．（16分）设函数f（x）＝x2+|x﹣a|（x∈R, a∈R）．（1）讨论f（x）的奇偶性；（2）当a＝1时, 求f（x）的单调区间；（3）若f（x）＜10对x∈（﹣1, 3）恒成立, 求实数a的取值范围．20．（16分）已知椭圆+y2＝1, A是它的上顶点, 点P n, Q n（n∈N*）各不相同且均在椭圆上（1）若P1, Q1恰为椭圆长轴的两个端点, 求△AP1Q1的面积；（2）若•＝0, 求证：直线P n Q n过一定点；（3）若＝＝1﹣, △AP n Q n的外接圆半径为R n, 求R n的值21．（18分）对任意正整数m, 若存在数列a1, a2, ……, a k, 满足m＝a1•1！+a2•2！+a3•3！+L+a k•k！, 其中a1∈N, a1≤i, a k＞0, i＝1, 2, ……, k, 则称数列a1, a2, ……, a k为正整数m的生成数列, 记为A[m]．（1）写出2018的生成数列A[2018]；（2）求证：对任意正整数m, 存在唯一的生成数列A[m]；（3）求生成数列A[2025！﹣1949！]的所有项的和．2018年上海市交大附中高考数学模拟试卷参考答案与试题解析一、填空题（本大题共12小题, 1-6题每题4分, 7～12题每题5分, 满分54分）1．（4分）函数f（x）＝的定义域为（﹣1, 0）∪（0, 1]．【解答】解：∵∴∴﹣1＜x≤1且x≠0,∴f（x）的定义域为（﹣1, 0）∪（0, 1]．故答案为：（﹣1, 0）∪（0, 1]．2．（4分）双曲线3x2﹣y2＝12的两渐近线的夹角大小为60°．【解答】解：由双曲线3x2﹣y2＝12, 可知双曲线的两条渐近线方程为y＝±x, ∴两条渐近线的倾斜角分别为60°, 120°,∴双曲线3x2﹣y2＝12的两条渐近线的夹角是60°,故答案为：60°．3．（4分）用行列式解线性方程组, 则D y的值为﹣9．【解答】解：行列式解线性方程组, 则D y＝＝2×（﹣1）﹣7×1＝﹣9,故答案为：﹣94．（4分）湖面上浮着一个球, 湖水结冰后将球取出, 冰上留下一个直径为24厘米, 深为8厘米的空穴, 则这个球的半径为13厘米．【解答】解：设球的半径为Rcm,由将球取出, 扭留下空穴的直径为24cm, 深8cm则截面圆的半径r＝12cm, 球心距d＝（R﹣8）cm,由R2＝r2+d2得：R2＝144+（R﹣8）2,即208﹣16R＝0解得R＝13故答案为：13cm5．（4分）直线2x+y﹣4＝0经过抛物线y2＝2px的焦点, 则抛物线的准线方程是x＝﹣2．【解答】解：抛物线y2＝2px的焦点坐标为（, 0）,又焦点在直线2x+y﹣4＝0上,∴p﹣4＝0,解得p＝4,则抛物线的准线方程是：x＝﹣2．故答案为：x＝﹣2．6．（4分）已知函数y＝sin（ωx+φ）（ω＞0, 0＜φ≤）的部分图象如图所示, 则点P （ω, φ）的坐标为（2, ）．【解答】解：∵T═﹣＝, ω＞0,∴T＝＝π,∴ω＝2；又曲线过（, 0）且为单调递减区间上的零点,∴ω+φ＝π+2kπ（k∈Z）,∴φ＝+2kπ（k∈Z）, 而0＜φ≤,∴φ＝,∴点P（ω, φ）的坐标为（2, ）．故答案为：（2, ）．7．（5分）设函数f（x）＝的反函数为f﹣1（x）, 若,则f（a+4）＝﹣2．【解答】解：∵,∴f（a）＝．当x≥6时, f（x）＝﹣log3（x+1）≤﹣log37＜0, 不符合条件, 舍去；当x＜6时, f（x）＝3x﹣6, 令＝3﹣2, ∴a﹣6＝﹣2, 解得a＝4, 满足条件．∴f（8）＝﹣log39＝﹣2．故答案为：﹣2．8．（5分）二项展开式（2x+3）7中, 在所有的项的系数、所有的二项式系数中随机选取一个, 恰好为奇数的概率为．【解答】解：二项展开式（2x+3）7中, 所有的项的系数为•3r•27﹣r, 其中, r＝0, 1, 2, 3, 4, 5, 6, 7,即所有的项的系数共有8个, 其中, r＝7时为奇数, 其余都为偶数．有的二项式系数为, 其中, r＝0, 1, 2, 3, 4, 5, 6, 7, 共有8个, 都是奇数．在所有的项的系数、所有的二项式系数中, 共有9个奇数, 7个偶数, 从中随机选取一个, 恰好为奇数的概率为＝,故答案为：．9．（5分）在平面直角坐标系xOy内, 曲线|x+1|+|x﹣3|+|y|＝7所围成的区域的面积为33．【解答】解：在平面直角坐标系xOy内, 曲线|x+1|+|x﹣3|+|y|＝7所围成的区域如下图所示：其面积为：2××6×+4×6＝33,故答案为：3310．（5分）已知梯形ABCD中, AD＝DC＝CB＝AB, P是BC边上一点, 且＝x+y, 当P在BC边上运动时, x+y的最大值是．【解答】解：设AB的中点为E, 则由题意可得△BCE为等边三角形, 且＝﹣＝﹣,再根据、共线, 可得＝λ＝λ（﹣）, λ∈[0, 1],∴＝+＝（1﹣）+λ．又＝x+y, ∴, ∴x+y＝1﹣+λ＝1+≤1+＝,故x+y的最大值是,故答案为：．11．（5分）求方程2sin x﹣sec x+tan x﹣1＝0在x∈[0, 2π]的解集{, , π}．【解答】解：方程2sin x﹣sec x+tan x﹣1＝0⇒．⇒2sin x cos x﹣1+sin x﹣cos x＝0．（cos x≠0）⇒sin x﹣cos x＝（sin x﹣cos x）2, （cos x≠0）．⇒sin x﹣cos x＝0或sin x﹣cos x＝1⇒tan x＝1⇒或cos x＝﹣1．x∈[0, 2π]∴x＝, , π．故答案为：{, , π}．12．（5分）已知底面为正方形且各侧棱长均相等的四棱锥V﹣ABCD可绕着AB任意旋转, AB⊂平面α, M是CD的中点, AB＝2, VA＝, 点V在平面α上的射影点为O, 则|OM|的最大值为1+．【解答】解：设∠VNO＝θ,则∵N、M分别是AB、CD的中点, AB＝2, VA＝,∴AN＝1, VN＝2,MN＝BC＝AB＝2, VN＝VM＝2,则三角形VNM为正三角形, 则∠MNV＝60°,则ON＝2cosθ,在三角形OMN中,OM2＝MN2+ON2﹣2MN•ON cos（60°+θ）＝4+4cos2θ﹣2×2×2cosθcos（60°+θ）＝4+4cos2θ﹣8cosθ（cosθ﹣sinθ）＝4+4cos2θ﹣4cos2θ+4sinθcosθ＝4+2sin2θ＝（2,∴要使OM最大, 则只需要sin2θ＝1, 即2θ＝90°即可, 则θ＝45°,此时OM＝+1．故答案为：1+．二、选择题（本大题共4道小题, 每小题只有一个正确选项, 答对得5分, 满分20分）13．（5分）下列以t为参数的方程所表示的曲线中, 与曲线xy＝1完全一致的是（）A．B．C．D．【解答】解：在A中, t＞0, 在xy＝1时, x, y∈（﹣∞, 0）∪（0, +∞）, 故A 错误；在B中, t≠0, 在xy＝1时, x, y∈（﹣∞, 0）∪（0, +∞）, 故B正确；在C中, t的终边不能在y轴上, 在xy＝1时, x, y∈（﹣∞, 0）∪（0, +∞）, 故C错误；在D中, t的终边不能在y轴上, x, y∈（﹣∞, 0）∪（0, +∞）, 故D错误．故选：B．14．（5分）已知无穷数列{a n}是公比为q的等比数列, S n为其前n项和, 则“0＜|q|＜1”是“存在M＞0, 使得|S n|＜M对一切n∈N*恒成立”的（）条件A．充分不必要B．必要不充分C．充要D．既不充分也不必要【解答】解：∵{a n}是公比为q的等比数列, 当0＜|q|＜1时, S n＝,|S n|＝||,即“存在M＞||, 使得|S n|＜M对一切n∈N*恒成立”,即“0＜|q|＜1”是“存在M＞0, 使得|S n|＜M对一切n∈N*恒成立”的充分条件,当q＝﹣1时, |S n|＝即取M＝2|a1|即可,即“0＜|q|＜1”是“存在M＞0, 使得|S n|＜M对一切n∈N*恒成立”的不必要条件, 综上可知：即“0＜|q|＜1”是“存在M＞0, 使得|S n|＜M对一切n∈N*恒成立”的充分不必要条件．故选：A．15．（5分）已知z均为复数, 则下列命题不正确的是（）A．若z＝, 则z为实数B．若z2＜0, 则z为纯虚数C．若|z+1|＝|z﹣1|, 则z为纯虚数D．若z3＝1, 则＝z2【解答】解：对于A, 设z＝a+bi（a, b∈R）, 由, 可得a+bi＝a﹣bi, 则2bi＝0, b＝0, ∴z为实数, 故A正确；对于B, 设z＝a+bi（a, b∈R）, 则z2＝a2+b2+2abi＜0, ∵z2＜0, ∴a＝0, 则z纯虚数, 故B正确；对于C, 当z＝0时, 有|z+1|＝|z﹣1|, 故C错误；对于D, 由z3＝1, 得z3﹣1＝0, 即（z﹣1）（z2+z+1）＝0, 可得z＝1或z＝, ∴, 故D正确．∴错误的是C．故选：C．16．（5分）直线l在平面α内, 直线m平行于平面α, 且与直线l异面, 动点P在平面α上, 且到直线l、m距离相等, 则点P的轨迹为（）A．直线B．椭圆C．抛物线D．双曲线【解答】解：设直线m在平面α的射影为直线n, 则l与n相交,不妨设l与n垂直, 设直线m与平面α的距离为d,在平面α内, 以l, n为x轴, y轴建立平面坐标系,则P到直线l的距离为|y|, P到直线n的距离为|x|,∴P到直线m的距离为,∴|y|＝, 即y2﹣x2＝d2,∴P点轨迹为双曲线．故选：D．三、解答题（本大题共5道小题, 每一问均需写出必要步骤, 满分共76分）17．（12分）如图, 在正三棱柱ABC﹣A1B1C1中, AA1＝A1B1＝4, D、E分别为AA1与A1B1的中点．（1）求异面直线C1D与BE所成角的大小；（2）求四面体BDEC1的体积．【解答】解：（1）以C为原点, 在平面ABC中过C作BC的垂线为x轴,以CB为y轴, CC1为z轴, 建立空间直角坐标系,则C1（0, 0, 4）, D（2, 2, 2）, B（0, 4, 0）,A1（2, 2, 4）, B1（0, 4, 4）, E（）,＝（2, 2, ﹣2）, ＝（）,设异面直线C1D与BE所成角的大小为θ,则cosθ＝＝＝,∴θ＝arccos．∴异面直线C1D与BE所成角的大小为arccos．（2）点C1到平面BDE的距离d＝＝2,S△BDE＝﹣﹣﹣S△ADB＝＝6,∴四面体BDEC1的体积：V＝＝＝4．18．（14分）某工厂在制造产品时需要用到长度为698m的A型和长度为518m的B型两种钢管．工厂利用长度为4000m的钢管原材料, 裁剪成若干A型和B型钢管．假设裁剪时损耗忽略不计, 裁剪后所剩废料与原材料的百分比称为废料率．（1）有两种裁剪方案的废料率小于4.5%, 请说明这两种方案并计算它们的废料率；（2）工厂现有100根原材料钢管, 一根A型和一根B型钢管为一套毛胚, 按（1）中的方案裁剪, 最多可裁剪多少套毛胚？最终的废料率为多少？（1）设每根原材料可裁剪a根A型和b根B型钢管, 则, 【解答】解：方案一：, 废料率最小为（1﹣）×100%＝0.35%,方案二：, 废料率最小为（1﹣）×100%＝4.3%,（2）设用方案一裁剪x根原材料, 用方案二裁剪y根原材料, 共裁剪得z套毛胚, 则, z＝2x+4y,得, z max＝320套,废料率为＝2.72%,答：最多可裁剪320套毛胚, 最终的废料率为2.72%19．（16分）设函数f（x）＝x2+|x﹣a|（x∈R, a∈R）．（1）讨论f（x）的奇偶性；（2）当a＝1时, 求f（x）的单调区间；（3）若f（x）＜10对x∈（﹣1, 3）恒成立, 求实数a的取值范围．【解答】解：（1）当a＝0时, f（x）为偶函数；当a≠0时, f（x）为非奇非偶函数（2）a＝1时, f（x）＝x2+|x﹣1|＝＝∴函数的单调减区间为（﹣∞, ）, 函数的单调增区间为（, +∞）（3）f（x）＝x2+|x﹣a|＜10对x∈（﹣1, 3）恒成立, 等价于x2﹣10＜x﹣a＜10﹣x2, 等价于对x∈（﹣1, 3）恒成立∴2≤a≤420．（16分）已知椭圆+y2＝1, A是它的上顶点, 点P n, Q n（n∈N*）各不相同且均在椭圆上（1）若P1, Q1恰为椭圆长轴的两个端点, 求△AP1Q1的面积；（2）若•＝0, 求证：直线P n Q n过一定点；（3）若＝＝1﹣, △AP n Q n的外接圆半径为R n, 求R n的值【解答】解：（1）椭圆+y2＝1, A（0, 1）, P1（﹣2, 0）, Q1（2, 0）,△AP1Q1的面积为×1×4＝2；（2）证明：设直线AP n的方程为y＝kx+1,由•＝0, 即⊥,可得直线AQ n的方程为y＝﹣x+1,由y＝kx+1和x2+4y2＝4联立, 可得（1+4k2）x2+8kx＝0,解得x1＝﹣, x2＝0,可得P n（﹣, ）,将k换为﹣可得Q n（, ）,直线P n Q n的斜率为＝,直线P n Q n的方程为y﹣＝（x+）,化简可得y＝x﹣,则直线P n Q n过一定点（0, ﹣）；（3）＝＝1﹣, △AP n Q n的外接圆半径为R n,由椭圆方程可得＝﹣, ＝,△AP n Q n为等腰三角形, △AP n Q n的外接圆圆心在y轴上,设为（0, t）, （0＜t＜1）,由圆的定义可得1﹣t＝,化为1﹣t＝4﹣,可得R n＝4﹣,可得R n＝（4﹣）＝4﹣0＝4．21．（18分）对任意正整数m, 若存在数列a1, a2, ……, a k, 满足m＝a1•1！+a2•2！+a3•3！+L+a k•k！, 其中a1∈N, a1≤i, a k＞0, i＝1, 2, ……, k, 则称数列a1,a2, ……, a k为正整数m的生成数列, 记为A[m]．（1）写出2018的生成数列A[2018]；（2）求证：对任意正整数m, 存在唯一的生成数列A[m]；（3）求生成数列A[2025！﹣1949！]的所有项的和．【解答】（1）解：2018＝1×2!+4×4!+4×5!+2×6!,故数列A[2018]为a1＝0, a2＝1, a3＝0, a4＝4, a5＝4, a6＝2；（2）证明：对于恰有k项生成数列, 其表示的正整数最小值为k!,表示的正整数最大值为1•1!+2•2!+3•3!+…+k•k!＝（k+1）!﹣1,即k项的不同生成数列共有2•3•4…k•k＝k•k!＝（k+1）!﹣k!,而满足k!≤n≤（k+1）!﹣1的正整数n恰好有（k+1）!﹣k!个．下面只需证明两个不同的k项生成数列表示的正整数不同．设生成数列a1, a2, a3, …, a k和b1, b2, b3, …, b k表示的数为A和B, 若a k＜b k,则a1•1!+a2•2!+…+a k•k!≤1•1!+2•2!+…+（k﹣1）•（k﹣1）!+a k•k!＝（a k+1）•k!﹣1＜b k•k!．即A＜B；同理若有a k＝b k, a k﹣1＜b k﹣1, 也可到A＜B；依此类推可得A＝B的充要条件是生成数列a1, a2, a3, …, a k和b1, b2, b3, …, b k相同．综上可得, 对任意正整数m, 存在唯一的生成数列A[m]；（3）解：∵（k+1）!﹣k!＝k•k!,∴2025！﹣1949！＝1949•1949!+1950•1950!+…+2024•2024!,即数列A[2025！﹣1949！]的通项为：．故所有项的和为：＝150974．。
上海市交大附中2019届高三上9月开学摸底考试数学试题(含参考答案)
上海市交大附中高三9月份开学考试一、填空题.1.方程组的增广矩阵是______．【答案】【解析】试题分析：根据增广矩阵的定义可知为.考点：本小题主要考查增广矩阵的定义和应用.点评：增广矩阵就是在系数矩阵的右边添上一列，这一列是线性方程组的等号右边的值。
2.若直线的参数方程为，则直线的倾斜角是_______．【答案】【解析】【分析】根据题意，将直线的参数方程变形为普通方程为y+2（x﹣3），求出其斜率，结合直线的斜率与倾斜角的关系可得tanθ，结合θ的范围，分析可得答案．【详解】根据题意，直线l的参数方程为，则其普通方程为y+2（x﹣3），其斜率k，则有tanθ，且0°≤θ＜180°，则θ＝120°；故答案为：120°．【点睛】本题考查直线的参数方程，关键是将直线的参数方程变形为普通方程,熟记斜率与倾斜角的关系是关键，是基础题3._______．【答案】【解析】【分析】利用二项式定理系数的性质，求解分子，然后利用数列极限的运算法则求解即可．【详解】由二项式定理系数的性质可得，．故答案为：．【点睛】本题考查二项式定理系数的性质，数列的极限的运算法则的应用，考查计算能力，是基础题4.已知数列的前项的和，则当为正偶数时，______．【答案】【解析】【分析】由已知求得，当n≥2且n为正偶数时，a n＝S n﹣S n﹣1＝2n﹣[2（n﹣1）﹣1]＝2n﹣2n+3，验证a2＝3适合，由此可得当n为正偶数时的a n．【详解】由，得＝1，；当n≥2且n为正偶数时，a n＝S n﹣S n﹣1＝2n﹣[2（n﹣1）﹣1]＝2n﹣2n+3．验证＝3适合上式，∴当n为正偶数时，．故答案为：2n﹣2n+3．【点睛】本题考查数列通项公式，考查利用数列的前n项和求数列的通项公式，是中档题．5.函数是奇函数，那么______．【答案】【解析】【分析】求f（﹣x）＝，再根据f（x）为奇函数，可得出＝-整理化简即可求出a的值．【详解】由题f（﹣x）＝函数是奇函数，∴-f（﹣x）＝，即-解得2,∴故答案为-1【点睛】本题考查奇函数的定义，多项式的运算，多项式相等的充要条件，准确利用定义计算是关键，是基础题6.若函数无最值，则的取值范围是______．【答案】a或a【解析】【分析】由题意函数f（x）＝lg（x2﹣ax+2）无最值，即f（x）的值域为R，那么（0，+∞）是y＝x2﹣ax+2的值域的子集，即△≥0，可得a的取值范围．【详解】由题意，函数f（x）＝lg（x2﹣ax+2）无最值，即f（x）的值域为R，那么（0，+∞）是y＝x2﹣ax+2的值域的子集，即△≥0，∴a2﹣8≥0，则a或a；故答案为：a或a．【点睛】本题考查对数型复合函数的值域，考查对数函数的性质，明确真数无最值是突破点，准确利用二次函数的△≥0解决问题是关键，是中档题7.△的内角，，的对边分别为，，，已知△的面积为，，则______．【答案】【解析】【分析】直接利用三角形的面积公式和正弦定理求出sinBsinC的值，进一步利用三角函数关系式的变换即可求出A的值．【详解】已知△ABC的面积为，则：S△ABC acsinB，整理得：3csinBsinA＝2a，由正弦定理得：3sinCsinBsinA＝2sinA，由于sinA≠0，故：sinBsinC，由于：6cosBcosC＝1，所以：cosBcosC，所以：cosBcosC﹣sinBsinC，所以：cos（B+C），故：cosA，A所以：A．故答案为：．【点睛】本题考查的知识要点：三角函数关系式的恒等变换，正弦定理和三角形面积公式的应用，主要考查学生的运算能力和转化能力，属于基础题型．8.设，是虚数单位，已知集合，，若，则的取值范围是________．【答案】【解析】【分析】根据复数的代数表示法及其几何意义可知集合A表示的点的轨迹是以（0，1）为圆心，半径为2的圆及内部；集合B表示圆的圆心移动到了（1，1+b）；两圆面有交点即可求解b的取值范围．【详解】由题意，集合A表示的点的轨迹是以（0，1）为圆心，半径为2的圆及内部；集合B表示点的轨迹为（1，1+b），半径为2的圆及内部∵A∩B≠∅，说明，两圆面有交点；∴．可得：，故答案：，【点睛】本题考查复数几何意义，圆与圆的位置关系，体现了数学转化思想方法，明确A.B集合的意义是关键，是中档题9.从双曲线（，）的左焦点引圆的切线，切点为，延长交双曲线右支于点，若是线段的中点，为坐标原点，则的值是____．【答案】【解析】试题分析：如图所示，设双曲线的右焦点为，连接，，，则，在中，，，所以，又是线段的中点，为中点，所以，所以即，故应填入.考点：1.双曲线的定义；2.直线与圆相切；3.数形结合的应用.10.胡涂涂同学用一颗均匀的骰子来定义递推数列，首先，他令，当时，他投一次骰子，若所得点数大于，即令，否则，令，则的概率为______（结果用最简分数表示）．【答案】【解析】【分析】胡涂涂同学掷了3轮，要使得，分两种情况讨论，再利用古典概型求的概率.【详解】胡涂涂同学掷了3轮，要使得，有两种情况，①一轮点数为1，二轮点数为1、2、3、4、5、6，三轮点数为1；②一轮点数为2、3、4、5、6，二轮点数为1、2，三轮点数为1；∴由古典概型得所求的概率为.故答案为：【点睛】本题主要考查排列组合的应用，考查古典概型，意在考查学生对这些知识的理解掌握水平和分析推理计算能力.11.关于的方程恰有3个实数根，，，则__________．【答案】2【解析】【分析】令f（x）＝x2+arcsin（cosx）+a，判断f（x）的奇偶性，由题意可得f（0）＝0，求得a，再由反三角函数的定义和性质，化简函数，求得f（x）＝0的解，即可得到所求和．【详解】令f（x）＝x2+arcsin（cosx）+a，可得f（﹣x）＝（﹣x）2+arcsin（cos（﹣x））+a＝f（x），则f（x）为偶函数，∵f（x）＝0有三个实数根，∴f（0）＝0，即0a＝0，故有a，关于x的方程即x2+arcsin（cosx）0，可设＝0，且2+arcsin（cos）0，2+arcsin（cos）0，＝﹣，由y＝x2和y arcsin（cosx），当x＞0，且0＜x＜π时，y arcsin（cosx）arcsin（sin（x））（x））＝x，则﹣π＜x＜0时，y arcsin（cosx）＝﹣x，由y＝x2和y arcsin（cosx）的图象可得：它们有三个交点，且为（0，0），（﹣1，1），（1，1），则2+2+2＝0+1+1＝2．故答案为：2．【点睛】本题考查函数与方程，函数的奇偶性，反三角函数的定义和性质，函数方程的转化思想，以及化简整理的运算能力，属于中档题．12.由无理数论引发的数字危机一直延续到19世纪，直到1872年，德国数学家戴德金从连续性的要求出发，用有理数的“分割”来定义无理数（史称戴德金分割），并把实数理论建立在严格的科学基础上，才结束了无理数被认为“无理”的时代，也结束了持续2000多年的数学史上的第一次大危机，所谓戴德金分割，是指将有理数集划分为两个非空的子集与，且满足，，中的每一个元素都小于中的每一个元素，则称为戴德金分割.试判断，对于任一戴德金分割，下列选项中，可能成立的是____．①没有最大元素，有一个最小元素；②没有最大元素，也没有最小元素；③有一个最大元素，有一个最小元素；④有一个最大元素，没有最小元素.【答案】①②④【解析】【分析】由题意依次举例对四个命题判断，从而确定答案．【详解】若M＝{x∈Q|x＜0}，N＝{x∈Q|x≥0}，则M没有最大元素，N有一个最小元素0，故①可能成立；若M＝{x∈Q|x}，N＝{x∈Q|x}；则M没有最大元素，N也没有最小元素，故②可能成立；若M＝{x∈Q|x≤0}，N＝{x∈Q|x＞0}；M有一个最大元素，N没有最小元素，故④可能成立；M有一个最大元素，N有一个最小元素不可能，因为这样就有一个有理数不存在M和N两个集合中，与M 和N的并集是所有的有理数矛盾，故③不可能成立．故答案为：①②④【点睛】本题考查新定义的理解和运用，考查列举法和推理能力，对每个选项举出反例说明是关键，属于基础题．二、选择题。
上海市交通大学附属中学2018届高三上学期开学摸底考试数学试题+Word版含解析
上海市交通大学附属中学2018届高三上学期开学摸底考试数学试题一、填空题1. 若集合，集合，则 __________．【答案】【解析】由题意得，或，所以.2. —个几何体的主视图、左视图、俯视图都是以为半径的圆，则该几何体的体积是__________．【答案】【解析】根据几何体的三视图的规则可知，该几何体表示半径为的球，所以该几何体的体积为.3. 已知是虚数单位，则的平方根是__________．【答案】【解析】设复数，则，即，解得，所以.4. 函数的反函数是__________．【答案】【解析】由，则，因为，则，所以函数的反函数.5. 设满足约束条件，则的最小值是__________．【答案】【解析】画出约束条件所表示的平面区域，如图所示，当经过可行域的点时，目标函数取得最小值，由，解得，则的最小值是.6. 如图，四个棱长为1的正方体排成一个正四棱柱，是一条侧棱，是上、下底面上其余十六个点，则的不同值的个数为__________．【答案】2【解析】由题意得，,则,因为，所以，所以的不同的值的个数为.7. 数列满足，其前项和记为，若，那么__________．【答案】3【解析】因为，所以，即，所以，即，即数列是周期为6的周期数列，因为，所以，所以，所以，又因为，解得，，且所以8. 若是展开式中项的系数，则__________．【答案】8【解析】试题分析：由题意，，∴.....................考点：二项展开式的通项与裂项相消法求和，极限．9. 设函数，其中，若，且的最小正周期大于，则__________．【答案】【解析】由的最小正周期大于，得，又，得，所以，则，所以，由，所以，取，得，所以.10. 已知函数，设，若关于的不等式在上恒成立，则的取值范围是__________．【答案】【解析】根据题意，函数的图象如图，令，其图象与x轴相交于点，在区间上我减函数，在上为增函数，若不等式在上恒成立，则函数的图象在上的上方或相交，则必有，即，可得.。
上海市交通大学附属中学2018-2019学年高三上期末考试数学试题(简略答案)
交大附中2018-2019学年度第一学期高三年级期末数学试卷2019.1一、填空题1.已知集合{}02A x =<≤，集合{}12B x x =-<<，则A B =U ______.2.若复数43z i =+，其中i 是虚数单位，则2z =______.3.函数()()4,43,4x x f x f x x -≥⎧⎪=⎨+<⎪⎩，则()1f f -=⎡⎤⎣⎦______. 4.已知1sin 43πα⎛⎫-= ⎪⎝⎭，则cos 4πα⎛⎫+ ⎪⎝⎭的值为______. 5.已知数列{}n a 的前n 项和()2*2n S n n n N =+∈，数列{}n a 的通项公式为n a =______.6.已知实数x 、y 满足约束条件222441x y x y x y +≥⎧⎪+≤⎨⎪-≥-⎩，则目标函数3z x y =+的取值范围为______.7.已知函数()()sin 2cos2,,0f x a x b x a b R ab =+∈≠，若其图像关于直线6x π=对称，则直线20ax by ++=的倾斜角α=______. 8.鲁班锁是中国传统的智力玩具，起源于古代汉族建筑中首创的椎卯结构，这种三维的拼插器具内部的凹凸部分（即榫卯结构）啮合，十分巧妙，外观看是严丝合缝的十字立方体，其上下、左右、前后完全对称，从外表上看，六根等长的正四棱柱分成三组，经90︒样卯起来，如图，若正四棱柱的高为6，底面正方形的边长为1，现将该鲁班锁放进一个球形容器内，则该球形容器的表面积的最小值为（容器壁的厚度忽略不计）______.14.某超市货架上摆放着某品牌红烧牛肉方便面，如图是它的三视图，则货架上的红烧牛肉便面至少有A.8桶B.9桶C.10桶D.11桶15.已知()23f x x x =+，若1x a -≤，则下列不等式一定成立的是A.()()33f x f a a -≤+B.()()5f x f a a -≤+C.()()24f x f a a -≤+D.()()()231f x f a a -≤+ 16.若2a b c ===&&&，且0a b ⋅=&&，()()0a c b c --≤&&&&，则a b c +-&&&的取值范围是A.0,2⎡⎤⎣⎦B.[]0,2C.2⎡⎤⎣⎦D.2,2⎡⎤⎣⎦三、解答题17.在ABC ∆中，a 、b 、c 分别为角A 、B 、C 的对边，已知()cos23cos 1A B C -+=.（1）求角A 的值；（2）若2a =，求ABC ∆周长的取值范围。
上海交通大学附属中学2018-2019学年高三第一学期摸底考试
上海交通大学附属中学2018-2019学年高三第一学期摸底考试英语试卷（考试时间120分钟；满分140分）第I卷Ⅰ. VocabularyDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.(A)1-5 BD A BC ABC D 6-10 B AB CB AC CTeenagers seem more likely to give in to peer influence than adults, but that may not be because teens are less 1 of making rational decisions themselves, studies on peer pressure suggest.Research finds people are strikingly susceptible to influences teenagers, but to what degree varies widely. In a growing body of work, scientists suggest that teens are more 2 to peer pressure than adults because they get greater pleasure from behavior they experience as rewarding. They tend to find being liked by other people very pleasing.Peer influence during adolescence is normal. During that time, people experience an increase in novelty-seeking, 3 by interest in exploring a new environment. This peer influence tends to 4 around age 15 and then decline. Teens get better at setting boundaries with peers by age 18, according to Laurence Steinberg, a psychology professor at Temple University.In years past, people thought teens didn’t have fully 5 frontal lobes (额叶), the part of the brain critical for decision-making and other more complex cognitive tasks. But an increasing amount of research seems to show that teens are able to make decisions as well as adults when they are not overwhelmed 6 with emotions.Peer pressure is often seen as negative, and indeed it can 7 teens into unhealthy behavior like smoking or speeding. But it can also lead to 8 in more useful social behavior. If peers value doing well in school or excelling at sports, for instance, it might encourage kids to study or train harder. And both peer pressure and learning to 9 it are important developmental steps to self-reliance, experts say.Facing the influence of friends represents an important developmental step for teens on their way to becoming independent-thinking adults, scientists say.In order for kids to develop the ability to stand up to peer pressure, parents have to let their children stand up to them, too, according to Dr. Steinberg. “If you’re the kind of parent that raises your children with the ‘do it because I said so’ 10 , you’re raising a child who’s going to be more susceptible to others saying, ‘Do this,’” he says.(B)1-5 CD C A D AC 6-10 ABC BC AB B BDStudents Honor 9/11 Through VolunteeringIn the days after Sept. 11, 2001, thousands poured into Ground Zero to lend their hands in one of the largest recovery efforts(恢复经济) in American history. Now, 12 years later, colleges are finding ways to channel（疏通）the same 1 into service projects in their own communities on September 11th."9/11 is such a(n) __2__ touchstone（试金石）for our country, I think people want to find ways to do something, and students as well," says Kevin Kruger, president of the student affairs group. "The idea of giving something back to the __3__ ties in well to（密切配合）the emotional significance of that."Though no corner of the country was untouched by the terrorist attacks, they especially rocked （晃动）New York University on 9/11. Less than two miles from Ground Zero, students__4__ the plane-on-tower impact from their dorm rooms. The university coordinated（协调）shelters and counseling（提供建议）in the days afterward.Because of this, NYU public affairs director Philip Lentz says the volunteer work the students do this week has a "special relevance（关联）." Students today __5__ at a rescue mission, wrote cards for soldiers and veterans（老兵）and made donations for the families of victims and first responders （现场目击者）on 9/11."NYU feels it has a special __6__ to offer this service opportunity to students that are here so they can give back to the community that was so devastated（毁灭）by the attacks," Lentz says.Similar deeds have been taking place in George Washington University in Washington, D.C. "For the past five years at GW, freshmen have boarded buses immediately after their official welcome __7__ in early September to head for the nation's financial capital and volunteer in __8__ that aid the environment, education, veterans and community organizations," says Amy Cohen, theuniversity's director for civic engagement and public service."We hope that students will learn that community __9__ is part of how we build strong communities and a vibrant __10__ society," Cohen wrote in an e-mail to USA TODAY. "We ask students to reflect on the tragedy of 9/11 and to use this day as a way to show civic engagement and community building."Ⅰ. Reading comprehension.Section ADirections:For each blank in the following passage there are four words or phrases marked A, B, C, D. Fill in each blank with the word or phrase that best fits the context.(A)Harvard LibraryIf we compare professors and students to the host of a university, then the library of a university can be compared to the hallway. The quality of a university, 1 , is in direct proportion to that of its library. At Harvard，the library is an essential part of everybody's life. Both the quantity and the 2 of the library make study a pleasant process.Harvard Library is not only the most ancient library in the United States, but the largest university library with the largest scale. In 1638 John Harvard 3 his whole library to the then Harvard College. After 300 years of development, the library now holds 10 million books and 4 more than 100 branch libraries. In addition to the libraries owned by each school, there are some branch libraries that are 5 in some aspects. While most of the branch libraries are on Harvard campus, some are as far as in Washington, D.C., or even in Florence of Italy. Yenching Library is famous for its 6 of East Asian literature. Lamont Library is the first library in the world that is 7 for undergraduates. Widener Library is the largest library in Harvard, only second to Library of Congress.What 8 to be mentioned is the system or rather theservice of the libraries. Usually the libraries are open from 9 a.m. to 5p.m.. The main libraries are open until 10 p.m.. The libraries forundergraduates will even be open all night during the 9period. The libraries also provide with students the service of 10 reading materials for all courses. At the beginning of a semester, each teacher will give a list of books to the librarians. The librarians are 11 to find out these books and put them at the places where students can easily find them.There is no limitation for the number of books that students can borrow. As the space for thelibrary is limited, many books are 12 in suburban library. Despite this, students can go to fetch the book at the 13 library within 24 hours after they submit request for that book. Even if there is only one book to be fetched from the suburban library, the libraries on campus will send someone to do the job. This kind of 14 which put readers in the first place is rare even in Ivy League. Therefore, study at Harvard will be a(n) 15 experience.A）1-5 BCADB 6-10 ACDAB 11-15 DCABD1. A. as a result B. to some extent C. on the contrary D. at all times2. A. influence B. discipline C. quality D. prospect3. A. donated B. assigned C. adapted D. distributed4. A. contains B. composes C. involves D. includes5. A. informative B. different C. secure D. peculiar6. A. collections B. documents C. phenomena D. exhibitions7. A. unusually B. formally C. specially D. especially8. A. remains B. happens C. appears D. deserves9. A. examination B. experiment C. vacation D. graduation10. A. confirming B. preparing C. selecting D. designing11. A. desperate B. willing C. reluctant D. responsible12. A. exhibited B. reserved C. stored D. classified13. A. appointed B. accepted C. expected D. restricted14. A. performance B. service C. activity D. response15. A. fortunate B. creative C. positive D. enjoyable(B)Instead of cruising in on a hoverboard, I rode my bike to the office today. The bicycle was invented in the 19th century. Instead of taking a pill for breakfast, I had a bacon roll, cooked using gas. Science fiction has 1 to us.Making predictions is tricky, especially about the future, as physicist Niels Bohr joked. In science fiction, you can't escape that 2 though. Since its birth in the 19th century, writers have 3 imagined the things to come: devices that humankind will invent to make life easier. But in so many instances, those promises have not come to pass. The biggest 4 are in travel--jet packs, hoverboards and flying cars are yet to fill the skies. Air travel has become significantly cheaper and wide-reaching, but only using distinctly 20th-century technology: commercial aero planes are much the same as they were 50 years ago.5 is what science fiction frequently delivers, but its arrival in the real world has been unpredictable. Domestic robots with a degree of intelligence are yet to6 , though robotic vacuum cleaners are commercially available--even if they are fairly hopeless. Video calls have now arrived--sort of--but conferencing on Skype is still dissatisfying. In mobiles, video call technology is now available, so when your dad rings to update you on his vegetable patch, he'll be able to7 your look of boredom.The truth is that we quickly 8 the astonishment of invention: our wonderment is soon replaced with the feeling of nothing new. We should ty to stay in that period of 9 . It is astonishing that the contents of every book ever written can be stored in a small box. Or that you can carry 10,000 albums on an object kept in your pocket. Or that almost all the information in the world can be accessed almost anywhere at any time. All these 10 are dependent on the emergence of the microchip and its place in computers. Yet sci-fi didn’t 11 the dominance of the computer in running our lives.But the real area where 12 far outstrips predictions is medicine. Sure, fiction would describe humans as ‘disease-free’ but without going into detail. “Disease-fee” humans are still absent, but the progress made in 13 life is breathtaking. With relative ease, we can sequence anyone's genome (基因组), giving a read-out of our entire genetic code. This means we can find out the underlying genetic cause of thousands of diseases in minutes.Photosensitive implants now exist that can replace damaged cells in the retina (视网膜) and can thus 14 sight to the blind. While the inventions of science fiction can show great ideas we’d like to happen, nothing 15 the inventiveness of people in the real world.B）1-5 BBDAC 6-10 ACCBD 11-15 ADACD1. A. tuned B. lied C. objected D. talked2. A. opportunity B. challenge C. Imagination D. conflict3. A. hesitantly B. critically C. temporarily D. tirelessly4. A. disappointments B. advancements C. enjoyments D. experiments5. A. Modernization B. Exploration C. Automation D. Transportation6. A. materialize B. identify C. honour D. liberate7. A. imagine B. feel C. see D. ignore8. A. arouse B. discover C. forget D. evaluate9. A. frustration B. amazement C. boredom D. limitation10. A. modes B. worries C. potentials D. actions11. A. predict B. overlook C. motivate D. prevent12. A. quality B. obstacle C. passion D. reality13. A. maintaining B. creating C. researching D. encountering14. A. show B. lend C. restore D. label15. A. guarantees B. overestimates C. releases D. OutperformsSection BDirections: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 you have just read.(A)In his book The Tipping Point Canadian author Malcolm Gladwell explains how a trend can take many forms. It can be a general change in social behaviour, an idea or a fashion. However, why do some trends catch on and others not? What makes one particular brand of training shoe suddenly become the must-have product? How do people find out about trends and what makes people want to buy into them? Is it simply a question of keeping up with other people?In his new work, Gladwell explores the moment when something becomes common and how products, ideas messages and forms of behaviour spread. He looks at the reasons why trends are similar in the way they develop to outbreaks of disease, or medical epidemics.Epidemics, like trends, start in a very small way, maybe from a single person with a virus, then spread very quickly until they take over the population and appear to be everywhere. Eventually, they will slowdown gradually or die out suddenly. Gladwell shows how these changes happen not gradually but at one dramatic moment.Gladwell identifies three types of people who are influential in the development of these kinds of social epidemics.Connectors are people in a community who have wide social circles. They know a lot of people and like to introduce people to catch other. The people they know also tend to come from a variety of social, cultural professional and economic circles.Mavens are people with a lot of knowledge or experts in a particular field. They wish to pass on their knowledge to others. Mavens collect and gather information so are the first to pick up on new trends.Salesmen are people with charisma and powerful negotiation skills. They have a soft influence over people rather than actual power. This means they are influential because people want to imitate them.Overall, Gladwell's book is a thought-provoking read for anyone interested in the origins oftrends. What's more, he writes in a clear style so even the most difficult ideas are easy to understand.71.According to Gladwell, which of the following is a similarity between trends and epidemics?A. Both of them start from nowhereB. Both of them die out in a short timeC. Both of them become popular overnightD. Both of them change he way people think72. What can be concluded from the passage about The Tipping Point?A. It is the writers first book to touch on social phenomenaB. It is intended to introduce the trends that once took overC. It deals with the ideas that most people identify withD. It explores How a trend originates and spreads73. The passage is most likely to beA. an adver sementB. a reviewC. a news reportD. a feature story74. Salesmen work in the development of trends by means ofA people intention to follow themB the way they gather informationD. their influence on the trendsC. their professional knowledge（B）(*Please purchase a ticket before boarding the coach75. On December 25th, 2014, the first coach from, Heathrow Terminal I to Reading departs at____.A.7:15B.7:00C.6:50D.6:3576. If you are to reach Terminal 2 of the Heathrow Airport before 10: 00 a.m., you need to get to Reading Station not later than____.A $. 30am B.8:45am C.9:00a,m. D.9:15a.m77. The information leaflet is produced by____.A. ReadingB. IntercityC. Railair LinkD. Heathrow Airport(C)As students are discussing their favorite colleges, there’s one characteristic they can’t control their race. That’s one reason voters, courts and politicians in six states have outlawed racial preferences in college admissions, while other colleges, fearful-of lawsuits, play down their affirmative action efforts these days. But make no mistake race still matters. How much depends on the school and the state.In: Texas, public universities have managed to reduce the effect of racial-preference bans by automatically admitting the top 10% of the graduating class of every high school, including those schools where most students are minorities. But Rice University in Houston, private and highly selective, has had to reinvent its admissions strategies to maintain the schools minority enrollment. Each February, 80 to 90 black, Hispanic(西班牙裔) and Native American kids visit Rice on an expenses-paid trip. Rice urges headmasters from high schools with large minority populations to recommend qualified students. And in the fall, Rice sends two recruiters on the road to find minority applicants; each recruiter visits about 80 mainly black or Hispanic high schools. Two weeks ago, Rice recruiter Tamara Siler dropped in on Westlake High in Atlanta, where 99% of the 1296 students are black. Siler went hearing literature and advice, and though only two kids showed up, she said: "Tm pleased I got two."Rice has also turned to some almost comical end-runs around the spirit of the law. The university used to award a yearly scholarship to a Mexican-American student; now it goes to a student who speaks Spanish really well. Admissions officers no longer know an applicant’s race. But a new essay question asks about each student "background" and "cultural traditions". When Rice officials read applications, they look for "diverse life experiences and what they awkwardly call "overcome students", who have triumphed over hardship.Last spring, admissions readers came across a student whose SAT score was lower than 1,200 and who did not rank in the top 10% of her class. Numerically speaking, she was far behind mostaccepted applicants: But her essay and recommendations indicated a strong interest in civil rights and personal experience with racial discrimination. She was admitted. All the newspapers say affirmative action is done, “says an experienced adviser at a large New York City high school. But nothing has changed. I have a(minority)kid at Yale with an SAT score in the high 900s.78. What does the word outlaw"(in Para. 1)most probably mean?A. supportB. consider D. hateC. ban79. What can we infer from the passage about affirmative action?A. It guarantees students, of different races to be admitted equally.B It discriminates against minority students in college admissions.C. It gives preference to minority students in college admissions.D. It is popular with American colleges but not with the American public80. Why does Rice University send two recruiters to find minority applicants?A. Rice has a large minority populationB: Rice wants to maintain its minority enrollmentC Minority students do not favor Rice very muchD. Minority students have better school performance81. Which of the following might be the best title of the passage?A: Here Comes Equality at LastB. Yes. Your Race Still MattersC. Well Done. Affirmative ActionD. Minorities Are Still a Minority in Universities(D)Direction: Read the following passage. Fill in each blank with a proper sentence given in the box.Sleep in a BoxHaving a newborn can be discouraging and troublesome. What do you need to buy? How will you get the baby to sleep? And how will you pay for everything? Could the answers to these worries lie in a “baby box"?The baby box has taken the world by storm, but what is it? Well, it originated in Finland in thel930s as a kind of basic items for parents-to-be. The box, provided by the government, contains supplies, such as bibs, nappies, and teething toys. It also comes with a small mattress, which is placed in the bottom of the box to create the baby’s first small bed. Now, the idea of putting a child to sleep inbox may shock some. 82_______ the box has been a huge success and is said to be one of the main reasons why the death rate of infants in Finland is so low.A few years ago, stories of the Finnish baby box went relating to viruses. Other countries around the world have since caught onto the idea and adapted it to suit local needs. For example, in South Africa, the box is made of plastic and can be used as a bath rather than a bed. A(83* The version tested in India includes other appropriate supplies, such as a mosquito het to protect babies from malaria(疟疾).In Canada, additions to the baby box include a bed-side assistance booklet “to help fathers connect with their babies, Written in the style of a car-repair manual, the booklet offers first-time dads helpful advice on matters such as how to burp(使打嗝)their child. 84________ I wanted to make that moment available to all parents,” said Morris, a father of a two-month baby boy.85_________ Many believe governments should focus on providing parents with adequate maternity and paternity Leaves(父母产假) and improving health care and education．Good things cone in small packages, though, and for such a simple idea the baby box really does do a lot of things!第II卷Ⅰ.GrammarDirections:After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)E-cigarettes, widely prompted as an alternative (1)_____to smoking, are actually attractingyoung people who might not otherwise (2)_____(smoke) tobacco products, a new US study suggested Monday.E-cigarettes (3)_____(think) by some to be responsible for a decline in American youth cigarette smoking, but researchers from the University of California, San Francisco (UCSF) indicated the reality is the opposite.The USCF researchers concluded that many kids who went on (4)_____(smoke) regular cigarettes may not have used nicotine(尼古丁) at all if e-cigarettes did not exist. "We didn't find any evidence that e-cigarettes are causing youth smoking to decline," said lead author Lauren Dutra of the UCSF.(5)_____ some of the kids using e-cigarettes were also smoking cigarettes, we found that kids who were at low risk of starting nicotine with cigarettes were using e-cigarettes," Dutra said. "Recent declines in youth smoking are likely (6)_____ tobacco control efforts, not to e-cigarettes."The findings, published in U.S. journal Pediatrics, built on a growing body of evidence that adolescents who start with e-cigarettes are (7)_____(likely) to subsequently smoke traditional cigarettes.They found that cigarette smoking among U.S. adolescents declined during that decade, but did not decline faster (8)_____ the advent of e-cigarettes in the U.S. between 2007 and 2009."E-cigarettes are encouraging -- not discouraging -- youth to smoke and to consume nicotine, and (9)_____(expend) the tobacco market," said senior author Stanton Glantz, UCSF professor of medicine and director of the UCSF Center for Tobacco Control Research and Education.In August 2016, the U.S. Food and Drug Administration restricted e-cigarette purchases to adults ages 18 and older. The FDA will also require a warning label on e-cigarettes, starting August 2018, regarding the addictive nature of nicotine. However, the FDA's (10)_____(rule) does not regulate advertising or flavors, and e-cigarettes continue to be sold in flavors that appeal to youth, the UCSF researchers said.(B)There seems never (11)______(be) a civilization without toys, but when and how they developed is unknown. They probably came about just to give children something to do.In the ancient world, (12)______is today, most boys played with some kinds of toys and most girls with another. In societies (13)______ social roles are rigidly determined, boys pattern their play after the activities of their fathers and girls are (14)______ the tasks of their mothers. This is true because boys and girls are being prepared, even in play, (15)______(step) into the roles and responsibilities of the adult world.What is remarkable about the history of toys is not so much (16)______ they changed over the centuries but how much they (17)______(remain) the same.The changes have been mostly in terms of craftsmanship, mechanics, and technology. It is the universality of toys with regard to their development in all parts of the world and (18)______ persistence to the present that is amazing. In Egypt, America, China, Japan and among the Arctic people, generally the same kinds of toys (19)______(appear). Variations depended on local customs and ways of life (20)______ toys imitate their surroundings. Nearly every civilization had dolls, little weapons, toy soldiers, tiny animals and vehicles.(C)In a shocking turn of events, the Academy Award for Best Picture was mistakenly awarded to "La La Land," (21)______ cast(剧组演员) and crew took the stage and began giving speeches－until a moment later the producers realized actually the award (22)_______(mean) for "Moonlight”.Presenter Warren Beatty explained afterwards that he was handed the wrong envelope, which contained the winner for best actress Emma Stone (23)______(star) in " La la land” After staring at the card for several moments, in (24)______ appeared to be an attempt to build suspense, his co-presenter Faye Dunaway announced that “La La Land" had won best picture.Realizing the mistake, representatives of Price Waterhouse Coopers raced onstage to stop the acceptance speeches from the "La La Land” team, and called the "Moonlight" cast and crew to the stage. Barry Jenkins’ "Moonlight" had actually won best picture in a historic Oscar upset."La La land" came in with 14 nominations, a record that tied (25)_____with “Titanic” and “All About Eve”. Barry Jenkins tender, coming-of-age drama, made for just $1.5 million, is an unusually small Oscar winner. "Even in my wildest dreams this cannot be true, " said the astonished Jenkins once he reached the stage.Host Jimmy Kimmel had come forward(26)______(inform) the cast that "Moonlight" had indeed won, showing the inside of the envelope(27)_______ proof. And Producer Jordan Horwitz then graciously passed his statue to the "Moonlight" producers.Barry Jenkins, the writer-director of "Moonlight" and Tare Alvin McCraney, whose play it was based on, won for(28)_______(adapt) screenplay. “For all you people out there (29)_______feel like there isn’t a mirror out there for you, the academy has your back, the ACLU has your back and for the next four years we will not leave you alone, we will not forget you, " said Jenkins.Backstage, Oscar winner Emma Stone said, "I think everyone’s in a state of confusion still. Later the actress, who pledged her deep love of "Moonlight" added, "Is that(30)______(crazy)Oscar moment of all time? Cool!Ⅰ.Translation1．你是否具备了成为一名演员的必要素质？(take)2．不可否认的是均衡的饮食是保持健康的关键之一。
上海交大附中高三(上)摸底数学试卷(解析版)
上海交大附中高三（上）摸底数学试卷参考答案与试题解析一、填空题（本大题共14题，每题4分，满分56分）1．已知全集U=R，集合A={x|x≤﹣2，x∈R}，B={x|x＜1，x∈R}，则（∁U A）∩B={x|﹣2＜x＜1}．【考点】交、并、补集的混合运算．【专题】集合．【分析】根据全集U及A求出A的补集，找出A补集与B的交集即可．【解答】解：∵全集U=R，集合A={x|x≤﹣2}，∴∁U A={x|x＞﹣2}，∵B={x|x＜1}，∴（∁U A）∩B={x|﹣2＜x＜1}．故答案为：{x|﹣2＜x＜1}【点评】此题考查了交、并、补集的混合运算，熟练掌握各自的定义是解本题的关键．2．已知互异的复数a，b满足ab≠0，集合{a，b}={a2，b2}，则a+b=﹣1．【考点】集合的相等．【专题】集合．【分析】根据集合相等的条件，得到元素关系，即可得到结论．【解答】解：根据集合相等的条件可知，若{a，b}={a2，b2}，则①或②，由①得，∵ab≠0，∴a≠0且b≠0，即a=1，b=1，此时集合{1，1}不满足条件．若b=a2，a=b2，则两式相减得a2﹣b2=b﹣a，∵互异的复数a，b，∴b﹣a≠0，即a+b=﹣1，故答案为：﹣1．【点评】本题主要考查集合相等的应用，根据集合相等得到元素相同是解决本题的关键，注意要进行分类讨论．3．若不等式ax2+5x﹣2＞0的解集是，则不等式ax2﹣5x+（a2﹣1）＞0的解集是．【考点】一元二次不等式的应用．【分析】先由二次不等式的解集形式，判断出，2是方程ax2+5x﹣2=0的两个根，利用韦达定理求出a的值，再代入不等式ax2﹣5x+a2﹣1＞0易解出其解集．【解答】解：∵ax2+5x﹣2＞0的解集是，∴a＜0，且，2是方程ax2+5x﹣2=0的两根韦达定理×2=，解得a=﹣2；则不等式ax2﹣5x+a2﹣1＞0即为﹣2x2﹣5x+3＞0，解得故不等式ax2﹣5x+a2﹣1＞0的解集．故答案为：【点评】本题考查的知识点是一元二次不等式的解法，及“三个二次”（三个二次指的是：二次函数，一元二次不等式，一元二次方程）之间的关系，“三个二次”之间的关系及应用是数形结合思想的典型代表．4．设等差数列{a n}的前n项和为S n，若S10=10，S20=30，则S30=60．【考点】等差数列的性质．【专题】计算题；等差数列与等比数列．【分析】由给出的数列是等差数列，可知数列的第一个10项和，第二个10项和，…仍然构成等差数列，结合S10=10，S20=30，列式求解S30的值．【解答】解：∵数列{a n}是等差数列，则S10，S20﹣S10，S30﹣S20仍然构成等差数列，由S10=10，S20=30，得2×20=10+S30﹣30，∴S30=60．故答案为：60．【点评】本题考查了等差数列的性质，考查了等差数列的前n项和，关键是对性质的理解与运用，是中档题．5．在△ABC中，已知角A，B，C所对的边分别为a，b，c，且c（acosB﹣bcosA）=2b2，则=．【考点】余弦定理；正弦定理．【专题】解三角形．【分析】由条件利用正弦定理和余弦定理代入进行化简即可．【解答】解：∵c（acosB﹣bcosA）=2b2，∴由余弦定理可得ac•﹣bc•=2b2，即a2+c2﹣b2﹣b2﹣c2+a2=4b2，即a2=3b2，则a=b，∴=．再利用正弦定理可得=，故答案为：【点评】本题主要考查正弦定理和余弦定理的应用，比较基础．要求熟练掌握相应的公式．6．若点A（2，3）与点B（1，y0）位于直线l：x﹣2y+5=0的两侧，则y0的取值范围是（3，+∞）．【考点】二元一次不等式（组）与平面区域；直线的斜率．【专题】直线与圆．【分析】由不等式与平面区域的关系可得y0的不等式，解不等式可得．【解答】解：∵点A（2，3）与点B（1，y0）位于直线l：x﹣2y+5=0的两侧，∴（2﹣2×3+5）（1﹣2y0+5）＜0，解得y0＞3故答案为：（3，+∞）【点评】本题考查不等式与平面区域，属基础题．7．将一枚质地均匀的一元硬币抛3次，恰好出现一次正面的概率是．【考点】相互独立事件的概率乘法公式．【专题】计算题．【分析】掷一枚硬币，正面向上的概率是，将一枚质地均匀的一元硬币抛3次，相当于做了三次独立重复试验，利用独立重复试验的概率公式写出结果．【解答】解：由题意知掷一枚硬币，正面向上的概率是，将一枚质地均匀的一元硬币抛3次，相当于做了三次独立重复试验，∴恰好出现一次正面的概率是故答案为：【点评】本题考查独立重复试验的概率公式，解题的关键是看出试验符合什么条件，注意应用概率的公式，本题是一个基础题．8．已知b∈R，若（1+bi）（2﹣i）为纯虚数，则|1+bi|=．【考点】复数求模．【专题】数系的扩充和复数．【分析】通过化简可知（1+bi）（2﹣i）=（2+b）+（2b﹣1）i，利用纯虚数的定义计算即可．【解答】解：∵（1+bi）（2﹣i）=（2+b）+（2b﹣1）i为纯虚数，∴，解得b=﹣2，∴|1+bi|===，故答案为：．【点评】本题考查复数求模，弄清纯虚数的概念是解决本题的关键，注意解题方法的积累，属于基础题．9．已知，且x+2y=1，则的最小值是．【考点】两向量的和或差的模的最值．【专题】计算题．【分析】根据要求的向量可以表示成两个向量的和的形式，把两个向量的系数用一个字母来表示，求向量的模长，利用二次函数的最值，做出结果．【解答】解：∵x+2y=1∴•===84y2﹣72y+16∴当y=时，原式=，故答案为：，【点评】本题考查向量的模长的最值，本题解题的关键是表示出向量的模长，再用函数求最值的方法来求解，这是这一类题目共同的特征．10．已知一圆锥的底面是半径为1cm的圆，若圆锥的侧面积是底面积的3倍，则该圆锥的体积是cm3．【考点】棱柱、棱锥、棱台的体积．【专题】空间位置关系与距离．【分析】由已知中，圆锥的底面半径为1，侧面积是底面积的3倍，分析圆锥的母线长，进而求出圆锥的高，结合圆锥的体积公式即可获得问题的解答．【解答】解：∵圆锥的底面半径r=1cm，侧面积是底面积的3倍，∴圆锥的母线长l=3cm，故圆锥的高h==2cm，故圆锥的体积V=Sh=πr2•h==cm3，故答案为：．【点评】本题考查的是圆锥的体积求解问题．在解答的过程当中充分体现了圆锥体积公式的应用以及转化思想的应用．值得同学们体会反思．11．抛物线y2=4x的焦点为F，过点P（2，0）的直线与该抛物线相交于A，B两点，直线AF，BF 分别交抛物线于点C，D．若直线AB，CD的斜率分别为k1，k2，则=．【考点】直线与圆锥曲线的关系．【专题】直线与圆；圆锥曲线的定义、性质与方程．【分析】设AF的方程是y=（x﹣1），与抛物线方程联立，求出C的坐标，同理求出D的坐标，可得k2，即可求出．【解答】解：设A（x1，y1），B（x2，y2），C（x3，y3），D（x4，y4），∴AF的方程是y=（x﹣1）设k0=，则AF：y=k0（x﹣1），与抛物线方程联立，可得k02x2﹣（2k02+4）x+k02=0，利用韦达定理x3x1=1∴x3=，∴y3=k0（x3﹣1）=﹣即C（，﹣）同理D（，﹣）∴k2==2k1，∴=．故答案为：．【点评】本题考查直线与抛物线的位置关系，考查斜率的计算，考查学生的计算能力，属于中档题．12．已知f（x）=x2﹣3x+4，若f（x）的定义域和值域都是[a，b]，则a+b=5．【考点】函数的值域；函数的定义域及其求法．【专题】函数的性质及应用．【分析】因为定义域和值域都是[a，b]，说明函数最大值和最小值分别是a和b，所以根据对称轴进行分类讨论即可．【解答】解：∵f（x）=x2﹣3x+4=+1，∴x=2是函数的对称轴，根据对称轴进行分类讨论：①当b＜2时，函数在区间[a，b]上递减，又∵值域也是[a，b]，∴得方程组即，两式相减得（a+b）（a﹣b）﹣3（a﹣b）=b﹣a，又∵a≠b，∴a+b=，由，得3a2﹣8a+4=0，∴a=∴b=2，但f（2）=1≠，故舍去．②当a＜2＜b时，得f（2）=1=a，又∵f（1）=＜2，∴f（b）=b，得，∴b=（舍）或b=4，∴a+b=5③当a＞2时，函数在区间[a，b]上递增，又∵值域是[a，b]，∴得方程组，即a，b是方程x2﹣3x+4=x的两根，即a，b是方程3x2﹣16x+16=0的两根，∴，但a＞2，故应舍去．故答案为：5【点评】本题考查了二次函数的单调区间以及最值问题，属于基础题．13．关于函数f（x）=cos（2x﹣）+cos（2x+），有下列说法（1）y=f（x）的最大值为；（2）y=f（x）是以π为最小正周期的函数；（3）y=f（x）在区间（，）上单调递减；（4）将函数y=cos2x的图象向左平移个单位后，将与已知函数的图象重合．其中正确说法的序号是（1）（2）（3）．【考点】两角和与差的余弦函数．【专题】三角函数的图像与性质．【分析】由三角函数公式化简可得f（x）=sin（2x+），由三角函数的性质逐个选项验证可得．【解答】解：化简可得f（x）=cos（2x﹣）+cos（2x+）=cos（2x+﹣）+cos（2x+）=sin（2x+）+cos（2x+）=sin（2x++）=sin（2x+）∴函数f（x）的最大值为，（1）正确；函数的周期T==π，（2）正确；由2kπ+＜2x+＜2kπ+可得kπ+＜x＜kπ+，当k=0时可得函数y=f（x）在区间（，）上单调递减，（3）正确；（4）y=cos2x的图象向左平移个单位后，可得y=cos2（x+）=cos（2x+）≠sin（2x+），错误；综上可知（1）（2）（3）正确，故答案为：（1）（2）（3）．【点评】本题考查三角函数的图象和性质，涉及两角和与差的三角函数公式和诱导公式，属中档题．14．定义在实数集R上的函数f（x），如果存在函数g（x）=Ax+B（A、B为常数），使得f（x）≥g（x）对一切实数x都成立，那么称g（x）为函数f（x）的一个承托函数．给出如下四个命题：①对于给定的函数f（x），其承托函数可能不存在，也可能有无数个；②定义域和值域都是R的函数f（x）不存在承托函数；③g（x）=2x为函数f（x）=|3x|的一个承托函数；④为函数f（x）=x2的一个承托函数．其中正确的命题有①③．【考点】函数最值的应用．【专题】压轴题；新定义．【分析】函数g（x）=Ax+B（A，B为常数）是函数f（x）的一个承托函数，即说明函数f（x）的图象恒在函数g（x）的上方（至多有一个交点）①举例可以说明，如f（x）=cosx，则g（x）=B （B＜﹣1）就是它的一个承托函数，且有无数个，反例如y=tanx或y=lgx就没有承托函数；②f（x）=2x+3的定义域和值域都是R，存在一个承托函数y=2x+1，故命题②不正确；③要说明g（x）=2x为函数f（x）=|3x|的一个承托函数；即证明F（x）=e x﹣2x 的图象恒在x轴上方；④举反例即可．【解答】解：①如f（x）=sinx，则g（x）=B（B＜﹣1）就是它的一个承托函数，且有无数个，再如y=tanx．y=lgx就没有承托函数，∴命题①正确；②f（x）=2x+3的定义域和值域都是R，存在一个承托函数y=2x+1，故命题②不正确；③令F（x）═|3x|﹣2x=，可见在x≥0时，函数F（x）单调递增，最小值F（0）=0，在x＜0时，函数F（x）单调递减，最小值大于F（0）=0，∴F（x）≥0在R上恒成立，符合定义∴命题③正确；④x=1时，g（1）=，f（1）=1，显然g（1）＜f（1），当x=时，g（）=，f（）=，显然g（）＞f（），命题④不正确．故答案为：①③【点评】本题是新定义题，考查对题意的理解和转化的能力，要说明一个命题是正确的，必须给出证明，如③，对于存在性命题的探讨，只需举例说明即可，如①，对于不正确的命题，举反例即可，如②③，属于中档题．二、选择题（本大题共4题，每题5分，满分20分）15．在（1+x）6（2+y）4的展开式中，含x4y3项的系数为（）A．210 B．120 C．80 D．60【考点】二项式定理的应用．【专题】二项式定理．【分析】利用二项展开式的通项公式求得（1+x）6（2+y）4的展开式中，含x4y3 的项，可得含x4y3项的系数．【解答】解：在（1+x）6（2+y）4的展开式中，含x4y3 的项为•x3••2•y3=120x4y3，故含x4y3项的系数为120，故选：B．【点评】本题主要考查二项式定理的应用，二项展开式的通项公式，二项式系数的性质，属于基础题．16．已知a、b、c是△ABC的三边长，且满足=0，则△ABC一定是（）A．等腰非等边三角形 B．等边三角形C．直角三角形D．等腰直角三角形【考点】高阶矩阵．【专题】选作题；矩阵和变换．【分析】方程化为2a2+2b2+2c2﹣2ab﹣2bc﹣2ca=0，配方可得结论．【解答】解：方程化为2a2+2b2+2c2﹣2ab﹣2bc﹣2ca=0，所以（a﹣b）2+（b﹣c）2+（a﹣c）2=0，所以a=b=c，故选：B．【点评】本题考查高阶矩阵，考查学生的计算能力，比较基础．17．甲乙两人进行相棋比赛，甲获胜的概率是0.4，两人下成和棋的概率是0.2，则甲不输的概率是（）A．0.6 B．0.8 C．0.2 D．0.4【考点】概率的基本性质．【专题】计算题．【分析】欲求甲不输的概率，利用等量关系：甲获胜的概率是0.4，两人下成和棋的概率是0.2，把相关数值代入即可求解．【解答】解，根据题意，甲获胜的概率是0.4，两人下成和棋的概率是0.2所以甲不输的概率为0.4+0.2=0.6．故选A．【点评】本题考查了等可能事件的概率，解答本题的关键是要判断出“甲获胜的概率，和棋的概率和乙获胜的概率的和是1”．18．圆ρ=（cosθ+sinθ）的圆心坐标是（）A．（1，）B．（，）C．（，）D．（2，）【考点】简单曲线的极坐标方程．【专题】坐标系和参数方程．【分析】利用化为直角坐标方程，进而得出．【解答】解：圆ρ=（cosθ+sinθ）即（cosθ+sinθ），∴，化为．∴圆心坐标是，∴=1，θ=arctan1=．极坐标为．【点评】本题考查了极坐标与直角坐标方程的互化，属于基础题．三．解答题（本大题共五题，满分74分，12+14+14+16+18=74）19．已知角α的终边经过点P（，﹣）．（1）求sinα的值．（2）求式•的值．【考点】任意角的三角函数的定义；运用诱导公式化简求值．【专题】计算题．【分析】（1）求出|OP|，利用三角函数的定义，直接求出sinα的值．（2）利用诱导公式化简表达式，根据角的终边所在象限，求出cosα=，可得结果．【解答】解：（1）∵|OP|=，∴点P在单位圆上．由正弦函数的定义得sinα=﹣（2）原式==..由余弦的定义可知，cosα=即所求式的值为【点评】本题考查任意角的三角函数的定义，运用诱导公式化简求值，考查计算能力，推理能力，是基础题．20．已知函数f（x）=|3x+2|．（Ⅰ）解不等式f（x）＜4﹣|x﹣1|；（Ⅱ）已知m+n=1（m，n＞0），若|x﹣a|﹣f（x）≤+（a＞0）恒成立，求实数a的取值范围．【考点】绝对值不等式的解法．【专题】不等式的解法及应用．【分析】（Ⅰ）把要解的不等式等价转化为与之等价的三个不等式组，求出每个不等式组的解集，再取并集，即得所求．（Ⅱ）由条件利用基本不等式求得+≥4，结合题意可得|x﹣a|﹣|3x+2|≤4恒成立．令g（x）=|x﹣a|﹣|3x+2|，利用单调性求得它的最大值，再由此最大值小于或等于4，求得a的范围．【解答】解：（Ⅰ）不等式f（x）＜4﹣|x﹣1|，即|3x+2|+|x﹣1|＜4，∴①，或②，或③．解①求得﹣＜x＜﹣，解②求得﹣≤x＜，解③求得x∈∅．综上可得，不等式的解集为（﹣，）．（Ⅱ）已知m+n=1（m，n＞0），∴+=（m+n）（+）=2++≥2+2=4，当且仅当m=n=时，取等号．再根据|x﹣a|﹣f（x）≤+（a＞0）恒成立，可得|x﹣a|﹣f（x）≤4，即|x﹣a|﹣|3x+2|≤4．设g（x）=|x﹣a|﹣|3x+2|=，故函数g（x）的最大值为g（﹣）=+a，再由+a≤4，求得0＜a≤．【点评】本题主要考查绝对值不等式的解法，函数的恒成立问题，基本不等式的应用，体现了转化、分类讨论的数学思想，属于中档题．21．如图，在直三棱柱ABC﹣A1B1C1中，∠ACB=90°，AC=BC=CC1=2．（1）证明：AB1⊥BC1；（2）求点B到平面AB1C1的距离；（3）求二面角C1﹣AB1﹣A1的大小．【考点】二面角的平面角及求法；点、线、面间的距离计算．【专题】综合题．【分析】（1）以C点为坐标原点，CA，CB，CC1为X，Y，Z轴正方向建立空间坐标系，分别求出AB1与BC1的方向向量，代入数量积公式，得到其数量积为0，即可得到AB1⊥BC1；（2）求出平面AB1C1的一个法向量，则AB的方向向量，代入到公式，即可求出点B到平面AB1C1的距离；（3）结合（2）的结合，再求出平面AB1A1的一个法向量，代入向量夹角公式，即可得到二面角C1﹣AB1﹣A1的大小．【解答】证明：（1）如图建立直角坐标系，其为C为坐标原点，题意A（2，0，0），B（0，2，0），A1（2，0，2），B1（0，2，2），C1（0，0，2）．∵，∴∴AB1⊥BC1解：（2）设的一个法向量，由得令∵，∴点B到平面AB1C1的距离．（3）解设是平面A1AB1的一个法向量由∴令∵，∴二面角C1﹣AB﹣A1的大小为60°．【点评】本题考查的知识点是二面角的平面角及求法，点到面的距离，异面直线的夹角，其中建立适当的空间坐标系，将问题转化为向量夹角及向量长度问题是解答本题的关键．22．已知F1，F2为椭圆E的左右焦点，点P（1，）为其上一点，且有|PF1|+|PF2|=4（Ⅰ）求椭圆C的标准方程；（Ⅱ）过F1的直线l1与椭圆E交于A，B两点，过F2与l1平行的直线l2与椭圆E交于C，D两点，求四边形ABCD的面积S ABCD的最大值．【考点】直线与圆锥曲线的综合问题．【专题】圆锥曲线中的最值与范围问题．【分析】（I）设椭圆E的标准方程为，由已知|PF1|+|PF2|=4，，由此能求出椭圆E的标准方程．（II）由题意可知，四边形ABCD为平行四边形，S△ABCD=4S△OAB，设直线AB的方程为x=my﹣1，且A（x1，y1），B（x2，y2），由，得（3m2+4）y2﹣6my﹣9=0，由此利用弦长公式能求出S△BCD的最大值．【解答】解：（I ）设椭圆E 的标准方程为，由已知|PF 1|+|PF 2|=4，得2a=4，∴a=2，…又点P （1，）在椭圆上，∴，∴b=， 椭圆E 的标准方程为=1．…（II ）由题意可知，四边形ABCD 为平行四边形，∴S ▱ABCD =4S △OAB ，设直线AB 的方程为x=my ﹣1，且A （x 1，y 1），B （x 2，y 2），由，得（3m 2+4）y 2﹣6my ﹣9=0， ∴y 1+y 2=，y 1y 2=﹣，… S △OAB =+=|OF 1||y 1﹣y 2|= ==6，…令m 2+1=t ，则t ≥1，S △OAB =6=6，…又∵g （t ）=9t+在[1，+∞）上单调递增 ∴g （t ）≥g （1）=10，∴S △OAB 的最大值为．∴S ▱ABCD 的最大值为6．…【点评】本题考查椭圆方程的求法，考查三角形面积的最大值的求法，解题时要认真审题，注意椭圆弦长公式的合理运用．23．给定数列{c n }，如果存在常数p 、q 使得c n+1=pc n +q 对任意n ∈N *都成立，则称{c n }为“M 类数列”． （1）若{a n }是公差为d 的等差数列，判断{a n }是否为“M 类数列”，并说明理由；（2）若{a n }是“M 类数列”且满足：a 1=2，a n +a n+1=3•2n ．①求a 2、a 3的值及{a n }的通项公式；②设数列{b n }满足：对任意的正整数n ，都有a 1b n +a 2b n ﹣1+a 3b n ﹣2+…+a n b 1=3•2n+1﹣4n ﹣6，且集合M={n|≥λ，n ∈N *}中有且仅有3个元素，试求实数λ的取值范围．【考点】数列的应用．【专题】点列、递归数列与数学归纳法．【分析】（1）通过a n+1=a n +d 与c n+1=pc n +q 比较可知p=1、q=d ，进而可得结论；（2）①通过a 1=2、a n +a n+1=3•2n 计算出a 2、a 3的值，进而利用数列{a n }是“M 类数列”代入计算可知数列{a n }是以首项、公比均为2的等比数列，计算可得结论；②通过①可知2b n +22b n ﹣1+23b n ﹣2+…+2n b 1=3•2n+1﹣4n ﹣6，利用2b n =（2b n +22b n ﹣1+23b n ﹣2+…+2n b 1）﹣（22b n ﹣1+23b n ﹣2+…+2n b 1）计算可知b n =2n ﹣1，从而M={n|≥λ，n ∈N *}，分别计算出当n=1、2、3时λ的值，进而可得结论．【解答】（1）结论：公差为d 的等差数列是“M 类数列”．理由如下：∵数列{a n }是公差为d 的等差数列，∴a n+1=a n +d ，此时p=1、q=d ，即公差为d 的等差数列是“M 类数列”；（2）①∵a 1=2，a n +a n+1=3•2n ，∴a 2=3•2﹣a 1=4， =8，又∵数列{a n }是“M 类数列”，∴，即，解得：p=2，q=0，即a n+1=2a n ， 又∵a 1=2，∴数列{a n }是以首项、公比均为2的等比数列，∴数列{a n }的通项公式a n =2n ； ②由①可知a 1b n +a 2b n ﹣1+a 3b n ﹣2+…+a n b 1=3•2n+1﹣4n ﹣6，即2b n +22b n ﹣1+23b n ﹣2+…+2n b 1=3•2n+1﹣4n ﹣6，∴2b n ﹣1+22b n ﹣2+23b n ﹣3+…+2n ﹣1b 1=3•2n ﹣4（n ﹣1）﹣6=3•2n ﹣4n ﹣2，∴22b n ﹣1+23b n ﹣2+…+2n b 1=3•2n+1﹣8n ﹣4，∴2b n =（2b n +22b n ﹣1+23b n ﹣2+…+2n b 1）﹣（22b n ﹣1+23b n ﹣2+…+2n b 1）=（3•2n+1﹣4n ﹣6）﹣（3•2n+1﹣8n ﹣4）=4n ﹣2，即b n =2n ﹣1，∴集合M={n|≥λ，n ∈N *}={n|≥λ，n ∈N *}，当n=1时，λ≤=；当n=2时，λ≤=；当n=3时，λ≤=；当n ≥4时，λ≤=；又∵集合M={n|≥λ，n ∈N *}中有且仅有3个元素，∴＜λ≤，故实数λ的取值范围是（，]．【点评】本题是一道关于数列与不等式的综合题，考查数列的通项，考查运算求解能力，注意解题方法的积累，属于难题．。
精品解析：上海市交通大学附属中学2018-2019学年高三上学期期末数学试题(解析版)
则
．
故答案为：25．
【点睛】本题考查了复数代数形式的乘法运算，考查了复数模的求法，是基础题．
3.函数
f
x
x 4，x 4
f
x
3，＜x
， 4
则
f
f
1
________.
【答案】0
【解析】
【分析】
先求出 f (1) 的值，再求 f f 1 的值.
交大附中 2018-2019 学年度第一学期高三年级期末数学试卷
一、填空题
A
1.已知集合
x 0 x2
B
，集合
x 1 x 2
，则 A B _____．
【答案】 1, 2
【解析】
A B x 0 x 2x 1 x 2 1, 2
2
，
∵
t2
2t
2
t
12
1
1
，∴
0
t2
1 2t
2
1
，
∴
1
t2
1 2t
2
0
1
，∴
2
t2
1 2t
2
2
，
∴
2 2
2 2
2 t2
1 2t 2
2 d
，即
2, 2
2
2 , 故答案为： 2
2
【点睛】本题考查直线与圆的位置关系，涉及圆的相交弦和点到直线的距离公式，以及不等式求函数的值
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上海市交大附中高三9月份开学考试
一、填空题：1-6题每题4分，第7-12题每题5分，共54分.
1.方程组21,
32x y x y 的增广矩阵是．
2.若直线l 的参数方程为33,23,x t y t t R ，则直线l 的倾斜角是．
3.0
2
222234lim n
n n n
n n C C C …．
4.已知数列n a 的前n 项的和2
1,2,n n n n S n 是正奇
是正偶，则当n 为正偶数时，n a ．
5.函数2
2()(1)(1)x ax
f x x x 是奇函数，那么a ．
6.若函数2()lg(2)f x x ax 无最值，则a 的取值范围是．
7.△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，已知△ABC 的面积为2
3sin a A ，6cos cos 1B C ，
则A ．
8.设b R ，i 是虚数单位，已知集合|||2A z z i ，11|1,B z z z bi z A ，若A B ，则b 的取值范围是．
9.从双曲线22
221x y a b （0a ，0b ）的左焦点F 引圆222x y a 的切线，切点为T ，延长FT 交双曲线右支于点P ，若M 是线段FP 的中点，O 为坐标原点，则||||MO MT 的值是．
10.胡涂涂同学用一颗均匀的骰子来定义递推数列n a ，首先，他令11a ，当1n 时，他投一次骰子，若所得点数大于n a ，即令11n n a a ，否则，令11n n a a ，则40a 的概率为（结果用最简分数表示）．
11.关于x 的方程2arcsin(cos )0x x a 恰有3个实数根1x ，2x ，3x ，则2
2
2
123x x x ．
12.由无理数论引发的数字危机一直延续到19世纪，直到1872年，德国数学家戴德金从连续性的要求出发，用有理数的“分割”来定义无理数（史称戴德金分割），并把实数理论建立在严格的科学基础上，才结束了
无理数被认为“无理”的时代，也结束了持续2000多年的数学史上的第一次大危机，所谓戴德金分割，是
指将有理数集Q 划分为两个非空的子集M 与N ，且满足M N Q ，M N ，M 中的每一个元素都小于N 中的每一个元素，则称(,)M N 为戴德金分割.试判断，对于任一戴德金分割(,)M N ，下列选项中，可能成立的是．
①M 没有最大元素，N 有一个最小元素；②M 没有最大元素，N 也没有最小元素；
③M 有一个最大元素，N 有一个最小元素；④M 有一个最大元素，N 没有最小元素.
二、选择题（每题5分，满分20分）
13.已知集合|1A x x ，则下列选项正确的是（）
A ．0A
B ．0A
C ．A
D ．0A
14.在空间直角坐标系O xyz 中，若点P 在第Ⅵ卦限，则与点P 关于y 轴对称的点在（）
A ．第Ⅰ卦限
B ．第Ⅲ卦限
C ．第Ⅴ卦限
D ．第Ⅶ卦限
15.设A ，B ，C 为实数，则实数“0ABC ”是“方程22Ax By C 表示的曲线为双曲线”的（）
A ．充分非必要条件
B ．必要非充分条件
C ．充要条件
D ．非充分非必要条件
16.已知O 、A 、B 、C 是同一平面上不共线的四点，若存在一组正实数1、2、3，使得
1230OA OB OC ，则三个角AOB 、BOC 、COA （）
A ．都是钝角
B ．至少有两个钝角
C ．恰有两个钝角
D ．至多有两个钝角
三、解答题（本大题共6小题，共76分.解答应写出文字说明、证明过程或演算步骤.）
17.如图所示，三棱柱111ABC A B C 的侧面11ABB A 是圆柱的轴截面，C 是圆柱底面圆周上不与A 、B 重合的一个点.
（1）若圆柱的轴截面是正方形，当点C 是弧AB 的中点时，求异面直线1AC 与AB 所成角的大小（结果用反三角函数值表示）；
（2）当点C 是弧AB 的中点时，求四棱锥111A BCC B 与圆柱的体积比.
18.（1）已知2()lg(2)lg f x x ax b 是定义在R 上的奇函数，求实数a 、b 的值；
（2）已知2()lg(2)lg f x x ax b 是定义在R 上的函数，求实数a 的取值范围.
19.某工厂在生产产品是需要用到长度为698mm 的A 型和长度为518mm 的B 型两种钢管.工厂利用长度为4000mm 的钢管原材料，建材成若干A 型和B 型钢管，假设裁剪时损耗忽略不计，裁剪后所剩废料与原材料的百分比称为废料率.
（1）要使裁剪的废料率小于 4.5%，共有几种方案剪裁？请写出每种方案中分别被裁剪A 型钢管和B 型钢管的根数；
（2）假设一根A 型钢管和一根B 型钢管能成为一套毛胚，假定只能按（1）中的那些方案裁剪，若工厂需要生产320套毛胚，则至少需要采购多少根长度为4000mm 的钢管原材料？最终的废料率为多少？
20.在平面上，给定非零向量b ，对任意向量a ，定义2()
'||a b a a b b .
（1）若(2,3)a ，(1,3)b ，求'a ；
（2）若(2,1)b ，证明：若位置向量a 的终点在直线0Ax By C 上，则位置向量'a 的终点也在一条直线上；
（3）已知存在单位向量b ，当位置向量a 的终点在抛物线C ：2x y 上时，位置向量'a 终点总在抛物线'C ：。