上海复旦大学附属中学2016届高三上学期期中考试数学(文)试题

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2016-2017学年上海师大附中高三(上)期中数学试卷

2016-2017学年上海师大附中高三（上）期中数学试卷一、填空题1．（3分）已知全集U=R，A={x|x2﹣2x＜0}，B={x|x≥1}，则A∩∁U B=．2．（3分）函数f（x）=的反函数f﹣1（x）= ．3．（3分）=．4．（3分）已知sin2θ+sinθ=0，θ∈（，π），则tan2θ=．5．（3分）方程log2（9x+7）=2+log2（3x+1）的解为．6．（3分）在△ABC中，角A，B，C所对的边分别为a，b，c，若其面积S=（b2+c2﹣a2），则∠A=．7．（3分）已知等比数列{a n}的各项均为正数，且满足：a1a7=4，则数列{log2a n}的前7项之和为．8．（3分）如果函数f（x）=sin2x+acos2x的图象关于直线x=对称，那么实数a=．9．（3分）若数列{a n}的通项公式是a n=，前n项和为S n，则S n的值为．10．（3分）已知f（x）=2sinωx（ω＞0）在[0，]单调递增，则实数ω的最大值为．11．（3分）函数y=arcsin（x2﹣x）的值域为．12．（3分）设函数y=f （x）的定义域为D，如果存在非零常数T，对于任意x ∈D，都有f（x+T）=T•f （x），则称函数y=f（x）是“似周期函数”，非零常数T 为函数y=f（x）的“似周期”．现有下面四个关于“似周期函数”的命题：①如果“似周期函数”y=f（x）的“似周期”为﹣1，那么它是周期为2的周期函数；②函数f（x）=x是“似周期函数”；③函数f（x）=2x是“似周期函数”；④如果函数f（x）=cosωx是“似周期函数”，那么“ω=kπ，k∈Z”．其中是真命题的序号是．（写出所有满足条件的命题序号）13．（3分）已知数列{a n}满足a1=81，a n=（k∈N*），则数列{a n}的前n项和S n的最大值为．14．（3分）已知函数f（x）是定义在[1，+∞）上的函数，且f（x）=，则函数y=2xf（x）﹣3在区间（1，2016）上的零点个数为．二、选择题15．（3分）“|x﹣1|＜2成立”是“x（x﹣3）＜0成立”的（）A．充分不必要条件 B．必要不充分条件C．充分必要条件D．既不充分也不必要条件16．（3分）函数y=2cos2（x﹣）﹣1是（）A．最小正周期为π的奇函数B．最小正周期为π的偶函数C．最小正周期为的奇函数D．最小正周期为的偶函数17．（3分）设变量x，y满足约束条件，则目标函数z=2x+5y的最小值为（）A．﹣4 B．6 C．10 D．1718．（3分）已知点列A n（a n，b n）（n∈N*）均为函数y=a x（a＞0，a≠1）的图象上，点列B n（n，0）满足|A n B n|=|A n B n+1|，若数列{b n}中任意连续三项能构成三角形的三边，则a的取值范围为（）A．（0，）∪（，+∞）B．（，1）∪（1，）C．（0，）∪（，+∞）D．（，1）∪（1，）三、解答题19．已知函数f（x）=2sin（x+）•cosx．（1）若0≤x≤，求函数f（x）的值域；（2）设△ABC的三个内角A，B，C所对的边分别为a，b，c，若A为锐角且f （A）=，b=2，c=3，求cos（A﹣B）的值．20．某公司生产的某批产品的销售量P万件（生产量与销售量相等）与促销费用x万元满足P=（其中0≤x≤a，a为正常数）．已知生产该产品还需投入成本6（P+）万元（不含促销费用），产品的销售价格定为（4+）元/件．（1）将该产品的利润y万元表示为促销费用x万元的函数；（2）促销费用投入多少万元时，该公司的利润最大？21．已知函数，其中a∈R．（1）根据a的不同取值，讨论f（x）的奇偶性，并说明理由；（2）已知a＞0，函数f（x）的反函数为f﹣1（x），若函数y=f（x）+f﹣1（x）在区间[1，2]上的最小值为1+log23，求函数f（x）在区间[1，2]上的最大值．22．设数列{a n}的前n项和为S n，且（S n﹣1）2=a n S n（n∈N*）．（1）求S1，S2，S3的值；（2）求出S n及数列{a n}的通项公式；（3）设b n=（﹣1）n﹣1（n+1）2a n a n+1（n∈N*），求数列{b n}的前n项和为T n．23．已知集合M是满足下列性制的函数f（x）的全体，存在实数a、k（k≠0），对于定义域内的任意x均有f（a+x）=kf（a﹣x）成立，称数对（a，k）为函数f （x）的“伴随数对”．（1）判断f（x）=x2是否属于集合M，并说明理由；（2）若函数f（x）=sinx∈M，求满足条件的函数f（x）的所有“伴随数对”；（3）若（1，1），（2，﹣1）都是函数f（x）的“伴随数对”，当1≤x＜2时，f（x）=cos（x）；当x=2时，f（x）=0，求当2014≤x≤2016时，函数y=f（x）的解析式和零点．2016-2017学年上海师大附中高三（上）期中数学试卷参考答案与试题解析一、填空题1．（3分）已知全集U=R，A={x|x2﹣2x＜0}，B={x|x≥1}，则A∩∁U B=（0，1）．【解答】解：全集U=R，A={x|x2﹣2x＜0}={x|0＜x＜2}=（0，2），B={x|x≥1}=[1，+∞），∴∁U B=（﹣∞，1），∴A∩∁U B=（0，1）．故答案为：（0，1）．2．（3分）函数f（x）=的反函数f﹣1（x）= x3+1．【解答】解：∵y=，∴x=y3+1，函数f（x）=的反函数为f﹣1（x）=x3+1．故答案为：x3+1．3．（3分）=﹣．【解答】解：原式===﹣，故答案为：﹣．4．（3分）已知sin2θ+sinθ=0，θ∈（，π），则tan2θ=．【解答】解：∵sin2θ+sinθ=0，⇒2sinθcosθ+sinθ=0，⇒sinθ（2cosθ+1）=0，∵θ∈（，π），sinθ≠0，∴2cosθ+1=0，解得：cosθ=﹣，∴tanθ=﹣=﹣，∴tan2θ==．故答案为：．5．（3分）方程log2（9x+7）=2+log2（3x+1）的解为x=0和x=1．【解答】解：由log2（9x+7）=2+log2（3x+1），得log2（9x+7）=log24（3x+1），即9x+7=4（3x+1），化为（3x）2﹣4•3x+3=0，解得：3x=1和3x=3，∴x=0和x=1．故答案为：x=0和x=1．6．（3分）在△ABC中，角A，B，C所对的边分别为a，b，c，若其面积S=（b2+c2﹣a2），则∠A=．【解答】解：由已知得：S=bcsinA=（b2+c2﹣a2）变形为：=sinA，由余弦定理可得：cosA=，所以cosA=sinA即tanA=1，又A∈（0，π），则A=．故答案为：7．（3分）已知等比数列{a n}的各项均为正数，且满足：a1a7=4，则数列{log2a n}的前7项之和为7．【解答】解：由等比数列的性质可得：a1a7=a2a6=a3a5=4=4，∴数列{log2a n}的前7项和=log2a1+log2a2+…+log2a7=log2（a1a2…a7）=log227=7，故答案为：7．8．（3分）如果函数f（x）=sin2x+acos2x的图象关于直线x=对称，那么实数a=1．【解答】解：f（x）=sin2x+acos2x=由正弦函数的对称轴方程，图象关于直线x=对称，即可得：，当k=0时，∵tanθ=a∴a=1故答案为1．9．（3分）若数列{a n}的通项公式是a n=，前n项和为S n，则S n的值为12．【解答】解：∵数列{a n}的通项公式是a n=，前n项和为S n，∴S n=4+8+=12．故答案为：12．10．（3分）已知f（x）=2sinωx（ω＞0）在[0，]单调递增，则实数ω的最大值为．【解答】解：∵f（x）=2sinωx（ω＞0）在[0，]单调递增，∴ω•≤，求得ω≤，则实数ω的最大值为，故答案为：．11．（3分）函数y=arcsin（x2﹣x）的值域为[﹣arcsin，] ．【解答】解：∵x2﹣x=（x﹣）2﹣≥﹣，∴函数y=arcsin（x2﹣x）的值域为[﹣arcsin，]．故答案为：[﹣arcsin，]．12．（3分）设函数y=f （x）的定义域为D，如果存在非零常数T，对于任意x ∈D，都有f（x+T）=T•f （x），则称函数y=f（x）是“似周期函数”，非零常数T 为函数y=f（x）的“似周期”．现有下面四个关于“似周期函数”的命题：①如果“似周期函数”y=f（x）的“似周期”为﹣1，那么它是周期为2的周期函数；②函数f（x）=x是“似周期函数”；③函数f（x）=2x是“似周期函数”；④如果函数f（x）=cosωx是“似周期函数”，那么“ω=kπ，k∈Z”．其中是真命题的序号是①④．（写出所有满足条件的命题序号）【解答】解：①∵似周期函数”y=f（x）的“似周期”为﹣1，∴f（x﹣1）=﹣f（x），∴f（x﹣2）=﹣f（x﹣1）=f（x），故它是周期为2的周期函数，故正确；②若函数f（x）=x是“似周期函数”，则f（x+T）=T•f （x），即x+T=Tx恒成立；故（T﹣1）x=T恒成立，上式不可能恒成立；③若函数f（x）=2x是“似周期函数”，则f（x+T）=T•f （x），即2x+T=T2x恒成立；故2T=T成立，无解；故错误；④若函数f（x）=cosωx是“似周期函数”，则f（x+T）=T•f （x），即cos（ω（x+T））=Tcosωx恒成立；故cos（ωx+ωT）=Tcosωx恒成立；即cosωxcosωT﹣sinωxsinωT=Tcosωx恒成立，故，故ω=kπ，k∈Z；故正确；故答案为：①④．13．（3分）已知数列{a n}满足a1=81，a n=（k∈N*），则数列{a n}的前n项和S n的最大值为127．【解答】解：∵数列{a n}满足a1=81，a n=（k∈N*），∴n=2k（k∈N*）时，a2k=﹣1+log3a2k﹣1，a2=3；n=2k+1时a2k+1=．==，a2k=﹣1+a2k﹣2．∴a2k+1∴数列{a n}的奇数项成等比数列，公比为；偶数项成等差数列，公差为﹣1．∴S n=S2k=（a1+a3+…+a2k﹣1）+（a2+a4+…+a2k）=+3k+=﹣+≤126．（k=5时取等号）．S n=S2k﹣1=S2k﹣2+a2k﹣1=﹣++≤127，k=5综上可得：数列{a n}的前n项和S n的最大值为127．故答案为：127．14．（3分）已知函数f（x）是定义在[1，+∞）上的函数，且f（x）=，则函数y=2xf（x）﹣3在区间（1，2016）上的零点个数为11．【解答】解：令函数y=2xf（x）﹣3=0，得到方程f（x）=，当x∈[1，2）时，函数f（x）先增后减，在x=时取得最大值1，而y=在x=时也有y=1；当x∈[2，22）时，f（x）=，在x=3处函数f（x）取得最大值，而y=在x=3时也有y=；当x∈[22，23）时，f（x）=，在x=6处函数f（x）取得最大值，而y=在x=6时也有y=；…；当x∈[210，211）时，f（x）=，在x=1536处函数f（x）取得最大值，而y=在x=1536时也有y=．∴函数y=2xf（x）﹣3在区间（1，2016）上的零点个数为11．故答案为：11．二、选择题15．（3分）“|x﹣1|＜2成立”是“x（x﹣3）＜0成立”的（）A．充分不必要条件 B．必要不充分条件C．充分必要条件D．既不充分也不必要条件【解答】解：由|x﹣1|＜2解得：﹣2+1＜x＜2+1，即﹣1＜x＜3．由x（x﹣3）＜0，解得0＜x＜3．“|x﹣1|＜2成立”是“x（x﹣3）＜0成立”必要不充分条件．故选：B．16．（3分）函数y=2cos2（x﹣）﹣1是（）A．最小正周期为π的奇函数B．最小正周期为π的偶函数C．最小正周期为的奇函数D．最小正周期为的偶函数【解答】解：由y=2cos2（x﹣）﹣1=cos（2x﹣）=sin2x，∴T=π，且y=sin2x奇函数，即函数y=2cos2（x﹣）﹣1是奇函数．故选：A．17．（3分）设变量x，y满足约束条件，则目标函数z=2x+5y的最小值为（）A．﹣4 B．6 C．10 D．17【解答】解：作出不等式组表示的可行域，如右图中三角形的区域，作出直线l0：2x+5y=0，图中的虚线，平移直线l0，可得经过点（3，0）时，z=2x+5y取得最小值6．故选：B．18．（3分）已知点列A n（a n，b n）（n∈N*）均为函数y=a x（a＞0，a≠1）的图象上，点列B n（n，0）满足|A n B n|=|A n B n+1|，若数列{b n}中任意连续三项能构成三角形的三边，则a的取值范围为（）A．（0，）∪（，+∞）B．（，1）∪（1，）C．（0，）∪（，+∞）D．（，1）∪（1，）【解答】解：由题意得，点B n（n，0），A n（a n，b n）满足|A n B n|=|A n B n+1|，由中点坐标公式，可得B n B n+1的中点为（n+，0），即a n=n+，b n=；，b n，b n+1为边长能构成一个三角形，当a＞1时，以b n﹣1只需b n+b n＞b n+1，﹣1b n﹣1＜b n＜b n+1，即+＞，即有1+a＜a2，解得1＜a＜；同理，0＜a＜1时，解得＜a＜1；综上，a的取值范围是1＜a＜或＜a＜1，故选：B．三、解答题19．已知函数f（x）=2sin（x+）•cosx．（1）若0≤x≤，求函数f（x）的值域；（2）设△ABC的三个内角A，B，C所对的边分别为a，b，c，若A为锐角且f （A）=，b=2，c=3，求cos（A﹣B）的值．【解答】解：（1）f（x）=2sin（x+）•cosx=（sinx+cosx）•cosx=sinxcosx+cos2x=sin2x+cos2x+=sin（2x+）+；…（2分）由得，，∴，…（4分）∴，即函数f（x）的值域为；…（6分）（2）由，得，又由，∴，∴，解得；…（8分）在△ABC中，由余弦定理a2=b2+c2﹣2bccosA=7，解得；…（10分）由正弦定理，得，…（12分）∵b＜a，∴B＜A，∴，∴cos（A﹣B）=cosAcosB+sinAsinB=．…（15分）20．某公司生产的某批产品的销售量P万件（生产量与销售量相等）与促销费用x万元满足P=（其中0≤x≤a，a为正常数）．已知生产该产品还需投入成本6（P+）万元（不含促销费用），产品的销售价格定为（4+）元/件．（1）将该产品的利润y万元表示为促销费用x万元的函数；（2）促销费用投入多少万元时，该公司的利润最大？【解答】解：（Ⅰ）由题意知，y=（4+）p﹣x﹣6（p+），将p=代入化简得：y=19﹣﹣x（0≤x≤a）；（Ⅱ）y=22﹣（+x+2）≤22﹣3=10，当且仅当=x+2，即x=2时，上式取等号；当a≥2时，促销费用投入2万元时，该公司的利润最大；y=19﹣﹣x，y′=﹣，∴a＜2时，函数在[0，a]上单调递增，∴x=a时，函数有最大值．即促销费用投入a万元时，该公司的利润最大．21．已知函数，其中a∈R．（1）根据a的不同取值，讨论f（x）的奇偶性，并说明理由；（2）已知a＞0，函数f（x）的反函数为f﹣1（x），若函数y=f（x）+f﹣1（x）在区间[1，2]上的最小值为1+log23，求函数f（x）在区间[1，2]上的最大值．【解答】解：（1）∵，∴f（﹣x）=﹣ax+log2（2﹣x+1）=﹣ax+log2（2x+1）﹣log22x=﹣ax+log2（2x+1）﹣x，∴f（﹣x）=f（x），即﹣ax﹣x=ax，故a=；此时函数为偶函数，若a≠﹣，函数为非奇非偶函数；（2）∵a＞0，∴单调递增，又∵函数f（x）的反函数为f﹣1（x），∴f﹣1（x）单调递增；∴f（1）+f﹣1（1）=1+log23，即a+log23+f﹣1（1）=1+log23，故f﹣1（1）=1﹣a，即a（1﹣a）+log2（2a﹣1+1）=1，解得，a=1；故f（2）=2+log25．22．设数列{a n}的前n项和为S n，且（S n﹣1）2=a n S n（n∈N*）．（1）求S1，S2，S3的值；（2）求出S n及数列{a n}的通项公式；（3）设b n=（﹣1）n﹣1（n+1）2a n a n+1（n∈N*），求数列{b n}的前n项和为T n．【解答】解：（1）∵（S n﹣1）2=a n S n（n∈N*），∴n≥2时，（S n﹣1）2=（S n﹣S n﹣1）S n（n∈N*）．∴n=1时，，解得a1==S1．n=2时，，解得S2=．同理可得：S3=．（2）由（1）可得：n≥2时，（S n﹣1）2=（S n﹣S n﹣1）S n（n∈N*）．化为：S n=．（*）猜想S n=．n≥2时，代入（*），左边=；右边==，∴左边=右边，猜想成立，n=1时也成立．∴n≥2时，a n=S n﹣S n﹣1=﹣=，n=1时也成立．∴S n=，a n=．（3）b n=（﹣1）n﹣1（n+1）2a n a n+1（n∈N*）=（﹣1）n﹣1=（﹣1）n﹣1，∴n=2k（k∈N*）时，数列{b n}的前n项和为T n=﹣++…+﹣==﹣．n=2k﹣1（k∈N*）时，数列{b n}的前n项和为T n=﹣++…﹣+==+．∴T n=×．23．已知集合M是满足下列性制的函数f（x）的全体，存在实数a、k（k≠0），对于定义域内的任意x均有f（a+x）=kf（a﹣x）成立，称数对（a，k）为函数f （x）的“伴随数对”．（1）判断f（x）=x2是否属于集合M，并说明理由；（2）若函数f（x）=sinx∈M，求满足条件的函数f（x）的所有“伴随数对”；（3）若（1，1），（2，﹣1）都是函数f（x）的“伴随数对”，当1≤x＜2时，f（x）=cos（x）；当x=2时，f（x）=0，求当2014≤x≤2016时，函数y=f（x）的解析式和零点．【解答】解：（1）f（x）=x2的定义域为R．假设存在实数a、k（k≠0），对于定义域内的任意x均有f（a+x）=kf（a﹣x）成立，则（a+x）2=k（a﹣x）2，化为：（k﹣1）x2﹣2a（k+1）x+a2（k﹣1）=0，由于上式对于任意实数x都成立，∴，解得k=1，a=0．∴（0，1）是函数f（x）的“伴随数对”，f（x）∈M．（2）∵函数f（x）=sinx∈M，∴sin（a+x）=ksin（a﹣x），∴（1+k）cosasinx+（1﹣k）sinacosx=0，∴sin（x+φ）=0，∵∀x∈R都成立，∴k2+2kcos2a+1=0，∴cos2a=，≥2，∴|cos2a|≥1，又|cos2a|≤1，故|cos2a|=1．当k=1时，cos2a=﹣1，a=nπ+，n∈Z．当k=﹣1时，cos2a=1，a=nπ，n∈Z．∴f（x）的“伴随数对”为（nπ+，1），（nπ，﹣1），n∈Z．（3）∵（1，1），（2，﹣1）都是函数f（x）的“伴随数对”，∴f（1+x）=f（1﹣x），f（2+x）=﹣f（2﹣x），∴f（x+4）=f（x），T=4．当0＜x＜1时，则1＜2﹣x＜2，此时f（x）=f（2﹣x）=﹣cos；当2＜x＜3时，则1＜4﹣x＜2，此时f（x）=﹣f（4﹣x）=﹣cos；当3＜x＜4时，则0＜4﹣x＜1，此时f（x）=﹣f（4﹣x）=cos．∴f（x）=．∴f（x）=．∴当2014≤x≤2016时，函数y=f（x）的零点为2014，2015，2016．。

上海市复旦大学附属中学2016届高三下学期5月月考数学试题

绝密★启用前上海市复旦大学附属中学2016届高三下学期5月月考数学试题注意事项：1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上第I 卷（选择题)请点击修改第I 卷的文字说明一、单选题1．等差数列{}n a 的前n 项和记为n S ，若2415a a a ++的值为一个确定的常数，则下列各数中也是常数的是（）A ．7SB ．8SC ．13SD ．15S 2．矩阵的一种运算a b x ax by c d y cx dy +⎛⎫⎛⎫⎛⎫= ⎪⎪ ⎪+⎝⎭⎝⎭⎝⎭，该运算的几何意义为平面上的点(,)x y 在矩阵a b c d ⎛⎫ ⎪⎝⎭作用下变换成点(,)ax by cx dy ++，若曲线22421x xy y ++=，在矩阵11a b ⎛⎫ ⎪⎝⎭的作用下变换成曲线2221x y -=，则+a b 的值为（） A ．2- B ．2 C ．2± D ．4-3．．函数()y f x =是R 上的增函数，则0()()()()a b f a f b f a f b +>+>-+-是的（） A ．充要条件B ．必要不充分条件C ．充分不必要条件D ．既不充分也不必要条件4．有一容积为33a cm 的正方体容器1111ABCD A B C D -，在棱AB 、1BB 和面对角线1BC 的中点各有一小孔E 、F 、G ，若此容器可以任意放置，则其可装水的最大容积是（） A ．331a cm B ．337a cm C ．3311a cm D ．3347a cm第II 卷（非选择题)请点击修改第II 卷的文字说明二、填空题5．已知集合{}|lg M x y x ==，{|N x y ==，则M N =I _____________.6．若复数z 满足()3443i z i -=+，则z 的虚部为________.7．在ABC ∆中，角A 、B 、C 所对的边分别为a 、b 、c ，若6a =，4c =，sin23B =，则b =_____.8．如图所示，一家面包销售店根据以往某种面包的销售记录，绘制了日销售量的频率分布直方图，若一个月以30天计算，估计这家面包店一个月内日销售量不少于150个的天数为________.9．现有5个女生和3个男生随机站成一排，则排头和排尾均为女生的概率是________（结果用分数表示）.10．在极坐标系中，圆2sin ρθ=的圆心到极轴的距离为________.11．无穷等比数列{}n a 的前n 项和为n S ，若12a =，且20152016201723S S S +=，则无穷等比数列{}n a 的各项和为________.12．如图，正三棱柱111ABC A B C -中，4AB =，16AA =，若E 、F 分别是棱1BB 、1CC 上的点，则三棱锥1A A EF -的体积是________.13．设直线2310x y ++=和圆22230x y x +--=相交于点A 、B ，则弦AB 的垂直平分线方程是____．14．在平面直角坐标系xOy 中，抛物线()220y px p =>的焦点为F ，双曲线22221x y a b-=()0,0a b >>的两条渐近线分别与抛物线交于A 、B 两点（A 、B 异于坐标原点O ），若直线AB 恰好过点F ，则双曲线的渐近线方程是________.15．在边长为6的等边△ABC 中，点M 满足2BM MA =u u u u r u u u r ，则CM CB ⋅u u u u r u u u r 等于．16．已知函数()2xf x =且()()()f xg xh x =+，其中()g x 为奇函数, ()h x 为偶函数，若不等式2()(2)0a g x h x ⋅+≥对任意[1,2]x ∈恒成立,则实数a 的取值范围是 .17．已知圆O ：221x y +=，O 为坐标原点，若正方形ABCD 的一边AB 为圆O 的一条弦，则线段OC 长度的最大值是 .18．如图，在正三棱锥P ABC -中，D 为线段BC 的中点，E 在线段PD 上，23PE PD =，AE l =为定长，则该棱锥的体积的最大值为________.三、解答题19．三角形的三个内角A 、B 、C 所对边的长分别为a 、b 、c ，设向量(,),(,)m c a b a n a b c =--=+r r ，若m r //n r ．（1）求角B 的大小；（2）求sin sin A C +的取值范围．20．如图，空间直角坐标系中，四棱锥P OABC -的正方形，且底面在xOy 平面内，点B 在y 轴正半轴上，PB ⊥平面OABC ，侧棱OP 与底面所成角为45°；（1）若(,,0)N x y 是顶点在原点，且过A 、C 两点的抛物线上的动点，试给出x 与y 满足的关系式；（2）若M 是棱OP 上的一个定点，它到平面OABC 的距离为a （02a <<），写出M 、N 两点之间的距离()d x ，并求()d x 的最小值；（3）是否存在一个实数a （02a <<），使得当()d x 取得最小值时，异面直线MN 与OB 互相垂直？请说明理由；21．已知k ∈R ，0a >且1a ≠，0b >且1b ≠，函数()x xf x a k b =+⋅. （1）设1a >，1ab =，若()f x 是奇函数，求k 的值；（2）设10>>>a b ，0k ≤，判断函数()f x 在R 上的单调性并加以证明；（3）设2a =，12b =，0k >，函数()f x 的图象是否关于某垂直于x 轴的直线对称？如果是，求出该对称轴，如果不是，请说明理由.22．已知A 、B 为椭圆22221x y a b+=（0a b >>）和双曲线22221x y a b -=的公共顶点，P 、Q 分为双曲线和椭圆上不同于A 、B 的动点，且满足()(),1AP BP AQ BQ R λλλ+=+∈>u u u r u u u r u u u r u u u r ，设直线AP 、BP 、AQ 、BQ 的斜率分别为1k 、2k 、3k 、4k .（1）求证：点P 、Q 、O 三点共线；（2）求1234k k k k +++的值；（3）若1F 、2F 分别为椭圆和双曲线的右焦点，且12//QF PF ，求22221234k k k k +++的值.23．已知n S 是数列{}n a 的前n 项和，对任意*n N ∈，都有()()140n n n m S ma m -=-+>；（1）若4m =，求证：数列14n n a -⎧⎫⎨⎬⎩⎭是等差数列，并求此时数列{}n a 的通项公式；（2）若4m ≠，求证：数列344n n a m ⎧⎫+⋅⎨⎬-⎩⎭是等比数列，并求此时数列{}n a 的通项公式；（3）设()4n n n a b n N *=∈，若2n b ≤，求实数m 的取值范围.参考答案1．C【解析】【分析】根据等差数列的性质，有()2415117318363a a a a d a d a ++=+=+=，故7a 为确定常数，再利用()11313713132a a S a +⋅==，可得出选项.【详解】由于题目所给数列为等差数列，根据等差数列的性质，有()2415117318363a a a a d a d a ++=+=+=，故7a 为确定常数，由等差数列前n 项和公式可知()11313713132a a S a +⋅==也为确定的常数，故选C. 【点睛】本小题主要考查等差数列的性质，考查对等差数列前n 项和公式的理解和运用.等差数列前n 项和公式有两个，在不同的已知条件下，要学会选择合适的公式的来求解.本小题属于中档题.2．B【解析】【分析】设点(),x y 是曲线22421x xy y ++=上的一点，在矩阵11a b ⎛⎫ ⎪⎝⎭的作用下的点为(),x y ''，可得出x x ay y bx y=+⎧⎨=+''⎩，代入方程2221x y ''-=化简后与方程22421x xy y ++=作比较，可得出关于a 、b 的方程组，解出这两个未知数的值，即可求出+a b 的值.【详解】设点(),x y 是曲线22421x xy y ++=上的一点，在矩阵11a b ⎛⎫ ⎪⎝⎭的作用下的点为(),x y ''，即11x a x x ay y b y bx y '+⎛⎫⎛⎫⎛⎫⎛⎫== ⎪ ⎪⎪ ⎪'+⎝⎭⎝⎭⎝⎭⎝⎭，代入方程2221x y ''-=，得()()2221x ay bx y +-+=，即()()()2222122421b x a b xy a y -+-+-=，得2212124422b a b a ⎧-=⎪-=⎨⎪-=⎩，解得20a b =⎧⎨=⎩，因此，2a b +=.故选：B.【点睛】本题主要考查几种特殊的矩阵变换、曲线与方程等基础知识，解题的关键就是利用待定系数法求出参数的值，考查运算求解能力，属于中等题.3．A【解析】【详解】又在R 上为增函数，则反之，若4．C【解析】【分析】分别讨论水面过直线EF 、FG 、EG时从正方体截去的几何体体积的最小值，即可得出此容器可装水的最大容积.【详解】当水面过直线FG 时，如下图所示，水面截去正方体1111ABCD A B C D -所得几何体为三棱柱MBF NCO -，当点E 在水面上方或水面上时，容器中的水不会漏，且当点M 与点E 重合时，截去的几何体体积最小为()233111228a a a cm ⎛⎫⨯⨯= ⎪⎝⎭；当水面过直线EF 时，如下图所示，水面截去正方体1111ABCD A B C D -所得几何体为三棱台EBF PCQ -，当点G 在水面上方或水面上时，容器中的水不会漏，且当点G 在直线FQ 上时，截去的几何体为三棱柱，且体积最小为()233111228a a a cm ⎛⎫⨯⨯= ⎪⎝⎭；当水面过直线EG 时，如下图所示，当点F 在水面上方或水面上时，容器中的水不会漏，此时水面截去正方体1111ABCD A B C D -所得几何体为EBR SCT -，且直线RT 过点G ，易知梯形BCTR 的面积为正方形11BCC B 面积的一半，此时，几何体EBR SCT -的体积为()233111132212EBR SCT E BCTR E SCT E BCTR V V V V a a a cm ----=+>=⨯⨯=. 当RT 与直线1B C 重合时，如下图所示，此时，点F 在水面上方，容器不会漏水，水面截去正方体1111ABCD A B C D -所得几何体为三棱锥1E BB C -，该三棱锥的体积为()1123311111332212E BB C BB C V BE S a a a cm -∆=⋅=⨯⨯=. 综上可知，水面截去截去正方体1111ABCD A B C D -所得几何体体积的最小值为()33112a cm . 因此，该容器可装水的最大容积是()33331111212a a a cm -=. 故选C.【点睛】本题主要考查几何体体积的计算，根据几何体的几何特征找出水面截去正方体所得几何体体积最小值时的位置是解题的关键，考查推理能力，属于难题. 5．(]0,1 【解析】【分析】求出集合M 、N ，然后利用交集的定义求出集合M N ⋂. 【详解】{}|lg (0,)M x y x ===+∞，{|[1,1]N x y ===-，(0,)[1,1](0,1].M N ⋂=+∞⋂-=故答案为(]0,1. 【点睛】本题考查集合的交集运算，同时与考查了具体函数的定义域，考查计算能力，属于基础题. 6．45【解析】【分析】求出复数43i +的模，然后在等式()3443i z i -=+两边同时除以34i -，利用复数的除法可求出复数z ，即可得出复数z 的虚部. 【详解】435i +==Q ，由()34435i z i -=+=，()()()()5345345343434342555i i z i i i i ++∴====+--+，因此，z 的虚部为45. 故答案为45. 【点睛】本题考查复数虚部的求解，同时也考查了复数模的计算以及复数的除法运算，解题的关键就是利用复数的四则运算将复数表示为一般形式，考查计算能力，属于基础题. 7．6 【解析】【分析】利用二倍角公式求出cos B ，然后利用余弦定理求出b 的值. 【详解】由二倍角的余弦公式可得221cos 12sin 1223B B =-=-⨯=⎝⎭. 由余弦定理得2222212cos 64264363b ac ac B =+-=+-⨯⨯⨯=，因此，6b =.故答案为：6. 【点睛】本题考查二倍角余弦公式的应用，同时也考查了利用余弦定理求三角形的边长，考查运算求解能力，属于中等题. 8．9 【解析】【分析】根据频率分布直方图计算出日销售量不少于150个的频率，然后乘以30即可. 【详解】根据频率分布直方图可知，一个月内日销售量不少于150个的频率为()0.0040.002500.3+⨯=，因此，这家面包店一个月内日销售量不少于150个的天数为300.39⨯=. 故答案为：9. 【点睛】本题考查频率分布直方图的应用，解题时要明确频数、频率和样本容量三者之间的关系，考查计算能力，属于基础题. 9．514【解析】【分析】计算出基本事件总数为8个学生的全排数，然后考虑排头和排尾均为女生的排法种数，利用古典概型的概率公式即可得出所求事件的概率. 【详解】将5个女生和3个男生随机站成一排，排法种数为88A ，现在考虑排头和排尾都是女生的情况，先从5个女生中选2人进行排列，剩余6个人全排，事件“排头和排尾均为女生”所包含的基本事件数为2656A A ，因此，排头和排尾均为女生的概率是265688514A A A =. 故答案为：514. 【点睛】本题考查利用古典概型概率公式的计算事件的概率，同时也考查了有限制的元素的排列问题，考查计算能力，属于中等题. 10．1 【解析】【分析】将圆的极坐标方程化为直角坐标方程，求出圆心的坐标，可求出圆心到极轴的距离. 【详解】在圆的极坐标方程两边同时乘以ρ，得22sin ρρθ=，化为普通方程得222x y y +=，标准方程为()2211x y +-=，圆心坐标为()0,1，因此，圆2sin ρθ=的圆心到极轴的距离为1. 故答案为：1. 【点睛】本题考查圆心到极轴的距离，解题的关键就是将圆的极坐标方程化为普通方程，考查计算能力，属于基础题. 11．32【解析】【分析】先求出等比数列{}n a 的公比，然后利用无穷等比数列的和可计算出结果. 【详解】设等比数列{}n a 的公比为q ，由20152016201723S S S +=，得()201720152017201620S S S S -+-=，即2017201630a a +=，310q ∴+=，解得13q =-，因此，无穷等比数列{}n a 的各项和为12311213a q ==-⎛⎫-- ⎪⎝⎭.故答案为：32. 【点睛】本题考查无穷等比数列各项和的计算，解题的关键就是求出等比数列的公比，考查计算能力，属于中等题. 12．【解析】【分析】用三棱柱111ABC A B C -的体积减去四棱锥111A EFC B -的体积和四棱锥A BCFE -的体积即可得出三棱锥1A A EF -的体积. 【详解】取BC 的中点D ，连接AD ，则AD BC ⊥.Q 平面ABC ⊥平面11BCC B ，平面ABC I 平面11BCC B BC =，AD ⊂平面ABC ，AD ∴⊥平面11BCC B .ABC ∆Q 是等边三角形，4AB =，AD =1//AA Q 平面11BCC B ，且E 、F 分别为1BB 、1CC 的中点.111114333A BCFE A EFCB BCFE V V S AD --∴==⋅=⨯⨯⨯=111122462A A EF ABC A B C A BCFE V V V ---∴=-=⨯-⨯=故答案为：【点睛】本题考查三棱锥的体积的计算，常用的方法有等体积法、间接法、割补法，解题时可充分选择合适的方法来进行计算，考查计算能力，属于中等题. 13．【解析】【详解】由22230x y x +--=得22(1)4x y -+=，所以圆的圆心为(1,0)，根据圆的相关性质，可知所求的直线过圆心，由直线垂直可得所求直线的斜率为32，根据直线的点斜式方程化简可得结果为．14．2y x =± 【解析】【分析】求出抛物线的焦点坐标与双曲线的渐近线方程，代入抛物线的方程可得出点A 、B 的坐标，再由A 、B 、F 三点共线，可得出2222pa pb =，可得出2b a =，由此可得出双曲线的渐近线方程. 【详解】抛物线的焦点F 的坐标为,02p ⎛⎫⎪⎝⎭，双曲线()222210,0x y a b a b-=>>的渐近线方程为b y x a =±，代入抛物线的方程，可得2222,pa pa A b b ⎛⎫ ⎪⎝⎭、2222,pa pa B b b ⎛⎫- ⎪⎝⎭，由A 、B 、F 三点共线可得2222pa pb =，2214a b ∴=，则2b a =.因此，双曲线的渐近线方程为2y x =±. 故答案为：2y x =±. 【点睛】本题考查抛物线焦点坐标以及双曲线渐近线方程的求解，解题的关键就是利用三点共线得出关系式，考查运算求解能力，属于中等题. 15． 24 【解析】试题分析：11·()?()?··2433CM CB CA AM CB CA AB CB CACB AB CB =+=+=+=u u u u r u u u r u u u r u u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r考点：本小题考查向量的线性运算及其向量的数量积．点评：解题关键在于向量CM 与AM 的转化,利用数量积定义求题． 16．17[,)12-+∞ 【解析】试题分析：∵h （x ）为定义在R 上的偶函数，g （x ）为定义在R 上的奇函数 ∴g （-x ）=-g （x ），h （-x ）=h （x ）, 又∵由h （x ）+g （x ）=2x， h （-x ）+g∵1≤x ≤2∴2x -2-x＞0,令t=2-x -2x ，整理得：考点：1.函数不等式的恒成立问题；2.换元法；3.基本不等式 17．【解析】试题分析：设,OBA θ∠=则2222cos (sin 2cos )14cos 2sin 23)34OC πθθθθθθ=++=++=++≤+，当且仅当8πθ=取等号，因此OC 长度的最大值是考点：直线与圆位置关系18．3932l【解析】【分析】设正三棱锥P ABC -的底面边长为a ，则该三棱锥的高为h =求出底面积，并代入三棱锥的体积公式，然后利用基本不等式可求出该棱锥体积的最大值. 【详解】如下图所示，设正三棱锥P ABC -的底面边长为a ，由正弦定理可知，底面ABC ∆的外接圆半径为2sin 60a r =o，连接AD ，过点P 作PO ⊥平面ABC ，垂足为点O ，过点E 作//EF PO 交AD 于点F ，23PE PD =Q ，22113323OF OD OA r ∴==⨯=，1433AF OA OF r r r ∴=+=+==，EF ∴===23PE PD =Q ，则13DE PD =，又//EF PO Q ，3PO EF ∴==，因此，该三棱锥的体积为221344P ABC V -=⨯⨯===339323232932l ≤===，当且仅当2221682727l a a -=，即当4a l =时，等号成立. 因此，该三棱锥体积的最大值为3932l . 故答案为3932l . 【点睛】本题考查正三棱锥体积最值的求法，建立关系式是解题的关键，考查数形结合思想的应用，属于中等题.19．（1）3B π=；（2）2⎛ ⎝．【解析】【详解】（1）由m r //n r知c a b aa b c--=+，即得222b a c ac =+-，据余弦定理知 1cos 2B =，得3B π= （2）()sin sin sin sin AC A A B +=++ sin sin 3A A π⎛⎫=++⎪⎝⎭13sin sin cos sin 2222A A A A A =++=+6A π⎛⎫=+ ⎪⎝⎭因为3B π=，所以23A C π+=，得20,3A π⎛⎫∈ ⎪⎝⎭所以5,666A πππ⎛⎫+∈ ⎪⎝⎭，得1sin ,162A π⎛⎫⎛⎤+∈ ⎪ ⎥⎝⎭⎝⎦，即得sin sin A C +的取值范围为⎝．点睛：本题关键是首先要得出向量平行的等式，再结合余弦定理即可得出B ，对于三角函数范围问题则通常需要将原式化简为sin)A x b ωφ++（的形式再求解答案（需注意范围的变化），此题属于基础题.20．（1）2y x =；（2）()min1,021222a d x a <≤=<<⎩；（3）a =. 【解析】【分析】（1）根据题意，求出点A 的坐标，代入抛物线方程，即可得出x 与y 的关系式；（2）设点M 和N 的坐标，根据两点间的距离公式，利用二次函数的基本性质，即可得出函数()y d x =的最小值；（3）由（2）可知，当1,22a ⎛⎫∈ ⎪⎝⎭时，当()y d x =取得最小值时，求得2212a x -=，由异面直线MN 与OB 垂直时，M N y y =，代入即可求出a 的值. 【详解】（1）由四棱锥P OABC -()1,1,0A ，可设x 与y 所满足的关系式为22x py =，将点A 横坐标和竖坐标代入该方程得12p =，解得12p =，因此，x 与y 所满足的关系式为2y x =；（2）设点()0,,M a a ，()2,,0N x x ，则()d x MN ===.令20t x =≥，设()()22122f t t a t a =+-+，对称轴为直线212a t -=. ①当022102a a <<⎧⎪⎨-≤⎪⎩时，即当102a <≤时，函数()()22122f t t a t a =+-+在[)0,+∞上单调递增，则()()2min 02f t f a ==，此时()min d x =；②当022102a a <<⎧⎪⎨->⎪⎩时，即当122a <<时，此时函数()()22122f t t a t a =+-+在212a t -=取得最小值，即()()2222min21214412224a a a a f t a --+-⎛⎫=-+=⎪⎝⎭，此时()mind x =. 因此，()min1,02122a d x a <≤=<<；（3）当10,2a ⎛⎤∈ ⎥⎝⎦时，此时点N 与原点重合，则直线MN 与OB 为相交直线，不符；当1,22a ⎛⎫∈ ⎪⎝⎭时，则当()y d x =取最小值时，2212a x -=，当异面直线MN 与OB 垂直时，M N y y =，即2212a a -⎛⎫= ⎪⎝⎭，化简得24810a a -+=.122a <<Q，解得a =【点睛】本题考查抛物线方程、异面直线所成的角、两点间的距离公式以及二次函数基本性质的应用，考查化归与转化思想的应用，属于中等题.21．（1）1k =-；（2）证明见解析；（3）对称轴为4log x k =，理由见解析. 【解析】【分析】（1）根据已知条件，将1b a=代入函数()y f x =的解析式，得出()x xf x a k a -=+⋅，利用奇函数的定义()()0f x f x +-=，可求出实数k 的值；（2）判断出函数1x y a =和函数2xy b =的单调性，然后利用函数单调性的运算法则，可判断出函数()xxf x a k b =+⋅的单调性，然后利用函数单调性的定义加以证明；（3）根据函数()22xxf x k -=+⋅图象的对称轴为直线x m =，得出()()f m x f m x -=+对任意的实数x 恒成立，即可求出实数m 的值. 【详解】（1）由已知，1b a=，()x xf x a k a -∴=+⋅，由于函数()y f x =为奇函数，则()()()()10xxx x x x f x f x a k aa k a k a a ---+-=+⋅++⋅=+⋅+=对任意的x ∈R 恒成立，10k ∴+=，因此，1k =-；（2）当10>>>a b 时，函数1x y a =为增函数，函数2xy b =为减函数，又0k ≤Q ，所以，函数()xxf x a k b =+⋅在R 上是增函数，下面利用定义来证明出函数()y f x =的单调性.任取12x x >，则()()()()()()1122121212x x x x x x x xf x f x a k b a k b a a k b b -=+⋅-+⋅=-+-，1a >Q ，12x x a a ∴>，即120x x a a ->，又01b <<Q ，12x x b b ∴<，0k ≤Q ，()120x xk b b ∴-≥，所以，()()120f x f x ->，即()()12f x f x >.因此，函数()xxf x a k b =+⋅在R 上是增函数；（3）()22xxf x k -=+⋅，若函数()y f x =的图象是轴对称图形，且对称轴为直线x m =，则()()f m x f m x -=+，2222m x x m m x x m k k --+--∴+⋅=+⋅，即()()22220m xm x x m x m k -+----+⋅-=，即()()2222220m x x m x x k ----+⋅-=，即()()22220mm x x k --⋅--=对任意的x ∈R 恒成立，22m m k -∴⋅=，即224m m k ==，因此，4log =m k . 【点睛】本题是一道函数综合问题，考查利用函数的奇偶性求参数、利用定义判定函数的单调性以及函数对称性问题，解题时要结合有关定义进行理解，考查分析问题与解决问题的能力，属于中等题.22．（1）见解析；（2）0；（3）8. 【解析】【分析】（1）由()AP BP AQ BQ λ+=+u u u r u u u r u u u r u u u r ，得到OP OQ λ=u u u r u u u r，由此可证明出点P 、Q 、O 三点共线；（2）设点()11,P x y 、()22,Q x y ，求出2112212x b k k a y +=⋅，2234222x b k k a y +=-⋅，由//OP OQ u u u r u u u r ，可得出1212x x y y =，从而可求出1234k k k k +++的值；（3）由OP OQ λ=u u u r u u u r ，可得222122211212x a y b λλ⎧+=⎪⎪⎨-⎪=⎪⎩，再由12//QF PF ，得出()2124k k +=，()2344k k +=，由此能求出22221234k k k k +++的值.【详解】（1）A Q 、B 为椭圆()222210x y a b a b +=>>和双曲线22221x y a b-=的公共顶点，P 、Q 分别为双曲线和椭圆上不同于A 、B 的动点，且()AP BP AQ BQ λ+=+u u u r u u u r u u u r u u u r ，即22OP OQ λ=⋅u u u r u u u r ，即OP OQ λ=u u u r u u u r，因此，点P 、Q 、O 三点共线；（2）设点()11,P x y 、()22,Q x y ，则21111111122222222111112222y y x y x y x b k k a x a x a x a a y y a a b +=+===⋅+--+-，同理可得2234222x b k k a y +=-⋅，//OP OQ u u u r u u u r Q ，1221x y x y ∴=，则1212x x y y =，因此，212123421220x x b k k k k a y y ⎛⎫+++=-= ⎪⎝⎭；（3）OP OQ λ=u u u r u u u r Q ，212111x x y y λλ⎧=⎪⎪∴⎨⎪=⎪⎩，2222221x y a b +=Q ，2221122x y a b λ∴+=，又2211221x y a b -=，解得222122211212x a y b λλ⎧+=⎪⎪⎨-⎪=⎪⎩，又12//QF PF Q ，21OF OF λ∴=⋅，则()22222a b a b λ+=-，则22222a b a b λ+=-. 222412224111x a a y b b λλ+∴=⋅=-，()2444211242441444x b b a k k a y a b∴+=⋅⋅=⋅⋅=，同理可得()2344k k +=，21111222111y y y k k x a x a x a =⋅=+--Q 且2211221x y a b -=，2222112a x a y b ∴-=，2122b k k a∴=，同理可得2342b k k a=-，因此，()()()22222212341234123428k k k k k k k k k k k k +++=+++-+=.【点睛】本题考查椭圆与双曲线的综合问题，考查整体代换与方程思想，在计算时应充分利用点在曲线上这一条件得出等式进行计算，考查运算求解能力，属于难题. 23．（1）证明见解析，()1314n n a n -=+⋅；（2）1443444n n n m a m m m --=⋅-⋅--；（3）50,2⎛⎤⎥⎝⎦. 【解析】【分析】（1）将4m =代入()14nn n m S ma -=-+，得344nn n S a =-，令1n =，求出1a ，然后令2n ≥，由344nn n S a =-得出111344n n n S a ---=-，两式作差可得出数列{}n a 的递推公式，然后利用定义证明出数列14n n a -⎧⎫⎨⎬⎩⎭是等差数列，确定该数列的首项，即可求出n a ；（2）令1n =求出14a =，然后令2n ≥，由()14nn n m S ma -=-+得出()11114n n n m S ma ----=-+，两式相减得出数列{}n a 的递推公式，然后利用定义证明出数列344n na m ⎧⎫+⋅⎨⎬-⎩⎭为等比数列，确定该数列的首项和公比，即可求出n a ；（3）结合（1）（2）中的结论，讨论4m =、4m >、01m <<、1m =、14m <<，结合条件2n b ≤，利用数列()41144nm m m m ⎧⎫-⎪⎪⎛⎫⋅⋅⎨⎬ ⎪-⎝⎭⎪⎪⎩⎭的单调性，即可得出实数m 的取值范围. 【详解】（1）将4m =代入()14nn n m S ma -=-+，得344n n n S a -=-+，即344nn n S a =-.当1n =时，则有11344a a =-，得14a =；当2n ≥时，由344nn n S a =-得出111344n n n S a ---=-，上述两式相减得()11113444444441n n n n n n n n a a a a a ----=---=---，整理得11434n n n a a --=+⋅，等式两边同时除以14n -得112344n n n n a a ---=+，即112344n n n n a a ----=，所以，数列14n n a -⎧⎫⎨⎬⎩⎭是以首项为1044a =为首项，以3为公差的等差数列，则()1431314n n a n n -=+-=+，因此，()1314n n a n -=+⋅；（2）对任意*n N ∈，都有()()140nn n m S ma m -=-+>.当1n =时，()1114m S ma -=-+，解得14a =；当2n ≥时，由()14nn n m S ma -=-+得出()11114n n n m S ma ----=-+，两式相减得()111114434nn n n n n n n m a ma ma ma ma -----=-++-=-++⋅，化简得1134n n n a ma --=+⋅，11111111133334344444444333444444n n n n nn n n n n n n n n a ma ma m m m a a a m m m ---------+⋅+⋅+⋅+⋅+⋅---==+⋅+⋅+⋅---Q 1111111111133334444444333444444n n n n n n n n n n n n m ma m a ma m m m m a a a m m m -----------⎛⎫⎛⎫++⋅+⋅+⋅ ⎪ ⎪--⎝⎭⎝⎭-====+⋅+⋅+⋅---，所以，数列344n n a m ⎧⎫+⋅⎨⎬-⎩⎭是以m 为公比，以()14131244444m a m m m -+⨯=+=---为首项的等比数列，则()1413444nn n m a m m m --+⋅=⋅--，因此，()141344n n n m m a m --⋅-⋅=-；（3）()4n n n a b n N *=∈Q ，且2n b ≤. 当4m =时，314n n b +=，当3n ≥时，2n b >，不满足条件；则4m ≠，可得()4113444nn m m b m m m -⎛⎫=⋅⋅-⎪--⎝⎭，可得()41313224444nm m m m m m -⎛⎫-+≤⋅⋅≤+⎪---⎝⎭，显然4m >时，数列()41144nm m m m ⎧⎫-⎪⎪⎛⎫⋅⋅⎨⎬ ⎪-⎝⎭⎪⎪⎩⎭单调递增，不满足条件，04m ∴<<. 当1m =时，则有301-<<显然成立；当01m <<时，若1n =，则数列()41144nm m m m ⎧⎫-⎪⎪⎛⎫⋅⋅⎨⎬ ⎪-⎝⎭⎪⎪⎩⎭的最大项为14m m --，31244m m m -∴+>--，即251m m -<-恒成立；当512m <≤时，数列()41144nm m m m ⎧⎫-⎪⎪⎛⎫⋅⋅⎨⎬ ⎪-⎝⎭⎪⎪⎩⎭的最大项为104m m -<-，则3204m +≥-满足条件；当542m <<时，3204m +<-，数列()41144nm m m m ⎧⎫-⎪⎪⎛⎫⋅⋅⎨⎬ ⎪-⎝⎭⎪⎪⎩⎭的最大项为104m m ->-，不满足条件；综上所述，实数m的取值范围是5 0,2⎛⎤ ⎥⎝⎦.【点睛】本题考查数列的递推公式，等差数列和等比数列的定义，同时也考查了数列不等式问题，考查分析问题和解决问题的能力，属于难题.。

上海市复旦大学附属中学2023-2024学年高三上学期期中考试数学试题(含答案)

复旦大学附属中学2023-2024学年高三上学期期中数学试题2023.11(满分 150分, 时间120分钟)学校:___________姓名：___________班级：___________考号：_________一、填空题(本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分)1.已知复数 (i是虚数单位)，则z的虚部是 .2.已知3sinα=cosα,则tan(π-α)的值是 .3.已知某班全体学生在某次数学考试中的成绩(单位：分)的频率分布直方图如图所示，则图中a所代表的数值是 .4.已知两点P(3,4),Q(-5,6),则以线段PQ为直径的圆的标准方程是 .5.已知向量则向量在向量上的投影向量的坐标为 .6 已知一个圆锥的轴截面是等边三角形，侧面积为8π，则该圆锥的体积等于 .7.齐王与田忌赛马，田忌的上等马优于齐王的中等马，劣于齐王的上等马；田忌的中等马优于齐王的下等马，劣于齐王的中等马；田忌的下等马劣于齐王的下等马.现各从双方的马匹中随机选一匹进行一场比赛，则田忌的马获胜的概率为 .8.已知一组数据: 10, 11, 12, 13, 13, 14, 15, 16, 记这组数据的第60百分位数为a, 众数为b,则a和b的大小关系是 . (用“<”,“>”,“=”连接)9.已知函数f(x)=3sinx+2cosx, 当f(x)取得最大值时,= .10.已知则abc的值为 .11. 如图在△ABC中, AB=2,AC=．5,∠BAC=60°,边BC、 AC上的中线AM、 BN相交于点P,则cos∠MPN= .l2.已知函数若有且仅有一个正整数使得不等式成立，则实数a的取值范围是 .二、选择题(本大题共有4题, 13、14每题4分, 15、16每题5分, 满分18分)13. 设ab>0, 则“a>b”是的 ( ) .A.充分非必要条件B.必要非充分条件C.充分必要条件D.既非充分也非必要条件14. 定义在区间(-∞,0)∪(0,+∞)的函数f(x), 如果对于任意给定的非常数等比数列{an},{f(an)}仍是等比数列，则称f(x)为“保等比数列函数”，下列函数是“保等比数列函数”的是( ) .B. f(x)=2x+1 D. f(x)=log ₃|x|15.《周髀算经》中“侧影探日行”一文有记载：“即取竹空，径一寸，长八尺，捕影而视之，空正掩目，而日应空之孔.”意为：“取竹空这一望筒，当望筒直径d 是一寸，筒长t 是八尺时(注：一尺等于十寸)，从筒中搜捕太阳的边缘观察，则筒的内孔正好覆盖太阳，而太阳的外缘恰好填满竹管的内孔.”如图所示， O 为竹空底面圆心，则太阳角∠AOB 的正切值为 ( ) . A.16. 已知F ₁、F ₂是椭圆的左、右焦点，Q 是Γ上一动点，记 ₁ ₂ ₁ ₂ ₁ ₂ 若 ₁ ₂ ₂ ₁ 则的值为( ) .三、解答题(本题共5道题，满分78分)17. (本题满分14分,第1题 6分, 第2题8分)如图，长方体 ₁ ₁ ₁ ₁的底面ABCD 是正方形, 点E 在棱AA ₁上, BE⊥EC ₁. (1)证明: BE⊥平面EB ₁C ₁;(2)若AA ₁=2,AB=1, 求四棱锥 ₁ ₁ 的体积.18. (本题满分14分,第1题6分,第2题8分)已知数列，若对于任意正整数n ， ₂ ₁仍为数列中的项，则称数列为“回归数列”. (1)已知判断数列是否为“回归数列”，并说明理由；(2)若数列为“回归数列”，且对于任意正整数n ，均有 ₁成立，证明：数列为等差数列.19. (本题满分14分,第1题6分, 第2题8分)“我将来要当一名麦田里的守望者，有那么一群孩子在一大块麦田里玩，几千几万的小孩子，附近没有一个大人，我是说，除了我.”《麦田里的守望者》中的主人公霍尔顿将自己的精神生活寄托于那广阔无垠的麦田.假设霍尔顿在一块平面四边形ABCD的麦田里成为守望者.如图所示，为了分割麦田，他将B、D连接,经测量知AB=BC=CD=1,AD=2.(1) 霍尔顿发现无论BD多长，2cosA-cosC都为一个定值. 请你证明霍尔顿的结论，并求出这个定值；(2) 霍尔顿发现小麦的生长和发育与分割土地面积的平方和呈正相关关系. 记△ABD与的面积分别为S₁和S₂，为了更好地规划麦田，请你帮助霍尔顿求出的最大值.20.(本题满分18分,第1题4分, 第2题6分,第3小题满分8分)已知椭圆过点且Γ的左焦点为直线l与Γ交于M,N两点.(1)求椭圆Γ的方程;(2)若且点P的坐标为(0，1)，求直线l的斜率；(3)若其中O为坐标原点，求△MON面积的最大值.21. (本题满分18分,第1题4分,第2题6分,第3小题满分8分)已知函数其中λ为实数.(1)若y=h(x)是定义域上的单调函数，求实数λ的取值范围；(2)若函数y=h(x)有两个不同的零点，求实数λ的取值范围；(3)记g(x)=h(x)-λx,若p,q(p<q)为g(x)的两个驻点,当λ在区间上变化时，求|g(p)-g(q)|的取值范围.复旦大学附属中学2023-2024学年高三上学期期中数学试题解析 2023.11(满分 150分, 时间120分钟)学校:___________姓名：___________班级：___________考号：_________一、填空题(本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分)1. 2.3.0.0154. =175.6.7.8.= 9.10. 或者11.12.10. 或者【详解】得 ,由得所以所以 1；即即所以或故答案为：10或11.【分析】根据题意建立直角坐标系，从而得到各点坐标，进而利用向量夹角余弦的坐标表示即可得解. 【详解】依题意，以A 为原点，AC 所在直线为x 轴，过A 作AC 的垂线为y 轴，如图所示，因为所以，，则，即为向量与的夹角，则, 则故答案为：12.【详解】函数有且仅有一个正整数使得不等式成立，① ②③恒成立，由②-①且③-②得：④ ⑤恒成立，恒成立,.二、选择题(本大题共有4题, 13、14每题4分, 15、16每题5分, 满分18分)13. C 14.A 15.B 16.A14.A.【分析】根据等比数列的定义及函数的性质计算即可一一选定.【详解】不妨设的公比为即；对于A项，仍是常数，即B项符合题意；对于B项，由题意可得，因为an非常数，则非常数，故非常数，即B项不符合题意；对于C项，,同B项可知，该比值非常数，即C项不符合题意；对于D项，，同B项可知，该比值非常数，即D项不符合题意.故选: A.16.A.【详解】中由勾股定理得：===即=,= ===,又 ₁ ₂ ₂ ₁，6=4,,=, c=a=, 故选A三、解答题(本题共5道题，满分78分)17.(1)证明：由长方体的性质可知，平面 ₁ ₁ 平面 ₁ ₁∴⊥平面E .(2)取棱 ₁的中点F, 连接EF、 ₁ 则由(1)知, ₁由题设可知， ₁ ₁₁ ₁ ₁∵在长方体 ₁ ₁ ₁ ₁中, ₁ 平面 ₁ ₁ ₁平面 ₁ ₁ ∴点E到平面 ₁ ₁ 的距离∴四棱锥 ₁ ₁ 的体积棱锥侧18.(1)对于任意仍为数列中的项，则称数列为“回归数列”.己知则 ₂ ₁显然不是数列中的项，故：数列不为“回归数列”.(2)由题意知:必存在使得： ₂ ₁由题意可知： ₁₂ ₁故 ₂ 因此即: ₂ ₁ ₁整理得： ₂ ₁ ₁则数列为等差数列.19.【详解】(1)在中,在中,,则为定值.(2)因为设则,所以，当时,取得最大值即时,的最大值为.20.【详解】(1)由题意得解得:∴椭圆C的标准方程为(2)设 ₁ ₁ ₂ ₂直线l的方程为代入椭圆的方程消去y得：①②，,=-2代入①和②得：③④③④得：=-2，解得：;(2)设 ₁ ₁ ₂ ₂直线l的方程为代入椭圆的方程消去y得：①②由解得=③把①和②代入③得：=4,④又+=16,又,,④中当且仅当即时，等号成立，的面积的最大值为21.已知函数其中λ为实数.(1)若y=h(x)是定义域上的单调函数，求实数λ的取值范围；(2)若函数y=h(x)有两个不同的零点，求实数λ的取值范围；(3)记g(x)=h(x)-λx,若p,q(p<q)为g(x)的两个驻点,当λ在区间上变化时，求|g(p)-g(q)|的取值范围.【详解】(1)易得定义域为，-=,①当且仅当0时，恒成立，y=h(x)是定义域上的单调递增函数，符合题意；而当0时，既不恒正，也不恒负，即y=h(x)不是定义域上的单调函数，不符合题意，舍去；所以，由题意得:实数λ的取值范围为，；(2)函数y=h(x)有两个不同的零点，y=h(x)不是定义域上的单调函数，即0；由①得：y=h(x)在上为单调递减函数，在，上为单调递增函数，函数y=h(x)有两个不同的零点=;(3)p,q(p<q)为g(x)=h(x)-λx=-λx的两个驻点,p,q(0p<q)为=--λ=0一元二次方程-x+的两个不同的正根，即, 又，===又=或者=, =- =,在p上为单调递增函数，=，.。

上海市复旦实验中学高三数学上学期期中试题(含解析)沪教版

上海市复旦实验中学高三数学上学期期中试题（含解析）沪教版一、填空题（本大题满分56分）本大题共有14题，每题4分，请在相应的空格内填上正确的答案，每个空格填对得5分，否则一律得0分.1. 已知集合{}03A x x =<<，401x B xx ⎧-⎫=<⎨⎬-⎩⎭，则A B ⋂= . 解析：()401,41x B xx ⎧-⎫=<=⎨⎬-⎩⎭，A B ⋂=()1,3. 2. 函数()2sin cos 44f x x x ππ⎛⎫⎛⎫=++ ⎪ ⎪⎝⎭⎝⎭的最小正周期为 .解析：()2sin cos sin 2cos2442f x x x x x πππ⎛⎫⎛⎫⎛⎫=++=+= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，所以最小正周期22T π==π.3. 已知()521px +的展开式中，6x 的系数为80，那么实数p = .解析：()22155rrrr r r T px p x C C +==，令3353802r p p C =⇒=⋅⇒=.4. 已知集合{}1,1A =-，{}10B x ax =+=，若B A ⊆，则实数a 的所有可能取值组成的集合为 .解析：分类讨论，不要忘了空集的情况：{}{}1,1,B B B a =-==∅⇒∈{}1,1,0-.5. 在ABC ∆中，角,,A B C 所对的边长分别为,,a b c .若sin :sin :sin 6:5:4A B C =，则最大角为 .解析：由正弦定理可得::6:5:4a b c =，有余弦定理即可得最大角A 的余弦值2225481cos 2548A +-==⋅⋅，即1arccos 8A =.6. 已知口袋里装有同样大小、同样质量的16个小球，其中8个白球、8个黑球. 现从口袋中任意摸出8个球恰好是4白4黑的概率为 .（结果精确到0.001）解析：44888160.381P C C C⋅=≈7. 在平面直角坐标系xOy 中，将点)A绕原点O 逆时针旋转90到点B ，若直线OB 的倾斜角为α，则tan2α的值为 .解析：很明显tan 30AOX AOX ∠=⇒∠=，所以120α=，即tan 2tan 240α==8. 若函数()()2lg 2f x x x a =-+在[)1,+∞上单调递增，则a 的取值范围是 . 解析：()()2lg 2f x x x a =-+在[)1,+∞上单调递增，内函数22y x x a =-+在[)1,+∞上递增且函数值大于0，所以2120a -+>⇒1a >.9. 若一个圆锥的轴截面是边长为4cm 的等边三角形，则这个圆锥的侧面积为 . 解析：轴截面是边长为4cm ，则底面半径2r =，母线4l =，所以侧面积为rl π=8π.10. 已知定义在02π⎛⎫⎪⎝⎭，上的函数()2sin 1y x =+与83y =的图像相交与点P ，过点P 作1PP x ⊥轴于1P ，直线1PP 与tan y x =的图像交于点2P ，则线段12P P 的长度为 .解析：()812sin 1sin 33P P x x +=⇒=,21221tan p y x PP y y ===-=11. 已知函数()()0,1x f x a a a =>≠满足()()23f f >，若()1f x -是()f x 的反函数，则关于x 的不等式()111f x -->的解集是 .解析：()()23f f >01a ⇒<<，所以()1log a f x x -=， ()111f x -->即()log 11a x x ->⇒∈()1,1a -.12. 设a 为非零实数，偶函数()()21f x x a x m x R =+-+∈在区间()2,3上存在唯一的零点，则实数a 的取值范围是 .解析：()()21f x x a x m x R =+-+∈为偶函数，()()0f x f x m =-⇒=，结合图形可知()()230f f a ⋅<⇒∈105,32⎛⎫-- ⎪⎝⎭.13. 设函数()()1f x x Q αα=+∈的定义域为[][],,b a a b --⋃，其中0a b <<.若函数()f x 在区间[],a b 上的最大值为6，最小值为3，则()f x 在区间[],b a --上的最大值与最小值之和为 .解析：令()g x x α=，定义域为[][],,b a a b --⋃，则有()g x 在区间[],a b 上的最大值为5，最小值为2，当()g x 为偶函数时，()g x 在区间[],b a --上的最大值为5，最小值为2，此时()f x 在区间[],b a --上的最大值与最小值之和为9；当()g x 为偶奇函数时，()g x 在区间[],b a --上的最大值为-2，最小值为-5，此时()f x 在区间[],b a --上的最大值与最小值之和为-5；综上，应填5-或914.已知命题“()22f x m x =，()22g x mx m =-，则集合()()1,12x f x g x x ⎧⎫<≤≤=∅⎨⎬⎩⎭”是假命题，则实数m 的取值范围是 .解析：原命题为假命题，即()()f x g x <在1,12⎡⎤⎢⎥⎣⎦上有解.显然0m ≠.当0m >时，结合函数图像可得()()11f g <2210m m m m ⇒<-⇒-<<，无解；当0m <时，结合函数图像可得1122f g ⎛⎫⎛⎫< ⎪ ⎪⎝⎭⎝⎭227044m m m m ⇒<-⇒-<<，所以，70m -<<.二、选择题（本大题满分20分）本大题共有4题，每题有且仅有一个正确答案，请在括号内填上正确的选项，选对得5分，否则一律得0分.15. 下列函数在其定义域内既是奇函数又是增函数的是（）解析：有各函数的基本性质即可知13y x =符合题意，选择D . 16.在钝角ABC ∆中，“sin A =”是“23A π∠=”的（）解析：23A π∠=能得到sin A =，反之不一定成立，A 还可以为3π.17. 已知函数()2a x f x x+=，其中0a >，(]0,x b ∈，则下列判断正确的是（）A.当b 时，()f x 的最小值为2a b b+B.当0b <≤()f x的最小值为C.当0b <≤()f x 的最小值为2a b b+D. 当0b >时，()f x的最小值为解析：()2a x a f x x x x +==+，令ax x x =⇒Nike函数图像，可得到当0b <≤时，()f x 取到最小值()2a b f b b+=，所以选择C.18. 给定方程：1sin 102xx ⎛⎫+-= ⎪⎝⎭，下列命题中：①该方程没有小于0的实数解；②该方程A.lg y x =B.tan y x =C.3x y =D.13y x =A.充分非必要条件B.必要非充分条件C.充要条件D.非充分非必要条件有无数个实数解；③该方程在(),0-∞内有且仅有一个实数解；④若x是该方程的实数解，则01x>-.其中正确的命题个数是（）A.1个B.2个C. 3个D.4个解析：1sin102xx⎛⎫+-=⎪⎝⎭1sin12xx⎛⎫⇒=-+⎪⎝⎭，1sin102xx⎛⎫+-=⎪⎝⎭的解就等价于函数siny x=与112xy⎛⎫=- ⎪⎝⎭的交点个数，作出图像即可判断只有①不对；所以选择C.三、解答题（本大题满分74分）本大题共5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19. （本题满分12分，第1小题6分，第2小题6分）如图，直三棱锥111ABC A B C -中，12AB AC AA ===，45ABC ∠=.⑴求直三棱锥111ABC A B C -的体积；⑵若D 是AC 的中点，求异面直线BD 与1A C 所成的角.解析：∵ 12AB AC ==且45ABC ∠=，∴ AB AC ⊥.⑴1111122242ABC A B C ABC V S AA -∆==⋅⋅⋅=；⑵如图，取1AA 中点E ，连接DE 、BE ，又D 是AC 的中点，所以1DE ∥A C ，所以EDB ∠即为异面直线BD 与1AC 所成的角. 计算可得112DE AC =BE BD ==，在EDB ∆中，由余弦定理可得cosEDB ∠==，即异面直线BD 与1A C 所成的角为.20.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分.已知函数()sin 2sin 22,33f x x x x m x R ππ⎛⎫⎛⎫=++-+-∈ ⎪ ⎪⎝⎭⎝⎭，且()f x 的最大值为1.⑴求m 的值，并求()f x 的单调递增区间；⑵在ABC ∆中，角A B C 、、的对边为a 、b 、c ，若()1f Bb c =+.试判断ABC ∆的形状.解析：⑴∵ ()sin 2sin 22sin 22332sin 23f x x x x m x x mx mπππ⎛⎫⎛⎫=++-+-=+- ⎪ ⎪⎝⎭⎝⎭⎛⎫=+- ⎪⎝⎭∴()max 21f x m =-=即1m =；C1B A 1CC1B 1C令222232k x k πππππ-≤+≤+，得()f x 的单调递增区间为()5,1212k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦；⑵()1sin 236f B B B ππ⎛⎫⇒+= ⎪⎝⎭，∴ 222cos 2a c b B ac +-==b c =+，∴ 21222a cb a bc c c ⇒=⇒=-=⇒=，即sin 111sin sin 2426B C C C π=⇒=<⇒<，故2A π>，所以ABC ∆为钝角三角形.21. （本题满分14分）本题共有2个小题，第1小题满分8分，第2小题满分6分.为保护环境，某单位采用新工艺，把二氧化碳转化为一种可利用的化工产品．已知该单位每月的处理量最多不超过300吨，月处理成本y （元）与月处理量x （吨）之间的函数关系式可近似的表示为：220040000y x x =-+，且每处理一吨二氧化碳得到可利用的化工产品价值为300元．⑴该单位每月处理量为多少吨时，才能使每吨的平均处理成本最低？ ⑵要保证该单位每月不亏损，则每月处理量应控制在什么范围？解析：⑴每吨的平均处理成本为22004000040000200400200200y x x x x x x-+==+-≥-= 当且仅当40000x x=即每月处理量为200吨时每吨的平均处理成本最低，最低为200元；⑵设该单位每月获利为S （元），则单位每月获利为处理二氧化碳得到可利用的化工产品价值减去月处理成本．()2230030020040000500400000S x y x x x x x =-=--+=-+-≥解之得：100400x ≤≤由题意可知0300x <≤，所以当100300x ≤≤时，该单位每月不亏损.小题满分6分.已知函数()()22,1,1x x f x a x -=+⋅∈-，其中常数0a ≠. ⑴1a =时，求()f x 的最小值. ⑵讨论函数的奇偶性.⑶若()()12f x f x +<恒成立，求实数a 的取值范围. 解析：⑴1a =时，()222x x f x -=+≥，当且仅当22x x -=即0x =时()f x 取最小值2. ⑵()22x x f x a --=+⋅，()22x x f x a --=--⋅，所以当1a =时()f x 为偶函数，因为此时有()()f x f x =-恒成立；当1a =-时()f x 为奇函数，因为此时有()()f x f x -=-恒成立. 当1a ≠±时()f x 为非奇非偶函数. ⑶由111121x x -<+<⎧⎨-<<⎩得102x -<<；()()1122122222x x x x f x f x a a +---+<⇒+⋅<+⋅，令2x t ⎫=∈⎪⎪⎝⎭，有234222122t a t t t a t t a t ⎛⎫+<+⇒->- ⎪⎝⎭，即()()332222a a t t t t ->-⇒>，所以2a ≥.小题满分8分.设函数()y f x =为定义在R 上的奇函数，()()2f x f x +=-. 当[]1,0x ∈-时，()()30f x f x x ==.⑴当[]1,3x ∈时，求()1y f x =的解析式；⑵记()y f x =，(]41,41,x k k k Z ∈-+∈为()k y f x =，求()k y f x =及其反函数()1k y f x -=的解析式；⑶定义()()()21,kg x k f x =+-其中[]21,21,1006,x k k k k Z ∈-+≤∈，探究方程()0g x b -=()0b >在区间[]2013,2013-上的解的个数.解析：⑴当[]0,1x ∈时，[]1,0x -∈-，()()3f x f x x =--=，即()()3,11f x x x =-≤≤；当[]1,3x ∈时，[]21,1x -∈-，有()()()322f x f x x =--=--.⑵()()()()()242f x f x f x f x f x +=-⇒+=-+=，则()f x 的周期为4T =；当(]41,41x k k ∈-+时，(]41,1x k -∈-，∴ ()()()(]3441,1k y f x f x k x k ==-=-∈-，4x k -=，即())1411k y f x k x -==+-<≤.⑶由()()()2f x f x f x +=-=-可得()f x 的对称轴为1x =，所以()f x 的图像如下：接下来求解()f x 在[]21,21k k -+上的解析式：①当k 为偶数时，2k 为其周期，[]21,1x k -∈-.所以()()()322f x f x k x k =-=-； ②当k 为奇数时，22k -为其周期，()[]221,3x k --∈. 所以()()()()()()3322222f x f x k x k f x k x k =-=--=--=--综上，()()()312k f x x k =--()()322g x k x k ⇒=+-，[]21,21,x k k k Z ∈-+∈，所以将3y x =向右移动2k 个单位，再向上移动2k 个单位即可得到()g x 的图像：显然，()g x 是连续的递增函数，∴ 当()020132013b g <≤=时，方程()0g x b -=在区间[]2013,2013-上有一解，当2013b >时，方程()0g x b -=在区间[]2013,2013-上无解.。

上海市复旦大学附中2016-2017学年高一下学期期中考试数学试题Word版缺答案

复旦大学附属中学2016学年第二学期高一年级数学期中考试试卷2017.4考试时间100分钟，满分120分一、填空题（每题4分，共48分）1.半径为2，圆心为︒300的圆弧的长为2.函数|tan |x y =的对称轴是3.在平面直角坐标系中，已知角θ的顶点在坐标原点，始边与x 轴正半轴重合，终边在直线上x y 3=，则=θ2sin4.求函数)22sin()(π+-=x x f 的单调递减区间 5.若锐角βα，满足=-=+=ββααcos ,135)cos(,53cos 则 6.已知函数,-91lg(tan )(2x x x f +-=），则)(x f 的定义域是 7.若长度为6,4,422++x x x 的三条线段可以构成一个锐角三角形，则x 取值范围是8.若函数]3,0[)10(sin 2)(πωω在区间<<=x x f 上的最大值是2，则ω＝ 9.如图所示，在塔底Ｂ测得山顶Ｃ的仰角为︒60，在山顶测得塔顶Ａ的仰角为︒45，已知塔高米20AB =，则山高=DC 米10.函数xx x x y cos sin 1cos sin ++=的值域为 11.已知,5)10(sin ),,(4cos sin )(333=︒∈++=f R b a x x b x a x f 且则=︒)100(cos f12.设,cos )(),sin ()(1,x b x g x b a x f b a +=+=的自然数，函数均为大于若存在实数m ，使得),()(m g m f =则=+b a二、选择题（每题４分，共１６分）13.若ＭＰ和ＯＭ分别是角67π的正选线和余弦线，则（） 0MP A <<OM 、Ｂ、ＯＭ＞０＞ＭＰＣ、ＯＭ＜ＭＰ＜０Ｄ、ＭＰ＞０＞ＯＭ14.已知),2,0(,πβα∈则下列不等式一定成立的是（）βαβαsin sin )sin(.A +<+βαβαs i n s i n )s i n (.B +>+ βαβαsin sin )cos(C.+<+ βαβαc o s c o s )c o s (.D +>+15.把函数x y 2sin =的图像沿着轴x 向左平移6π个单位，纵坐标伸长到原来的２倍（横坐标不变）后得到函数)(x f y =的图像，对于函数)(x f y =有以下四个判断：）（1该函数的解析式为)62sin(2π+=x y ；）（2该函数图像关于点）（0,3π对称；）（3该函数在]6,0[π上是增函数；）（4若函数a x f y +=)(在]2,0[π上的最小值为3，则32=a 其中正确的判断有（）个1.A 个2.B 个3.C 个4.D16.定义在区间]3,3[ππ-上的函数图像与的图像的交点个数为（）个12.A 个14.B 个16.C 个18.D三、解答题（本题共５大题，满分５６分） 17.是第四象限角且分）已经（θπθ,257)32cos(10=-，的值。

《解析》上海师大附中2016届高三上学期期中数学试卷(文科)Word版含解析

2015-2016学年上海师大附中高三（上）期中数学试卷（文科）一、填空题（本大题共14小题，每小题4分，共56分）1．不等式logx≥2的解集为__________．2．已知复数z=，则z=__________．3．已知sin（﹣α）=，则cos（π﹣α）=__________．4．若，则行列式=__________．5．函数f（x）=x+的值域__________．6．设g（x）=，则g（g（））=__________．7．若偶函数f（x）在（﹣∞，0]上为增函数，则不等式f（2x+1）＞f（2﹣x）的解集__________．8．函数y=f（x）的图象与y=2x的图象关于y轴对称，若y=f﹣1（x）是y=f（x）的反函数，则y=f﹣1（x2﹣2x）的单调递增区间是__________．9．将函数y=log2x的图象上每一点的纵坐标不变，横坐标变为原来的m（m＞0）倍，得到图象C，若将y=log2x的图象向上平移2个单位，也得到图象C，则m=__________．10．如果，那么函数f（x）=cos2x+sinx的最小值是__________．11．函数y=|x2﹣1|的图象与函数y=x+k的图象交点恰为3个，则实数k=__________．12．数列{a n}满足（n∈N*）．①存在a1可以生成的数列{a n}是常数数列；②“数列{a n}中存在某一项”是“数列{a n}为有穷数列”的充要条件；③若{a n}为单调递增数列，则a1的取值范围是（﹣∞，﹣1）∪（1，2）；④只要，其中k∈N*，则一定存在；其中正确命题的序号为__________．13．已知函数f（x）=，x∈，函数g（x）=ax+2，x∈．若对任意x1∈，总存在x2∈，使f（x1）=g（x2）成立．则实数a的取值范围是__________．14．设表示不超过x的最大整数，如：=1，=﹣2．若集合A={x|x2﹣﹣1=0}，，则A∩B=__________．二、选择题（本大题共4小题，每小题5分，共20分，在每小题给出的四个选项中，只有一项是符合要求的.）15．若a∈R，则“关于x的方程x2+ax+1=0无实根”是“z=（2a﹣1）+（a﹣1）i（其中i表示虚数单位）在复平面上对应的点位于第四象限”的( )A．充分非必要条件B．必要非充分条件C．充要条件 D．既非充分又非必要条件16．如果一个函数f（x）满足：（1）定义域为R；（2）任意x1，x2∈R，若x1+x2=0，则f（x1）+f（x2）=0；（3）任意x∈R，若t＞0，f（x+t）＞f（x）．则f（x）可以是( )A．y=﹣x B．y=3x C．y=x3D．y=log3x17．已知函数f（x）满足f（1）=，4f（x）f（y）=f（x+y）+f（x﹣y）（x，y∈R），则f=( ) A． B． C．﹣D．018．若数列{a n}满足：对任意的n∈N*，只有有限个正整数m使得a m＜n成立，记这样的m的个数为（a n）*，则得到一个新数列{（a n）*}．例如，若数列{a n}是1，2，3，…n，…，则数列{（a n）*}是0，1，2，…，n﹣1，…已知对任意的n∈N*，a n=n2，则（（a4）*）*=( ) A．8 B．20 C．32 D．16三、解答题（12+14+14+16+18=74分）19．在△ABC中，角A、B、C的所对边的长分别为a、b、c，且a=，b=3，sinC=2sinA．（Ⅰ）求c的值；（Ⅱ）求的值．20．（14分）设P表示幂函数在（0，+∞）上是增函数的c的集合；Q表示不等式|x﹣1|+|x ﹣4|≥c对任意x∈R恒成立的c的集合．（1）求P∪Q；（2）试写出一个解集为P∪Q的不等式．21．（14分）已知复数z0满足|2z0+15|=|+10|，（1）求证：|z0|为定值；（2）设x=，z n=z0x n，若a n=|z n﹣z n﹣1|，n∈N*，求（a1+a2+…+a n）．22．（16分）已知函数f（x）=log2（2x+1）．（1）求证：函数f（x）在（﹣∞，+∞）内单调递增；（2）记f﹣1（x）为函数f（x）的反函数．若关于x的方程f﹣1（x）=m+f（x）在上有解，求m的取值范围；（3）若f（x+t）＞2x对于x∈恒成立，求t的取值范围．23．（18分）已知递增的等差数列{a n}的首项a1=1，且a1、a2、a4成等比数列．（1）求数列{a n}的通项公式a n；（2）设数列{c n}对任意n∈N*，都有++…+=a n+1成立，求c1+c2+…+c2014的值（3）若b n=（n∈N*），求证：数列{b n}中的任意一项总可以表示成其他两项之积．2015-2016学年上海师大附中高三（上）期中数学试卷（文科）参考答案一、填空题（本大题共14小题，每小题4分，共56分）1．（0，]；2． -1-i；3．-；4．；5．（-∞，-3]∪∪[2，+∞）；14．；二、选择题（本大题共4小题，每小题5分，共20分，在每小题给出的四个选项中，只有一项是符合要求的.）15．B； 16．C； 17．B； 18．D；三、解答题（12+14+14+16+18=74分）19．；20．；21．；22．；23．；。

【全国百强校】上海市复旦大学附属中学2016届高三上学期期中考试英语试题

复旦大学附属中学2015学年第一学期高三年级英语期中考试试卷第I卷（共103分）I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversation and the question will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. This afternoon. B. Tomorrow. C. Next week. D. Next month.2. A. She doesn’t play tennis well. B. She likes other sports as well.C. She is an enthusiastic tennis player.D. She is a professional athlete.3. A. At a paint store. B. At an oil market.C. At a science museum.D. At a gallery.4. A. Work in the yard. B. Buy some wood.C. Go to the bookstore.D. Take a walk.5. A. A taxi driver. B. A passenger. C. A car cleaner. D. A mechanic.6. A. Call a repairman. B. Get out the paper stuck.C. Turn to her colleague for help.D. Restart the machine.7. A. There are not enough gardens. B. Parking areas are full before 10:00.C. Parking areas are closed after 10:00.D. All classes begin at 10:00.8. A. The presentation will begin at noon.B. She’ll present her work to the man.C. She’d like to invite the man for lunch.D. She suggests working on the presentation at 12:00.9. A. The dormitory hours. B. The problem with the rules.C. The door number of the dormitory.D. The time to open the dormitory.10. A. The chairs didn’t need to be painted.B. He doesn’t like the color of the chairs.C. The park could have avoided the problem.D. The woman should have been more careful.Section BDirections: In section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Worried B. Surprised. C. Satisfied. D. Uninterested.12. A. It spoiled Juana’s reputation. B. It copied her ideas without permission.C. It bought Juana’s dishwashers.D. It wanted to share the dishwasher market.13. A. A successful business case. B. Juana’s waterless laundry.C. A case against a global company.D. The worldwide dishwasher market.Questions 14 through 16 are based on the following passage.14. A. Footprints. B. Food. C. Living insects. D. Orange seeds.15. A. Don’t touch animals under any circumstances.B. Don’t take away any natural objects from the park.C. Don’t leave litter in the park or throw any off the boat.D. Don’t transport animals from one island to another.16. A. To protect the guide’s interest. B. To improve the unique environment.C. To ensure a trouble-free visit.D. To get rid of illegal behaviors.Section CDirections: In section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. (本题做在答题纸上)Write onlyII. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages 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)Libraries are my world. I've been a patron all my life, and for the past nine years I (25) ______ (work) at multiple libraries and archives in and around Detroit. The library (26) ______ an institution has many roles, but as our country struggles through an economic crisis, I have watched the library where I work evolve into a career and business center, a community gathering place and a bastion for hope.In the spring of 2007 I got a library internship at the Southfield Public Library, just north of Detroit. Summers at SPL (27) ______ (be) usually slow, but that year, we experienced a library that hustled and bustled like science-fair project week, midterms or tax season. Yet patrons weren't looking for Mosby's Nursing Drug Reference or 1040 forms. They were coming for information on entrepreneurship and growing their small business.I interpreted people's interest in our business collection as the first step to (28) ______ (pursue) their dreams, but these patrons were not motivated by dreams. They were responding to reality, and they were looking for Plan B.Things worsened in 2008, and in 2009 the economic crisis continues to suffocate Michigan. Last year, we put up a display with a variety of job resources that we restocked every hour. Each night the library closed, the display was bare. (29) ______ we normally keep displays up for a week, we kept the job resources display up for months.Then there's the tightening credit market. People see the writing on the wall and they want to get educated. They can't afford a financial adviser, but checking books out is free. Some of (30) ______(popular) titles now are "Rich Dad, Poor Dad," "Think and Grow Rich," and "Suze Orman's 2009 Action Plan."The economic downturn affects us all. I have had to work long hours and don't get to see (31) ______ of my boyfriend or experience any kind of social life lately, but I am thankful to be in a position where I can help people overcome this struggle. In Michigan, we haven't lost hope. (32) ______ ______ ______ there are libraries here, there will always be hope.(B)It’s estimated that 300 million people in China are studying, or (33) ______ (study), English. That’s an impressive number and I can’t think of any other country in the world where one quarter of the population is so dedicated to (34) ______ (learn) a second language. But some people are questioning whether this “craze” for studying English is worthwhile.Professor Zhang Shuhua of the Chinese Academy of Social Sciences says that too much emphasis is placed on learning English and that it is a waste of education resources as well as a threat to the study of Chinese. He says that having English as a compulsory course in university “has distracted much of students’attention (35) ______ specialized subjects,”and that some students have been denied access to postgraduate education because they failed English. Others have admitted that studying so much English has made them (36) ______ (poor) Chinese speakers.Both of these criticisms are legitimate, but they beg the question of why so many Chinese still want to learn English. English, (37) ______ recognize, is the lingua franca of the modern world. It is the language of business and has become the language of international relations and culture. When people from different countries get together, they frequently speak in English rather than try to translate their native languages. It seems that everyone everywhere can speak at least some English.For China to be part of that international conversation, it is necessary that some level ofEnglish proficiency (38) ______ be achieved. But what, you may ask, about those who will never speak a word of English once they leave school? Well, for good or ill, they will still be surrounded by English. It is there in signs, in music, in movies and in the casual conversations they overhear of the increasing number of foreigners on the city streets. To know English is to be included in the rest of the world, (39) ______ ______ your world is limited to China.I agree with Professor Zhang on one point, (40) ______. English should not be a compulsory subject in university. For most, passing the CET is just the endless drudgery of memorizing word lists. There is little emphasis placed on communication. And if you can’t communicate in English after years of study in primary school, middle school and high school, a few more years in university probably won’t help.Section BDirections: 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.For writers in Western cultures, autumn is a difficult season to describe. On the one hand, it is the end of the summer, and therefore a little sad. The nights draw in, and when you wake in the morning, there’s mist and it’s cooler: Winter is around the ___41___. American writer Ernest Hemingway wrote in his book A Moveable Feast: “You expected to be sad in the fall. Part of you died each year when the leaves fell from the trees and their branches were ___42___against the wind and th e cold, wintry light.”On the other hand, autumn has its good side. There are so many changes in nature at this time of year, such as the reds and browns that the leaves change to, and the ___43___ they fall from the trees. French writer Albert Camus even though autumn was a second spring: “Autumn is a second spring when every leaf is a flower.” It’s a view you can also find in the most famous autumn poem in English literature, To Autumn by John Keats. In that poem Keats says that the autumn has its own songs, just like spring.Another autumn theme is wisdom. The arrival of the season is thought to be similar to a person becoming ___44___. Their summer peak may have been and gone, but old age has not yet come. At this time it’s thought that people have ___45___ a thing or two about life. The great Irish poet W.B. Yeats takes up this theme in his poem The Wild Swans at Coole. Yeats puts together a picture for the reader out of the ___46___ of the changing seasons in Coole Park in the west of Ireland, a place he knew well. Seeing and counting 59 swans, he remembers first making the count 19 years ago. He ___47___ whether he can still love like the lover swans do.Of course, many other themes and subject matters can play a part in the literature of autumn. For ex ample, it’s the beginning of a new term of the school year. As you would expect, autumncan___48___ in writing for children and young people. But autumn writing usually ___49___ on the changes in nature that we see, which writers often use as a ___50___ for changes in human life.III. Reading ComprehensionSection ADirections: 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.After the college-board examinations in June, Basil Duke Lee and five other boys from St. Regis School ___51___ the train for the West. Two got out at Pittsburgh, one slanted south toward St. Louis and two stayed in Chicago; from then on Basil was alone. It was the first time in his life that he had ever felt the need of tranquility, but now he took long breaths of it; for, though things had gone better toward the end, he had had a / an ___52___ year at school.He wore one of those extremely flat derbies (常礼帽) in vogue during the twelfth year of the century, and a blue business suit became a little too short for his constantly ___53___ body. Within he was by turns a disembodied (空洞的) spirit, almost ___54___ of his person and moving in a mist of impressions and emotions, and a fiercely competitive individual trying ___55___ to control the rush of events that were the steps in his own ___56___ from child to man. He believed that everything was a matter of ___57___ — the current principle of American education — and his fantastic ___58___ was continually leading him to expect too much. He wanted to be a great athlete, popular, brilliant and always happy. During this year at school, where he had been punished for his “freshness,” for fifteen years of thorough spoiling at home, he had grown uselessly introspective, and this ___59___ with that observation of others which is the beginning of wisdom. It was apparent that before he obtained much success in dealing with the world he would know that he’d been in a fight.Fifteen is of all ages the most difficult to ___60___———to put one’s fingers on and say, “That’s the way I was.” And all one can know is that somewhere between thirteen, boyhood’s___61___, and seventeen, when one is a sort of counterfeit young man, there is a time when youth ___62___ hourly between one world and another —— pushed ceaselessly forward into unprecedented experiences and ___63___ trying to struggle back to the days when nothing had to be ___64___ for. Fortunately none of our contemporaries remember much more than we do of how we behaved in those days; nevertheless the ___65___ is about to be drawn aside for an inspection of Basil’s madness that summer.51. A. boarded B. missed C. jumped D.followed52. A. happy B. unhappy C.memorableD.favourable53. A. swelling B. bending C. lengthening D.strengthening54. A. aware B.fond C. critical D.unconscious55. A. randomly B. desperately C. particularly D.indifferently56. A. evolution B. revolution C. solutionD.introduction57. A. fact B. opinion C. course D. effort58. A. fashion B. ambition C. character D.treasure59. A. contacted B. associated C. interfered D.smashed60. A. digest B.describe C. deal D. locate61. A. majority B. minority C. senior D. junior62. A. floats B. varies C. fluctuates D. ranges63. A. successfully B. vainly C. wildly D. gently64. A. hunted B. provided C. compensated D. paid65. A.curtain B. adolescence C. portrait D. ceilingSection 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)There are people in Italy who can’t stand soccer. Not all Canadians love hockey. A similar situation exists in America, where there are those individuals you may be one of them who yawn or even frown when somebody mentions baseball. Baseball to them means boring hours watching grown men in funny tight outfits standing around in a field staring away while very little of anythin g happens. They tell you it’s a game better suited to the 19th century, slow, quiet, gentlemanly. These are the same people you may be one of them who love football because there’s the sport that glorifies “the hit”.By contrast, baseball seems abstract, cool, silent, still.On TV the game is fractured into a dozen perspectives, replays, close ups. The geometry of the game, however, is essential to understanding it. You will contemplate the game from one point as a painter does his subject; you may, of course, project yourself into the game. It is in this projection that the game affords so much space and time for involvement. The TV won’t do it for you.Take, for example, the third baseman. You sit behind the third base dugout and you watch him watching home plate. His legs are apart, knees flexed. His arms hang loose. He does a lot of this. The skeptic still cannot think of any other sports so still, so passive. But watch what happens every time the pitcher throws: the third baseman goes up on his toes, flexes his arms or brings the glove to a point in front of him, takes a step right or left, backward or forward, perhaps he glances across the field to check his first baseman’s position. Suppose the pitch is a ball. “Nothing happened,” you say. “I could have had my eyes closed.”The skeptic and the innocent must play the game. And this involvement in the stands is no more intellectual than listening to music is. Watch the third baseman. Smooth the dirt in front of you with one foot; smooth the pocket in your glove; watch the eyes of the batter, the speed of the bat, the sound of horsehide on wood. If football is a symphony of movement and theatre, baseball is chamber music, a spacious interlocking of notes, chores and responses.66. Those who don’t like baseball may complain that ______.A. it is only to the taste of the oldB. it involves fewer players than footballC. it is not exciting enoughD. it is pretentious and looks funny67. The author admits that ______.A. baseball is too peaceful for the youngB. baseball may seem boring when watched on TVC. football is more attracting than baseballD. baseball is more interesting than football68. By stating ‘I could have had my eyes closed.’ the author means (4th paragraph last sentence):A. The third baseman would rather sleep than play the game.B. Even if the third baseman closed his eyes a moment ago, it could make no difference to theresult.C. The third baseman is so good at baseball that he could finish the game with eyes closed allthe time and do his work well.D. The consequence was too bad he could not bear to see it.69. We can safely conclude that the author ______.A. likes footballB. hates footballC. hates baseballD. likes baseball(B)Some of the world’s most significant problems never hit headlines. One example comes from agriculture. Food riots and hunger make news. But the trend lying behind these matters is rarely talked about. This is the decline in the growth in yields of some of the world’s major crops. A new study by the University of Minnesota and McGill University in Montreal looks at where, and how far, this decline is occurring.The authors take a vast number of data points for the four most important crops: rice, wheat corn and soybeans. They find that on between 24% and 39% of all harvested areas, the improvement in yields that took place before the 1980s slowed down in the 1990s and 2000s.There are two worrying features of the slowdown. One is that it has been particularly sharp in the world’s most populo us countries, India and China. Their ability to feed themselves has been an important source of relative stability both within the countries and on world food markets. That self-sufficiency cannot be taken for granted if yields continue to slow down or reverse.Second, yield growth has been lower in wheat and rice than in corn and soya beans. This is problematic because wheat and rice are more important as foods, accounting for around half of all calories consumed. Corn and soybeans are more important as feed grains. The authors note that “we have preferentially focused our crop improvement efforts on feeding animals and cars rather than on crops that feed people and are the basis of food security in much of the world.”The report qualifies the more optimistic findings of another new paper which suggests that the world will not have to dig up a lot more land for farming in order to feed 9 billion people in 2050, as the Food and Agriculture Organization has argued.Instead, it says, thanks to slowing population growth, land currently ploughed up for crops might be able to revert to forest or wilderness. This could happen. The trouble is that the forecast assumes continued improvements in yields, which may not actually happen.70. What does the author try to draw attention to?A. Food riots and hunger in the world.B. News headlines in the leading media.C. The decline of the grain yield growth.D. The food supply in populous countries.71. Why does the author mention India and China in particular?A. Their self-sufficiency is vital to the stability of world food markets.B. Their food yields have begun to decrease sharply in recent years.C. Their big populations are causing worldwide concerns.D. Their food self-sufficiency has been taken for granted.72. What does the new study by the two universities say about recent crop improvement efforts?A. They fail to produce the same remarkable results as before the 1980s.B. They contribute a lot to the improvement of human food production.C. They play a major role in guaranteeing the food security of the world.D. They focus more on the increase of animal feed than human food grains.73. What does the Food and Agriculture Organization say about world food production in the coming decades?A. The growing population will greatly increase the pressure on world food supplies.B. The optimistic prediction about food production should be viewed with caution.C. The slowdown of the growth in yields of major food crops will be reversed.D. The world will be able to feed its population without increasing farmland.(C)Among the more colorful characters of Leadville’s golden age were H.A.W. Tabor and his second wife, Elizabeth McCourt, better known as “Baby Doe”. Their history is fast becoming one of the legends of the Old West. Horace Austin Warner Tabor was a school teacher in Vermont. With his first wife and two children he left Vermont by covered wagon in 1855 to homestead in Kansas. Perhaps he did not find farming to his liking, or perhaps he was tempted by rumors of fortunes to be made in Colorado mines. At any rate, a few years later he moved west to the small Colorado mining camp known as California Gulch, which he later renamed Leadville when he became its leading citizen. “Great deposits of lead are sure to be found here.” he said.As it turned out, it was silver, not lead, that was to make Leadville’s fortune and wealth. Tabor knew little about mining himself, so he opened a general store, which sold everything from boots to salt, flour, and tobacco. It was his c ustom to “grubstake” prospective miners, in other words, to supply them with food and supplies, or “grub”, while they looked for ore, in return for which he would get a share in the mine if one was discovered. He did this for a number of years, but no one that he aided ever found anything of value.Finally one day in the year 1878, so the story goes, two miners came in and asked for “grub”. Tabor had decided to quit supplying it because he had lost too much money that way. These werepersistent, however, and Tabor was too busy to argue with them. “Oh help yourself. One more time won’t make any difference,” He said and went on selling shoes and hats to other customers. The two miners took $17 worth of supplies, in return for which they gave Tabor a one-third interest in their findings. They picked a barren place on the mountain side and began to dig. After nine days they struck a rich vein of silver. Tabor bought the shares of the other two men, and so the mine belonged to him alone. This mine, known as the “Pittsburgh Mine,” made 1 300 000 for Tabor in return for his $17 investment.Later Tabor bought the Matchless Mine on another barren hillside just outside the town for $117 000. This turned out to be even more fabulous than the Pittsburgh, yielding $35 000 worth of silver per day at one time. Leadville grew. Tabor became its first mayor, and later became lieutenant governor of the state.74. Leadville got its name for the following reasons EXCEPT ______.A. because Tabor became its leading citizenB. because great deposits of lead is expected to be found thereC. because it could bring good fortune to TaborD. because Tabor renamed it so75. The word “grubstake” in paragraph 2 means ______.A. to supply miners with food and suppliesB. to open a general storeC. to do one’s contribution to the development of the mineD. to supply miners with food and supplies and in return get a share in the mine, if one wasdiscovered76. Tabor made his first fortune ______.A. by supplying two prospective miners and getting in return a one-third interest in the findingsB. because he was persuaded by the two miners to quit supplyingC. by buying the shares of the otherD. as a land speculator77. The underlying reason for Tabor’s life career is ______.A. purely accidentalB.based on the analysis of miner’s being very poor and their possibility of discoveringprofitable mining siteC. through the help from his second wifeD. he planned well and accomplished targets step by stepSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.When the Internet powerhouse Yahoo wanted to teach ethics to its employees, it faced a challenge familiar to multinational companies.Yahoo employs nearly 14,000 people at 25 sites worldwide. They would feel bored at sitting down in front of a dated video in which actors with 1980s haircuts tell them what to do. So it hired a company called The Network to design a game. In the game, the truck where Yahoo wasfounded traveled the world, turning into a boat and a helicopter along the way as it visited some of Yahoo's foreign offices. Participants play in game show-like scenarios that quiz them about conflicts of interest and doing business fairly. And employees note: Yahoo is tracking how well they do.Such activities draw more enthusiastic participation and teach more effectively than traditional methods. They are described as alternative-reality games (ARGs), involving both interactive and real-world elements. Besides teaching employees, ARGs have also been used in many areas for a number of different purposes.From a marketing perspective, a number of very successful ARGs have been written as a way to build product awareness. A very popular ARG called I Love Bees was produced to market the 2004 video game Halo 2. At its height, I Love Bees received between two to three million unique visitors over the course of three months.ARGs are more than just a fun way to learn. They have also been used to solve real world problems. An ARG called World Without Oil was created to obtain collective input from players about dealing with the world's dependency on oil. World without Oil simulates the first 32 days of a global oil crisis and anybody could play by creating a personal story that recorded the imagined reality of their life in the crisis. World Without Oil's success on a small budget has opened the door for similar games to engage mainstream Internet users with climate change, education reform, governmental policy and other timely, vital issues.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)78.What challenge did yahoo face in teaching ethics to its employees?79.In the game designed for yahoo, participants had to answer questions about ________.80.What are the three major functions of ARG mentioned in the passage?81.The success of World Without Oil suggests that ARGs can ________.第II卷（共47分）I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.没有人不希望和平。

2016-2017学年上海市复旦大学附属中学高一上学期期中考试数学试卷含详解

复旦附中高一期中数学卷一.填空题1.集合{1,2,3,,2015,2016}⋅⋅⋅的子集个数为________2.已知全集U =R ，集合{|1}A x x =≤，集合{|2}B x x =≥，则()U C A B =________3.已知集合{|12}A x x =≤≤，集合{|}B x x a =≤，若A B ⋂≠∅，则实数a 的取值范围是________4.如果全集{,,,,,}U a b c d e f =，{,,,}A a b c d =，{}A B a = ，(){}U C A B f = ，则B =________5.已知210a a >>，210b b >>，且12121a a b b +=+=，记1122A a b a b =+，1221B a b a b =+，12C =，则、、A B C按从小到大的顺序排列是________.6.已知Rt ABC ∆的周长为定值2，则它的面积最大值为__________.7.我们将b a -称为集合{|}M x a x b =≤≤的“长度”，若集合2{|}3M x m x m =≤≤+，{|0.5}N x n x n =-≤≤，且集合M 和集合N 都是集合{|01}x x ≤≤的子集，则集合M N ⋂的“长度”的最小值是________8.已知{}A x x =>，{|(3)(3)0}B x x x x =-+>，则A B = ________9.对任意两个集合X 与Y ，定义①{X Y x x X-=∈且}x Y ∉，②()()X Y X Y Y X∆=-- ，已知{}2,A y y x x R==∈，{}22B y y =-≤≤，则A B ∆=_________.10.已知常数a 是正整数，集合1{|||,}2A x x a a x Z =-<+∈，{|||2,}B x x a x Z =<∈，则集合A B ⋃中所有元素之和为________11.非空集合G 关于运算*满足：①对任意,a b G ∈，都有a b G *∈；②存在e G ∈使对一切a G ∈都有a e e a a *=*=，则称G 是关于运算*的融洽集，现有下列集合及运算：①G 是非负整数集，*运算：实数的加法；②G 是偶数集，*运算：实数的乘法；③G 是所有二次三项式组成的集合，*运算：多项式的乘法；④{|,}G x x a a b Q ==+∈，*运算：实数的乘法；其中为融洽集的是________12.集合(){},,R A x y y a x x ==∈，(){},,R B x y y x a x ==+∈，已知集合A B ⋂中有且仅有一个元素，则常数a 的取值范围是______________．二.选择题13.已知集合{1,2,3,,2015,2016}A =⋅⋅⋅，集合{|31,}B x x k k Z ==+∈，则A B ⋂中的最大元素是（）A.2014B.2015C.2016D.以上答案都不对14.已知全集U =A B ⋃中有m 个元素，()()U U A B ⋃痧中有n 个元素．若A B ⋂非空，则A B ⋂的元素个数为A.mnB.m n+ C.n m- D.m n-15.命题“已知,x y R ∈，若220x y +=，则0x =且0y =”的逆否命题是（）A.已知,x y R ∈，若220x y +≠，则0x ≠且0y ≠B.已知,x y R ∈，若220x y +≠，则0x ≠或0y ≠C.已知,x y R ∈，若0x ≠且0y ≠，则220x y +≠D.已知,x y R ∈，若0x ≠或0y ≠，则22x y +≠16.对任意实数,,a b c ，给出下列命题：①“a b =”是“ac bc =”的充要条件；②“5a +是无理数”是“a 是无理数”的充要条件；③“a b >”是“22a b >”的充分条件；④“4a <”是“3a <”的必要条件；其中真命题的个数是（）A.1个B.2个C.3个D.4个三.解答题17.已知集合{1,2,3}A =，2{|(1)0,}B x x a x a x R =-++=∈，若A B A ⋃=，求实数a ；18.已知,,a b c R +∈，求证：3332222222()a b c ab a b bc b c ac a c ++≥+++++；19.设正有理数1a21211a a =++，求证：（11a 与2a 之间；（2）2a 比1a20.已知对任意实数x ，不等式2(3)10mx m x --+>成立或不等式0mx >成立，求实数m 的取值范围；21.已知关于x 的不等式2(4129)(211)0kx k k x ---->，其中R k ∈；（1）试求不等式的解集A ；（2）对于不等式的解集A ，记B A Z = （其中Z 为整数集），若集合B 为有限集，求实数k 的取值范围，使得集合B中元素个数最少，并用列举法表示集合B；复旦附中高一期中数学卷一.填空题1.集合{1,2,3,,2015,2016}⋅⋅⋅的子集个数为________【答案】20162【分析】若集合中有n 个元素,则该集合有2n 个子集,显然,集合中的元素有2016个,即2016n =,代入2n 中即可【详解】由题,集合中有2016个元素,所以该集合有20162个子集,故答案为:20162【点睛】本题考查集合的子集个数,属于基础题2.已知全集U =R ，集合{|1}A x x =≤，集合{|2}B x x =≥，则()U C A B = ________【答案】{|12}x x <<【分析】先求的A B ⋃,再求得补集即可【详解】由题,{|1A B x x ⋃=≤或}2x ≥,所以(){}U |12A B x x ⋃=<<ð,故答案为:{|12}x x <<【点睛】本题考查集合的并集、补集运算,属于基础题3.已知集合{|12}A x x =≤≤，集合{|}B x x a =≤，若A B ⋂≠∅，则实数a 的取值范围是________【答案】1a ≥【分析】由A B ⋂≠∅,画出数轴,表示出集合,即可求解【详解】因为A B ⋂≠∅,则画出数轴,并表示出集合,如下:可得1a ≥,故答案为:1a ≥【点睛】本题考查已知交集结果求参数范围,属于基础题4.如果全集{,,,,,}U a b c d e f =，{,,,}A a b c d =，{}A B a = ，(){}U C A B f = ，则B =________【答案】{,}a e 【分析】由题,用维恩图来表示集合,由图即可得到B 集合【详解】由题,将集合用维恩图表示,则{},B a e =,故答案为:{,}a e 【点睛】本题考查图示法处理集合问题,属于基础题5.已知210a a >>，210b b >>，且12121a a b b +=+=，记1122A a b a b =+，1221B a b a b =+，12C =，则、、A B C 按从小到大的顺序排列是________.【答案】B ＜C ＜A【分析】根据题设，取符合题设的特殊值即可快速判断，或者采用排序原理也可判断.【详解】方法一：212112120,0,1a a b b a a b b >>>>+=+= ，不妨令12121212,,,3333a ab b ====，11221221145224,999999A a b a bB a b a b =+=+==+=+=，1 4.529C == ，B C A \<<，故答案为：B ＜C ＜A .方法二：∵210a a >>，210b b >>，∴由排序原理可知：22112112a b a b a b a b +>+，∵12121,1a a b b +=+=，()()1212111221221a a b b a b a b a b a b ∴=++=+++()()()2211211222112a b a b a b a b a b a b =+++<+221112a b a b ∴+>，∴A ＞C ＞B ﹒故答案为：B ＜C ＜A .6.已知Rt ABC ∆的周长为定值2，则它的面积最大值为__________.【答案】3-.【分析】设出三角形的边长，根据周长和勾股定理列方程组，利用基本不等式求得ab 的最大值，进而求得三角形面积的最大值.【详解】设Rt ABC ∆三条边长分别为,,a b c ，其中c 为斜边长，所以2222a b c c a b++=⎧⎨=+⎩，2a b +=，2≥，2≤=-，所以6ab ≤-则三角形的面积132ABC S ab ∆=≤-.故答案为3-.【点睛】本小题主要考查利用基本不等式求三角形面积的最大值，考查直角三角形的性质，考查化归与转化的数学思想方法，属于中档题.7.我们将b a -称为集合{|}M x a x b =≤≤的“长度”，若集合2{|}3M x m x m =≤≤+，{|0.5}N x n x n =-≤≤，且集合M 和集合N 都是集合{|01}x x ≤≤的子集，则集合M N ⋂的“长度”的最小值是________【答案】16【分析】当集合M N ⋂的“长度”的最小值时,M 与N 应分别在区间[]0,1的左右两端,由此能求出M N ⋂的“长度”的最小值【详解】由题,M 的“长度”为23,N 的“长度”为12,当集合M N ⋂的“长度”的最小值时,M 与N 应分别在区间[]0,1的左右两端,故M N ⋂的“长度”的最小值是2111326+-=,故答案为:16【点睛】本题考查交集的“长度”的最小值的求法,考查新定义的合理运用8.已知{}A x x =>，{|(3)(3)0}B x x x x =-+>，则A B = ________【答案】{|30}-<<x x【分析】先分别求解集合中元素的所满足的不等式,再由交集的定义求解即可【详解】由题,因为20xx >-≥⎪⎩,解得1x <,则{}|1A x x =<,因为()()330x x x -+>,解得30x -<<或3x >,则{|30B x x =-<<或}3x >,所以{}|30A B x x ⋂=-<<,故答案为:{|30}-<<x x 【点睛】本题考查集合的交集运算,考查含根式的不等式的运算,考查解高次不等式9.对任意两个集合X 与Y ，定义①{X Y x x X -=∈且}x Y ∉，②()()X Y X Y Y X ∆=-- ，已知{}2,A y y x x R ==∈，{}22B y y =-≤≤，则A B ∆=_________.【答案】[)()2,02-+∞ ，【分析】由A ＝{y |y ＝x 2，x ∈R }＝{y |y ≥0}，B ＝{y |﹣2≤y ≤2}，先求出A ﹣B ＝{y |y ＞2}，B ﹣A ＝{y |﹣2≤y ＜0}，再求A △B 的值．【详解】∵A ＝{y |y ＝x 2，x ∈R }＝{y |y ≥0}，B ＝{y |﹣2≤y ≤2}，∴A ﹣B ＝{y |y ＞2}，B ﹣A ＝{y |﹣2≤y ＜0}，∴A △B ＝{y |y ＞2}∪{y |﹣2≤y ＜0}，故答案为[﹣2，0）∪（2，+∞）．【点睛】本题考查集合的交、并、补集的运算，解题时要认真审题，仔细解答，注意正确理解X ﹣Y ＝{x |x ∈X 且x ∉Y }、X △Y ＝（X ﹣Y ）∪（Y ﹣X ）．10.已知常数a 是正整数，集合1{|||,}2A x x a a x Z =-<+∈，{|||2,}B x x a x Z =<∈，则集合A B ⋃中所有元素之和为________【答案】2a【分析】分别求出集合A 、B 中的元素,再求出集合A 、B 的并集,即可求解【详解】由题,因为12x a a -<+,所以11222x a -<<+,则11|2,22A x x a x Z ⎧⎫=-<<+∈⎨⎬⎩⎭；因为2x a <,所以22a x a -<<,则{}|22,B x a x a x Z =-<<∈,因为常数a 是正整数,所以{}0,,,,2A a a = ,{}21,,0,,21B a a =-+- ,所以{}21,,0,,21,2A B a a a ⋃=-+- ,所以A B ⋃中所有元素之和是2a ,故答案为:2a【点睛】本题考查集合的并集,考查解含绝对值的不等式11.非空集合G 关于运算*满足：①对任意,a b G ∈，都有a b G *∈；②存在e G ∈使对一切a G ∈都有a e e a a *=*=，则称G 是关于运算*的融洽集，现有下列集合及运算：①G 是非负整数集，*运算：实数的加法；②G 是偶数集，*运算：实数的乘法；③G 是所有二次三项式组成的集合，*运算：多项式的乘法；④{|,}G x x a a b Q ==+∈，*运算：实数的乘法；其中为融洽集的是________【答案】①④【分析】逐一验证几个选项是否分别满足“融洽集”的两个条件,若两个条件都满足,是“融洽集”,有一个不满足,则不是“融洽集”【详解】①对于任意非负整数,a b ,则a b +仍为非负整数,即a b G +∈；取0e =,则00a a a +=+=,故①符合题意；②对于任意偶数,a b ,则ab 仍为偶数,即ab G ∈；但是不存在e G ∈,使对一切a G ∈都有ae ea a ==,故②不符合题意；③对于G 是所有二次三项式组成的集合,若,a b G ∈,ab 不再是二次三项式,故③不符合题意；④对于{|,}G x x a a b Q ==+∈,设1x a =+2x c =+,则()(122x x ac bd ad bc ⋅=+++,即12x x G ⋅∈；取1e =,则11a a a ⨯=⨯=,故④符合题意,故答案为:①④【点睛】本题考查对新定义“融洽集”的理解,考查理解分析能力12.集合(){},,R A x y y a x x ==∈，(){},,R B x y y x a x ==+∈，已知集合A B ⋂中有且仅有一个元素，则常数a 的取值范围是______________．【答案】[]1,1-【分析】将A B ⋂中有且仅有一个元素，转化为方程只有一个解，分情况讨论，确定参数范围.【详解】由集合(){},,R A x y y a x x ==∈，(){},,R B x y y x a x ==+∈，且A B ⋂中有且仅有一个元素，a x x a ∴=+只有1个解，若0x ≥，则ax x a =+，1a x a =-，若0x <，则ax x a -=+，1ax a =-+，所以0101a a a a ⎧≥⎪⎪-⎨⎪-≥⎪+⎩或0101aa a a ⎧≤⎪⎪-⎨⎪-≤⎪+⎩或101a a a =⎧⎪⎨-<⎪+⎩或011a a a ⎧≥⎪-⎨⎪=-⎩，解得11a -≤≤，故答案为：[]1,1-.二.选择题13.已知集合{1,2,3,,2015,2016}A =⋅⋅⋅，集合{|31,}B x x k k Z ==+∈，则A B ⋂中的最大元素是（）A.2014B.2015C.2016D.以上答案都不对【答案】A【分析】由题意可知集合B 表示整数的3倍且大1的数的集合,则找到集合A 中符合条件的最大元素即可【详解】由题,因为{|31,}B x x k k Z ==+∈,即为整数的3倍且大1的数的集合,则A B ⋂中的最大元素为2014,故选:A【点睛】本题考查集合的交集定义,属于基础题14.已知全集U =A B ⋃中有m 个元素，()()U U A B ⋃痧中有n 个元素．若A B ⋂非空，则A B ⋂的元素个数为A.mnB.m n+ C.n m- D.m n-【答案】D【详解】因为()()()U UUB A B A ⋃=⋂痧所以()()U UU A B A B ⋂=⋃⎡⎤⎣⎦痧，所以A B ⋂共有m n -个元素，故选D ．15.命题“已知,x y R ∈，若220x y +=，则0x =且0y =”的逆否命题是（）A.已知,x y R ∈，若220x y +≠，则0x ≠且0y ≠B.已知,x y R ∈，若220x y +≠，则0x ≠或0y ≠C.已知,x y R ∈，若0x ≠且0y ≠，则220x y +≠D.已知,x y R ∈，若0x ≠或0y ≠，则22x y +≠【答案】D【分析】直接利用逆否命题的定义得到答案.【详解】己知,x y R ∈，若220x y +=，则0x =且0y =”的逆否命题是：己知,x y R ∈，若0x ≠或0y ≠，则220x y +≠故选D【点睛】本题考查了命题的逆否命题，意在考查学生对于命题基础知识的掌握情况.16.对任意实数,,a b c ，给出下列命题：①“a b =”是“ac bc =”的充要条件；②“5a +是无理数”是“a 是无理数”的充要条件；③“a b >”是“22a b >”的充分条件；④“4a <”是“3a <”的必要条件；其中真命题的个数是（）A.1个B.2个C.3个D.4个【答案】B【分析】利用等式与不等式的性质逐一验证命题的真假即可【详解】①“a b =”⇒“ac bc =”,但当0c =时,“ac bc =”无法推出“a b =”,则“a b =”是“ac bc =”的充分不必要条件,故①是假命题；②“5a +是无理数”⇒“a 是无理数”,且“a 是无理数”⇒“5a +是无理数”,则“5a +是无理数”是“a 是无理数”的充要条件,故②是真命题；③当12a b =>-=时,2214a b =<=,即“a b >”无法推出“22a b >”,且当2241a b =>=时,21a b =-<=,即“22a b >”无法推出“a b >”,则“a b >”是“22a b >”的既不充分也不必要条件,故③是假命题；④因为{}|3a a <{}|4a a <,所以“4a <”是“3a <”的必要条件,故④是真命题；综上,真命题有2个,故选:B【点睛】本题考查命题的真假的判断,考查两命题的充分性和必要性的判断,考查等式与不等式的性质的应用三.解答题17.已知集合{1,2,3}A =，2{|(1)0,}B x x a x a x R =-++=∈，若A B A ⋃=，求实数a ；【答案】1a =或2或3【分析】由A B A ⋃=可得B A ⊆,分别讨论B =∅与B ≠∅的情况,进而求解即可【详解】由A B A ⋃=可得B A ⊆,若B =∅,则()2140a a ∆=+-<,解得a ∈∅；若B ≠∅,则()()10x a x --=,解得1x a =,21x =,①当1a =,则{}1B =,符合题意；②当2a =,则{}1,2B =,符合题意；③当3a =,则{}1,3B =,符合题意；综上,1a =或2或3【点睛】本题考查已知集合的包含关系求参数,考查分类讨论思想18.已知,,a b c R +∈，求证：3332222222()a b c ab a b bc b c ac a c ++≥+++++；【答案】证明见解析【分析】先对33+a b 与22a b ab +作差证明3322a b a b ab +≥+,同理证明3322a c a c ac +≥+,3322b c b c bc +≥+,再求和即可得证【详解】证明:()()()()()()()()233222222a b a b ab a a b b b a a b a b a b a b +-+=-+-=--=+-,因为,,a b c R +∈,所以0a b +>,()20a b -≥,所以()()33220a b a b ab +-+≥,即3322a b a b ab +≥+,同理,3322a c a c ac +≥+,3322b c b c bc +≥+,所以333333222222a b b c a c a b ab b c bc a c ac +++++≥+++++,即3332222222()a b c ab a b bc b c ac a c++≥+++++【点睛】本题考查作差法证明不等式,考查推理论证能力19.设正有理数1a21211a a =++，求证：（11a 与2a 之间；（2）2a 比1a【答案】（1）证明见解析（2）证明见解析【分析】（1）作差(12111a a a -=+,讨论1a2a （2）整理问题为21a a <-,进而求证即可【详解】证明:（1）(121112111a a a a --=+-++,因为若1a >,则10a >,又10<,则2a <；若1a <则10a <,又10-<,则2a >,介于1a 与2a 之间（2）12111121a a a a a a ----=--+,因为10a >20-<,10a>,所以210a a -<,所以21a a -<-所以2a 比1a 【点睛】本题考查不等式的证明,考查运算能力与分类讨论思想20.已知对任意实数x ，不等式2(3)10mx m x --+>成立或不等式0mx >成立，求实数m 的取值范围；【答案】19m <<【分析】①对任意实数x ,不等式2(3)10mx m x --+>成立,讨论0m =与0m ≠的情况,进而求解；②对任意实数x ,不等式0mx >成立,则m ∈∅,二者求并集即可【详解】解:①由题,对任意实数x ,不等式2(3)10mx m x --+>成立,当0m =时,不等式为310x -+>不成立,舍去；当0m ≠时,()20340m m m >⎧⎪⎨∆=--<⎪⎩,解得19m <<；②对任意实数x ,不等式0mx >成立,则m ∈∅,综上,19m <<【点睛】本题考查含参的一元二次不等式恒成立问题,考查分类讨论思想21.已知关于x 的不等式2(4129)(211)0kx k k x ---->，其中R k ∈；（1）试求不等式的解集A ；（2）对于不等式的解集A ，记B A Z = （其中Z 为整数集），若集合B 为有限集，求实数k 的取值范围，使得集合B 中元素个数最少，并用列举法表示集合B ；【答案】（1）答案见解析（2）[44k ∈--+，{2,3,4,5}B =【分析】（1）对k 进行分类讨论,分别讨论0k =,0k <,01k <<或9k >,19k ≤≤的情况,进而求解即可；（2）由（1）可知当0k <时,集合B 为有限集,利用对勾函数可知933442k k ++≤,当且仅当3k =-时等号成立,进而求解即可【详解】（1）当0k =,11{|}2A x x =<；当0k ≠时,令21291142k k k ++=,解得1k =或9k =,则当1k <或9k >时,9113442k k ++<,当19k <<时,9113442k k ++>,①当0k <,911{|3}442k A x x k =++<<；②当01k <<或9k >,11{|2A x x =<或93}44k x k >++；③当19k ≤≤,9{|344k A x x k =<++或11}2x >；（2）因为B A Z = （其中Z 为整数集）,由（1）,当0k ≥时,集合B 中的元素的个数无限；当0k <时,集合B 中的元素的个数有限,此时集合B 为有限集,因为0k <,所以9933333444422k k k k ⎛⎫++=---+≤-+= ⎪⎝⎭,当且仅当944k k -=-,即3k =-时等号成立,所以{2,3,4,5}B =且93144k k++≥，所以2890k k ++≤，所以[44k ∈--+【点睛】本题考查解含参的不等式,考查交集的定义的应用,考查分类讨论思想。

2016-2017学年上海市复旦大学附中浦东分校高三(上)第二次月考数学试卷(解析版)

2016-2017学年上海市复旦大学附中浦东分校高三（上）第二次月考数学试卷一.填空题1．函数f（x）=的定义域为．2．已知复数z满足z+i=1﹣iz（i是虚数单位），则z=．3．以抛物线y2=4x的焦点F为圆心，与抛物线的准线相切的圆的标准方程为．4．已知二元一次方程组的增广矩阵是（），若该方程组无解，则实数m的值为．5．已知定义域为R的函数y=f（x）的图象关于点（﹣1，0）对称，y=g（x）是y=f（x）的反函数，若x1+x2=0，则g（x1）+g（x2）=．6．已知x，y∈R+，且4x+y=1，则的最小值是．7．若二项式（x+）n展开式中只有第四项的系数最大，则这个展开式中任取一项为有理项的概率是．8．等比数列{a n}前n项和，n∈N*，则=．9．把一个大金属球表面涂漆，共需油漆2.4公斤．若把这个大金属球熔化制成64个大小都相同的小金属球，不计损耗，将这些小金属球表面都涂漆，需要用漆公斤．10．已知函数f（x）=，记a n=f（n）（n∈N*），若{a n}是递减数列，则实数t的取值范围是．11．已知f（x）=asin2x+bcos2x（a，b为常数），若对于任意x∈R都有f（x）≥f（），则方程f（x）=0在区间[0，π]内的解为．12．对于具有相同定义域D的函数f（x）和g（x），若存在函数h（x）=kx+b（k，b为常数），对任给的正数m，存在相应的x0∈D，使得当x∈D且x＞x0时，总有，则称直线l：y=kx+b为曲线y=f（x）和y=g（x）的“分渐近线”．给出定义域均为D={x|x＞1}的四组函数如下：①f（x）=x2，g（x）=；②f（x）10﹣x+2，g（x）=；③f（x）=，g（x）=；④f（x）=，g（x）=2（x﹣1﹣e﹣x）其中，曲线y=f（x）和y=g（x）存在“分渐近线”的是．二.选择题13．设全集U=R，已知A={x|＞0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A．（﹣，﹣1）B．（﹣1，﹣2]C．（2，3] D．[2，3）14．已知a、b为实数，命题甲：ab＞b2，命题乙：，则甲是乙的（）条件．A．充分不必要B．必要不充分C．充要D．非充分非必要15．下列命题中，正确的个数是（1）直线上有两个点到平面的距离相等，则这条直线和这个平面平行；（2）a，b为异面直线，则过a且与b平行的平面有且仅有一个；（3）直四棱柱是直平行六面体（4）两相邻侧面所成角相等的棱锥是正棱锥（）A．0 B．1 C．2 D．316．已知符号函数sgnx=，f（x）是R上的增函数，g（x）=f（x）﹣f（ax）（a＞1），则（）A．sgn[g（x）]=sgnx B．sgn[g（x）]=﹣sgnx C．sgn[g（x）]=sgn[f（x）] D．sgn[g（x）]=﹣sgn[f（x）]17．已知f（x）=Asin（wx+θ），（w＞0），若两个不等的实数x1，x2∈，且|x1﹣x2|min=π，则f（x）的最小正周期是（）A．3πB．2πC．πD．18．已知O是正三角形△ABC内部的一点，+2+3=，则△OAC的面积与△OAB的面积之比是（）A．B．C．2 D．1三.解答题19．如图，在直三棱柱ABC﹣A1B1C1中，已知AA1=BC=AB=2，AB⊥BC．（1）求四棱锥A1﹣BCC1B1的体积；（2）求二面角B1﹣A1C﹣C1的大小．20．已知△ABC的三个内角分别为A，B，C，且．（Ⅰ）求A的度数；（Ⅱ）若BC=7，AC=5，求△ABC的面积S．21．平面直角坐标系中，点A（﹣2，0）、B（2，0），平面内任意一点P满足：直线PA的斜率k1，直线PB的斜率k2，k1k2=﹣，点P的轨迹为曲线C1．双曲线C2以曲线C1的上下两顶点M，N为顶点，Q是双曲线C2上不同于顶点的任意一点，直线QM的斜率k3，直线QN的斜率k4．（1）求曲线C1的方程；（2）如果k1k2+k3k4≥0，求双曲线C2的焦距的取值范围．22．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n=+（∈N*），试求（b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．23．已知函数f（x）=log2（x+a）．（1）若0＜f（1﹣2x）﹣f（x）＜，当a=1时，求x的取值范围；（2）若定义在R上奇函数g（x）满足g（x+2）=﹣g（x），且当0≤x≤1时，g （x）=f（x），求g（x）在[﹣3，﹣1]上的反函数h（x）；（3）对于（2）中的g（x），若关于x的不等式g（）≥1﹣log23在R 上恒成立，求实数t的取值范围．2016-2017学年上海市复旦大学附中浦东分校高三（上）第二次月考数学试卷参考答案与试题解析一.填空题1．函数f（x）=的定义域为[﹣2，0）∪（0，2] ．【分析】由根式内部的代数式大于等于0，且分母不为0联立不等式组求解．解：由，解得：﹣2≤x≤2且x≠0．∴函数f（x）=的定义域为[﹣2，0）∪（0，2]．故答案为：[﹣2，0）∪（0，2]．【点评】本题考查函数的定义域及其求法，是基础题．2．已知复数z满足z+i=1﹣iz（i是虚数单位），则z=﹣i．【分析】根据复数z满足z+i=1﹣iz，移项得到z+zi=1﹣i，提出公因式z（1+i）=1﹣i，两边同除以1+i，进行复数的除法运算，分子和分母同乘以分母的共轭复数，得到结果．解：复数z满足z+i=1﹣iz，∴z+zi=1﹣iz（1+i）=1﹣i∴z===﹣i故答案为：﹣i【点评】本题考查复数的代数形式的运算，本题解题的关键是整理出复数的表示式，再进行复数的除法运算，或者设出复数的代数形式，根据复数相等的充要条件来解题．3．以抛物线y2=4x的焦点F为圆心，与抛物线的准线相切的圆的标准方程为（x ﹣1）2+y2=4．【分析】求出抛物线的焦点坐标，焦点到准线的距离就是所求圆的半径，然后写出圆的方程即可．解：因为抛物线y2=4x的焦点为圆心即（1，0），与抛物线的准线相切的圆的半径为：2．所求圆的方程为：（x﹣1）2+y2=4．故答案为：（x﹣1）2+y2=4．【点评】本题考查圆的方程的求法，抛物线的简单性质的应用，考查计算能力．4．已知二元一次方程组的增广矩阵是（），若该方程组无解，则实数m的值为﹣2．【分析】根据二元一次方程组的增广矩阵是（），该方程组无解，可得且，从而可求实数m的值．解：∵二元一次方程组的增广矩阵是（），该方程组无解，∴且，∴m2﹣4=0且4m﹣m（m+2）≠0，∴m=﹣2．故答案为：﹣2．【点评】本题考查二元一次方程组的增广矩阵．考查行列式，解答的关键是二元线性方程组的增广矩阵的涵义．5．已知定义域为R的函数y=f（x）的图象关于点（﹣1，0）对称，y=g（x）是y=f（x）的反函数，若x1+x2=0，则g（x1）+g（x2）=﹣2．【分析】由题意可得y=g（x）是y=f（x）的反函数，得到函数y=g（x）的图象关于（0，﹣1）点中心对称图形，结合x1+x2=0，可得g（x1）+g（x2）的值．解：∵定义域为R的函数y=f（x）的图象关于点（﹣1，0）对称，且y=g（x）是y=f（x）的反函数，∴函数y=g（x）的图象与函数y=f（x）的图象关于直线x﹣y=0对称，故函数y=g（x）的图象关于（0，﹣1）点中心对称图形，∴点（x1，g（x1））和点（x2，g（x2））是关于点（0，﹣1）中心对称，∴，，∵x1+x2=0，∴g（x1）+g（x2）=﹣2．故答案为：﹣2．【点评】本题考查了函数的性质，考查了互为反函数的两个函数图象间的关系，是中档题．6．已知x，y∈R+，且4x+y=1，则的最小值是25．【分析】利用“乘1法”与基本不等式的性质即可得出．解：∵x，y∈R+，且4x+y=1，则=（4x+y）=13++≥13+2=25．故答案为：25．【点评】本题考查了“乘1法”与基本不等式的性质，考查了推理能力与计算能力，属于基础题．7．若二项式（x+）n展开式中只有第四项的系数最大，则这个展开式中任取一项为有理项的概率是．【分析】先利用展开式中只有第四项的二项式系数最大求出n=6，再求出其通项公式，求出r=0，2，4，6时，为有理项，即可求出概率．解：因为二项式（x+）n展开式中只有第四项的系数最大，所以n=6．=所以其通项为T r+1所以r=0，2，4，6时，为有理项，所以所求概率为，故答案为：．【点评】本题主要考查二项式定理中的常用结论：如果n为奇数，那么是正中间两项的二项式系数最大；如果n为偶数，那么是正中间一项的二项式系数最大．8．等比数列{a n}前n项和，n∈N*，则=﹣．【分析】根据等比数列{a n}的前n项和推知a1和q，然后根据求和公式进行计算并求极限．解：∵等比数列{a n}前n项和为S n=a+（﹣）n，n∈N*，∴a n=S n﹣S n﹣1=a+（﹣）n﹣a+（﹣）n﹣1=﹣•（﹣）n﹣1，∴a1=﹣，q=﹣∴a1，a3，a5，…，a2n﹣1，为首项﹣，公比为的等比数列，∴a1+a3+a5+…+a2n﹣1==﹣（1﹣），∴=（﹣（1﹣）=﹣故答案为：【点评】本题考查数列的前2n项中奇数项和的极限的求法，是基础题，解题时要认真审题，注意等比数列的性质的合理运用．9．把一个大金属球表面涂漆，共需油漆2.4公斤．若把这个大金属球熔化制成64个大小都相同的小金属球，不计损耗，将这些小金属球表面都涂漆，需要用漆9.6公斤．【分析】设大金属球的半径为R，小金属球的半径为r，根据体积相等建立等式关系，然后求出64个小球球面的总面积，从而求出所求．解：设大金属球的半径为R，小金属球的半径为r，依题意得知：面积为4πR2需要要用油漆2.4kg．由=64×，可得r=R64个小球球面的总面积为：64×4πr2=4×（4πR2）∴4×2.4=9.6（kg）故答案为：9.6．【点评】本题是基础题，考查球的体积的求法，考查计算能力，送分题．10．已知函数f（x）=，记a n=f（n）（n∈N*），若{a n}是递减数列，则实数t的取值范围是．【分析】要使函数f（x）=x2﹣3tx+18在x≤3（x∈N*）时单调递减，则＞，解得t，解得t；要使函数f（x）=在x＞3单调递减，则必须满足t ﹣13＜0，解得t；又函数f（x）在x∈N*时单调递减，则f（3）＞f（4），解得t．联立解得即可．解：要使函数f（x）=x2﹣3tx+18在x≤3（x∈N*）时单调递减，则＞，解得t；要使函数f（x）=在x＞3单调递减，则必须满足t﹣13＜0，解得t ＜13．又函数f（x）在x∈N*时单调递减，则f（3）=27﹣9t＞f（4）=（t﹣13）•，解得t＜4．故t的取值范围是．故答案为：．【点评】本题考查了利用函数的单调性研究数列的单调性、二次函数的单调性、一次函数的单调性，属于难题．11．已知f（x）=asin2x+bcos2x（a，b为常数），若对于任意x∈R都有f（x）≥f（），则方程f（x）=0在区间[0，π]内的解为或．【分析】由f（x）≥f（），可知f（）是函数f（x）的最小值，利用辅助角公式求出a，b的关系，然后利用三角函数的图象和性质进求解即可．解：∵f（x）=asin2x+bcos2x=sin（2x+θ）其中tan，由f（x）≥f（），则f（）是函数f（x）的最小值，即f（）=，∴f（）=，即，平方得，，即，∴，解得b=﹣，∵tan=，不妨设，则f（x）=asin2x+bcos2x=sin（2x﹣），由f（x）=sin（2x﹣）=0，解得2x﹣=kπ，即x=，k∈Z，∵x∈[0，π]，∴当k=0时，x=，当k=1时，x=，故x=或=．故答案为：或．【点评】本题主要考查三角函数的图象和性质，利用三角函数的辅助角公式是解决本题的关键，考查学生的计算能力．12．对于具有相同定义域D的函数f（x）和g（x），若存在函数h（x）=kx+b （k，b为常数），对任给的正数m，存在相应的x0∈D，使得当x∈D且x＞x0时，总有，则称直线l：y=kx+b为曲线y=f（x）和y=g（x）的“分渐近线”．给出定义域均为D={x|x＞1}的四组函数如下：①f（x）=x2，g（x）=；②f（x）10﹣x+2，g（x）=；③f（x）=，g（x）=；④f（x）=，g（x）=2（x﹣1﹣e﹣x）其中，曲线y=f（x）和y=g（x）存在“分渐近线”的是②④．【分析】题目给出了具有相同定义域D的函数f（x）和g（x），若存在函数h （x）=kx+b（k，b为常数），对任给的正数m，存在相应的x0∈D，使得当x∈D 且x＞x0时，总有，则称直线l：y=kx+b为曲线y=f（x）和y=g （x）的“分渐近线”．当给定的正数m无限小的时候，函数f（x）的图象在函数h（x）=kx+b的图象的上方且无限靠近直线，函数g（x）的图象在函数h（x）=kx+b 的图象的下方且无限靠近直线，说明f（x）和g（x）存在分渐近线的充要条件是x→∞时，f（x）﹣g（x）→0．对于第一组函数，通过构造辅助函数F（x）=f （x）﹣g（x）=，对该函数求导后说明函数F（x）在（1，+∞）上是增函数，不满足x→∞时，f（x）﹣g（x）→0；对于第二组函数，直接作差后可看出满足x→∞时，f（x）﹣g（x）→0；对于第三组函数，作差后得到差式为，结合函数y=x和y=lnx图象的上升的快慢，说明当x＞1时，为为负值且逐渐减小；第四组函数作差后，可直接看出满足x→∞时，f（x）﹣g（x）→0．由以上分析可以得到正确答案．解：f（x）和g（x）存在分渐近线的充要条件是x→∞时，f（x）﹣g（x）→0．对于①f（x）=x2，g（x）=，当x＞1时，令F（x）=f（x）﹣g（x）=由于，所以h（x）为增函数，不符合x→∞时，f（x）﹣g （x）→0，所以①不存在；对于②f（x）=10﹣x+2，g（x）=f（x）﹣g（x）==，因为当x＞1且x→∞时，f（x）﹣g（x）→0，所以存在分渐近线；对于③f（x）=，g（x）=，f（x）﹣g（x）==当x＞1且x→∞时，与均单调递减，但的递减速度比快，所以当x→∞时f（x）﹣g（x）会越来越小，不会趋近于0，所以不存在分渐近线；对于④f（x）=，g（x）=2（x﹣1﹣e﹣x），当x→∞时，f（x）﹣g（x）===→0，因此存在分渐近线．故存在分渐近线的是②④．故答案为②④．【点评】本题从大学数列极限定义的角度出发，仿造构造了分渐近线函数，目的是考查学生分析问题、解决问题的能力，考生需要抓住本质：存在分渐近线的充要条件是x→∞时，f（x）﹣g（x）→0进行作答，是一道好题，思维灵活，要透过现象看本质．二.选择题13．设全集U=R，已知A={x|＞0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A．（﹣，﹣1）B．（﹣1，﹣2]C．（2，3] D．[2，3）【分析】求出A与B中不等式的解集确定出A与B，找出A补集与B的交集即可．解：由A中不等式变形得：（2x+3）（x﹣2）＞0，解得：x＜﹣或x＞2，即A=（﹣∞，﹣）∪（2，+∞），∴∁U A=[﹣，2]，由B中不等式变形得：﹣2＜x﹣1＜2，解得：﹣1＜x＜3，即B=（﹣1，3），∴（∁U A）∩B=（﹣1，2]，故选：B．【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．14．已知a、b为实数，命题甲：ab＞b2，命题乙：，则甲是乙的（）条件．A．充分不必要B．必要不充分C．充要D．非充分非必要【分析】利用不等式的性质与解法分别化简命题甲、乙，即可判断出关系．解：由命题乙：，可得：a＜b＜0．命题甲：ab＞b2，化为：b（a﹣b）＞0，∴，或，解得a＞b＞0，或a＜b＜0．∴甲是乙的必要不充分条件．故选：B．【点评】本题考查了不等式的解法、简易逻辑的判定方法，考查了推理能力与计算能力，属于基础题．15．下列命题中，正确的个数是（1）直线上有两个点到平面的距离相等，则这条直线和这个平面平行；（2）a，b为异面直线，则过a且与b平行的平面有且仅有一个；（3）直四棱柱是直平行六面体（4）两相邻侧面所成角相等的棱锥是正棱锥（）A．0 B．1 C．2 D．3【分析】（1）利用线面平行的判定定理即可判断出正误；（2）利用异面直线的性质与线面平行的判定定理即可断出正误；（3）利用直四棱柱与直平行六面体的定义，即可判断出正误；（4）两相邻侧面所成角相等的棱锥不一定是正棱锥，例如把如图所示的正方形折叠成三棱锥不是正棱锥．解：（1）直线上有两个点到平面的距离相等，则这条直线和这个平面不一定平行，因此不正确；（2）a，b为异面直线，则过a且与b平行的平面有且仅有一个，正确；（3）直四棱柱不是直平行六面体，因此不正确；（4）两相邻侧面所成角相等的棱锥不一定是正棱锥，例如把如图所示的正方形折叠成三棱锥不是正棱锥．综上正确的有1个．故选：B．【点评】本题考查了简易逻辑的判定方法、线面平行的判定定理、直四棱柱与直平行六面体的定义等基础知识，考查了推理能力与计算能力，属于中档题．16．已知符号函数sgnx=，f（x）是R上的增函数，g（x）=f（x）﹣f（ax）（a＞1），则（）A．sgn[g（x）]=sgnx B．sgn[g（x）]=﹣sgnx C．sgn[g（x）]=sgn[f（x）] D．sgn[g（x）]=﹣sgn[f（x）]【分析】直接利用特殊法，设出函数f（x），以及a的值，判断选项即可．解：由于本题是选择题，可以采用特殊法，符号函数sgnx=，f（x）是R上的增函数，g（x）=f（x）﹣f（ax）（a＞1），不妨令f（x）=x，a=2，则g（x）=f（x）﹣f（ax）=﹣x，sgn[g（x）]=﹣sgnx．所以A不正确，B正确，sgn[f（x）]=sgnx，C不正确；D正确；对于D，令f（x）=x+1，a=2，则g（x）=f（x）﹣f（ax）=﹣x，sgn[f（x）]=sgn（x+1）=；sgn[g（x）]=sgn（﹣x）=，﹣sgn[f（x）]=﹣sgn（x+1）=；所以D不正确；故选：B．【点评】本题考查函数表达式的比较，选取特殊值法是解决本题的关键，注意解题方法的积累，属于中档题．17．已知f（x）=Asin（wx+θ），（w＞0），若两个不等的实数x1，x2∈，且|x1﹣x2|min=π，则f（x）的最小正周期是（）A．3πB．2πC．πD．【分析】由题意可得•=π，求得ω 的值，可得f（x）的最小正周期是的值．解：由题意可得sin（wx+θ）=的解为两个不等的实数x1，x2，且•=π，求得ω=，故f（x）的最小正周期是=3π，故选：A．【点评】本题主要考查正弦函数的图象特征，正弦函数的周期性，属于中档题．18．已知O是正三角形△ABC内部的一点，+2+3=，则△OAC的面积与△OAB的面积之比是（）A．B．C．2 D．1【分析】对所给的向量等式进行变形，根据变化后的条件对两个三角形的面积进行探究即可，解：+2+3=，变为，设D，E分别是对应边的中点，由平行四边形法则知，2，故，由于正三角形ABC，故S△AOC =S△ADC=××S△ABC=S△ABC，又D，E是中点，故O到AB的距离是正三角形ABC高的一半，所以S△AOB=×S△ABC∴△OAC的面积与△OAB的面积之比为．故选：B．【点评】本题考查向量的加法与减法，及向量共线的几何意义，本题中把两个三角形的面积都用三角形ABC的面积表示出来，这是求比值问题时常采用的思路，统一标准，属于中档题．三.解答题19．如图，在直三棱柱ABC﹣A1B1C1中，已知AA1=BC=AB=2，AB⊥BC．（1）求四棱锥A1﹣BCC1B1的体积；（2）求二面角B1﹣A1C﹣C1的大小．【分析】（1）证明AB⊥BCC1B1，说明A1B1是四棱锥A1﹣BCC1B1的高，然后求解四棱锥A1﹣BCC1B1的体积．（2）建立空间直角坐标系．求出相关点的坐标，求出=（1，1，0）是平面A1C1C 的一个法向量．平面A1B1C的一个法向量利用向量的数量积求解二面角B1﹣A1C ﹣C1的大小．【解答】（本题满分12分）本题共2小题，第（1）小题，第（2）小题．解：（1）因为AB⊥BC，三棱柱ABC﹣A1B1C1是直三棱柱，所以AB⊥BCC1B1，从而A1B1是四棱锥A1﹣BCC1B1的高．…四棱锥A1﹣BCC1B1的体积为V=×2×2×2=…（2）如图（图略），建立空间直角坐标系．则A（2，0，0），C（0，2，0），A1（2，0，2），B1（0，0，2），C1（0，2，2），…设AC的中点为M，∵BM⊥AC，NM⊥CC1，∴BM⊥平面A1C1C，即=（1，1，0）是平面A1C1C的一个法向量．设平面A1B1C的一个法向量是=（x，y，z），=（﹣2，2，﹣2），=（﹣2，0，0）…∴=﹣2x=0，，令z=1，解得x=0，y=1.=（0，1，1），…设法向量与的夹角为β，二面角B1﹣A1C﹣C1的大小为θ，显然θ为锐角．∵cosθ=|cosβ|=，∴θ=．二面角B1﹣A1C﹣C1的大小为…【点评】本题考查二面角的平面角的求法，几何体的体积的求法，考查空间想象能力以及逻辑推理能力．20．已知△ABC的三个内角分别为A，B，C，且．（Ⅰ）求A的度数；（Ⅱ）若BC=7，AC=5，求△ABC的面积S．【分析】（Ⅰ）利用二倍角公式、诱导公式化简已知的等式求得，可得A=60°．（Ⅱ）在△ABC中，利用余弦定理求得AB的值，再由，运算求得结果．解：（Ⅰ）∵．∴，…．∵sinA≠0，∴，∴，…．∵0°＜A＜180°，∴A=60°．…（Ⅱ）在△ABC中，∵BC2=AB2+AC2﹣2AB×AC×cos60°，BC=7，AC=5，∴49=AB2+25﹣5AB，∴AB2﹣5AB﹣24=0，解得AB=8或AB=﹣3（舍），…．∴．…【点评】本题主要考查二倍角公式、诱导公式、余弦定理的应用，属于中档题．21．平面直角坐标系中，点A（﹣2，0）、B（2，0），平面内任意一点P满足：直线PA的斜率k1，直线PB的斜率k2，k1k2=﹣，点P的轨迹为曲线C1．双曲线C2以曲线C1的上下两顶点M，N为顶点，Q是双曲线C2上不同于顶点的任意一点，直线QM的斜率k3，直线QN的斜率k4．（1）求曲线C1的方程；（2）如果k1k2+k3k4≥0，求双曲线C2的焦距的取值范围．【分析】（1）设P（x，y），运用直线的斜率公式，化简整理，即可得到曲线C1的方程；（2）设双曲线方程为，Q（x0，y0）在双曲线上，再由直线的斜率公式，结合条件，得到b的范围，即可得到双曲线C2的焦距的取值范围．解：（1）设P（x，y），则，∴曲线C1的方程为；（2）设双曲线方程为，Q（x0，y0）在双曲线上，所以，∵，∴，∴0＜b≤2，由双曲线C2的焦距为2，故双曲线C2的焦距的取值范围∈（2，2]．【点评】本题考查轨迹方程的求法，主要考查椭圆和双曲线的方程和性质，同时考查直线的斜率公式的运用，属于中档题．22．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n=+（∈N*），试求（b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．【分析】（1）通过4a n+1=4S n+1﹣4S n得（a n+1+a n）（a n+1﹣a n﹣2）=0，进而可得结论；（2）通过分离b n的分母可得b n=2+2（﹣），累加后取极限即可；（3）假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，通过（1）可得300=（2m+k﹣1）（k+1），利用2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，即得结论．解：（1）∵4S n=（a n+1）2，∴4S n+1=（a n+1+1）2，两式相减，得4a n+1=4S n+1﹣4S n=（a n+1）2﹣（a n+1+1）2=﹣+2a n+1﹣2a n，化简得（a n+1+a n）（a n+1﹣a n﹣2）=0，又∵数列{a n}各项均为正数，∴a n+1﹣a n=2 （n∈N*），∴数列{a n}是以1为首项，2为公差的等差数列，∴a n=2n﹣1 （n∈N*）．（2）因为b n=+=+=2+2（﹣），故b1+b2+…+b n=2n+2[（1﹣）+（﹣）+…+（﹣）]=2n+2（1﹣），于是（b1+b2+…+b n﹣2n）=[2（1﹣）]=2；（3）结论：存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300．理由如下：假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，由（1），可得a m+a m+1+…+a m+k=（2m+k﹣1）（k+1），从而（2m+k﹣1）（k+1）=300，由于正整数m、k均大于2，知2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，故由300=22×3×52，得或，解得或，因此，存在大于2的正整数m、k：或，使得a m+a m+1+…+a m+k=300．【点评】本题考查求数列的通项，涉及到极限等知识，注意解题方法的积累，属于中档题．23．已知函数f（x）=log2（x+a）．（1）若0＜f（1﹣2x）﹣f（x）＜，当a=1时，求x的取值范围；（2）若定义在R上奇函数g（x）满足g（x+2）=﹣g（x），且当0≤x≤1时，g （x）=f（x），求g（x）在[﹣3，﹣1]上的反函数h（x）；（3）对于（2）中的g（x），若关于x的不等式g（）≥1﹣log23在R 上恒成立，求实数t的取值范围．【分析】（1）根据对数函数的真数部分大于0，及对数的运算性质，可将不等式化为1＜，且2﹣2x＞0且x+1＞0，解不等式组可得x的取值范围；（2）函数g（x）满足g（x+2）=﹣g（x），表示函数的周期为4，结合函数g（x）为奇函数，可求出x∈[﹣3，﹣1]时，函数g（x）的解析式，进而得到其反函数；（3）关于x的不等式关于x的不等式g（）≥1﹣log23在R上恒成立，等价于g（）≥g（﹣）在R上恒成立，即u==﹣，∈[﹣，]，分类讨论后，综合讨论结果，可得实数t的取值范围．解：（1）原不等式可化为，∴1＜，且2﹣2x＞0，且x+1＞0，得3﹣2．（2）∵g（x）是奇函数，∴g（0）=0，得a=1，当x∈[﹣3，﹣2]时，﹣x﹣2∈[0，1]，g（x）=﹣g（x+2）=g（﹣x﹣2）=log2（﹣x﹣1），此时g（x）∈[0，1]，x=﹣2g（x）﹣1，h（x）=﹣2x﹣1（x∈[0，1]）．当x∈（﹣2，﹣1]时，﹣x﹣2∈[﹣1，0），x+2∈（0，1]，g（x）=﹣g（x+2）=﹣log2（x+3），此时，﹣g（x）∈[﹣1，0），x=2﹣g（x）﹣3，h（x）=2﹣x﹣3．（x∈[﹣1，0））．∴．（3）∵关于x的不等式g（）≥1﹣log23在R上恒成立，∴记u==﹣，∵关于x的不等式g（）≥1﹣log23在R上恒成立，∴g（）≥=﹣log2=﹣log2（1+）=﹣g（）=g（﹣）在R上恒成立，当t+1≥0时，u∈（﹣，﹣）=（﹣），∴（﹣，）∈[﹣，]，解得t∈[﹣1，20]．当t+1＜0时，u∈（﹣，﹣）=（），由g（）≥=﹣log2=﹣log2（1+）=﹣g（）=g（﹣）在R 上恒成立，得，解得t∈[﹣4，﹣1）．综上所述，实数t的取值范围是[﹣4，20]．【点评】本题考查的知识点是对数函数的图象和性质，函数的奇偶性，函数的周期性，函数的单调性，反函数，对数的运算性质，存在性问题，函数的最值，是函数图象和性质较为综合的应用，难度较大．。

上海市复旦大学附属中学2018-2019学年高三上学期期中数学试题

绝密★启用前上海市复旦大学附属中学2018-2019学年高三上学期期中数学试题试卷副标题注意事项：1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上第I 卷（选择题)请点击修改第I 卷的文字说明一、单选题1．若平面中两条直线12,l l 的方向向量分别是,a b ，则12l l //是//a b 的（）. A ．充分不必要条件 B ．必要不充分条件 C ．充要条件 D ．既不充分又不必要2．已知定义域为R 的奇函数()y f x =有反函数()-1y fx =，那么必在函数()11y f x -=+图像上的点是（）.A ．()(),1f t t ---B ．()()1,f t t -+- C ．((t)1,)f t ---D ．()()1,f t t -+-3．已知数列{}n a 的通项公式为()()*11n a n N n n =∈+，其前n 项和910n S =，则双曲线2211x y n n-=+的渐近线方程为（） A ．y x = B ．y x = C ．y = D ．y x = 4．已知定义在0,+∞上的函数f x 满足2f x f x x +=+，且当0,2x ∈时,()8f x x =-，则()93f =（）.A ．2019B ．2109C ．2190D ．2901第II 卷（非选择题)请点击修改第II 卷的文字说明二、填空题5．20191lim 12019n n →∞⎛⎫+= ⎪⎝⎭____________.6．若复数z 满足:()()211z i i ⋅+=-(i 为虚数单位)，则z =____________.7．已知向量()()2,1,3,1a b =-=-，()2,c y =，且()a b c -⊥，则y =____________. 8．若集合12A x y lg x ⎧⎫⎛⎫==-⎨⎬ ⎪⎝⎭⎩⎭，{}B y y arcsinx ==，则A B =____________. 9．在622x x ⎛⎫- ⎪⎝⎭的二项展开式中，3x 的系数是____________. 10．在ABC △中，角,,A B C 所对的边分别是,,a b c ，若3,3a b π===，则角C的大小为____________.11．若圆锥侧面积为20π，且母线与底面所成角为34arctan ，则该圆锥的体积为____________.12．若无穷等比数列{}n a 的各项和为n S ，首项11a = ，公比为32a -，且l i m n x S a →∞= ，则a =_____．13．某学生选择物理、化学、地理三门学科参加等级考，已知每门学科考A +得70分，考A 得67分，考B +得64分，该生每门学科均不低于64分，则其总分至少为207分的概率为________14．已知,a b ∈R ，且22425a b ≤+≤，则22a b ab ++的取值范围是____________. 15．已知函数()f x asinx bcosx c =++的图像经过()0,5,,52A B π⎛⎫⎪⎝⎭两点，当0,2x π⎡⎤∈⎢⎥⎣⎦时，()10f x ≤，则实数c 的取值范围是____________.16．定义在[0,)+∞上的函数()f x 满足:①当[)1,2x ∈时，()1222f x sin x ππ⎛⎫=+ ⎪⎝⎭；②对任意[0,)x ∈+∞都有()()22f x f x =.设关于x 的函数()()F x f x a =-的零点从小到大依次为123,x ,,,x x ⋅⋅⋅若1,12a ⎛⎫∈ ⎪⎝⎭，则122n x x x ++⋅⋅⋅+=____________. 三、解答题17．已知递增的等差数列{}n a 的首项11a =，且124a a a 、、成等比数列. (1) 求数列{}n a 的通项公式n a ；(2) 设数列{}n b 满足()21n na n nb a =+-，n T 为数列{}n b 的前n 项和，求2n T . 18．已知函数()22f x sinxcosx x x R =+∈. (1)求函数()31y f x =-+的最小正周期和单调递减区间；(2)已知ABC △中的三个内角,,A B C 所对的边分别为,,a b c ，若锐角A 满足26A f π⎛⎫-= ⎪⎝⎭7a =， sinB sinC +=,b c 的长. 19．某饮料生产企业为了占有更多的市场份额，拟在2017年度进行一系列促销活动，经过市场调查和测算，饮料的年销售量x 万件与年促销费t 万元间满足311t x t +=+.已知2017年生产饮料的设备折旧，维修等固定费用为3万元，每生产1万件饮料需再投入32万元的生产费用，若将每件饮料的售价定为其生产成本的150%与平均每件促销费的一半之和，则该年生产的饮料正好能销售完．(1)将2017年的利润y (万元)表示为促销费t (万元)的函数； (2)该企业2017年的促销费投入多少万元时，企业的年利润最大？(注：利润＝销售收入－生产成本－促销费，生产成本＝固定费用＋生产费用) 20．设数列{}n a 的前n 项和为n S ，对任意*n N ∈，点,n S n n⎛⎫⎪⎝⎭都在函数()2n a f x x x =+的图象上.(1)求123,,a a a ，归纳数列{}n a 的通项公式(不必证明).()78910,,,,a a a a (){}111213141515,,,,,a a a a a a ，()1718,a a ，()192021,,a a a ，各个括号内各数之和，设由这些和按原来括号的前后顺序构成的数列为{}n b ，求6100b b +的值.(3)设n A 为数列1n n a a ⎧⎫-⎨⎬⎩⎭的前n 项积，若不等式()32n a A f a a +<-对一切*n N ∈都成立，其中0a >，求a 的取值范围.21．已知函数()2327mx n h x x +=+为奇函数，()13x mk x -⎛⎫ ⎪⎝⎭=，其中m n R ∈、.(1)若函数()h x 的图像过点()1,1A ，求实数m 和n 的值；(2)若3m =，试判断函数()()()11f x h x k x =+在[3,)x ∈+∞上的单调性并证明； (3)设函数()()(),39,3h x x g x k x x ⎧≥⎪=⎨<⎪⎩若对每一个不小于3的实数1x ，都恰有一个小于3的实数2x ，使得()()12g x g x =成立，求实数m 的取值范围.参考答案1．A 【解析】【分析】平面中两条直线1l ，2l 的方向向量分别是a ，b ，可得12l l //⇒//a b ，反之不成立，可能重合．【详解】平面中两条直线1l ，2l 的方向向量分别是a ，b ，则12l l //⇒//a b ，反之不成立，可能重合．12//l l ∴是//a b 的充分不必要条件．故选：A ．【点睛】本题考查了线面位置关系、平面向量的性质、简易逻辑的判定方法，考查了推理能力与计算能力，属于基础题 2．C 【解析】【分析】由()()f t f t -=-得1(())f f t t --=-，再由函数图象的平移规律得出答案．【详解】()f x 定义在R 上的奇函数，()()f t f t ∴-=-，1(())f f t t -∴-=-，即(()f t -，)t -在1()y f x -=的图象上，1(1)y f x -=+图象是由1()y f x -=的图象向左平移1个单位得到的， (()1f t ∴--，)t -在1(1)y f x -=+图象上．故选：C ．【点睛】本题考查了奇函数、反函数的性质及函数图象变换，利用互为反函数的函数图象关系是关键． 3．C【解析】试题分析：根据数列的通项公式为*1()(1)n a n N n n =∈+,其前项和910n S =，那么可知n 1111n S n n =-=++，可知n=9,那么根据2211x y n n-=+可知a=10,b= 3,故可知双曲线2211x y n n -=+的渐近线方程为31010y x =±，选C. 考点：数列的求和，双曲线的性质点评：主要是考查了数列的通项公式和双曲线的性质的运用，属于基础题。

2016年上海市复旦大学附中高二下学期数学期中试卷

2016年上海市复旦大学附中高二下学期数学期中试卷一、填空题（共12小题；共60分）1. 复数的虚部是．2. 若两个球的表面积之比是，则它们的体积之比是．3. 已知平面平面，直线，，点，点，记点，之间的距离为，点到直线的距离为，直线和的距离，则，，的大小关系是．4. 设，是平面同侧的两点，点，，是平面的斜线，射线，在内的射影分别是射线，，若，则是（锐角、直角或钝角）．5. 在复平面内，到点的距离与到直线：的距离相等的点的轨迹是．6. 在正方体中，为棱的中点，则异面直线与所成的角的大小为（结果用反三角函数值表示）．7. 已知实数和复数满足，则的最小值是．8. 正四棱锥底面边长为，侧棱长为，则其体积为．9. 在半径为的球面上有，，三点，如果，，则球心到平面的距离为．10. 在地球表面上，地点位于东经，北纬，地点位于西经，南纬，则，两点的球面距离是（设地球的半径为）．11. 在三棱锥中，三条侧棱，，两两垂直，且，，又是底面内一点，则到三个侧面的距离的平方和的最小值是．12. 小明在研究三棱锥的时候，发现下面一个真命题，在三棱锥中，已知，，（如图），设二面角的大小为，则，其中是一个与有关的代数式，请写出符合条件的．二、选择题（共4小题；共20分）13. 从正方体的八个顶点中任取四个点连线，在能构成的一对异面直线中，其所成的角的度数不可能是A. B. C. D.14. 对于复数（，为虚数单位），定义，给出下列命题：对任何复数，都有，等号成立的充要条件是；；，则；对任何复数，，，不等式恒成立，其中真命题的个数是A. B. C. D.15. 下列四个命题：①任意两条直线都可以确定一个平面；②若两个平面有个不同的公共点，则这两个平面重合；③直线，，，若与共面，与共面，则与共面；④若直线上有一点在平面外，则在平面外．其中错误命题的个数是A. B. C. D.16. 两个相同的正四棱锥底面重合组成一个八面体，可放于棱长为的正方体中，重合的底面与正方体的某一个面平行，各顶点均在正方体的表面上（如图），该八面体的体积可能值有A. 个B. 个C. 个D. 无数个三、解答题（共4小题；共52分）17. 在复数范围内解方程：．18. 如图，是圆柱的一条母线，已知过底面圆的圆心，是圆上不与点，重合的任意一点，，，．（1）求直线与平面所成角的大小；（2）求点到平面的距离；（3）将绕母线旋转一周，求由旋转而成的封闭几何体的体积．19. 如图，在四棱柱中，侧棱底面，，，，，，．（1）求证：平面；（2）现将与四棱柱形状和大小完全相同的两个四棱柱拼接成一个新的四棱柱．规定：若拼接成的新四棱柱形状和大小完全相同，则视为同一种拼接方案．问：共有几种不同的拼接方案?在这些拼接成的新四棱柱中，记其中最小的表面积为，写出的解析式．（直接写出答案，不必说明理由）20. 在四面体中，有两条棱的长为，其余棱的长度为．（1）若，且，求二面角的余弦值；（2）求的取值范围，使得这样的四面体是存在的．答案第一部分1.【解析】复数故其虚部为．2.【解析】由已知两个球的表面积之比是，所以两个球的半径之比是，所以两个球的体积之比．3.【解析】由于平面平面，直线和又分别是两平面的直线，则即是平面之间的距离，即两个平面内直线的最短距离．而由于两直线不一定在同一平面内，则一定大于，判断和时，因为是上任意一点，则大于．4. 锐角【解析】在，上取点，，使得，则射影长等于，设，，则，所以，所以是锐角．5.【解析】设（），则直线：化为：．因为点在直线上，所以在复平面内，到点的距离与到直线：的距离相等的点的轨迹是．6.【解析】以为原点，为轴，为轴，为轴，建立空间直角坐标系，设正方体棱长为，则，，，，，，设异面直线与所成的角为，．所以．所以异面直线与所成的角为．7.【解析】设，因为，所以，所以所以，，所以当且仅当时“”成立．8.9.【解析】设外接圆半径为，球半径为，则在中，，即：，所以，所以球心到面的距离．10.【解析】由题意，在大圆上．因为地点位于东经，北纬，地点位于西经，南纬，所以纬度差为，因为地球半径为，所以，两地的球面距离是．11.【解析】以为原点，为轴，为轴，为轴，建立空间直角坐标系，由已知得，，，所以平面为：，所以，解得．又是底面内一点，所以到三棱锥三个侧面的距离的平方和的最小值是．12.【解析】如图，在平面内，作于，在平面内作于，连接，则为二面角的平面角，大小为，设，，则，，，所以在中，在中，，在中，，所以，所以，所以，所以．第二部分13. A 【解析】从正方体的八个顶点中任取四个点连线中，在能构成的一对异面直线中，其所成的角的度数可能有以下几种情况：①若两异面直线为和，此时两直线所成的角为．②若两异面直线为和，此时两直线所成的角为．③若两异面直线为和，此时两直线所成的角为．所以在能构成的一对异面直线中，其所成的角的度数不可能是．14. C 【解析】由复数（，为虚数单位），定义，知：在中，对任何复数，都有，当时，；反之，当时，，所以等号成立的充要条件是，故成立；在中，因为，，所以，故成立；在中，当，时，，但，故错误；对任何复数，，，设，，，则，，，，所以恒成立．故成立．15. C16. D 【解析】设与正方体的截面四边形为，设，则，故，所以该八面体的体积可能值有无数个．第三部分17. 设（），则原方程变成或或或所以原方程的解为，，．18. （1）因为平面，平面，所以，因为是圆的直径，又平面，平面，，所以平面．所以是与平面所成的角．因为，所以，所以．所以直线与平面所成角的大小为．（2）过作，垂足为，由（）得平面，平面，所以平面平面，又平面平面，平面，，所以平面．因为，所以．所以，即到平面的距离为．（3）线段绕旋转一周所得几何体为以为底面半径，以为高的圆锥，线段绕旋转一周所得几何体为以为底面半径，以为高的圆锥，所以绕旋转一周而成的封闭几何体的体积．19. （1）取的中点，连接，如图．因为，，所以四边形为平行四边形，所以且．在中，因为，，．所以，所以，即．所以．因为平面，平面，所以．又，所以平面．（2）共有种不同的拼接方案．．20. （1）如图，过作，垂足为，连接，则为二面角的平面角，在等边三角形中，因为，所以，在等腰三角形中，因为，，所以．在中，由余弦定理得．（2）当两条长为的棱相交时，不妨设，，因为面与平面重合且，在异侧时，，此时，面与平面重合且，在同侧时，，此时，所以；当两条长为的棱互为对棱时，不妨设，，，可以无限趋近于，当为平面四边形时，所以．综上，若四面体存在，则．。

上海市复旦大学附属中学2016-2017学年高二上学期期中考试英语试题 Word版含答案

复旦大学附属中学2016学年第一学期高二年级期中考试试卷2016-11-8II. Grammar and vocabulary (26%)Section ADirections: Read the following two passages. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper from of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.(A)As a student, I get so many assignments every day. I have to stay up late in order to finish all my homework. I used to complain about all this pressure (25) ______ school with my classmates. We did not appreciate our teachers for their hard work. We only (26) _______ (know) that we got a lot of homework.After a few months, we did not complain about homework anymore (27) _______ we knew that our teachers worked (28) ________ (hard) than we did. We had no right to complain. Sometimes, we said, “I didn’t go to bed until 12:00 o’clock last night. Now I just want to sleep.” Our teacher would answer us, “I go to bed at 1:00 a.m. every day.” Since we knew how hard teachers work, we started to appreciate them. To give our thanks, we wrote a big card to the teachers (29) _______ it was teachers’ day. When they got our card, they (30) _________ (touch) because their students finally knew the teachers’ effort.After (31) _________ (give) the card, I realized (31) powerful the sentence “thank you” is. When we give our thanks to somebody, the world is full of love.I say “thank you” to my friends, fami ly, classmates, teachers, and even strangers.I like to see the smiles on their faces, so (32) ________ (say) “thank you” every day is the way I make the world a better place.(B)Phyllis Rawlins’ house was destroyed after a tornado(龙卷风) swept through her town of Kokomo, Ind., on Sunday. Last summer, she lost (33)_______ husband of over 40 years, Edgar. In the tornado’s rubble(瓦砾), Rawlins searched for Edgar’s wedding ring. “Digging and praying. Digging and praying,” she told local station Fox 59.“It was everything to me, because that’s one thing that I had,” she said.Rawlins had been visiting family in Kentucky when the storm came through. She returned (34) ________ find her home completely in pieces. “This was the house that love built,” she told W THR.Without her husband or her house, she was determined to find the ring. But (35) ________ (locate) it among the piles of rubble seemed to be hopeless.Somehow, her brother spotted something under (36) _______ piece of the roof and called her over. The ring, (37) _______ (bury) in the rubble, had turned up.“It was a miracle,” Rawlins said. “We both just hugged each other, (38) _______ (cry). That was (39) _______ I had searched and searched for,”When all was lost, the special ring he left was finally found.“I’m very strong with my faith, and I know that God is in control of (40) _________, the good and the bad,” Rawlins said.Section BDirections: 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．Zhou Yeling couldn’t wait until 7am for a long-awaited date with her favourite Englishman.The 19-year-old from the city of Shanghai dragged herself out of bed at 5am to watch the third season premiere of Sherlock on the BBC’s website. Two hours later, the episode started showing with Chinese subtitles on , a video website.Youku says it was viewed more than 5 million times in the first 24 hours, becoming the site’s most popular programme to 41.“I was excited beyond words,” said Zhou, a student in the central Chinese city of Changsha.Sherlock has become a global 42, but nowhere more than in China, which was one of the first countries where the new season was shown.Online fan clubs have attracted thousands of members. Chinese fans write their own stories about the modern version of author Arthur Conan Doyle’s prickly, Victorian detective and his 43, Dr Watson, to fill the time between the brief, three-episode seasons.“The Sherlock production team shoot something more like a movie, not just a TV drama,” said YuFei, a veteran writer of TV crime dramas for Chinese television. Scenes in which Holmes44 clues in a suspect’s clothes or picks apart an alibi are so richly detailed that “it seems like a wasteful luxury,” Yu said.Even the Communist Party newspaper People’s Daily is a fan.“45 plot, bizarre story, exquisite production, excellent performances,” it said of the third season’s first episode.With its mix of odd villains, eccentric aristocrats and fashionable London settings, Sherlock can draw on a Chinese fondness for a storybook version of Britain. Wealthy Chinese send their children to local branches of British schools such as Eton and Dulwich.On the outskirts of Shanghai, a developer has built Thames Town, modelled on an English village with 46Tudor houses and classic red phone booths.“The whole drama has the rich scent of British culture and47,” Yu said. “Our drama doesn’t have that.”The series has given a48to , part of a fast-growing Chinese online video industry. Dozens of sites, some independent and others run by Chinese television stations, show local and imported programmes such as The Good Wife and The Big Bang Theory. says that after two weeks, total 49 for the Sherlock third season premiere had risen to 14.5 million people. That compares with the 8 to 9 million people who the BBC says watch first-run episodes in Britain. The total in China is bumped up by viewers on pay TV service BesTV, which also has rights to the programme.Appearing online gives Sherlock an unusual 50over Chinese dramas. To support a fledgling industry, communist authorities have exempted video websites from most censorship and limits on showing foreign programming that apply to traditional TV stations. That allows outlets such as Youku to show series that might be deemed too violent or political for state TV and to release them faster.III. Cloze (15points)Section A (15分)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.Summer is an ideal time to get the jump on your college admissions essay. These less hurried months before the onslaught of a highly pressured fall offer the chance for students to think, 51and connect with a writing topic of your admissions essay.So how can you use the summer to your best 52?First，you’d better clear your head. 53 like TV, texting, video games and Internet surfing can seriously inhibit 54. Once your school term is over, 55some time away from those electronic diversions and find a park bench, rooftop, library carrel or some other quiet place where you can hear your thoughts bubbling up from deep down in your 56Next，ask yourself questions. In looking for an essay topic, an excellent way to begin is by asking questions that can 57some juicy conflict. But don’t forget write it down. carry around a pad and pencil or some kind of wireless 58to record your thoughts. if you don’t write it down, you’re 59 to lose them.Certainly,you are required to familiarize yourself with the narrative form. It is far better to think of the college admissions essay as your chance to tell a good story. Stories are narratives. Be conscious of their narrative 60Last but not least.you should enjoy yourself. These feel-good months make it easier to relax, and61the college admissions essay with less anxiety is a good thing. It would be extremely 62to view this assignment as a creative act. You’ll want to 63yourself to the work, think that your essay will 64 through a series of drafts and allow yourself to take some 65in the process.51. A. renew B. reflect C. reserve D. resign52. A.advantage B.gain C favor D. profit53. A.Discussions B. Distributions C. Distinctions D. Distractions54. A.motive B. awareness C.inspiration D.shelter55. A.schedule B. program C. draft D. enclose56. A. consequence B.consensus C. conscience D. consciousness.57. A. turn B. dress C.catch D. run58. A. dignity B.devil C. dialect D.device59. A. free B. bound C. obliged D.possible60. A. technologies B.negotiations C.discussions D. techniques61. A. moving B.accessing C.approaching D. entering62. A. Irresistible B. beneficial C.discreet D.inevitable63. A.attach B.reply C. commit D. appeal64. A.emerge B.flutter C.stoop D. evolve65. A. pleasure B. worth C.literacy D. courageIV. Reading Comprehension (15 points)Section ADirections: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,andD.choose the one that fits best according to the information given in the passage you have read.(A)Texting has long been bemoaned (哀叹）as the downfall of the written word,“penmanship for illiter,”as one critic called it. To which the proper response is LOL.Texting properly isn′t writing at all. It′s a “spoken” language that is getting richer and more complex by the year.First,some historical perspective. Writing was only invented 5,500 years ago. whereas Ianguage probably traces back at least 80.000 years. Thus talking came first; writing is just a craft that came along later. As such, the first writing was based on the way people talk，with short sentences.However, while talking is largely subconscious and rapid, writing is deliberate and slow, Over time,writers took advantage of this and started cratting long-winded sentences such as this one:The whole engagement lasted above 12 hours, till the gradual retreat of the Per trsians was changed into a disorderly flight, of which the shameful example was given by the principal leaders and……”No one talks like that casually — or should. But it is natural to desire to do so for special occasions. In the old days, we didn’t much wr ite like talking because there was no mechanism to reproduce the speed of conversation. But texting and instant messaging do —and a revolution has begun. It involves the crude mechanics of writing, but in its economy, spontaneity and even vulgaritv. texting is actually a new kind of talking, with its own kind of grammar and conventions.Take LOL. It doesn’t actually mean “laughing out loud” in a literal sense anymore. LOL has evolved into something much subtler and sophisticated and is used even when not hing is remotely amusing. Jocelyn texts “Where have you been?” and Annabelle texts back ，LOL at the library studying for two hours.” LOL signals basic empath)’（同感)between tcxters. easing tension and creating a sense of equality. Instead of having a literal meaning, it docs something - conveying an attitude —just like the -cd ending conveys past tense rather than “meaning.，anything. LOL.of all things, is grammar.Of course no one thinks about that consciously. But then most of communication operates without being noticed. Over time, the meaning of a word or an expression drifts meat used to mean any kind of food, silly used to mean, believe it or not，blessed.Civilization, then，is fine 一 people banging away on their smartphones are fluently using a code separate from the one they use in actual writing, and there is no evidence that texting is ruining composition skills. Worldwide people speak differently from the w ay they write, and texting -quick, casual and only intended to be read once — is actually a way of talking with your fingers.66.In what way does the author say writing is different from talking?A) It is crafted with specific skills.B) It expresses ideas more deeply.C) It does not have as long a history.D) It is not as easy to comprehend.67.Why is LOL much used in texting?A) It brings texters closer to each other.B) It shows the texter's sophistication.C) It is a trendy way to communicateD) It adds to the humor of the text68.Examples like meat and silly are cited to showA) the difference between writing and talkingB) how different words are used in textingC) why people use the words the way they doD) the gradual change of word meaning(B)Mark Twain has been called the inventor of the American novel. And he surely deserves additional praise: the man who popularized the clever literary attack on racism.I say clever because anti-slavery fiction had been the important part of the literature in the years before the Civil War. H. B. Stowe’s Uncle Tom’s Cabin is only the most famous example. These early stories dealt directly with slavery. With minor exceptions, Twain planted his attacks on slavery and prejudice into tales that were on the surface about something else entirely. He drew his readers into the argument by drawing them into the story.Again and again, in the postwar years, Twain seemed forced to deal with the challenge of race. Consider the most controversial, at least today, of Twain’s novels, Adventures of Huckleberry Finn. Only a few books have been kicked off the shelves as o ften as Huckleberry Finn, Twain’s most widely read tale. Once upon a time, people hated the book because it struckthemas rude. Twain himself wrote that those who banned the book considered the novel “trash and suitable only for the slums(贫民窟).” More recently the book has been attacked because of the character Jim, the escaped slave, and many occurrences of the word nigger. (The term Nigger Jim, for which the novel is often severely criticized, never appears in it.)But the attacks were and are silly—and miss the point. The novel is strongly anti-slavery. Jim’s search through the slave states for the family from whom he has been forcibly parted is heroic. As J. Chadwick has pointed out, the character of Jim was a first in American fiction—a recognition that the slave had two personalities, “the voice of survival within a white slave culture and the voice of the individual: Jim, the father and the man.”There is much more. Twain’s mystery novel Pudd’nhead Wilson stood as a challenge to the racial beliefs of even many of the liberals of his day. Written at a time when the accepted wisdom held Negroes to be inferior (低等的) to whites, especially in intelligence, Twain’s tale centered in part around two babies switched at birth.A slave gave birth to her master’s bab y and, for fear that the child should be sold South, switched him for the master’s baby by his wife. The slave’s light-skinned child was taken to be white and grew up with both the attitudes and the education of the slave-holding class. The master’s wife’s baby was taken for black and grew up with the attitudes and intonations of the slave.The point was difficult to miss: nurture (养育), not nature, was the key to social status. The features of the black man that provided the stuff of prejudice—manner of speech, for example—were, to Twain, indicative of nothing other than the conditioning that slavery forced on its victims.Twain’s racial tone was not perfect. One is left uneasy, for example, by the lengthy passage in his autobiography (自传) about how much he loved what were called “nigger shows” in his youth—mostly with white men performing in black-face—and his delight in getting his mother to laugh at them. Yet there is no reason to think Twain saw the shows as representing reality. His frequent attacks on slavery and prejudice suggest his keen awareness that they did not.Was Twain a racist? Asking the questioning the 21 stcentury is as wise as asking the same of Lincoln. If we read the words and attitudes of the past through the “wisdom” of the considered m oral judgments of the present, we will find nothing but error. Lincoln, who believed the black man the inferior of the white, fought and won a war to free him. And Twain, raised in a slave state, briefly a soldier, and inventor of Jim, may have done more to anger the nation over racial injustice and awaken its collective conscience than any other novelist in the past century.69. How do Twain’s novels on slavery differ from Stowes?A. Twain was more willing to deal with racism.B. Twain’s attack on racism w as much less open.C. Twain’s themes seemed to agree with plots.D. Twain was openly concerned with racism.70. What best proves Twain’s anti slavery stand according to the author?A. Jim’s search for his family was described in detail.B. The slave’s voic e was first heard in American novels.C. Jim grew up into a man and a father in the white culture.D. Twain suspected that the slaves were less intelligent.71. The story of two babies switched mainly indicates that .A. slaves were forced to give up their babies to their mastersB. slaves babies could pickup slave holders‵ way of speakingC. blacks‵ social position was shaped by how they were brought upD. blacks were born with certain features of prejudice72. What does the author mainly argue for?A. Twain had done more than his contemporary writers to attack racism.B. Twain was an admirable figure comparable to Abraham Lincoln.C. Twain’s works had been banned on unreasonable grounds.D. Twain s works should be read from a historical point of view.Section BDirections:complete the following passage by using the sentences in the box. Each sentence can only be used once. Note that there are two sentences more than you need.Scientists have identified the clear biological advantages that give the world's sporting champions a head start in life before they have even begun their rigorous training programs.The coach for the French Olympic team says: “We measure special attributes between the ages of 16 and 18. But only one in 10, 000 people has the physical aspects needed to compete at the very top level in sporting events.___73___.We take into account the height, strength and endurance of a person. We also regard mental application as important, how an individual reacts when the competition gets really tou gh.”Scientists say that medical evidence is playing an increasingly important role in the selection of athletes. A study of the 40-year dominance of Kenyan runners in long distance athletic events has revealed that 45 per cent of them come from the Nandi tribe. What is remarkable is that this tribe makes up only 3 per cent of the Kenyan population.___74___. Athletic organizations consider these genetic factors a good indicator when selecting athletes to produce superior running performances.___75___.For example, David Beckham's bandy legs have been partly credited with helping to put a spin on the football when he takes a free kick for England. Other biological characteristics are more measurable. The American tennis player, AndyRoddick, has the fastest serve in the game. He is able to arch his back so much that it increases the rotation of his arm to 130 degrees. This is 44 per cent better than the average professional player and this allows him to drive the ball over the net at 240 kilometers per hour. Michael Phelps, the fourteen-times Olympic swimming champion, has over-size feet which act like flippers to propel him through the water.___76___.Mia Hamm, probably the best all-round woman footballer in the world, produces less than one liter of sweat an hour when doing vigorous exercise, which is half the human average. When it comes to speed, take the example of woman racing driver, Liz Halliday. A normal person would take 300 milliseconds to make a reactive decision. She can do it in 260 milliseconds. It may not sound much quicker but at top race speeds this makes a difference of three car lengths.The difference between success and failure is very small and all these biological factors are crucial in finding future champions.V. Productive Grammar (10 points )Directions:After reading the passage below, fill in the blanks to make the passages 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.Golden Rules of Good DesignWhat makes good design? Over the years, designers and artists __77_____(try) to capture the essentials of good design. They have found that some sayings can help people understand the ideas of good design. There are four____78_____follows.Less is more. This saying is associated with the German-born architect Mies van der Rohe. In his Modernist view, beauty lies in simplicity and elegance, and the aim of the designer is ____79_____(create) solutions to problems through the most efficient means. Design should avoid unnecessary decorations.More is not a bore. The American-born architect Robert Venturi concluded that____80___ simplicity is done badly, the result is soulless design. Post-Modernist designers began to experiment with decoration and color again. Product design was heavily influenced by this view and can be seen in kitchen appliances ____81____ ____81____ovens and kettles.Fitness for purpose. Successful product design takes into consideration a product’s function, purpose, shape, form, color, and so on.The most important result for the user is that the product does____82____is needed. For example, think of a(n) adjustable desk lamp. It needs to be constructed from materials that will stand the heat of the lamp and regular adjustments by the user. It also needs to be stable. ____83____(importantly), it needs to direct light where it is needed.From follows emotion. This phrase is associated with the German designer Hartmut Esslinger.He believes design____84____take into account the sensory side of____85____nature—sight, smell, touch and taste. These are as important as rational(理性的). When ____86____(choose)everyday products such as toothpaste, we appreciate a cool-looking device that allows us to easily squeeze the toothpaste onto our brush.VI. Translation ( 15 points )Directions: Translate the following sentences into English, using the words given in the brackets.86.不可否认鼓励学生的最好方法之一是颁发奖学金。

上海市复旦附中2016届高三(上)期中英语试卷(精校版含题目材料来源衔接)

复旦大学附属中学2015-2016学年第一学期高三年级英语期中考试试卷第I卷（共103分）I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversation and the question will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. This afternoon. B. Tomorrow. C. Next week. D. Next month.2. A. She doesn’t play tennis well. B. She likes other sports as well.C. She is an enthusiastic tennis player.D. She is a professional athlete.3. A. At a paint store. B. At an oil market.C. At a science museum.D. At a gallery.4. A. Work in the yard. B. Buy some wood.C. Go to the bookstore.D. Take a walk.5. A. A taxi driver. B. A passenger. C. A car cleaner. D. A mechanic.6. A. Call a repairman. B. Get out the paper stuck.C. Turn to her colleague for help.D. Restart the machine.7. A. There are not enough gardens. B. Parking areas are full before 10:00.C. Parking areas are closed after 10:00.D. All classes begin at 10:00.8. A. The presentation will begin at noon.B. She’ll present her work to the man.C. She’d like to invite the man for lunch.D. She suggests working on the presentation at 12:00.9. A. The dormitory hours. B. The problem with the rules.C. The door number of the dormitory.D. The time to open the dormitory.10. A. The chairs didn’t need to be painted.B. He doesn’t like the color of the chairs.C. The park could have avoided the problem.D. The woman should have been more careful.Section BDirections: In section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Worried B. Surprised. C. Satisfied. D. Uninterested.12. A. It spoiled Juana’s reputation. B. It copied her ideas without permission.C. It bought Juana’s dishwashers.D. It wanted to share the dishwasher market.13. A. A successful business case. B. Juana’s waterless laundry.C. A case against a global company.D. The worldwide dishwasher market.Questions 14 through 16 are based on the following passage.14. A. Footprints. B. Food. C. Living insects. D. Orange seeds.15. A. Don’t touch animals under any circumstances.B. Don’t take away any natural objects from the park.C. Don’t leave litter in the park or throw any off the boat.D. Don’t transport animals from one island to another.16. A. To protect the guide’s interest. B. To improve the unique environment.C. To ensure a trouble-free visit.D. To get rid of illegal behaviors.Section CDirections: In section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. (本题做在答题纸上)Write only ONE WORD for each blank.Write no more than THREE WORDS for each blank.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages 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)Libraries are my world. I've been a patron all my life, and for the past nine years I (25) ______ (work) at multiple libraries and archives in and around Detroit. The library (26) ______ an institution has many roles, but as our country struggles through an economic crisis, I have watched the library where I work evolve into a career and business center, a community gathering place and a bastion for hope.In the spring of 2007 I got a library internship at the Southfield Public Library, just north of Detroit. Summers at SPL (27) ______ (be) usually slow, but that year, we experienced a library that hustled and bustled like science-fair project week, midterms or tax season. Yet patrons weren't looking for Mosby's Nursing Drug Reference or 1040 forms. They were coming for information on entrepreneurship and growing their small business.I interpreted people's interest in our business collection as the first step to (28) ______ (pursue) their dreams, but these patrons were not motivated by dreams. They were responding to reality, and they were looking for Plan B.Things worsened in 2008, and in 2009 the economic crisis continues to suffocate Michigan. Last year, we put up a display with a variety of job resources that we restocked every hour. Each night the library closed, the display was bare. (29) ______ we normally keep displays up for a week, we kept the job resources display up for months.Then there's the tightening credit market. People see the writing on the wall and they want to get educated. They can't afford a financial adviser, but checking books out is free. Some of (30) ______(popular) titles now are "Rich Dad, Poor Dad," "Think and Grow Rich," and "Suze Orman's 2009 Action Plan."The economic downturn affects us all. I have had to work long hours and don't get to see (31) ______ of my boyfriend or experience any kind of social life lately, but I am thankful to be in a position where I can help people overcome this struggle. In Michigan, we haven't lost hope. (32) ______ ______ ______ there are libraries here, there will always be hope. （349）(源自201306CET4)(B)It’s estimated that 300 million people in China are studying, or (33) ______ (study), English. That’s an impressive number and I can’t think of any other country in the world where one quarter of the population is so dedicated to (34) ______ (learn) a second language. But some people are questioning whether this “craze” for studying English is worthwhile.Professor Zhang Shuhua of the Chinese Academy of Social Sciences says that too much emphasis is placed on learning English and that it is a waste of education resources as well as a threat to the study of Chinese. He says that having English as a compulsory course in university “has distracted much of students’attention (35) ______ specialized subjects,”and that some students have been denied access to postgraduate education because they failed English. Others have admitted that studying so much English has made them (36) ______ (poor) Chinese speakers.Both of these criticisms are legitimate, but they beg the question of why so many Chinese still want to learn English. English, (37) ______ recognize, is the lingua franca of the modern world. It is the language of business and has become the language of international relations and culture. When people from different countries get together, they frequently speak in English rather than try to translate their native languages. It seems that everyone everywhere can speak at least some English.For China to be part of that international conversation, it is necessary that some level of English proficiency (38) ______ be achieved. But what, you may ask, about those who will never speak aword of English once they leave school? Well, for good or ill, they will still be surrounded by English. It is there in signs, in music, in movies and in the casual conversations they overhear of the increasing number of foreigners on the city streets. To know English is to be included in the rest of the world, (39) ______ ______ your world is limited to China.I agree with Professor Zhang on one point, (40) ______. English should not be a compulsory subject in university. For most, passing the CET is just the endless drudgery of memorizing word lists. There is little emphasis placed on communication. And if you can’t communicate in English after years of study in primary school, middle school and high school, a few more years in university probably won’t help. (390)Section BDirections: 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.For writers in Western cultures, autumn is a difficult season to describe. On the one hand, it is the end of the summer, and therefore a little sad. The nights draw in, and when you wake in the morni ng, there’s mist and it’s cooler: Winter is around the ___41___. American writer Ernest Hemingway wrote in his book A Moveable Feast: “You expected to be sad in the fall. Part of you died each year when the leaves fell from the trees and their branches were ___42___against the wind and the cold, wintry light.”On the other hand, autumn has its good side. There are so many changes in nature at this time of year, such as the reds and browns that the leaves change to, and the ___43___ they fall from the trees. French writer Albert Camus even though autumn was a second spring: “Autumn is a second spring when every leaf is a flower.” It’s a view you can also find in the most famous autumn poem in English literature, To Autumn by John Keats. In that poem Keats says that the autumn has its own songs, just like spring.Another autumn theme is wisdom. The arrival of the season is thought to be similar to a person becoming ___44___. Their summer peak may have been and gone, but old age has not yet come. At this time it’s thought that people have ___45___ a thing or two about life. The great Irish poet W.B. Yeats takes up this theme in his poem The Wild Swans at Coole. Yeats puts together a picture for the reader out of the ___46___ of the changing seasons in Coole Park in the west of Ireland, a place he knew well. Seeing and counting 59 swans, he remembers first making the count 19 years ago. He ___47___ whether he can still love like the lover swans do.Of course, many other themes and subject matters can play a part in the literature of autumn. For example, it’s the beginning of a new term of the school year. As you would expect, autumn can___48___ in writing for children and young people. But autumn writing usually ___49___ on the changes in nature that we see, which writers often use as a ___50___ for changes in human life. (372)III. Reading ComprehensionSection ADirections: 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.After the college-board examinations in June, Basil Duke Lee and five other boys from St. Regis School ___51___ the train for the West. Two got out at Pittsburgh, one slanted south toward St. Louis and two stayed in Chicago; from then on Basil was alone. It was the first time in his life that he had ever felt the need of tranquility, but now he took long breaths of it; for, though things had gone better toward the end, he had had a / an ___52___ year at school.He wore one of those extremely flat derbies (常礼帽) in vogue during the twelfth year of the century, and a blue business suit became a little too short for his constantly ___53___ body. Within he was by turns a disembodied (空洞的) spirit, almost ___54___ of his person and moving in a mist of impressions and emotions, and a fiercely competitive individual trying ___55___ to control the rush of events that were the steps in his own ___56___ from child to man. He believed that everything was a matter of ___57___ —the current principle of American education —and his fantastic ___58___ was continually leading him to expect too much. He wanted to be a great athlete, popular, brilliant and always happy. During this year at school, where he had been punished for his “freshness,” for fifteen years of thorough spoiling at home, he had grown uselessly introspective, and this ___59___ with that observation of others which is the beginning of wisdom. It was apparent that before he obtained much success in dealing with the world he would know that he’d been in a fight.Fifteen is of all ages the most difficult to ___60___———to put one’s fingers on and say, “That’s the way I was.” And all one can know is that somewhere between thirteen, boyhood’s ___61___, and seventeen, when one is a sort of counterfeit young man, there is a time when youth ___62___ hourly between one world and another —— pushed ceaselessly forward into unprecedented experiences and ___63___ trying to struggle back to the days when nothing had to be ___64___ for. Fortunately none of our contemporaries remember much more than we do of how we behaved in those days;nevertheless the ___65___ is about to be drawn aside for an inspection of Basil’s madness that summer. (380) （源自Taps at Reveille F. Scott Fitzgerald The Scandal Detectives The SaturdayEvening Post (28 April, 1928) ）51. A. boarded B. missed C. jumped D.followed52. A. happy B. unhappy C.memorable D.favourable53. A. swelling B. bending C. lengthening D. strengthening54. A. aware B.fond C. critical D. unconscious55. A. randomly B. desperately C. particularly D. indifferently56. A. evolution B. revolution C. solution D.introduction57. A. fact B. opinion C. course D. effort58. A. fashion B. ambition C. character D. treasure59. A. contacted B. associated C. interfered D. smashed60. A. digest B.describe C. deal D. locate61. A. majority B. minority C. senior D. junior62. A. floats B. varies C. fluctuates D. ranges63. A. successfully B. vainly C. wildly D. gently64. A. hunted B. provided C. compensated D. paid65. A.curtain B. adolescence C. portrait D. ceilingSection 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)There are people in Italy who can’t stand soccer. Not all Canadians love hockey. A similar situation exists in America, where there are those individuals you may be one of them who yawn or even frown when somebody mentions baseball. Baseball to them means boring hours watching grown men in funny tight outfits standing around in a field staring away while very little of anything happens. They tell you it’s a game better suited to the 19th century, slow, quiet, gentlemanly. These are the same people you may be one of them who love football because there’s the sport that glorifies “the hit”.By contrast, baseball seems abstract, cool, silent, still.On TV the game is fractured into a dozen perspectives, replays, close ups. The geometry of the game, however, is essential to understanding it. You will contemplate the game from one point as a painter does his subject; you may, of course, project yourself into the game. It is in this projection that the game affords so much space and tim e for involvement. The TV won’t do it for you.Take, for example, the third baseman. You sit behind the third base dugout and you watch him watching home plate. His legs are apart, knees flexed. His arms hang loose. He does a lot of this. The skeptic still cannot think of any other sports so still, so passive. But watch what happens every time the pitcher throws: the third baseman goes up on his toes, flexes his arms or brings the glove to a point in front of him, takes a step right or left, backward or forward, perhaps he glances across the field to check his first baseman’s position. Suppose the pitch is a ball. “Nothing happened,” you say. “I could have had my eyes closed.”The skeptic and the innocent must play the game. And this involvement in the stands is no more intellectual than listening to music is. Watch the third baseman. Smooth the dirt in front of you with one foot; smooth the pocket in your glove; watch the eyes of the batter, the speed of the bat, the sound of horsehide on wood. If football is a symphony of movement and theatre, baseball is chamber music, a spacious interlocking of notes, chores and responses. (381) （201312CET4）66. Those who don’t like baseball may complain that ______.A. it is only to the taste of the oldB. it involves fewer players than footballC. it is not exciting enoughD. it is pretentious and looks funny67. The author admits that ______.A. baseball is too peaceful for the youngB. baseball may seem boring when watched on TVC. football is more attracting than baseballD. baseball is more interesting than football68. By stating ‘I could have had my eyes closed.’ the author means (4th paragraph last sentence):A. The third baseman would rather sleep than play the game.B. Even if the third baseman closed his eyes a moment ago, it could make no difference to theresult.C. The third baseman is so good at baseball that he could finish the game with eyes closed all thetime and do his work well.D. The consequence was too bad he could not bear to see it.69. We can safely conclude that the author ______.A. likes footballB. hates footballC. hates baseballD. likes baseball(B)Some of the world’s most significant problems never hit headlines. One example comes from agriculture. Food riots and hunger make news. But the trend lying behind these matters is rarely talked about. This is the decline in the growth in yields of some of the world’s major crops. A new study by the University of Minnesota and McGill University in Montreal looks at where, and how far, this decline is occurring.The authors take a vast number of data points for the four most important crops: rice, wheat corn and soybeans. They find that on between 24% and 39% of all harvested areas, the improvement in yields that took place before the 1980s slowed down in the 1990s and 2000s.There are two worrying features of the slowdown. One is that it has been particularly sharp in the world’s most populous countries, India and China. Their ability to feed themselves has been an important source of relative stability both within the countries and on world food markets. That self-sufficiency cannot be taken for granted if yields continue to slow down or reverse.Second, yield growth has been lower in wheat and rice than in corn and soya beans. This is problematic because wheat and rice are more important as foods, accounting for around half of all calories consumed. Corn and soybeans are more important as feed grains. The authors note that “we have preferentially focused our crop improvement efforts on feeding animals and cars rather than on crops that feed people and are the basis of food security in much of the world.”The report qualifies the more optimistic findings of another new paper which suggests that the world will not have to dig up a lot more land for farming in order to feed 9 billion people in 2050, as the Food and Agriculture Organization has argued.Instead, it says, thanks to slowing population growth, land currently ploughed up for crops might be able to revert to forest or wilderness. This could happen. The trouble is that the forecast assumes continued improvements in yields, which may not actually happen. (350) 201506CET4）70. What does the author try to draw attention to?A. Food riots and hunger in the world.B. News headlines in the leading media.C. The decline of the grain yield growth.D. The food supply in populous countries.71. Why does the author mention India and China in particular?A. Their self-sufficiency is vital to the stability of world food markets.B. Their food yields have begun to decrease sharply in recent years.C. Their big populations are causing worldwide concerns.D. Their food self-sufficiency has been taken for granted.72. What does the new study by the two universities say about recent crop improvement efforts?A. They fail to produce the same remarkable results as before the 1980s.B. They contribute a lot to the improvement of human food production.C. They play a major role in guaranteeing the food security of the world.D. They focus more on the increase of animal feed than human food grains.73. What does the Food and Agriculture Organization say about world food production in the coming decades?A. The growing population will greatly increase the pressure on world food supplies.B. The optimistic prediction about food production should be viewed with caution.C. The slowdown of the growth in yields of major food crops will be reversed.D. The world will be able to feed its population without increasing farmland.(C)Among the more colorfu l characters of Leadville’s golden age were H.A.W. Tabor and his second wife, Elizabeth McCourt, better known as “Baby Doe”. Their history is fast becoming one of the legends of the Old West. Horace Austin Warner Tabor was a school teacher in Vermont. With his first wife and two children he left Vermont by covered wagon in 1855 to homestead in Kansas. Perhaps he did not find farming to his liking, or perhaps he was tempted by rumors of fortunes to be made in Colorado mines. At any rate, a few years later he moved west to the small Colorado mining camp known as California Gulch, which he later renamed Leadville when he became its leading citizen. “Great deposits of lead are sure to be found here.” he said.As it turned out, it was silver, not lead, that was to make Leadville’s fortune and wealth. Tabor knew little about mining himself, so he opened a general store, which sold everything from boots to salt, flour, and tobacco. It was his custom to “grubstake” prospective miners, in other words, to supply them with food and supplies, or “grub”, while they looked for ore, in return for which he would get a share in the mine if one was discovered. He did this for a number of years, but no one that he aided ever found anything of value.Finally one day in the ye ar 1878, so the story goes, two miners came in and asked for “grub”. Tabor had decided to quit supplying it because he had lost too much money that way. These were persistent, however, and Tabor was too busy to argue with them. “Oh help yourself. One more time won’t make any difference,” He said and went on selling shoes and hats to other customers. The two miners took $17 worth of supplies, in return for which they gave Tabor a one-third interest in their findings. They picked a barren place on the mountain side and began to dig. After nine days they struck a rich vein of silver. Tabor bought the shares of the other two men, and so the mine belonged to him alone. This mine, known as the “Pittsburgh Mine,” made 1 300 000 for Tabor in return for his $17 investment.Later Tabor bought the Matchless Mine on another barren hillside just outside the town for $117 000. This turned out to be even more fabulous than the Pittsburgh, yielding $35 000 worth of silver per day at one time. Leadville grew. Tabor became its first mayor, and later became lieutenant governor of the state. (435) （201112CET4）74. Leadville got its name for the following reasons EXCEPT ______.A. because Tabor became its leading citizenB. because great deposits of lead is expected to be found thereC. because it could bring good fortune to TaborD. because Tabor renamed it so75. The word “grubstake” in paragraph 2 means ______.A. to supply miners with food and suppliesB. to open a general storeC. to do one’s contribution to the develo pment of the mineD. to supply miners with food and supplies and in return get a share in the mine, if one wasdiscovered76. Tabor made his first fortune ______.A. by supplying two prospective miners and getting in return a one-third interest in the findingsB. because he was persuaded by the two miners to quit supplyingC. by buying the shares of the otherD. as a land speculator77. The underlying reason for Tabor’s life career is ______.A. purely accidentalB.based on the analysis of miner’s be ing very poor and their possibility of discovering profitablemining siteC. through the help from his second wifeD. he planned well and accomplished targets step by stepSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.When the Internet powerhouse Yahoo wanted to teach ethics to its employees, it faced a challenge familiar to multinational companies.Yahoo employs nearly 14,000 people at 25 sites worldwide. They would feel bored at sitting down in front of a dated video in which actors with 1980s haircuts tell them what to do. So it hired a company called The Network to design a game. In the game, the truck where Yahoo was founded traveled the world, turning into a boat and a helicopter along the way as it visited some of Yahoo's foreign offices. Participants play in game show-like scenarios that quiz them about conflicts of interest and doing business fairly. And employees note: Yahoo is tracking how well they do.Such activities draw more enthusiastic participation and teach more effectively than traditional methods. They are described as alternative-reality games (ARGs), involving both interactive and real-world elements. Besides teaching employees, ARGs have also been used inmany areas for a number of different purposes.From a marketing perspective, a number of very successful ARGs have been written as a way to build product awareness. A very popular ARG called I Love Bees was produced to market the 2004 video game Halo 2. At its height, I Love Bees received between two to three million unique visitors over the course of three months.ARGs are more than just a fun way to learn. They have also been used to solve real world problems. An ARG called World Without Oil was created to obtain collective input from players about dealing with the world's dependency on oil. World without Oil simulates the first 32 days of a global oil crisis and anybody could play by creating a personal story that recorded the imagined reality of their life in the crisis. World Without Oil's success on a small budget has opened the door for similar games to engage mainstream Internet users with climate change, education reform, governmental policy and other timely, vital issues. (330) （2012徐汇二模）(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)78. What challenge did yahoo face in teaching ethics to its employees?79. In the game designed for yahoo, participants had to answer questions about ________.80. What are the three major functions of ARG mentioned in the passage?81. The success of World Without Oil suggests that ARGs can ________.第II卷（共47分）I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.没有人不希望和平。

上海市复旦大学附中2016年10月2016～2017学年度高一第一学期期中考试数学试卷及参考答案教师专用

2016年10月2016～2017学年度上海市复旦大学附中高一上学期期中数学试卷一.填空题1.集合{1,2,3,…,2015,2016}的子集个数为.2.已知全集U＝R,集合A＝{x|x≤1},集合B＝{x|x≥2},则∁U(A∪B)＝.3.已知集合A＝{x|1≤x≤2},集合B＝{x|x≤a},若A∩B≠∅,则实数a的取值范围是.4.己知集合U＝{a,b,c,d,e,f},集合A＝{a,b,c,d},A∩B＝{b},∁U(A∪B)＝{f},求集合B.5.已知a2＞a1＞0,b2＞b1＞0,且a1＋a2＝b1＋b2＝1,记A＝a1b1＋a2b2,B＝a1b2＋a2b1,C＝,则按A、B、C从小到大的顺序排列是.6.已知Rt△ABC的周长为定值2,则它的面积最大值为.7.我们将b﹣a称为集合M＝{x|a≤x≤b}的“长度”,若集合M＝{x|m≤x≤m＋},N＝{x|n﹣0.5≤x≤n},且集合M和集合N都是集合{x|0≤x≤1}的子集,则集合M∩N的“长度”的最小值是.8.已知A＝{x|＞x},B＝{x|x(x﹣3)(x＋3)＞0},则A∩B＝.9.对于任意集合X与Y,定义:①X﹣Y＝{x|x∈X且x∉Y},②X△Y＝(X﹣Y)∪(Y﹣X),已知A ＝{y|y＝x2,x∈R},B＝{y|﹣2≤y≤2},则A△B＝.10.已知常数a是正整数,集合A＝{x||x﹣a|＜a＋,x∈Z},B＝{x||x|＜2a,x∈Z},则集合A∪B中所有元素之和为.11.非空集合G关于运算⊕满足:(1)对任意a,b∈G,都有a＋b∈G;(2)存在e∈G使得对于一切a∈G都有a⊕e＝e⊕a＝a,则称G是关于运算⊕的融洽集,现有下列集合与运算:①G是非负整数集,⊕:实数的加法;②G是偶数集,⊕:实数的乘法;③G是所有二次三项式构成的集合,⊕:多项式的乘法;④G＝{x|x＝a＋b,a,b∈Q},⊕:实数的乘法;其中属于融洽集的是(请填写编号)12.集合A＝{(x,y)|y＝a|x|,x∈R},B＝{(x,y)|y＝x＋a,x∈R},已知集合A∩B中有且仅有一个元素,则常数a的取值范围是.二.选择题13.已知集合A＝{1,2,3,…,2105,2016},集合B＝{x|x＝3k＋1,k∈Z},则A∩B中的最大元素是()A.2014B.2015C.2016D.以上答案都不对14.已知全集U＝A∪B中有m个元素,(∁U A)∪(∁U B)中有n个元素.若A∩B非空,则A∩B的元素个数为()A.mnB.m＋nC.n﹣mD.m﹣n15.命题“已知x,y∈R,如果x2＋y2＝0,那么x＝0且y＝0”的逆否命题是()A.已知x,y∈R,如果x2＋y2≠0,那么x≠0且y≠0B.已知x,y∈R,如果x2＋y2≠0,那么x≠0或y≠0C.已知x,y∈R,如果x≠0或y≠0,那么x2＋y2≠0D.已知x,y∈R,如果x≠0且y≠0,那么x2＋y2≠016.对任意实数a,b,c,给出下列命题:①“a＝b”是“ac＝bc”的充要条件;②“a＋5是无理数”是“a是无理数”的充要条件;③“a＞b”是“a2＞b2”的充分条件;④“a＜4”是“a＜3”的必要条件;其中真命题的个数是()A.1个B.2个C.3个D.4个三.解答题17.已知集合A＝{1,2,3},B＝{x|x2﹣(a＋1)x＋a＝0,x∈R},若A∪B＝A,求实数a.18.已知a,b,c∈R＋,求证:2(a3＋b3＋c3)≥ab2＋a2b＋bc2＋b2c＋ac2＋a2c.19.设正有理数a1是的一个近似值,令a2＝1＋,求证:(1)介于a1与a2之间;(2)a2比a1更接近于.20.已知对任意实数x,不等式mx2﹣(3﹣m)x＋1＞0成立或不等式mx＞0成立,求实数m的取值范围.21.已知关于x的不等式(4kx﹣k2﹣12k﹣9)(2x﹣11)＞0,其中k∈R;(1)试求不等式的解集A;(2)对于不等式的解集A,记B＝A∩Z(其中Z为整数集),若集合B为有限集,求实数k的取值范围,使得集合B中元素个数最少,并用列举法表示集合B.2016年10月2016～2017学年度上海市复旦大学附中高一(上)期中数学试卷参考答案与试题解析一.填空题1.(2016秋•杨浦区校级期中)集合{1,2,3,…,2015,2016}的子集个数为22016.【知识考查点】子集与真子集.【专题】集合思想;集合.【试题分析】对于有限集合,我们有以下结论:若一个集合中有n个元素,则它有2n个子集. 【试题解答】解:∵集合{1,2,3,…,2015,2016}中有2016个元素,∴集合M{1,2,3,…,2015,2016}的子集的个数为22016;故答案为:22016.【点评】本题考查了集合的子集个数,若一个集合中有n个元素,则它有2n个子集,有(2n﹣1)个真子集,属于基础题.2.(2016秋•杨浦区校级期中)已知全集U＝R,集合A＝{x|x≤1},集合B＝{x|x≥2},则∁U(A∪B)＝{x|1＜x＜2} .【知识考查点】交、并、补集的混合运算.【专题】集合思想;定义法;集合.【试题分析】根据并集与补集的定义,进行计算即可.【试题解答】解:全集U＝R,集合A＝{x|x≤1},集合B＝{x|x≥2},所以A∪B＝{x|x≤1或x≥2},所以∁U(A∪B)＝{x|1＜x＜2}.故答案为:{x|1＜x＜2}.【点评】本题考查了并集与补集的定义与应用问题,是基础题目.3.(2016秋•杨浦区校级期中)已知集合A＝{x|1≤x≤2},集合B＝{x|x≤a},若A∩B≠∅,则实数a的取值范围是[1,＋∞).【知识考查点】交集及其运算.【专题】计算题;集合思想;定义法;集合.【试题分析】题中条件:“A∩B≠∅,”表示两个集合的交集的结果不是空集,即可求解实数a的取值范围.【试题解答】解:集合A＝{x|1≤x≤2},集合B＝{x|x≤a},因为A∩B≠∅,所以a≥1故答案为:[1,＋∞)【点评】本题考查集合的关系、一元二次不等式的解法,考查运算能力,是基础题.4.(2016秋•杨浦区校级期中)己知集合U＝{a,b,c,d,e,f},集合A＝{a,b,c,d},A∩B＝{b},∁U(A∪B)＝{f},求集合B.【知识考查点】交、并、补集的混合运算.【专题】集合思想;综合法;集合.【试题分析】根据全集U,以及A与B并集的补集确定出A与B的并集,再根据A与B的交集及A,确定出B即可.【试题解答】解:∵U＝{a,b,c,d,e,f},∁U(A∪B)＝{f},∴A∪B＝{a,b,c,d,e},∵A∩B＝{b};A＝{a,b,c,d},∴b∈B,e∈B,b∉B,c∉B,d∉B,∴B＝{b,e}.【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.5.(2016秋•杨浦区校级期中)已知a2＞a1＞0,b2＞b1＞0,且a1＋a2＝b1＋b2＝1,记A＝a1b1＋a2b2,B＝a1b2＋a2b1,C＝,则按A、B、C从小到大的顺序排列是B＜C＜A.【知识考查点】不等式比较大小.【专题】计算题;转化思想;转化法;不等式.【试题分析】不妨令a1＝,a2＝,b1＝,b2＝,分别求出A,B,比较即可【试题解答】解:∵a2＞a1＞0,b2＞b1＞0,且a1＋a2＝b1＋b2＝1,不妨令a1＝,a2＝,b1＝,b2＝,A＝a1b1＋a2b2＝＋＝,B＝a1b2＋a2b1＝＋＝,∵C＝＝∴B＜C＜A故答案为:B＜C＜A.【点评】本题主要考查不等式与不等关系,利用特殊值代入法比较几个式子在限定条件下的大小关系,是一种简单有效的方法,属于基础题.6.(2016秋•杨浦区校级期中)已知Rt△ABC的周长为定值2,则它的面积最大值为3﹣2.【知识考查点】正弦定理.【专题】计算题;转化思想;综合法;解三角形;不等式的解法及应用.【试题分析】设直角边长为a,b,则斜边长为,利用直角三角形ABC的三边之和为2,可得a＋b＋＝2,利用基本不等式,即可求△ABC的面积的最大值.【试题解答】解:设直角边长为a,b,则斜边长为,∵直角三角形ABC的三边之和为2,∴a＋b＋＝2,∴2≥2＋,∴≤＝2﹣,∴ab≤6﹣4,∴S＝ba≤3﹣2,∴△ABC的面积的最大值为3﹣2.故答案为:3﹣2.【点评】本题考查基本不等式的运用,考查学生的计算能力,正确运用基本不等式是关键,属于中档题.7.(2016秋•杨浦区校级期中)我们将b﹣a称为集合M＝{x|a≤x≤b}的“长度”,若集合M＝{x|m≤x≤m＋},N＝{x|n﹣0.5≤x≤n},且集合M和集合N都是集合{x|0≤x≤1}的子集,则集合M∩N的“长度”的最小值是.【知识考查点】交集及其运算.【专题】计算题;新定义;转化思想;转化法;集合.【试题分析】当集合M∩N的长度的最小值时,M与N应分别在区间[0,1]的左右两端,由此能求出M∩N的长度的最小值.【试题解答】解:根据题意,M的长度为,N的长度为,当集合M∩N的长度的最小值时,M与N应分别在区间[0,1]的左右两端,故M∩N的长度的最小值是＝.故答案为:.【点评】本题考查交集的“长度”的最小值的求法,是基础题,解题时要认真审题,注意新定义的合理运用.8.(2016秋•杨浦区校级期中)已知A＝{x|＞x},B＝{x|x(x﹣3)(x＋3)＞0},则A∩B＝{x|﹣3＜x＜0} .【知识考查点】交集及其运算.【专题】计算题;方程思想;定义法;集合.【试题分析】先利用不等式的性质分别求出集合A和B,由此利用交集的性质能求出A∩B. 【试题解答】解:∵A＝{x|＞x}＝{x|﹣2≤x≤1,或x＜0},B＝{x|x(x﹣3)(x＋3)＞0}＝{x|﹣3＜x＜0或x＞3},∴A∩B＝{x|﹣3＜x＜0}.故答案为:{x|﹣3＜x＜0}.【点评】本题考查交集的求法,是中档题,解题时要认真审题,注意无理不等式和高次不等式性质的合理运用.9.(2016秋•杨浦区校级期中)对于任意集合X与Y,定义:①X﹣Y＝{x|x∈X且x∉Y},②X△Y ＝(X﹣Y)∪(Y﹣X),已知A＝{y|y＝x2,x∈R},B＝{y|﹣2≤y≤2},则A△B＝[﹣3,0)∪(3,＋∞).【知识考查点】子集与交集、并集运算的转换.【专题】综合题;方程思想;演绎法;集合.【试题分析】由A＝{y|y＝x2,x∈R}＝{y|y≥0},B＝{y|﹣2≤y≤2},先求出A﹣B＝{y|y＞2},B﹣A＝{y|﹣2≤y＜0},再求A△B的值.【试题解答】解:∵A＝{y|y＝x2,x∈R}＝{y|y≥0},B＝{y|﹣2≤y≤2},∴A﹣B＝{y|y＞2},B﹣A＝{y|﹣2≤y＜0},∴A△B＝{y|y＞2}∪{y|﹣2≤y＜0},故答案为:[﹣3,0)∪(3,＋∞).【点评】本题考查集合的交、并、补集的运算,解题时要认真审题,仔细解答,注意正确理解X ﹣Y＝{x|x∈X且x∉Y}、X△Y＝(X﹣Y)∪(Y﹣X).10.(2016秋•杨浦区校级期中)已知常数a是正整数,集合A＝{x||x﹣a|＜a＋,x∈Z},B＝{x||x|＜2a,x∈Z},则集合A∪B中所有元素之和为2a.【知识考查点】并集及其运算.【专题】集合思想;转化法;集合.【试题分析】分别求出集合A、B中的元素,从而求出A、B的并集,求和即可.【试题解答】解:A＝{x||x﹣a|＜a＋,x∈Z}＝{0,a,2a},B＝{x||x|＜2a,x∈Z}＝{﹣a,0,a},则集合A∪B＝{﹣a,0,a,2a},故集合A∪B中所有元素之和是2a,故答案为:2a.【点评】本题考查了集合的运算,考查解绝对值不等式问题,是一道基础题.11.(2016秋•杨浦区校级期中)非空集合G关于运算⊕满足:(1)对任意a,b∈G,都有a＋b∈G;(2)存在e∈G使得对于一切a∈G都有a⊕e＝e⊕a＝a,则称G是关于运算⊕的融洽集,现有下列集合与运算:①G是非负整数集,⊕:实数的加法;②G是偶数集,⊕:实数的乘法;③G是所有二次三项式构成的集合,⊕:多项式的乘法;④G＝{x|x＝a＋b,a,b∈Q},⊕:实数的乘法;其中属于融洽集的是①④(请填写编号)【知识考查点】元素与集合关系的判断.【专题】新定义;集合思想;集合.【试题分析】逐一验证几个选项是否分别满足“融洽集”的两个条件,若两个条件都满足,是“融洽集”,有一个不满足,则不是“融洽集”.【试题解答】解:①对于任意非负整数a,b知道:a＋b仍为非负整数,所以a⊕b∈G;取e＝0,及任意非负整数a,则a＋0＝0＋a＝a,因此G对于⊕为整数的加法运算来说是“融洽集”;②对于任意偶数a,b知道:a＋b仍为偶数,故有a＋b∈G;但是不存在e∈G,使对一切a∈G都有a⊕e＝e⊕a＝a,故②的G不是“融洽集”.③对于G＝{二次三项式},若a、b∈G时,a,b的两个同类项系数,则其积不再为二次三项式,故G不是和谐集,故③不正确;④G＝{x|x＝a＋b,a,b∈Q},设x1＝a＋b,x2＝c＋d,则设x1＋x2＝(a＋c)＋(b＋d),属于集合G,取e＝1,a×1＝1×a＝a,因此G对于⊕实数的乘法运算来说是“融洽集”,故④中的G是“融洽集”.故答案为①④.【点评】本题考查了对新定义“融洽集”理解能力,及对有关知识的掌握情况.关键是看所给的数集是否满足“融洽集”的两个条件.12.(2016秋•杨浦区校级期中)集合A＝{(x,y)|y＝a|x|,x∈R},B＝{(x,y)|y＝x＋a,x∈R},已知集合A∩B中有且仅有一个元素,则常数a的取值范围是[﹣1,1] .【知识考查点】交集及其运算.【专题】计算题;转化思想;转化法;集合.【试题分析】由已知得a|x|＝x＋a有1个解,由此能求出常数a的取值范围.【试题解答】解:∵集合A＝{(x,y)|y＝a|x|,x∈R},B＝{(x,y)|y＝x＋a,x∈R},集合A∩B中有且仅有一个元素,∴a|x|＝x＋a有1个解,若x≥0,ax＝x＋a,x＝,若x＜0,﹣ax＝x＋a,x＝﹣,由已知得或或或,解得﹣1≤a≤1.∴常数a的取值范围是[﹣1,1].故答案为:[﹣1,1].【点评】本题考查常数的取值范围的求法,是基础题,解题时要认真审题,是基础题,解题时要认真审题,注意交集性质的合理运用.二.选择题13.(2016秋•杨浦区校级期中)已知集合A＝{1,2,3,…,2105,2016},集合B＝{x|x＝3k＋1,k∈Z},则A∩B中的最大元素是()A.2014B.2015C.2016D.以上答案都不对【知识考查点】交集及其运算.【专题】计算题;集合思想;定义法;集合.【试题分析】由题意求出A与B的交集,即可作出判断.【试题解答】解:∵A＝{1,2,3,…,2105,2016},集合B＝{x|x＝3k＋1,k∈Z}∴则A∩B中的最大元素是2014.故选:A.【点评】此题考查了交集及其运算,熟练掌握交集的定义是解本题的关键.14.(2009•江西)已知全集U＝A∪B中有m个元素,(∁U A)∪(∁U B)中有n个元素.若A∩B非空,则A∩B的元素个数为()A.mnB.m＋nC.n﹣mD.m﹣n【知识考查点】Venn图表达集合的关系及运算.【专题】数形结合.【试题分析】要求A∩B的元素个数,可以根据已知绘制出满足条件的韦恩图,根据图来分析(如解法一),也可以利用德摩根定理解决(如解法二).【试题解答】解法一:∵(C U A)∪(C U B)中有n个元素,如图所示阴影部分,又∵U＝A∪B中有m个元素,故A∩B中有m﹣n个元素.解法二:∵(C U A)∪(C U B)＝C U(A∩B)有n个元素,又∵全集U＝A∪B中有m个元素,由card(A)＋card(C U A)＝card(U)得,card(A∩B)＋card(C U(A∩B))＝card(U)得,card(A∩B)＝m﹣n,故选D.【点评】解答此类型题目时,要求对集合的性质及运算非常熟悉,除教材上的定义,性质,运算律外,还应熟练掌握:①(C U A)∪(C U B)＝C U(A∩B)②(C U A)∩(C U B)＝C U(A∪B)③card(A∪B)＝card(A)＋card(B)﹣card(A∩B)等.15.(2016秋•杨浦区校级期中)命题“已知x,y∈R,如果x2＋y2＝0,那么x＝0且y＝0”的逆否命题是()A.已知x,y∈R,如果x2＋y2≠0,那么x≠0且y≠0B.已知x,y∈R,如果x2＋y2≠0,那么x≠0或y≠0C.已知x,y∈R,如果x≠0或y≠0,那么x2＋y2≠0D.已知x,y∈R,如果x≠0且y≠0,那么x2＋y2≠0【知识考查点】四种命题间的逆否关系.【专题】定义法;简易逻辑.【试题分析】根据已知中原命题,写出逆否命题,可得答案.【试题解答】解:命题“已知x,y∈R,如果x2＋y2＝0,那么x＝0且y＝0”的逆否命题是“已知x,y∈R,如果x≠0或y≠0,那么x2＋y2≠0”故选:C【点评】本题考查的知识点是四种命题,难度不大,属于基础题.16.(2016秋•杨浦区校级期中)对任意实数a,b,c,给出下列命题:①“a＝b”是“ac＝bc”的充要条件;②“a＋5是无理数”是“a是无理数”的充要条件;③“a＞b”是“a2＞b2”的充分条件;④“a＜4”是“a＜3”的必要条件;其中真命题的个数是()A.1个B.2个C.3个D.4个【知识考查点】命题的真假判断与应用;必要条件、充分条件与充要条件的判断.【专题】综合法;简易逻辑.【试题分析】逐项判断即可.①由ac＝bc不能推出a＝b;②由5是有理数易判断;③根据不等式的性质可得;④根据充分必要条件的定义易得.【试题解答】解:①由“a＝b“可得ac＝bc,但当ac＝bc时,不能得到a＝b,故“a＝b”是“ac＝bc”的充分不必要条件,故①错误;②因为5是有理数,所以当a＋5是无理数时,a必为无理数,反之也成立,故②正确;③取a＝1,b＝﹣2,此时a2＜b2,故③错误;④当a＜4时,不能推出a＜3;当a＜3时,有a＜4成立,故“a＜4”是“a＜3”的必要不充分条件,故④正确.综上可得正确的命题有2个.故选:B.【点评】本题考查充分必要条件的判断,掌握充分必要条件的定义是关键.属于基础题.三.解答题17.(2016秋•杨浦区校级期中)已知集合A＝{1,2,3},B＝{x|x2﹣(a＋1)x＋a＝0,x∈R},若A∪B ＝A,求实数a.【知识考查点】并集及其运算.【专题】计算题;分类讨论;集合.【试题分析】根据A∪B＝A,得到B⊆A,然后分B为空集和不是空集讨论,A为空集时,只要二次方程的判别式小于0即可,不是空集时,分别把1和2代入二次方程求解a的范围,注意求出a后需要验证.【试题解答】解:由A∪B＝A,得B⊆A.①若B＝∅,则△＝(a＋1)2﹣4a＜0,解得:a∈∅;②若1∈B,△＝(a＋1)2﹣4a＝0,此时a＝1,满足12﹣a﹣1＋a＝0,此时B＝{1},符合题意;③若2∈B,则22﹣2a﹣2＋a＝0,解得:a＝2,此时A＝{2,1},满足题意.④若3∈B,则32﹣3a﹣3＋a＝0,解得:a＝3,此时A＝{3,1},满足题意.综上所述,实数a的值为:1,2,3.【点评】本题考查了并集及其运算,考查了分类讨论的数学思想,求出a值后的验证是解答此题的关键,是基础题.18.(2016秋•杨浦区校级期中)已知a,b,c∈R＋,求证:2(a3＋b3＋c3)≥ab2＋a2b＋bc2＋b2c＋ac2＋a2c.【知识考查点】不等式的证明.【专题】证明题;转化思想;演绎法;不等式的解法及应用.【试题分析】作差,因式分解,即可得到结论.【试题解答】证明:(a3＋b3)﹣(a2b＋ab2)＝a2(a﹣b)＋b2(b﹣a)＝(a﹣b)(a2﹣b2)＝(a﹣b)2(a＋b)∵a＞0,b＞0,∴(a3＋b3)﹣(a2b＋ab2)≥0∴a3＋b3≥a2b＋ab2.同理b3＋c3≥bc2＋b2c,a3＋c3≥ac2＋a2c,三式相加,可得2(a3＋b3＋c3)≥ab2＋a2b＋bc2＋b2c＋ac2＋a2c.【点评】本题考查不等式的证明,考查作差法的运用,考查学生分析解决问题的能力,属于中档题.19.(2016秋•杨浦区校级期中)设正有理数a1是的一个近似值,令a2＝1＋,求证:(1)介于a1与a2之间;(2)a2比a1更接近于.【知识考查点】二分法求方程的近似解.【专题】证明题;转化思想;作差法;不等式.【试题分析】(1)利用作差法,再因式分解,确定其符号,即可得到结论;(2)利用作差法,判断|a2﹣|﹣|a1﹣|＜0,即可得到结论【试题解答】证明:(1)a2﹣＝1＋﹣＝,∵若a1＞,∴a1﹣＞0,而1﹣＜0,∴a2＜∵若a1＜,∴a1﹣＜0,而1﹣＜0,∴a2＞,故介于a1与a2之间;(2)|a2﹣|﹣|a1﹣|＝﹣|a1﹣|＝|a1﹣|×,∵a1＞0,﹣2＜0,|a1﹣|＞0,∴|a2﹣|﹣|a1﹣|＜0∴|a2﹣|＜|a1﹣|∴a2比a1更接近于.【点评】本题考查不等式的证明,考查作差法的运用,确定差的符号是关键.20.(2016秋•杨浦区校级期中)已知对任意实数x,不等式mx2﹣(3﹣m)x＋1＞0成立或不等式mx＞0成立,求实数m的取值范围.【知识考查点】一元二次不等式的解法.【专题】分类讨论;转化思想;不等式的解法及应用.【试题分析】①对任意实数x,不等式mx2﹣(3﹣m)x＋1＞0成立,对m分类讨论,m＝0时,易判断出.m≠0时,,解出即可得出.②对任意实数x,不等式mx＞0成立,m∈∅.【试题解答】解:①对任意实数x,不等式mx2﹣(3﹣m)x＋1＞0成立,m＝0时化为:﹣3x＋1＞0,不成立,舍去.m≠0时,,解得.②对任意实数x,不等式mx＞0成立,m∈∅.综上可得:.∴实数m的取值范围是.【点评】本题考查了一元二次不等式的解法,考查了分类讨论方法、推理能力与计算能力,属于中档题.21.(2016秋•杨浦区校级期中)已知关于x的不等式(4kx﹣k2﹣12k﹣9)(2x﹣11)＞0,其中k∈R;(1)试求不等式的解集A;(2)对于不等式的解集A,记B＝A∩Z(其中Z为整数集),若集合B为有限集,求实数k的取值范围,使得集合B中元素个数最少,并用列举法表示集合B.【知识考查点】一元二次不等式的解法.【专题】分类讨论;不等式的解法及应用;不等式.【试题分析】(1)对k分类讨论,利用一元二次不等式的解法即可得出.(2)根据B＝A∩Z(其中Z为整数集),集合B为有限集,即可得出.【试题解答】解:(1)①当k＜0,A＝{x|};②当k＝0,A＝{x|x};③当0＜k＜1或k＞9,A＝{x|x,或x＞};④当1≤k≤9,A＝{x|x＜,或x＞};(2)B＝A∩Z(其中Z为整数集),集合B为有限集,只有k＜0,B＝{2,3,4,5}.【点评】本题考查了一元二次不等式的解法,考查了分类讨论方法、推理能力与计算能力,属于中档题.11。

上海市复旦大学附属中学2024届高三上学期期中数学试题

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