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上海交大附中2017-2018学年高二(下)期中数学试卷 Word版含解析
2017-2018学年上海交大附中高二（下）期中数学试卷一、填空题（本大题满分56分）1．抛物线y2=x的准线方程为______．2．计算i+2i2+3i3+…+2016i2016=______．3．异面直线a，b成60°，直线c⊥a，则直线b与c所成的角的范围为______．4．如图，在棱长为1的正方体ABCD﹣A1B1C1D1中，M、N分别是A1B1和BB1的中点，那么直线AM和CN所成角的余弦值为______．5．已知△AOB内接于抛物线y2=4x，焦点F是△AOB的垂心，则点A，B的坐标______．6．在图中，G、H、M、N分别是正三棱柱的顶点或所在棱的中点，则表示直线GH、MN 是异面直线的图形有______．（填上所有正确答案的序号）7．已知复数z1，z2满足|z1|=|z2|=1，|z1﹣z2|=，则|z1+z2|等于______．8．三个平面能把空间分为______部分．（填上所有可能结果）9．已知复数Z1，Z2满足|Z1|=2，|Z2|=3，若它们所对应向量的夹角为60°，则=______．10．已知复数z1，z2满足|z1|=|z2|=1，，则复数|z1+z2|=______．11．二面角α﹣l﹣β的平面角为120°，在面α内，AB⊥l于B，AB=2在平面β内，CD⊥l 于D，CD=3，BD=1，M是棱l上的一个动点，则AM+CM的最小值为______．12．已知虚数z=（x﹣2）+yi（x，y∈R），若|z|=1，则的取值范围是______．13．已知F是抛物线C：y2=4x的焦点，A，B是C上的两个点，线段AB的中点为M（2，2），则△ABF的面积等于______．14．如图，直线y=x与抛物线y=x2﹣4交于A，B两点，线段AB的垂直平分线与直线y=﹣5交于Q点，当P为抛物线上位于线段AB下方（含A，B）的动点时，则△OPQ面积的最大值为______．二、选择题（本大题满分20分，共计4小题，每题5分）15．在正方体AC1中，E，F分别是线段BC，CD1的中点，则直线A1B与直线EF的位置关系是（）A．相交 B．异面 C．平行 D．垂直16．（1）两个共轭复数的差是纯虚数；（2）两个共轭复数的和不一定是实数；（3）若复数a+bi（a，b∈R）是某一元二次方程的根，则a﹣bi是也一定是这个方程的根；（4）若z为虚数，则z的平方根为虚数，其中正确的个数为（）A．3 B．2 C．1 D．017．如图所示，在正方体ABCD﹣A1B1C1D1的侧面ABB1A1内有一动点P到直线A1B1和直线BC的距离相等，则动点P所在曲线形状为（）A．B．C．D．18．设P表示一个点，a，b表示两条直线，α，β表示两个平面，给出下列四个，其中正确的是（）①P∈a，P∈α⇒a⊂α②a∩b=P，b⊂β⇒a⊂β③a∥b，a⊂α，P∈b，P∈α⇒b⊂α④α∩β=b，P∈α，P∈β⇒P∈b．A．①②B．②③C．①④D．③④三、解答题（满分74分）19．已知复数z1=+（a2﹣3）i，z2=2+（3a+1）i（a∈R，i是虚数单位）．（1）若复数z1﹣z2在复平面上对应点落在第一象限，求实数a的取值范围；（2）若虚数z1是实系数一元二次方程x2﹣6x+m=0的根，求实数m值．20．如图，已知直四棱柱ABCD﹣A1B1C1D1，DD1⊥底面ABCD，底面ABCD为平行四边形，∠DAB=45°，且AD，AB，AA1三条棱的长组成公比为的等比数列，（1）求异面直线AD1与BD所成角的大小；（2）求二面角B﹣AD1﹣D的大小．21．已知z为复数，ω=z+为实数，（1）当﹣2＜ω＜10，求点Z的轨迹方程；（2）当﹣4＜ω＜2时，若u=（α＞0）为纯虚数，求：α的值和|u|的取值范围．22．动圆M与圆（x﹣1）2+y2=1相外切且与y轴相切，则动圆M的圆心的轨迹记C，（1）求轨迹C的方程；（2）定点A（3，0）到轨迹C上任意一点的距离|MA|的最小值；（3）经过定点B（﹣2，1）的直线m，试分析直线m与轨迹C的公共点个数，并指明相应的直线m的斜率k是否存在，若存在求k的取值或取值范围情况[要有解题过程，没解题方程只有结论的只得结论分]．23．已知复数z1=m+ni（m，n∈R），z=x+yi（x，y∈R），z2=2+4i且．（1）若复数z1对应的点M（m，n）在曲线上运动，求复数z所对应的点P（x，y）的轨迹方程；（2）将（1）中的轨迹上每一点按向量方向平移个单位，得到新的轨迹C，求C的轨迹方程；（3）过轨迹C上任意一点A（异于顶点）作其切线，交y轴于点B，求证：以线段AB为直径的圆恒过一定点，并求出此定点的坐标．2015-2016学年上海交大附中高二（下）期中数学试卷参考答案与试题解析一、填空题（本大题满分56分）1．抛物线y2=x的准线方程为x=﹣．【考点】抛物线的简单性质．【分析】抛物线y2=x的焦点在x轴上，且开口向右，2p=1，由此可得抛物线y2=x的准线方程．【解答】解：抛物线y2=x的焦点在x轴上，且开口向右，2p=1∴∴抛物线y2=x的准线方程为x=﹣故答案为：x=﹣2．计算i+2i2+3i3+…+2016i2016=1008﹣1008i．【考点】复数代数形式的混合运算．【分析】利用复数单位的幂运算，化简求解即可．【解答】解：i+2i2+3i3+…+2016i2016=（i﹣2﹣3i+4）+（5i﹣6﹣7i+8）+…+2016=504（2﹣2i）=1008﹣1008i．故答案为：1008﹣1008i．3．异面直线a，b成60°，直线c⊥a，则直线b与c所成的角的范围为[30°，90°] ．【考点】异面直线及其所成的角．【分析】作b的平行线b′，交a于O点，所有与a垂直的直线平移到O点组成一个与直线a垂直的平面α，O点是直线a与平面α的交点，在直线b′上取一点P，作垂线PP'⊥平面α，交平面α于P'，∠POP'是b′与面α的线面夹角，在平面α所有与OP'垂直的线，由此能求出直线b与c所成的角的范围．【解答】解：如图作b的平行线b′，交a于O点，所有与a垂直的直线平移到O点组成一个与直线a垂直的平面α，O点是直线a与平面α的交点，在直线b′上取一点P，作垂线PP'⊥平面α，交平面α于P'，∠POP'是b′与面α的线面夹角，∠POP'=30°．在平面α中，所有与OP'平行的线与b′的夹角都是30°．在平面α所有与OP'垂直的线∵PP'⊥平面α，∴该线⊥PP′，则该线⊥平面OPP'，∴该线⊥b'，与b'的夹角为90°，与OP'夹角大于0°，小于90°的线，与b'的夹角为锐角且大于30°．∴直线b与c所成的角的范围[30°，90°]．故答案为：[30°，90°]．4．如图，在棱长为1的正方体ABCD﹣A1B1C1D1中，M、N分别是A1B1和BB1的中点，那么直线AM和CN所成角的余弦值为．【考点】异面直线及其所成的角．【分析】先通过平移将两条异面直线平移到同一个起点B1，得到的锐角或直角就是异面直线所成的角，在三角形中再利用余弦定理求出此角即可．【解答】解：如图，将AM平移到B1E，NC平移到B1F，则∠EB1F为直线AM与CN所成角设边长为1，则B1E=B1F=，EF=∴cos∠EB1F=，故答案为5．已知△AOB内接于抛物线y2=4x，焦点F是△AOB的垂心，则点A，B的坐标A（5，2），B（5，﹣2）．【考点】抛物线的简单性质．【分析】根据垂心的性质可得A，B关于x轴对称，且AF⊥OB，设A（，y1）（y1＞0），则B（，﹣y1）．求出AF，OB的斜率，令k OB•k AF=﹣1解出y1即可得出A，B的坐标．【解答】解：抛物线焦点F（1，0），∵焦点F是△AOB的垂心，∴直线AB⊥x轴．∴A，B关于x轴对称．设A（，y1）（y1＞0），则B（，﹣y1）．∴k OB==﹣．k AF==．∵焦点F是△AOB的垂心，∴AF⊥OB．∴k OB•k AF=﹣1，即﹣•=﹣1，解得y1=2．∴A（5，2），B（5，﹣2）．故答案为：A（5，2），B（5，﹣2）．6．在图中，G、H、M、N分别是正三棱柱的顶点或所在棱的中点，则表示直线GH、MN 是异面直线的图形有（2）、（4）．（填上所有正确答案的序号）【考点】异面直线的判定．【分析】图（1）中，直线GH∥MN，图（2）中M∉面GHN，图（3）中GM∥HN，图（4）中，H∉面GMN．【解答】解析：如题干图（1）中，直线GH∥MN；图（2）中，G、H、N三点共面，但M∉面GHN，因此直线GH与MN异面；图（3）中，连接MG，GM∥HN，因此，GH与MN共面；图（4）中，G、M、N共面，但H∉面GMN，∴GH与MN异面．所以图（2）、（4）中GH与MN异面．故答案为：（2）、（4）7．已知复数z1，z2满足|z1|=|z2|=1，|z1﹣z2|=，则|z1+z2|等于1．【考点】复数求模；复数的代数表示法及其几何意义．【分析】复数z1，z2满足|z1|=|z2|=1，故可令z1=cosA+isinA，z2=cosB+isinB，代入，|z1﹣z2|=，及|z1+z2|，比较即可求得所求的答案【解答】解：∵复数z1，z2满足|z1|=1，|z2|=1，可令z1=cosA+isinA，z2=cosB+isinB∵|z1﹣z2|=，故有（cosA﹣cosB）2+（sinA﹣sinB）2=3，整理得2cosAcosB+2sinAsinB=﹣1又|z1+z2|2=（cosA+cosB）2+（sinA+sinB）2=2+2cosAcosB+2sinAsinB=1∴|z1+z2|=1故答案为：1．8．三个平面能把空间分为4，或6，或7，或8部分．（填上所有可能结果）【考点】平面的基本性质及推论．【分析】此类问题可以借助实物模型来研究，用房屋的结构来研究就行．【解答】解：若三个平面两两平行，则把空间分成4部分；若三个平面两两相交，且共线，则把空间分成6部分；若三个平面两两相交，且有三条交线，则把空间分成7部分；当两个平面相交，第三个平面同时与两个平面相交时，把空间分成8部分，故答案为：4，或6，或7，或8．9．已知复数Z1，Z2满足|Z1|=2，|Z2|=3，若它们所对应向量的夹角为60°，则=．【考点】余弦定理的应用；复数求模．【分析】由余弦定理可得Z1+Z2|=，|Z1﹣Z2|=，故==【解答】解：如图在三角形OBC中由余弦定理得|Z1+Z2|=|OB|==，同理可得|Z1﹣Z2|=|CA=|=，∴===10．已知复数z1，z2满足|z1|=|z2|=1，，则复数|z1+z2|=．【考点】复数求模．【分析】复数z1，z2满足|z1|=|z2|=1，，判断三角形是直接三角形，即可求得所求的答案．【解答】解：因为|z1|=|z2|=1，，所以复数z1，z2，构成的三角形是直角三角形，|z1+z2|是平行四边形的对角线，则|z1+z2|=．故答案为：．11．二面角α﹣l﹣β的平面角为120°，在面α内，AB⊥l于B，AB=2在平面β内，CD⊥l于D，CD=3，BD=1，M是棱l上的一个动点，则AM+CM的最小值为．【考点】点、线、面间的距离计算．【分析】要求出AM+CM的最小值，可将空间问题转化成平面问题，将二面角展开成平面中在BD上找一点使AM+CM即可，而当A、M、C在一条直线时AM+CM的最小值，从而求出对角线的长即可．【解答】解：将二面角α﹣l﹣β平摊开来，即为图形当A、M、C在一条直线时AM+CM的最小值，最小值即为对角线AC而AE=5，EC=1故AC=故答案为：12．已知虚数z=（x﹣2）+yi（x，y∈R），若|z|=1，则的取值范围是[﹣，]．【考点】复数求模；复数的代数表示法及其几何意义．【分析】根据复数的模，利用模长公式得：（x﹣2）2+y2=1，根据表示动点（x，y）与原点（0，0）连线的斜率．根据直线与圆相切的性质得到结果．【解答】解：∵复数（x﹣2）+yi（x，y∈R）的模为1，∴（x﹣2）2+y2=1根据表示动点（x，y）到定点（0，0）的斜率知：的最大值是，同理求得最小值是﹣，如图示：∴的取值范围是[﹣，]故答案为：[﹣，]．13．已知F是抛物线C：y2=4x的焦点，A，B是C上的两个点，线段AB的中点为M（2，2），则△ABF的面积等于2．【考点】抛物线的简单性质．【分析】设A（x1，y1），B（x2，y2），则，=4x2，两式相减可得：（y1+y2）（y1﹣y2）=4（x1﹣x2），利用中点坐标公式、斜率计算公式可得k AB，可得直线AB的方程为：y﹣2=x﹣2，化为y=x，与抛物线方程联立可得A，B的坐标，利用弦长公式可得|AB|，再利用点到直线的距离公式可得点F到直线AB的距离d，利用三角形面积公式求得答案．【解答】解：∵F是抛物线C：y2=4x的焦点，∴F（1，0）．设A（x1，y1），B（x2，y2），则，=4x2，两式相减可得：（y1+y2）（y1﹣y2）=4（x1﹣x2），∵线段AB的中点为M（2，2），∴y1+y2=2×2=4，又=k AB，4k AB=4，解得k AB=1，∴直线AB的方程为：y﹣2=x﹣2，化为y=x，联立，解得，，∴|AB|==4．点F到直线AB的距离d=，∴S△ABF===2，故答案为：2．14．如图，直线y=x与抛物线y=x2﹣4交于A，B两点，线段AB的垂直平分线与直线y=﹣5交于Q点，当P为抛物线上位于线段AB下方（含A，B）的动点时，则△OPQ面积的最大值为30．【考点】二次函数的性质．【分析】把直线方程抛物线方程联立求得交点A，B的坐标，则AB中点M的坐标可得，利用AB的斜率推断出AB垂直平分线的斜率，进而求得AB垂直平分线的方程，把y=﹣5代入求得Q的坐标；设出P的坐标，利用P到直线0Q的距离求得三角形的高，利用两点间的距离公式求得QO的长，最后利用三角形面积公式表示出三角形OPQ，利用x的范围和二次函数的单调性求得三角形面积的最大值．【解答】解：直线y=x与抛物线y=x2﹣4联立，得到A（﹣4，﹣2），B（8，4），从而AB的中点为M（2，1），由k AB═，直线AB的垂直平分线方程y﹣1=﹣2（x﹣2）．令y=﹣5，得x=5，∴Q（5，﹣5）．∴直线OQ的方程为x+y=0，设P（x，x2﹣4）．∵点P到直线OQ的距离d==|x2+8x﹣32|，|OQ|=5，∴S△OPQ=|OQ|d=|x2+8x﹣32|，|∵P为抛物线上位于线段AB下方的点，且P不在直线OQ上，∴﹣4≤x＜4﹣4或4﹣4＜x≤8．∵函数y=x2+8x﹣32在区间[﹣4，8]上单调递增，∴当x=8时，△OPQ的面积取到最大值30．故答案为：30．二、选择题（本大题满分20分，共计4小题，每题5分）15．在正方体AC1中，E，F分别是线段BC，CD1的中点，则直线A1B与直线EF的位置关系是（）A．相交 B．异面 C．平行 D．垂直【考点】空间中直线与直线之间的位置关系．【分析】直线AB与直线外一点E确定的平面为A1BCD1，EF⊂平面A1BCD1，且两直线不平行，故两直线相交，可得结论．【解答】解：如图，在正方体AC1中：∵A1B∥D1C∴A1B与D1C可以确定平面A1BCD1，又∵EF⊂平面A1BCD1，且两直线不平行，∴直线A1B与直线EF的位置关系是相交，故选A．16．（1）两个共轭复数的差是纯虚数；（2）两个共轭复数的和不一定是实数；（3）若复数a+bi（a，b∈R）是某一元二次方程的根，则a﹣bi是也一定是这个方程的根；（4）若z为虚数，则z的平方根为虚数，其中正确的个数为（）A．3 B．2 C．1 D．0【考点】的真假判断与应用；复数的基本概念．【分析】直接利用复数的基本概念频道的真假即可．【解答】解：（1）两个共轭复数的差是纯虚数；如果两个复数是实数，差值也是实数，所以（1）不正确；（2）两个共轭复数的和不一定是实数；不正确，和一定是实数；（3）若复数a+bi（a，b∈R）是某一元二次方程的根，则a﹣bi是也一定是这个方程的根；不正确，因为实系数方程的虚根是共轭复数，所以（3）不正确；（4）若z为虚数，则z的平方根为虚数，如果虚数为i，则设z=x+yi（x，y∈R），由z2=（x+yi）2=i，得x2﹣y2+2xyi=i，∴，解得：或．∴z=+i或z=﹣﹣i．所以正确．故选：C．17．如图所示，在正方体ABCD﹣A1B1C1D1的侧面ABB1A1内有一动点P到直线A1B1和直线BC的距离相等，则动点P所在曲线形状为（）A．B．C．D．【考点】轨迹方程．【分析】点P到BC的距离就是当P点到B的距离，它等于到直线A1B1的距离，满足抛物线的定义，推断出P的轨迹是以B为焦点，以A1B1为准线的过A的抛物线的一部分．从而得出正确选项．【解答】解：依题意可知点P到BC的距离就是当P点B的距离，P到点B的距离等于到直线A1B1的距离，根据抛物线的定义可知，动点P的轨迹是以B为焦点，以A1B1为准线的过A的抛物线的一部分．A的图象为直线的图象，排除A．B项中B不是抛物线的焦点，排除B．D项不过A点，D排除．故选C．18．设P表示一个点，a，b表示两条直线，α，β表示两个平面，给出下列四个，其中正确的是（）①P∈a，P∈α⇒a⊂α②a∩b=P，b⊂β⇒a⊂β③a∥b，a⊂α，P∈b，P∈α⇒b⊂α④α∩β=b，P∈α，P∈β⇒P∈b．A．①②B．②③C．①④D．③④【考点】空间中直线与平面之间的位置关系．【分析】根据公理1及直线在面内的定义，逐一对四个结论进行分析，即可求解．【解答】解：当a∩α=P时，P∈a，P∈α，但a⊄α，∴①错；当a∩β=P时，②错；如图∵a∥b，P∈b，∴P∉a，∴由直线a与点P确定唯一平面α，又a∥b，由a与b确定唯一平面β，但β经过直线a与点P，∴β与α重合，∴b⊂α，故③正确；两个平面的公共点必在其交线上，故④正确．故选D三、解答题（满分74分）19．已知复数z1=+（a2﹣3）i，z2=2+（3a+1）i（a∈R，i是虚数单位）．（1）若复数z1﹣z2在复平面上对应点落在第一象限，求实数a的取值范围；（2）若虚数z1是实系数一元二次方程x2﹣6x+m=0的根，求实数m值．【考点】复数代数形式的混合运算；复数的基本概念；复数的代数表示法及其几何意义．【分析】（1）由题设条件，可先通过复数的运算求出的代数形式的表示，再由其几何意义得出实部与虚部的符号，转化出实数a所满足的不等式，解出其取值范围；（2）实系数一元二次方程x2﹣6x+m=0的两个根互为共轭复数，利用根与系数的关系求出a 的值，从而求出m的值．【解答】解：（1）由条件得，z1﹣z2=（）+（a2﹣3a﹣4）i…因为z1﹣z2在复平面上对应点落在第一象限，故有…∴解得﹣2＜a＜﹣1…（2）因为虚数z1是实系数一元二次方程x2﹣6x+m=0的根所以z1+==6，即a=﹣1，…把a=﹣1代入，则z1=3﹣2i，=3+2i，…所以m=z1•=13…20．如图，已知直四棱柱ABCD﹣A1B1C1D1，DD1⊥底面ABCD，底面ABCD为平行四边形，∠DAB=45°，且AD，AB，AA1三条棱的长组成公比为的等比数列，（1）求异面直线AD1与BD所成角的大小；（2）求二面角B﹣AD1﹣D的大小．【考点】二面角的平面角及求法；异面直线及其所成的角．【分析】（1）不妨设AD=1，由AD，AB，AA1三条棱的长组成公比为的等比数列，可得AB=，AA1=2．在△ABD中，利用余弦定理可得：DB=1．利用勾股定理的逆定理可得∠ADB=90°．由DD1⊥底面ABCD，可得DD1⊥DB，可得DB⊥平面ADD1，即可得出异面直线AD1与BD所成角．（2）由（1）可得：DB⊥平面ADD1．在Rt△ADD1中，经过点D作DO⊥AD1，垂足为O，连接OB，可得OB⊥AD1．∠BOD即为二面角B﹣AD1﹣D的平面角．利用直角三角形的边角关系即可得出．【解答】解：（1）不妨设AD=1，∵AD，AB，AA1三条棱的长组成公比为的等比数列，∴AB=，AA1=2．在△ABD中，DB2==1，解得DB=1．∴AD2+DB2=AB2，∠ADB=90°．∴AD⊥DB．∵DD1⊥底面ABCD，DB⊂平面ABCD，∴DD1⊥DB，又AD∩DD1=D，∴DB⊥平面ADD1，∴DB⊥AD1，∴异面直线AD1与BD所成角为90°．（2）由（1）可得：DB⊥平面ADD1．在Rt△ADD1中，经过点D作DO⊥AD1，垂足为O，连接OB，则OB⊥AD1．∴∠BOD即为二面角B﹣AD1﹣D的平面角．在Rt△ADD1中，OD===．在Rt△ODB中，tan∠BOD===．∴∠BOD=arctan．21．已知z为复数，ω=z+为实数，（1）当﹣2＜ω＜10，求点Z的轨迹方程；（2）当﹣4＜ω＜2时，若u=（α＞0）为纯虚数，求：α的值和|u|的取值范围．【考点】复数代数形式的乘除运算；复数的代数表示法及其几何意义．【分析】（1）设z=x+yi，x，y∈R，则ω=+i为实数，可得y﹣=0，因此y=0，或x2+y2=9．通过分类讨论即可得出．（2）由（1）可得：①y=0时，ω=x+，由﹣4＜ω＜2，可得﹣4＜＜2，利用基本不等式的性质即可得出．②x2+y2=9时．ω=2x，由于﹣4＜ω＜2，即可得出x的取值范围．由u=（α＞0）为纯虚数，化简可得α，再利用模的计算公式、函数的单调性即可得出．【解答】解：（1）设z=x+yi，x，y∈R，则ω=z+=x+yi+=x+yi+=+i为实数，∴y﹣=0，∴y=0，或x2+y2=9．①y=0时，ω=x+∵﹣2＜ω＜10，∴﹣2＜＜10，x＞0时，解得1＜x＜9．x＜0时，x∈∅．综上可得：y=0时，点Z的轨迹方程是．②x2+y2=9时．ω=2x，∵﹣2＜ω＜10，∴﹣2＜2x＜10，解得﹣1＜x＜5．因此x2+y2=9时．可得：点Z的轨迹方程是x2+y2=9（﹣1＜x＜5）．（2）由（1）可得：①y=0时，ω=x+∵﹣4＜ω＜2，∴﹣4＜＜2，∵x＜0时，≤﹣6；x＞0时，≥6．综上可得：y=0时，x∈∅，点Z的轨迹无方程．②x2+y2=9时．ω=2x，∵﹣4＜ω＜2，∴﹣4＜2x＜2，解得﹣2＜x＜1．∵u=（α＞0）为纯虚数，u==，∴α2﹣9=0，2yα≠0，解得α=3，y≠0．∴u==，∵x∈（﹣2，1），∴|u|===∈．∴α=3，|u|∈．22．动圆M与圆（x﹣1）2+y2=1相外切且与y轴相切，则动圆M的圆心的轨迹记C，（1）求轨迹C的方程；（2）定点A（3，0）到轨迹C上任意一点的距离|MA|的最小值；（3）经过定点B（﹣2，1）的直线m，试分析直线m与轨迹C的公共点个数，并指明相应的直线m的斜率k是否存在，若存在求k的取值或取值范围情况[要有解题过程，没解题方程只有结论的只得结论分]．【考点】轨迹方程．【分析】（1）设出动圆圆心M的坐标，利用动圆M与y轴相切且与圆（x﹣1）2+y2=1外切建立方程，化简得答案；（2）设M的坐标，利用两点间的距离公式结合配方法求得定点A（3，0）到轨迹C上任意一点的距离|MA|的最小值；（3）写出过B斜率存在的直线方程，联立直线方程与抛物线方程，由判别式等于0求得k 值，再结合图形求得直线m与轨迹C的公共点个数，并分析对应的斜率情况．【解答】解：（1）设动圆圆心M的坐标为（x，y），则，∴（x﹣1）2+y2=x2+2|x|+1，当x＜0时，y=0；当x≥0时，y2=4x；（2）如图，由图可知，M到轨迹C上的点与A的距离最小，则M在抛物线y2=4x上，设M（x，y），则|MA|===．∴当x=1，即M（1，±2）时，|MA|的最小值为；（3）设过B与抛物线y2=4x相切的直线方程为y﹣1=k（x+2），即y=kx+2k+1，联立，得k2x2+（4k2+2k﹣4）x+4k2+4k+1=0．由△=（4k2+2k﹣4）2﹣4k2（4k2+4k+1）=0，解得：k=﹣1或k=．∴当直线m的斜率k不存在时或斜率存在为0时或直线m的斜率k∈（，+∞）∪（﹣∞，﹣1）时，m与C有1个交点；当直线m的斜率为k=﹣1或k=或k∈[﹣，0）时，m与C有2个交点；当直线m的斜率k∈（0，）∪（﹣1，﹣）时，m与C有3个交点．23．已知复数z1=m+ni（m，n∈R），z=x+yi（x，y∈R），z2=2+4i且．（1）若复数z1对应的点M（m，n）在曲线上运动，求复数z所对应的点P（x，y）的轨迹方程；（2）将（1）中的轨迹上每一点按向量方向平移个单位，得到新的轨迹C，求C的轨迹方程；（3）过轨迹C上任意一点A（异于顶点）作其切线，交y轴于点B，求证：以线段AB为直径的圆恒过一定点，并求出此定点的坐标．【考点】抛物线的简单性质．【分析】（1）根据复数条件求出关系式，结合复数z1对应的点M（m，n）在曲线上运动即可得出复数z所对应的点P（x，y）的轨迹方程；（2）先按向量方向平移个单位得到即为向 x 方向移动 1×=个单位，向 y 方向移动 1×1=1 个单位，再进行函数式的变换即可得出C 的轨迹方程； （3）设A （x 0，y 0），斜率为k ，切线y ﹣y 0=k （x ﹣x 0） 代入（y +6）2=﹣2x ﹣3消去x 得到关于y 的一元二次方程，再结合根的判别式为0利用向量的数量即可求得定点，从而解决问题．【解答】解：（1）∵i ﹣z 2=（m ﹣ni ）•i ﹣（2+4i ）=（n ﹣2）+（m ﹣4）i ；∴⇒．∵复数z 1对应的点M （m ，n ）在曲线上运动∴x +2=﹣（y +7）2﹣1⇒（y +7）2=﹣2（x +3）．复数z 所对应的点P （x ，y ）的轨迹方程：（y +7）2=﹣2（x +3）．（2）∵按向量方向平移个单位，==1×．即为向 x 方向移动 1×=个单位，向 y 方向移动 1×1=1 个单位（y +7）2=﹣2（x +3）⇒y +7=±．得轨迹方程 y +7=±⇒（y +6）2=﹣2（x +）=﹣2x ﹣3．C 的轨迹方程为：（y +6）2=﹣2x ﹣3． （3）设A （x 0，y 0），斜率为k ，切线y ﹣y 0=k （x ﹣x 0） （k ≠0）， 代入（y +6）2=﹣2x ﹣3整理得：（y +6）2=﹣2（）﹣3，△=0⇒k=，设定点M （1，0），且．∴以线段AB 为直径的圆恒过一定点M ，M 点的坐标（1，0）．2016年9月14日。
2017-2018交附高二下期中化学卷(含解析)
2017-2018交附高二下期中化学卷（含解析）上海交通大学附属中学 2017-2018 学年度第二学期高二化学期中试卷（满分 100 分，60 分钟完成，答案一律写在答题纸上）相对原子质量：H-1C-12 O-16 S-32 Ba-137一、选择题（本题共 40 分，每小题 2 分，只有一个正确选项）1. 化学与生产和生活密切相关，下列过程中没有发生化学变化的是（）A. 用热碱水清除炊具上的油污C. 用洁厕净去除马桶污垢 B. 用食醋除去水壶内的水垢D. 用汽油擦洗衣物上的油污2. 下列将“地沟油”变废为宝的各方案中一定不能实现的是（）A. 发酵生产乙醇、沼气C. 水解制造肥皂、甘油 B. 分馏获得汽油、煤油D. 转化成工业润滑油3. 化学的发展离不开科学家的贡献，下列科学家与贡献不匹配的是（）A. 门捷列夫~元素周期律C. 汤姆逊~原子结构的行星模型 B. 诺贝尔~硝化甘油D. 凯库勒~苯的结构4. 下列有关氮原子核外 p 亚层中的各电子的说法中错误的是（）A. 能量相同C. 自旋方向相同 B. 电子云形状相同D. 电子云伸展方向相同5. 126C 质量的 1/12 相对原子质量的标准，一个 12 6 C 原子的质量是（）A. 12 克B. 1.2×10-10 克C. 2.0×10-23 克D. 1.2×10-27 克6. 元素性质周期性变化的根本原因是（）A. 原子半径的周期性变化C. 金属性和非金属性的周期性变化 B. 化合价的周期性变化D. 核外电子排布的周期性变化7. 下列关于物质间相互关系的判断正确的是（）A. C 60、N 60 属于同分异构体B. 235U 与 238U 属于同素异形体C. 甘油和 1,2,3-丙三醇属于同系物D. 1H 、H 2、2H +属于同一元素的不同形态8.物质分离提纯是利用物质某些性质差异而进行的行为，下列分离方法利用的物质的主要性质差异不正确的是（）A.实验室模拟石油分馏：沸点差异C.分液：溶解性及密度差异B.结晶：温度差异D.过滤：溶解度差异9.下图中小球可表示分子、原子或原子团，其中表示取代反应的是（）10.实验室制取乙酸乙酯和乙酸乙酯的水解的有关说法中不正确的是（）A.二者都需催化剂C.二者都有冷凝措施B.二者都需要水浴加热D.二者都用右图装置完成11.下列各组物质所含化学键类型相同、形成晶体类型也相同的是（）A.乙酸和乙醇B.CO2和SiO2C.NH4Cl和KClD.HCl和NaCl12.下列微粒半径大小关系错误的是（）A.Na＞Mg＞SC.K+＞Na+＞S2-B.P＞Cl＞FD.O2-＞Na+＞Al3+13.下列事实能判断短周期元素甲的金属性一定比乙强的有（）①甲单质能与乙的硫酸盐水溶液反应②甲、乙两元素原子的最外层电子数相同，且甲的电子层数小于乙③甲、乙两原子的电子层数相同，且甲的原子半径大于乙④甲、乙金属片插入稀硫酸中，乙单质表面生成氢气⑤两单质分别与氯气反应时生成的阳离子，甲失去的电子数比乙多A.仅③可以B.仅②不可以C.仅①③⑤可以D.全部可以14.常温下下列离子在溶液中能共存，加NaOH有沉淀析出，加稀HCl能放出气体的是（）A.Na+、Mg2+、Cl—、SO32-C.Al3+、Na+、SO42-、HCO3—B.Fe2+、K+、Cl—、NO3—D.Fe3+、Ca2+、Cl—、SCN—15.某同学给出了下列有机物的名称，其中正确的是（）A.3,3,4-三甲基戊烷C.3,5-二甲基-2-己炔B.2,3-二甲基-1,3-丁二烯D.1,3,5-三硝基甲苯K16. 有机化合物常会混有杂质，以下除杂方法正确的是（）A. 溴苯（杂质：溴），加入苯，分液B. 乙烷（杂质：乙烯），催化条件下通入 H 2 C. 甲酸乙酯（杂质：甲酸），加入饱和碳酸钠溶液振荡、分液 D. 乙酸（杂质：乙醛），加入适量新制氢氧化铜悬浊液，蒸馏17. 实验室将化学式为 C 8H 16O 2 的酯水解,得 到 A 和 B 两种物质,A 氧化可转变为B,符合上述性质的酯的结构种类有（）A. 1 种B. 2 种C. 3 种D. 4 种18. 胡椒酚是植物挥发油中的一种成分。
上海市上海交通大学附属中学2017-2018学年高二下学期3月月考数学试题
（2）求到两定点 、 的“直角距离”和为定值 的动点轨迹方程，并在直角坐标系内作出该动点的轨迹.（在以下三个条件中任选一个做答）
① ， ， ；
② ， ， ；
③ ， ， .
（3）写出同时满足以下两个条件的“格点”的坐标，并说明理由（格点指横、纵坐标均为整数的点）.
（2）答案不唯一，见解析
（3） 、 、 、 、 、 、 、 、 ，理由见解析
【解析】
【分析】
（1）由“直角距离”的定义知 ，进而得到所求点坐标；
（2）根据“直角距离”的定义，分别结合条件①②③，得到动点轨迹方程；利用分类讨论的方式去掉绝对值符号即可得到不同区间内动点的轨迹，从而做出图形；
（3）由条件①可得： ；由条件②可得： ，在平面直角坐标系中做出两个条件下的点构成的区域，取交集，结合图形得到最终结果.
11．已知椭圆 的焦距为 ，则实数 __________．
12．已知 ， 是实系数一元二次方程 的两根，则 的值为__________．
13．若 为非零实数，则下列四个命题都成立：
① ② ③若 ，则
④若 ，则 。则对于任意非零复数 ，上述命题仍然成立的序号是 。
14．如图， 是三角形 所在平面外的一点， ，且 ， 、 分别是 和 的中点，则异面直线 与 所成角的大小为__________（用反三角函数表示）．
【详解】
由双曲线的相关性质可知，双曲线 的焦点为 ，顶点为 ，
所以椭圆的顶点为 ，焦点为 ，
因为 ，所以椭圆的方程为 ，
故答案为 。
【点睛】
本题考查圆锥曲线的相关性质，主要考查椭圆、双曲线的几何性质，考查椭圆的标准方程，正确运用椭圆、双曲线的几何性质是关键．
2017-2018学年上海交大附中高二(下)期末数学试卷(J)
2017-2018学年上海交大附中高二（下）期末数学试卷(J)副标题题号一二三总分得分一、选择题（本大题共4小题，共4.0分）1.设地球的半径为R，地球上A，B两地都在北纬45∘的纬度线上去，且其经度差为90∘，则A，B两地的球面距离是()A. πRB. πR2C. πR3D. πR6【答案】C【解析】解：设在北纬45∘的纬圆的圆心为C，球心为O，连结OA、OB、OC、AC、BC，则OC⊥平面ABCRt△ACO中，AC=OAcos45∘=√22R，同理BC=√22R，∵A、B两地经度差为90∘，∴∠ACB=90∘，Rt△ABC中，AB=√AC2+BC2=R由此可得△AOB是边长为R的等边三角形，得∠AOB=π3∴A、B两地的球面距离是π3R.故选：C．设在北纬45∘的纬圆的圆心为C，球心为O，连结OA、OB、OC、AC、BC.根据地球纬度的定义，算出小圆半径AC=BC=√22R.由A、B两地经度差为90∘，在Rt△ABC中算出AB=√AC2+BC2=R，从而得到∠AOB=π3，利用球面距离的公式加以计算，即可得到A、B两地的球面距离．本题求地球上北纬45度圈上两点的球面距离，着重考查了球面距离及相关计算、经纬度等基础知识，考查运算求解能力，考查空间想象能力，属于基础题．2.对于不重合的两个平面α与β，给定下列条件：①存在平面γ，使得α，β都平行于γ②存在平面γ，使得α，β都垂直于γ；③α内有不共线的三点到β的距离相等；④存在异面直线l，m，使得l//α，l//β，m//α，m//β．其中，可以判定α与β平行的条件有()A. 1个B. 2个C. 3个D. 4个【答案】B【解析】解：①α与β平行.此时能够判断①存在平面γ，使得α，β都平行于γ；两个平面平行，所以正确．②存在平面γ，使得α，β都垂直于γ；可以判定α与β平行，如正方体的底面与相对的侧面.也可能α与β不平行.②不正确．③不能判定α与β平行.如α面内不共线的三点不在β面的同一侧时，此时α与β相交；④可以判定α与β平行．∵可在α面内作l′//l，m′//m，则l′与m′必相交．又∵l//β，m//β，∴l′//β，m′//β，∴α//β．故选：B．直线与平面的位置关系，平面与平面的位置关系，对选项进行逐一判断，确定正确选项即可．本题考查平面与平面平行的判定与性质，平面与平面垂直的判定，考查空间想象能力，逻辑思维能力，是基础题．3.一个正方体的展开图如图所示，B，C，D为原正方体的顶点，A为原正方体一条棱的中点.在原来的正方体中，CD与AB所成角的余弦值为()A. √510B. √105C. √55D. √1010【答案】D【解析】解：还原正方体如右图所示设AD=1，则AB=√5，AF=1，BE=EF=2√2，AE=3，CD与AB所成角等于BE与AB所成角，所以余弦值为cos∠ABE=2×√5×2√2=√1010，故选：D．先还原正方体，将对应的字母标出，CD与AB所成角等于BE与AB所成角，在三角形ABE中再利用余弦定理求出此角的余弦值即可．本小题主要考查异面直线所成的角，考查空间想象能力、运算能力和推理论证能力，属于基础题．4.已知函数y=f(x)的图象是一条连续不断的曲线，若f(0)=A，f(1)=B，那么下列四个命题中①必存在x∈[0,1]，使得f(x)=A+B2；②必存在x∈[0,1]，使得f(x)=√AB；③必存在x∈[0,1]，使得f(x)=√A2+B22；④必存在x∈[0,1]，使得f(x)=21A+1B．真命题的个数是()A. 1个B. 2个C. 3个D. 4个【答案】A【解析】解：函数y=f(x)的图象是一条连续不断的曲线，若f(0)=A，f(1)=B，对于①，由y=f(x)−A+B2，[f(0)−A+B2]⋅[f(1)−A+B2]=−(A−B)22≤0，可得函数y存在零点，即①成立；对于②，由y=f(x)−√AB，[f(0)−√AB]⋅[f(1)−√AB]=(A−√AB)(B−√AB)，若A>0，B>0，则上式为−√AB(√A−√B)2≤0，可得函数y存在零点；若A<0，B<0，则上式>0，可得函数y不一定存在零点；即有②不成立；对于③，由y=f(x)−√A2+B22，[f(0)−√A2+B22]⋅[f(1)−√A2+B22]=[A−√A2+B22][B−√A2+B22]，若A<0，B<0，则上式>0，可得函数y不一定存在零点；即有③不成立；对于④，由y=f(x)−21A+1B，[f(0)−21A+1B]⋅[f(1)−21A+1B]=(A−21A+1B]⋅[B−21A+1B]=−ABA+B⋅(A−B)2，若AB(A+B)<0，则上式>0，可得函数y不一定存在零点；即有④不成立．故选：A．对于①，由y=f(x)−A+B2；对于②，由y=f(x)−√AB；对于③，由y=f(x)−√A2+B22；对于④，由y=f(x)−21A+1B，运用函数零点存在定理，即可判断是否成立．本题考查命题的真假判断，注意运用函数的零点存在定理，考查运算能力，属于中档题．二、填空题（本大题共12小题，共12.0分）5.函数f(x)=√x+1+12−x的定义域为______．【答案】[−1,2)U(2,+∞)【解析】解：根据题意：{2−x≠0x+1≥0解得：x≥−1且x≠2∴定义域是：[−1,2)∪(2,+∞)故答案为：[−1,2)∪(2,+∞)根据负数不能开偶次方根和分母不能为零来求解，两者求解的结果取交集．本题主要考查定义域的求法，这里主要考查了分式函数和根式函数两类．6.表面积为π的球的体积为______．【答案】16π【解析】解：由S=4πR2=π得R=12，所以V=43πR3=π6．则该球的体积为π6．故答案为：π6．先根据球的表面积，就可以利用公式得到半径，再求解该球的体积即可．本题考查球的体积和表面积，主要考查学生对公式的利用，是基础题．7.(2x−1x)7的二项展开式中，x项的系数是______.(用数字作答)【答案】−448【解析】解：(2x−1x )7的二项展开式的通项为T r+1=C7r(2x)7−r(−1x)r=(−1)r⋅27−r⋅C7r⋅x7−2r．由7−2r=1，得r=3．∴(2x−1x)7的二项展开式中，x项的系数是−24×C73=−448．故答案为：−448．写出二项展开式的通项，由x的指数为1求得r值，则答案可求．本题考查二项式定理的应用，关键是熟悉二项展开式的通项，是基础题．8.高一(10)班有男生36人，女生12人，若用分层抽样的方法从该班的全体同学中抽取一个容量为8的样本，则抽取男生的人数为______人.【答案】6【解析】解：由分层抽样的定义得抽取男生的人数为3636+12×8=3648×8=6人，故答案为：6根据分层抽样的定义建立比例关系即可得到结论．本题主要考查分层抽样的应用，根据条件建立比例关系是解决本题的关键.比较基础．9.6个人排成一行，其中甲、乙两人不相邻的不同排法共有______种.(用数字作答)【答案】480【解析】解：6个人排成一行，其中甲、乙两人不相邻的不同排法：排列好甲、乙两人外的4人，有A44中方法，然后把甲、乙两人插入4个人的5个空位，有A52种方法，所以共有：A44⋅A52=480．故答案为：480．排列好甲、乙两人外的4人，然后把甲、乙两人插入4个人的5个空位中即可．本题考查了乘法原理，以及排列的简单应用，插空法解答不相邻问题．10.若交大附中共有400名教职工，那么其中至少有两人生日在同一天的概率为______．【答案】1【解析】解：∵交大附中共有400名教职工，∴其中至少有两人生日在同一天是必然事件，∴其中至少有两人生日在同一天的概率为1．故答案为：1．交大附中共有400名教职工，其中至少有两人生日在同一天是必然事件，由此能求出其中至少有两人生日在同一天的概率．本题考查概率的求法，考查必然事件等基础知识，考查运算求解能力，考查函数与方程思想，是基础题．11.设函数f(x)=ln(1+|x|)−11+x2，则使得f(x)>f(2x−1)成立的x的取值范围为______．【答案】13<x<1【解析】解：由函数的解析式可得函数f(x)是定义域上的偶函数，且x>0时函数单调递增，则不等式等价于：f(|x|)>f(|2x−1|)，脱去f 符号有：|x|>|2x −1|，求解关于实数x 的不等式可得使得f(x)>f(2x −1)成立的x 的取值范围为13<x <1． 故答案为：13<x <1．首先确定函数的单调性和函数的奇偶性，然后脱去f 符号求解自变量的取值范围即可． 本题考查函数的单调性，函数的奇偶性，不等式的解法等，重点考查学生对基础概念的理解和计算能力，属于中等题．12. 在长方体ABCD −A 1B 1C 1D 1中，AB =BC =4，AA 1=2，则直线BC 1与平面BB 1D 1D所成角的正弦值为______． 【答案】√105【解析】解：以D 为原点，DA 为x 轴，DC 为y 轴，DD 1为z 轴，建立空间直角坐标系，则B(2,2，0)，C 1(0,2，1)，D(0,0，0)，D 1(0,0，1)， BC 1⃗⃗⃗⃗⃗⃗⃗ =(−2,0，1)，DB ⃗⃗⃗⃗⃗⃗ =(2,2，0)，DD 1⃗⃗⃗⃗⃗⃗⃗⃗ =(0,0，1)，设平面BB 1D 1D 的法向量n⃗ =(x,y ，z)， 则{n⃗ ⋅DB ⃗⃗⃗⃗⃗⃗ =2x +2y =0n ⃗ ⋅DD 1⃗⃗⃗⃗⃗⃗⃗⃗ =z =0，取x =1，得n⃗ =(1,−1,0)，设BC 1与平面BB 1D 1D 所成的角为θ， 则sinθ=|n ⃗⃗ ⋅BC 1⃗⃗⃗⃗⃗⃗⃗⃗ ||BC 1⃗⃗⃗⃗⃗⃗⃗⃗ ||n⃗⃗ |=√105． ∴BC 1与平面BB 1D 1D 所成的角的正弦值为：√105．故答案为：√105．以D 为原点，DA 为x 轴，DC 为y 轴，DD 1为z 轴，建立空间直角坐标系，利用向量法能求出BC 1与平面BB 1D 1D 所成的角的正弦值．本题考查线面角的正弦值的求法，是中档题，解题时要认真审题，注意向量法的合理运用．13. 一个正方体的8个顶点可以组成______个非等边三角形． 【答案】48【解析】解：一个正方体的8个顶点可以组成C 83=56个三角形， 其中等边三角形有8个，如图所示；所以非等边三角形有56−8=48个．故答案为：48．找出一个正方体的8个顶点可以组成三角形的个数，去掉等边三角形的个数，即得所求．本题考查了空间几何体的结构特征应用问题，是基础题，14.将集合M={1,2，…，12}的元素分成互不相交的三个子集：M=A∪B∪C，其中A={a1,a2,a3,a4}，B={b1,b2,b3,b4}，C={c1,c2,c3,c4}，且a k+b k=c k，k=1，2，3，4，则满足条件的集合C有______个.【答案】3【解析】解：若A={1,2，3，4}，B={5,8，7，9}，则C={6,10，12，11}，若A={1,2，3，4}，B={5,6，8，10}，则C={7,9，12，11}，若A={1,2，3，4}，B={5,6，7，11}，则C={8,10，12，9}，故满足条件的集合C为3个，故答案为：3．讨论集合A与集合B，根据完并集合的概念知集合C本题考查集合的交、并、补的混合运算，是中档题.解题时要认真审题，仔细解答，注意合理地进行等价转化．15.设非空集合A为实数集的子集，若A满足下列两个条件：(1)0∈A，1∈A；(2)对任意x，y∈A，都有x+y∈A，x−y∈A，xy∈A，xy∈A(y≠0)则称A为一个数域，那么命题：①有理数集Q是一个数域；②若A为一个数域，则Q⊆A；③若A，B都是数域，那么A∩B也是一个数域；④若A，B都是数域，那么A∪B也是一个数域．其中真命题的序号为______．【答案】①②③④【解析】解：由已知中数域的定义可得：则有理数集Q满足定义，是一个数域，故①正确；若A为一个数域，则A中包含任意整数和分数，故Q⊆A，故②正确；若A，B都是数域，那么Q⊆A∩B，故A∩B中的元素均满足定义，故A∩B也是一个数域，故③正确；若A，B都是数域，那么Q⊆A∪B，故A∪B中的元素均满足定义，故A∪B也是一个数域，故④正确；故真命题的序号为①②③④，故答案为：①②③④根据已知中数域的定义，逐一分析给定四个答案的真假，可得答案．本题考查的知识点是命题的真假判断与应用，正确理解数域的定义，是解答的关键．16.已知函数f(x)=−2x2+bx+c在x=1时有最大值1，0<m<n，并且x∈[m,n]时，f(x)的取值范围为[1n ,1m]，则m+n=______．【答案】3+√32【解析】解：根据题意，函数f(x)=−2x2+bx+c在x=1时有最大值1，则有−b−4=b4=1，即b=4，且−2+4+c=1，解可得c=−1，则f(x)=−2x2+4x−1，又有x ∈[m,n]时，f(x)的取值范围为[1n ,1m ]， 则1m ≤1，解可得m ≥1， f(x)在[m,n]上单调递减， 则有f(m)=1m ，f(n)=1n ，即有m 、n 是方程−2x 2+4x −1=1x 的两个根， −2x 2+4x −1=1x ⇒(x −1)(2x 2−2x −1)=0， 其根为1、1+√32、1−√32，又有1≤m <n ， 则m =1，n =1+√32，则m +n =3+√32； 故答案为：3+√32．根据题意，结合二次函数的性质分析可得b 、c 的值，即可得f(x)=−2x 2+4x −1，进而可得1m ≤1，解可得m ≥1，分析可得f(x)在[m,n]上单调递减，据此可得f(m)=1m ，f(n)=1n ，即有m 、n 是方程−2x 2+4x −1=1x 的两个根，又有−2x 2+4x −1=1x ⇒(x −1)(2x 2−2x −1)=0，求出方程的根，分析可得m 、n 的值，相加即可得答案． 本题考查二次函数的性质以及应用，关键是求出m 、n 的值，属于基础题．三、解答题（本大题共5小题，共5.0分）17. 某公司生产一种产品，每年投入固定成本0.5万元，此外，每生产1件这种产品还需要增加投入25元，经测算，市场对该产品的年需求量为500件，且当出售的这种产品的数量为t(单位：百件)时，销售所得的收入约为5t −12t 2(万元)． (1)若该公司这种产品的年产量为x(单位：百件).试把该公司生产并销售这种产品所得的年利润y 表示为年产量x 的函数；(2)当该公司的年产量x 多大时，当年所得利润y 最大？ 【答案】解：(1)由题意得：y ={(5x −12x 2)−0.5−0.25x,0<x ≤5(5×5−12×52)−0.5−0.25x,x >5={−12x 2+194x −12,0<x ≤5−14x +12,x >5(6分) (2)当0<x ≤5时，函数对称轴为x =194=4.75∈(0,5)，故x =4.75时y 最大值为34532. (3分) 当x >5时，函数单调递减，故y <−54+12=434<34532，(3分)所以当年产量为475件时所得利润最大. (2分)【解析】(1)由已知中某公司生产一种产品，每年投入固定成本0.5万元，此外，每生产1件这种产品还需要增加投入25元，经测算，市场对该产品的年需求量为500件，且当t2(万元).根据年出售的这种产品的数量为t(单位：百件)时，销售所得的收入约为5t−12利润=销售额−成立，构造出该公司生产并销售这种产品所得的年利润y表示为年产量x 的函数．(2)根据(1)的分段函数解析式，我们分别求出各段上函数的最大值，进而得到该公司当年所得利润y的最大值，及相应的生产量．本题考查的知识点是函数模型的选择与应用，函数的值域，分段函数的解析式求法，二次函数的性质，其中(1)中要注意由于市场对该产品的年需求量为500件，故要分0<x≤5，x>5两种情况将问题转化为分段函数模型，(2)要注意分段函数最值，分段处理．18.解关于x的不等式ax2+ax−1>x.(a∈R)【答案】解：关于x的不等式ax2+ax−1>x，a∈R；①当a=0时，解不等式得x<−1；②当a≠0时：(i)若a>0，则不等式化为ax2+(a−1)x−1>0，因为△=(a−1)2+4a=(a+1)2>0，；所以不等式化为：(ax−1)(x+1)>0，解得x<−1或x>1a<x<−1；(ii)当−1<a<0时，不等式化为(−ax+1)(x+1)<0，解得1a(iii)当a=−1时，不等式化为x2+2x+1<0，此时解集为空集；(iv)当a<−1时，不等式化为(−ax+1)(x+1)<0，解得−1<x<1；a综上，a=0时，不等式的解集为(−∞,−1)；,+∞)；a>0时，不等式的解集为(−∞,−1)∪(1a,−1)；−1<a<0时，不等式的解集为(1aa=−1时，不等式的解集为空集；).a<−1时，不等式的解集为(−1,1a【解析】讨论a=0以及a>0和−1<a<0、a=−1以及a<−1时，求出对应不等式的解集．本题考查了含有字母系数的不等式的解法与应用问题，是中档题．19.如图，二面角D−AB−E的大小为π，四边形ABCD是边长2为2的正方形，AE=EB，F为CE上的点，且BF⊥平面ACE．(1)求证：AE⊥BE；(2)求二面角B−AC−E的大小；(3)求点D到平面ACE的距离．【答案】(1)证明：∵BF ⊥平面ACE ，∴BF ⊥AE ，∵二面角D −AB −E 为直二面角，且CB ⊥AB ，∴CB ⊥平面ABE ， ∴CB ⊥AE ， ∵BF ∩CB =B ，∴AE ⊥平面BCE ，则AE ⊥BE ；(2)解：设二面角B −AC −E 的大小为θ， 由(1)知，AE ⊥EB ，AE ⊥EC ，在Rt △AEB 中，由AB =2，可得AE =EB =√2， 则S △AEB =12×√2×√2=1，在Rt △CBE 中，由BE =√2，BC =2，可得EC =√6， ∴S △AEC =12×√2×√6=√3， ∴cosθ=S △AEB S △AEC=1√3=√33，即θ=arccos√33； (3)解：设点D 到平面ACE 的距离为h ， 则V E−ADC =V D−ACE ，即12×13×2×2×1=13×√3ℎ，则ℎ=2√33， 即点D 到平面ACE 的距离为2√33． 【解析】(1)由BF ⊥平面ACE ，得BF ⊥AE ，再由二面角D −AB −E 为直二面角，且CB ⊥AB ，可得CB ⊥平面ABE ，则CB ⊥AE ，由线面垂直的判断可得AE ⊥平面BCE ，从而得到AE ⊥BE ；(2)设二面角B −AC −E 的大小为θ，分别求出三角形AEB 与三角形AEC 的面积，由两三角形面积比为二面角B −AC −E 的余弦值求解；(3)设点D 到平面ACE 的距离为h ，由V E−ADC =V D−ACE 列式求解点D 到平面ACE 的距离．本题考查空间中直线与直线、直线与平面间位置关系的判定，考查二面角平面角的求法，训练了利用等积法求多面体的体积，是中档题．20. 设全体空间向量组成的集合为V ，a⃗ =(a 1,a 2,a 3)为V 中的一个单位向量，建立一个“自变量”为向量，“应变量”也是向量的“向量函数”f(x ⃗ )：f(x ⃗ )=−x ⃗ +2(x ⃗ ⋅a ⃗ )a ⃗ (x ⃗ ∈V)． (1)设u ⃗ =(1,0,0)，v ⃗ =(0,0,1)，若f(u ⃗ )=v ⃗ ，求向量a ⃗ ； (2)对于V 中的任意两个向量x ⃗ ，y ⃗ ，证明：f(x ⃗ )⋅f(y ⃗ )=x ⃗ ⋅y ⃗ ； (3)对于V 中的任意单位向量x ⃗ ，求|f(x ⃗ )−x ⃗ |的最大值．【答案】解：(1)依题意得：f(u⃗ )=−u ⃗ +2(u ⃗ ⋅a ⃗ )a ⃗ =v ⃗ ， 设a⃗ =(x,y,z)， 代入运算得：{2x 2−1=02xy =02xz =1，解得a ⃗ =(√22,0，√22)或a ⃗ =(−√22,0,−√22)．证明：(2)设x⃗ =(a,b,c)，y ⃗ =(m,n,t)，a ⃗ =(a 1,a 2,a 3)， 则f(x⃗ )⋅f(y ⃗ )=[−x ⃗ +2(x ⃗ ⋅a ⃗ )a ⃗ ]⋅[−y ⃗ +2(y ⃗ ⋅a ⃗ )a ⃗ ]=x ⃗ ⋅y ⃗ −4(y ⃗ ⋅a ⃗ )(x ⃗ ⋅a ⃗ )+4(y ⃗ ⋅a ⃗ )(x ⃗ ⋅a ⃗ )(a ⃗ )2=x ⃗ ⋅y ⃗ −4(y ⃗ ⋅a ⃗ )(x ⃗ ⋅a ⃗ )+4(y ⃗ ⋅a ⃗ )(x ⃗ ⋅a ⃗ )=x ⃗ ⋅y ⃗ ． ∴f(x ⃗ )⋅f(y ⃗ )=x ⃗ ⋅y ⃗ ．解：(3)设x⃗ 与a ⃗ 的夹角为α， 则x ⃗ ⋅a ⃗ =|x ⃗ |⋅|a ⃗ |cosα=cosα， 则|f(x ⃗ )−x ⃗ |=|2x ⃗ −2(x ⃗ ⋅a ⃗ )a ⃗ |=√(2x ⃗ −2cosαa ⃗ )2=√4−4cos 2α≤2， ∴|f(x ⃗ )−x ⃗ |的最大值为2．【解析】(1)f(u⃗ )=−u ⃗ +2(u ⃗ ⋅a ⃗ )a ⃗ =v ⃗ ，设a ⃗ =(x,y,z)，列方程组能求出向量a ⃗ ． (2)设x ⃗ =(a,b,c)，y ⃗ =(m,n,t)，a ⃗ =(a 1,a 2,a 3)，由此能证明f(x ⃗ )⋅f(y ⃗ )=x ⃗ ⋅y ⃗ ． (3)设x ⃗ 与a ⃗ 的夹角为α，则x ⃗ ⋅a ⃗ =|x ⃗ |⋅|a ⃗ |cosα=cosα，由此能求出|f(x ⃗ )−x ⃗ |的最大值为2．本题考查向量的求法，考查等式的证明，考查向量的模的最大值的求法，考查向量、向量的模、向量的数量积公式等基础知识，考查推理能力与计算能力，考查函数与方程思想，是中档题．21. 对于函数y =f(x)，若关系式t =f(x +t)中变量t 是变量x 的函数，则称函数y =f(x)为可变换函数.例如：对于函数f(x)=2x ，若t =2(x +t)，则t =−2x ，所以变量t 是变量x 的函数，所以f(x)=2x 是可变换函数．(1)求证：反比例函数g(x)=kx (k >0)不是可变换函数； (2)试判断函数y =−x 3是否是可变换函数并说明理由；(3)若函数ℎ(x)=log b x 为可变换函数，求实数b 的取值范围．【答案】(1)证明：假设g(x)是可变换函数，则t =g(x +t)=kx+t ⇒t 2+tx −k =0， ∵变量x 是任意的，故当△=x 2+4k <0时，此时有关变量t 的一元二次方程无解， 与假设矛盾，故原结论正确，∴反比例函数g(x)=kx (k >0)不是可变换函数； (2)解：若y =−x 3是可变换函数，则t =−(x +t)3， 则有关t 的两个函数：{ℎ(t)=(t +x)3ϕ(t)=−t必须有交点，而φ(t)连续且单调递减，值域为R ，ℎ(t)连续且单调递增，值域为R ， ∴这两个函数φ(t)与ℎ(t)必定有交点，即变量t 是变量x 的函数，故y =−x 3是可变换函数；(3)解：函数ℎ(x)=log b x 为可变换函数，则t =ℎ(x +t)⇒t =log b (x +t)，若b >1，则t 恒大于log b (x +t)，即函数y =t 与y =log b (t +x)无交点，不满足题意； 若0<b <1，则{y =log b (t +x)y=t必定有交点，即方程t =log b (x +t)有解，从而满足题意，∴实数b 的取值范围为(0,1)．【解析】(1)利用可变换函数的定义结合反证法证明；(2)由题意可得t =−(x +t)3，结合关于t 的两函数y =−t 与y =(x +t)3有交点可得函数y =−x 3是可变换函数；(3)由题意可得t =log b (x +t)，若b >1，则t 恒大于log b (x +t)，函数y =t 与y =log b (t +x)无交点；若0<b <1，则{y =log b (t +x)y=t必定有交点，从而得到实数b 的取值范围．本题是新定义题，考查函数解析式的求解及常用方法，考查逻辑思维能力与推理运算能力，属中档题．第11页，共11页。
2017-2018交附高二第二学期期末化学卷
2017-2018交附高二第二学期期末化学卷上海交通大学附属中学2017-2018 学年度第二学期高二化学等级考期末试卷（考试时间：60 分钟）2018 年6 月相对原子质量：Si-28 Fe-56一、选择题(本题共40 分，每小题 2 分，只有一个答案)1.我国歼-20 将含20984Po 涂于表面，以屏蔽雷达，起到隐身作用。
该Po（钋）原子核内中子数是A. 42B. 84C. 125D. 2092.符号“3p x”没有给出的信息是A.电子层B. 电子亚层C. 电子云在空间的伸展方向D. 电子的自旋方向3.除去NaCl 固体中少量KNO3的操作不包括A.溶解B. 加热浓缩C. 冷却结晶D. 趁热过滤4.对储氢材料CaH2描述错误的是A.离子半径：H—>Li+B. H—有强还原性C. Ca2+在水溶液中不能够大量存在D. Ca2+最外层电子排布式3s23p65.下列关于物质结构的叙述中正确的是A.分子晶体中一定含有共价键B.离子晶体中一定含有离子键C.含有极性键的分子一定是极性分子D.含有非极性键的分子一定是非极性分子6.下列各组中的离子，能在溶液中大量共存的是A. K+、H+、SO42-、AlO2-B. H+、Al3+、CO32-、Cl-C. Mg2+、Na+、Cl-、SO42-D. Na+、K+、OH-、HCO3-7.从下列叙述中能肯定判断某化学平衡发生移动的是A.混合物中各组分的浓度改变B. 混合体系中气体密度发生变化C. 正、逆反应速率改变D. 反应物的转化率改变8.N A为阿伏伽德罗常数的值，下列说法正确的是A.1L1mol·L-1 Na2CO3溶液中含有的CO32-数目为N AB.0.1mol 钠在空气中完全燃烧生成Na2O2，转移的电子数为0.1N AC.0.5mol 苯中含有碳碳双键的数目为1.5N AD.标准状况下，2.24LCl2通入足量NaOH 溶液中，转移电子的数目为0.2N A个9.下列离子方程式正确的是A.NH3通入稀H2SO4中：NH3+H+→NH4+B.过量的CO2通入澄清石灰水中：CO2+Ca2++2OH-→CaCO3↓+H2OC.醋酸与氨水混合：CH3COOH+OH-→CH3COO-+ H2OD.足量NaHSO4溶液与Ba(OH)2溶液混合：H++SO42-+Ba2++OH-→BaSO4↓+H2O10.有关物质结构的下列说法中正确的是A.碘升华时破坏了共价键B.含极性键的共价化合物一定是电解质C.氯化钠固体中的离子键在溶于水时被破坏D.HCl 的分子间作用力大于HBr，故HCl 比HBr 更稳定11.X、Y、Z、W 均为短周期元素，它们在周期表中的位置如右下图所示。
最新-上海交大附中2018学年度第二学期高二期终 精品
上海交通大学附属中学2018-2018学年度第二学期高二相关物理期终试卷（满分100分，90分钟完成，答案请写在答题纸上）一、选择题1．物体从粗糙斜面的顶端由静止开始下滑，下列叙述正确的是： A ．物体滑到斜面中点时的速度与滑到斜面底端时的速度之比1：2 B ．物体滑到斜面中点的时间与滑到斜面底端的时间之比)12(:1+C ．把斜面分成三等分，则物体依次通过这三段位移所用的时间之比)23(:)12(:1--D ．把时间分成三等分，则物体在这三段时间内通过的位移之比为1：3：5 2．三个在同一平面的共点力大小分别为F 1=5N 、F 2=8N 、F 3=12N ，则下列说法正确的是：A ．三个力合力的最大值为25NB ．三个力的合力可能等于9NC ．三个力的合力最小值为1ND ．三个力的合力等于0时，F 1与F 3的合力大小一定等于8N3、如图所示，A 、B 、C 三个质量相同的木块，迭放在水平桌面上，在施于B 的水平恒力F 作用下，此三木块保持相对静止，一起沿桌面向右作匀速运动，则：A ．B 对A 的摩擦力为零B ．B 对A 的摩擦力为F/3C ．C 对桌面的摩擦力为0D ．B 对C 的摩擦力为F4、一艘小船沿一定的航向渡河，由于水流作用，此时小船恰能沿垂直河岸方向抵达河岸，今保持小船的航向和速度的大小不变，则：A ．若水流速度减小，则小船将抵达上游B ．若水流速度减小，则小船的合速度增大C ．若水流速度增大，则小船的抵达对岸时间减小D ．若水流速度增大，则小船合速度不变5．关于自由落体运动，下列说法正确的是：A．下落的瞬时速度与下落的位移平方成正比B．每秒钟瞬时速度的增量为9.8米/秒C．每秒钟位移的增量为9.8米D．自由落体运动可以看成是一个竖直上抛运动和一个匀速直线运动的合运动6．如图所示，一根木棒AB在O点被悬挂起来，AO＝OC，在A、C两点分别挂有两个和三个钩码，木棒处于平衡状态。
如在木棒的A、C点各增加一个同样的钩码，则木棒：A．绕O点顺时针方向转动B．绕O点逆时针方向转动C．平衡可能被破坏，转动方向不定D．仍能保持平衡状态7．一轻杆AB，A端铰于墙上，B端用细线挂于墙上的C点，（C点位置不变）并在B端挂一重物，细线较长使轻杆位置如图（甲）所示时杆所受的压力大小为N1，细线较短使轻杆位置如图（乙）所示时杆所受的压力大小为N2，则有：A．N1>N2B．N1<N2C．N1=N2D．无法比较8．如图所示，AB是一段质量分布不均匀的棒．两次将棒靠在光滑的竖直墙壁和粗糙的水平地面之间，棒处于静止状态，一次是A端在上，一次是B端在上，两次棒与地面的夹角可视为相同，两次相比较，则有：A．地面对棒的弹力相同B．地面对棒的摩擦力相同C．墙壁对棒的弹力相同D．棒的下端受到地面的作用力相同．二、填充题9．一轻绳跨过两个等高的轻定滑轮，两端分别挂上质量m1=4千克和m2=2千克的物体，如图所示，在滑轮之间的绳上悬挂物体M，为使三个物体能保持平衡。
最新上海交通大学附属中学2017-2018高二下学期期末考试数学试题含答案
2018年交附高二下数学期末试卷第Ⅰ卷（共54分）一、填空题（本大题共12题，1-6题每题4分，7-12题每题5分，满分54分，将答案填在答题纸上） 1.函数()112f x x x=+-的定义域为 ． 2.表面积为π的球的体积为 ．3.712x x ⎛⎫- ⎪⎝⎭的二项展开式中，x 项的系数是 ．（用数字作答）4.高一（10）班有男生36人，女生12人，若用分层抽样的方法从该班的全体同学中抽取一个容量为8的样本，则抽取男生的人数为 人．5.6人并排站成一行，其中甲、乙两人必须相邻，那么不同的排法有 种.（用数学作答）6.若交大附中共有400名教职工，那么其中至少有两人生日在同一天的概率为 ．7.设函数()()21ln 11f x x x =+-+，则使得()()21f x f x >-成立的x 的取值范围是 ．8.在长方体1111ABCD A B C D -中，4AB BC ==，12AA =，则直线1BC 与平面11BB D D 所成角的正弦值为 ．9.一个正方体的8个顶点可以组成 个非等边三角形. 10.将集合{}1,2,,12M =的元素分成互不相交的三个子集：M A B C =，其中{}1234,,,A a a a a =,{}1234,,,B b b b b =，{}1234,,,C c c c c =，且k k k a b c +=，1,2,3,4k =，则满足条件的集合C 有 个．11.设非空集合A 为实数集的子集，若A 满足下列两个条件： （1）0A ∈，1A ∈；（2）对任意,x y A ∈，都有x y A +∈，x y A -∈，xy A ∈，()0xA y y∈≠ 则称A 为一个数域，那么命题：①有理数集Q 是一个数域；②若A 为一个数域，则Q A ⊆；③若A ，B 都是数域，那么A B也是一个数域；④若A ，B 都是数域，那么AB 也是一个数域.其中真命题的序号为 ．12.已知函数()22f x x bx c =-++在1x =时有最大值1，0m n <<，并且[],x m n ∈时，()f x 的取值范围为11,n m ⎡⎤⎢⎥⎣⎦，则m n += ．第Ⅱ卷（共96分）二、选择题：本大题共4个小题,每小题5分,共20分.在每小题给出的四个选项中，只有一项是符合题目要求的.13.设地球的半径为R ，地球上A ，B 两地都在北纬45的纬度线上去，且其经度差为90，则A ，B 两地的球面距离是（ ） A ．R π B ．2R π C.3R π D ．6Rπ 14.对于不重合的两个平面α与β，给定下列条件： ①存在平面γ，使得α、β都垂直于γ； ②存在平面γ，使得α、β都平行于γ； ③α内有不共线的三点到β的距离相等；④存在异面直线l ，m ，使得//l α，//l β，//m α，//m β 其中，可以判定α与β平行的条件有（ ）A ．1个B ．2个 C. 3个 D ．4个15.一个正方体的展开如图所示，点B ，C ，D 为原正方体的顶点，点A 为原正方体一条棱的中点，那么在原来的正方体中，直线CD 与AB 所成角的余弦值为（ ）A ．10 B ．5 C.5 D ．1016.已知函数()y f x =的图像是一条连续不断的曲线，若()0f A =，()1f B =，那么下列四个命题中①必存在[]0,1x ∈，使得()2A Bf x +=；②必存在[]0,1x ∈，使得()f x =；③必存在[]0,1x ∈，使得()222A B f x +=； ④必存在[]0,1x ∈，使得()211f x A B=+.真命题的个数是（ ）A ．1个B ．2个 C. 3个 D ．4个三、解答题 （本大题共5小题，共76分.解答应写出文字说明、证明过程或演算步骤.） 17. 某公司生产一种产品，每年投入固定成本0.5万元.此外，每生产1件这种产品还需要增加投入25万元.经测算，市场对该产品的年需求量为500件，且当出售的这种产品的数量为t （单位：百件）时，销售所得的收入约为2152t t -（万元）. （1）若该公司这种产品的年产量为x （单位：百件），试把该公司生产并销售这种产品所得的年利润y 表示为年产量()x x R +∈的函数；（2）当该公司的年产量x 为多少时，当年所得利润y 最大？最大为多少？ 18. 解关于x 的不等式21ax ax x +->.（a R ∈） 19. 如图，二面角D AB E --的大小为2π，四边形ABCD 是边长为2的正方形，AE EB =，F 为CE 上的点，且BF ⊥平面ACE .（1）求证：AE BE ⊥；（2）求二面角B AC E --的大小； （3）求点D 到平面ACE 的距离.20. 设全体空间向量组成的集合为V ，()123,,a a a a =为V 中的一个单位向量，建立一个“自变量”为向量，“应变量”也是向量的“向量函数”()()()():2f x f x x x a a x V =-+⋅∈.（1）设()1,0,0u =，()0,0,1v =，若()f u v =，求向量a ； （2）对于V 中的任意两个向量x ，y ，证明：()()f x f y x y ⋅=⋅； （3）对于V 中的任意单位向量x ，求()f x x -的最大值.21. 对于函数()y f x =，若关系式()t f x t =+中变量t 是变量x 的函数，则称函数()y f x =为可变换函数.例如：对于函数()2f x x =，若()2t x t =+，则2t x =-，所以变量t 是变量x 的函数，所以()2f x x =是可变换函数. （1）求证：反比例函数()()0kg x k x=>不是可变换函数； （2）试判断函数3y x =-是否是可变换函数并说明理由； （3）若函数()log b h x x =为可变换函数，求实数b 的取值范围.试卷答案一、填空题 1.[)()1,22,-+∞ 2.16π 3.448- 4.6 5.480 6.1 7.113x <<8.5 9.48 10.3 11. ①②③④12.32+ 二、选择题 13-16:CBDA 三、解答题17.解析：（1）由题意得：2221119150.50.25,05,0522421112,55550.50.25,542x x x x x x x y x x x x ⎧⎛⎫⎧---<≤-+-<≤ ⎪⎪⎪⎪⎝⎭⎪=⇒⎨⎨⎛⎫⎪⎪-+>⨯-⨯--> ⎪⎪⎪⎩⎝⎭⎩；（2）当05x <≤时，函数对称轴为(]190,54x =∈， 故当194x =时，max 34532y =； 当5x >时，函数单调递减，故543345124432y <-+=<， 所以当年产量为475件时，所得利润最大. 18.解析：讨论法！ ①当0a =时，1x <-； ②当0a ≠时：1 0a >，()2110ax a x +-->，因为()()221410a a a ∆=-+=+>，故等式左边因式分解得：()()()1110,1,ax x x a ⎛⎫-+>⇒∈-∞-+∞ ⎪⎝⎭； 2当10a -<<时，()()11101ax x x a-++<⇒<<-； 3当1a =-时，2210x x ++<，此时解集为空集；4当1a <-时，()()11101ax x x a-++<⇒-<<； 19.解析：（1）证明：∵BF ⊥平面ACE ，∴BF AE ⊥，∵二面角D AB E --为直二面角，且CB AB ⊥，∴CB ⊥平面ABE ， ∴CB AE ⊥，∴AE ⊥平面BCE . （2）arcsin3；（3）3. 20.解析：（1）依题意得：()()2f u u u a a v =-+⋅=，设(),,a x y z =，代入运算得：2210220,0,21x xy a xz ⎧-=⎛⎪=⇒= ⎨ ⎝⎭⎪=⎩或2,0,a ⎛=- ⎝⎭；（2）设(),,x a b c =，(),,y m n t =，()123,,a a a a =，则()()()()22f x f y x x a a y y a a ⎡⎤⎡⎤⋅=-+⋅⋅-+⋅⎣⎦⎣⎦()()()()()()()()()24444x y y a x a y a x a ax y y a x a y a x a x y =⋅-⋅⋅+⋅⋅=⋅-⋅⋅+⋅⋅=⋅从而得证；（3）设x 与a 的夹角为α，则cos cos x a x a αα⋅=⋅=， 则()()()22222cos 44cos 2f x x x x a a x a α-=-⋅=-=-≤，故最大值为2.21.解析：（1）证明：假设()g x 是可变换函数，则()20kt g x t t tx k x t=+=⇒+-=+， 因为变量x 是任意的，故当240x k ∆=+<时，此时有关变量t 的一元二次方程无解， 则与假设矛盾，故原结论正确，得证；（2）若3y x =-是可变换函数，则()3t x t =-+，则有关t 的两个函数：()()()3t t h t t x ϕ=-⎧⎪⎨=+⎪⎩必须有交点，而()t ϕ连续且单调递减，值域为R ， ()h t 连续且单调递增，值域为R ，所以这两个函数()t ϕ与()h t 必定有交点，即：变量t 是变量x 的函数，所以3y x =-是可变换函数；（3）函数()log b h x x =为可变换函数，则()()log b t h x t t x t =+⇒=+，若1b >，则t 恒大于()log b x t +，即无交点，不满足题意；()log b tt x ==+必定有交点，即方程()log b t x t =+有解，从而满足题意.单独统一招生考试语文冲刺模拟试题（五）总分：__________一、语文知识（每小题4分，共40分）】内讧．（h òng ） 呼号．（h ào ） 循规蹈矩． （j ǔ） 押解．（ji è） 贿赂．（l ù） 脍．（ku ài ）炙人 埋．（m án ）怨 勉强．（qi ǎng ） 含情脉脉．（m ò） 剽．（pi āo ）悍 拘泥．（n ì） 拈．（ni ān ）轻怕 】磐竹难书 临渊羡鱼，不如退而结网 并行不背 功欲善其事，必先利其器 一诺千金 城门失火，殃及池鱼自立更生 穷则独善其身，达则兼济天下 】_________这个成语的含义通常不很好。
上海交通大学附属中学2017-2018学年高二10月月考英语试题(word版无听力部分附答案)
上海交通大学附属中学 2017-2018学年度第一学期高二英语月考一试卷出卷人：张睿燕审稿人：朱雪艳（满分150分，130分钟完成答案一律写在网上阅卷答题纸上）II. GrammarandVocabulary(32')SectionADirections：BeneatheachofthefollowingsentencestherearefourchoicesmarkedA,B,C,and D. Choose the one answer that best completesthe sentence.21.T he local people have doubt this carpenter can make of such material.A.w hetherB.whatC.whichD. if22.thegoodsdisplayedinthesupermarket sosurprising thatIdon’t knowwhat to choose.A.variety of, areB.The variety of, is isC.Avarietyof, isD.The variety of, are23.The population ofthe earth increasing fast. one third of the populationhere workers.A.i s; areB. is; isC. are; isD. are; are24.A snightfelluponthetown,theassistants thelastordersso as to close the shop early,A.digested,fewB. rushed across, pieces ofC. rushed through, fewD. rushed into,pieces of25.theimportanceofholdingthe2017MovieFestivalsuccessfully,allcitizensaremaking every effort to work for the beauty of thecity of Shanghai.ARealize B.Aware of C. Encouraged by D. Beingaware of26.inamisunderstanding,abitterquarrelbetweenthetwoneighborshaslasted for two hours.A.OriginatedB. OriginatingC. BeingoriginatedD.Havingbeen originated27.There is only onething I'mnot in favor of in this restaurant-----thecoffee .A.serves coldB. is served coldlyC. serves coldlyD.is servedcold28.Itwasonlywhenthereservedtroopswerebroughtintoplaythatthebattleturned.A.to favor usB. in our favorC. to our favorD.in favor of us29.T he littleboy still needs the 30 dollars to do withsomethings .A.remaining;remained to be settledB. remaining;remainingto be settledC.remained;remainedto settleD. remained; remaining to settle30.Roughlyhalfthenation'scitizensliveoverseas,mostlyintheU.S.andAustralia,where theywork folks back home.A.helping to supportB. to helpsupportingC.to help support D. helpingsupporting31.Theinvestmentmoneywaswithdrawnfromthecountry,leavingsomeofitscitieswith abandoned constructionsA. litteringB. to litterC.to be litteredD. littered32. They searched for hours, but befound.A.at no time could the missingspoon.A. nowhere could the missingspoonC.nowhere the missing spooncouldD. nowhere can the missing spoonSection BDirections:Completethefollowingpassagebyusingthewordsinthebox.Eachwordcan only beused once.Note thatthere is one wordmore than you need.Junkfoodiseverywhere.We’re eatingwaytoomuchofit.Mostofusknowwhat we’re doing and yet we do it anyway.Sohere'sasuggestion_ 33bytworesearchersattheRandCorporation:Whynottakealess onfromalcoholcontrolpoliciesandapplythemtowherefoodissold and how it’s 34 ?"Manypolicymeasurestocontrolobesity （肥胖症）assumethatpeopleconsciouslyandrationallychoosewhatandhowmuchtheyeatandtherefo refocusonproviding informationand moreaccess to 35 foods,"note the tworesearchers."Incontrast,"theresearcherscontinue,"manyregulationsthatdon'tassumepeoplemak erationalchoiceshavebeensuccessfullyappliedtocontrolalcohol,asubstance--likefood--ofwhichimmoderate36 leadstoserioushealthproblems."Theresearchreferencesstudiesofp eople’s behaviorwithfoodandalcoholandresultsofa lcoholrestrictions,andthenlistsfiveregulationsthattheresearchersthinkmight be promising if appliedto junkfoods.Amongthem:Density _37 :licensestosellalcoholaren'thandedout 38toallcomers but are allotted （分配）based on the number of places in an area that alreadysell alcohol.These makealcohol less easy to getandreduce thenumberofpsychological cuesto drink.Similarly,theresearcherssay,beingpresentedwithjunkfood39 ourA.healthierB.stimulatesC.particularlyD.consumptionE.offeredF.displayedG.primarily K.peculiar H.special I.restrictions J.unplanneddesiretoeatit.Sowhynotlimitthedensityoffoodoutlets, 40onesthatsellfoodrichinemptycalories?Andwhynotlimitsaleoffoodinplacesthat aren’t41 food stores?Displayandsalesrestrictions:Californiahasaruleprohibitingalcoholdisplaysnearthecashre gistersingasstations,andinmostplaces youcan’tbuy alcoholatdrive-throughfacilities.Atsupermarkets,foodcompaniespaytohavetheirwaresinplaceswherethe y'reeasilyseen.Onecouldremovejunkfoodtothebackofthestoreandbanthemfromtheshelv esatcheckoutlines.Theothermeasuresincluderestrictingportionsizes,taxingandprohibiting42 price dealsforjunkfoods, and placing warning labels onthe products.(B)Tea —ABrief Historyof theNation'sFavoriteBeverageThe birth of tea in ChinaTeaisoftenthoughtofasbeingatypicallyBritishdrink,andwehavebeendrinkingit for over 350years, Butinfact the history of tea goesmuchfurther back.ThestoryofteabeginsinChina.Accordingtolegend,in2737BC,theChineseemperorSh enNungwassittingbeneathawhilehisservant43drinkingwater,whensomeleavesfromthetreeblewintothewater.ShenNung,arenownedh erbalist,decidedtotheinfusionthathissavanthad 44_created.Thetreewas Camellia sinensis drink and theresulting drink was what we now calltea.Itisimpossibletoknowwhetherthereisanytruthinthisstory.Butteadrinkingcertainlybe cameestablishedinChinamanycenturiesbeforeithadevenbeenheardofinthewest.Containe rsforteahavebeenfoundintombs45fromtheHandynasty(206BC220AD)'butitwasundertheTangdynastyAD),thatteabecamefirmly 46 as the nationaldrink of China. It becamesuch47thatduringthelateeighthcenturyawritercalledYuwrotethefirstbookentirelyAbouttea,theC h'aChing,orTeaClassic.Itwasshortlyafterthisthatteawag first 48 toJapan,by JapaneseBuddhistmonkswho had travelledtoChina to studyThegrowthof tea in EuropeSoatthisstageinthehistoryoftea,Europewagratherlaggingbehind.Inthelatterhalfofth esixteenthcenturytherearethefirstbriefmentionsofteaasadrinkamongEuropeans.Theseare 49 fromPortuguesewhowerelivingintheEastastradersandmissionaries(,Butalthoughsomeof theseindividualsmayhaveA.accidentallyB.establishedmercialD.datingE.introducedF.favoriteG.boiledH.turnJ.mostly K.originbroughtbacksamplesofteatotheirnativecountry,itwasnotthePortuguesewhowerethefirstt oshipbackt:asa 50 import.ThiswasdonebytheDutch,whointhelastyearsofthesixteenthcenturybegantoencro achonPortuguesetradingroutesintheEast,By'the 51 ofthecenturytheyhadestablishedatradingposton the islandof Java,anditwag viaJava thatin1606 the firstconsignmentof teawasshippedfromChinatoHolland,TeasoonbecameafashionabledrinkamongtheDutch, andfromtherespreadtoothercountriesincontinentalwesternEope,butbecause of its high priceit 52 drink for the wealthy.Ill.ReadingComprehensionSectionA ClozeDirections:ForeachblankinthefollowingpassagetherearefourwordsorphrasesmarkedA, B,C and D. Fill i)each blank with theword or phrase that bestfits the context.Inspiration"Mama,whenIgrowup,I'mgoingtobeoneofthose!"IsaidthisaftergeeingtheCapitalDa ncingCompanyperformwhenIwagthree.Itwasthefirsttimethatmydreamtookonavividfor mandactedassomethingimportanttostartmytraining.AsI grewolder and was 53 to more,myinterestsintheworldofdance54variedbutthatlittlegirl'sdreamofsomedaybecomingadancerinthecompanyneverleftme.In thesummerof2005whenIwas18,Ireceivedthephonecallwhichmadethatdreama55 ;Ibecameamemberofthecompany56 back to 1925.AsIlookbackonthatdaynow,itsurely57__anysenseofreality,IbelieveIstayedinastate ofpleasantdisbeliefuntilIwashalfwaythroughrehearsalsonmyfirstday.Ineveractually58 togetthejob,Afterbeingofferedtheposition,Iwag completely 59 . I remember shaking withexcitement.ThoughIwasabsolutelythrilledwiththechange,itdidnotcomewithoutitsfairshareof60.Throughthestrictrehearsalperiodofdancingsixdaysaweek;Ifounditvitalto61_upthematerialfastwitheverylastbitofconcentration.Itisthat62__todetailandonpractic ethatsetus_63_.Tothen follow thosehigh-energyrehearsalswithabusyshowscheduleofuptofiveperformancesaday,Idiscoveredanew64ofthe words "hardwork."WhatIthoughtweremyphysical _65 werepushedmuchfurtherthan I thought 66I learned to make each performancebetter than thelast,Today,whenIlookattheunbelievablecompanythatIhavethegreat 67 ofbeingapartof.notBillyamember,butdancecaptain,Iseeatraditionthathasinspirednotonly generationsoflittlegirlsbutasplendidcompanythatcontinuestodevelop andgrow-and inspires people every dayfollow their dreams.(B)Starbucksliterallychangedthedefinitionof"agoodcupofcoffee".ForStarbucks,thebra ndhadthreeelements:coffee,68_andstores.Strictcontroloverthequalityandprocessingofth ebeans69_thatthecoffeewouldbeofthehighestpossiblequality.Outstandingstorepersonnel wereemployedandtrainedincoffeeknowledge and70_ service. Store design,aroma all71 _ the"StarbucksExperience".AlmostallStarbucks storeswerecorporatelyowned andcontrolled.Starbuckspridedonthe"StarbucksExperience",72_coffeetoprovideauniqu eexperienceforits customers.73_,yougetmorethancoffeewhenyouvisitStarbucks;Yougetgreatpeople,first-ratemusic,acomfortableandupbeatmeetingplace,onbrewingexcellentcoffeeathome.Atho meyouarepartofafamily.Atworkyouarepartofacompany.Andsomewhereinbetweenisapla cewhereyoucansitbackandbeyourself.ThatiswhataStarbucksstoreistomanyofitscustome rs —akindof"thirdplace"wheretheycanescape, read,75_ or listen.ThegreenStarbuckslogoisamermaidthatlooksliketheendofthedoubleimageofthesea .ItwasdesignedbyTerryHeckler,whogotthe76_fromthewoodenstatueofthesea.Mermaidl ogoalso77_originalandmodemmeanings:herfaceisvery simple, but with modern abstract forms of packaging.StarbucksmakesthetypicalAmericanculturegraduallybrokendownintoelements of 78_:thevisualwarmth,hearingtheway,smellingthearomaofcoffeeandsoon.Justthink,thro ughthehugeglasswindows,watchingthecrowdedstreets,__79_sippingacoffeeflavor,whic hisinlinewiththe"Yapi",thefeelingofexperienceinthe_80_life.ButtheessenceofStarbucksisnotaboutthecoffee,althoughit's 81_.coffee. 53. A.connected 54. A.rarely 55. A.symbol 56. A.bouncing 57. cks 58.A. cared 59.A. motivated60.A, challenges 61.A. put 62,A. attention 63. A. apart 64. A, function 65. A. boundaries 66. A.necessary 67. A. talent B. expanded B. certainly B. memory B. dating B. adds B. expected B.relaxed B. profits B. mix B. association B. aside B.eating B. problems B. perfect B. honor C. exposed C. probably C. truth C. turning C. makesC.aske d C. convinced C. advantage C.build C. attraction C. off C. expression C. barriers C. proper C. potential D. extended D. consistently D.reality D.tracking D.brings D. decided D. astonishedD.ad ventures D. pick D. adaptation D. back D usage D. efforts D. possible D. responsibilityCoffee is only a carrier. Coffeeconsumption,_82_, is and cultural.Section BDirections:Readthefollowingfourpassages.Eachpassageisfollowedbyseveralquestionso runfinishedstatements.ForeachofthemtherearefourchoicesmarkedA,B,CandD.Chooset heonethatfitsbestaccordingtotheinformationgiveninthepassage youhave justread.(A)Foodandideasaboutcookingithavebeenpassingfromonepartoftheworldtoanothersin cetheStoneAgeRevolutionbeganintheMiddleEast.Theywerepartofthespreadofcivilizati on,thoughsincepeoplechangetheirtastesinpaintingandarchitecturemuchfasterthantheirta stesinfood,knowledgeofwhatwaseatenisfarlessthanknowledgeofthehousesthatwerelive dinortheclothesthatwereworn.Cookerybookswerefew before the 17th century, andhow closeare the generaleatinghabitsat anyperiodto the cookery books published?Changesowedmoretothemovementofpeople,ofanimals,ofmerchants,ofrichlandow ners,thantobooks.Beforecanals,railways,goodroads,mostpeopleatewhatcouldbeproduce dwithinathirty-milearea.Portsdidbetter,ofcourse,iftheywereonaroute.Formostpeoplefoodwasbasicallyr egionalfoodandtherewasnotalwaysenoughofiteither.Eveningoodareas,poorcountrypeop lehadlittletoeatsincemostofwhattheyproducedwentforsaleatlocalmarkets.Onlyrichmenc ouldbuy expensive seedsto grow unusual vegetables,oremploygardenerswho68. A. customers69. A. ensured70. A. employment71. A. resulted from72. A. coming across73. A.Actually74.A. dependence75.A. negotiate76.A. imagination77.A.conveys78.A.brand79.A. greedily80.A. easy81.A. great82.A. for instance B. managers B. assured B.customer B. contributedB.m aking upB.Gradually B. effect B. perform B. inspiration B. creates B. logo B. gently B. busy B. rareB. to large extentC.people C.predicted C. environment C. benefited from C. going beyond C.Luckily C. advice C. quarrel C. patent C. credits C. possession C. reluctantly C. miserable C. expensive C. in additionD. clients D. confessed D. greeting D. stuck to D. depending on D. extremelyD.remark D. chat D. illustration D. cultivates D. experience D. indifferently D. energetic D. bitter D. by contrastunderstoodhowtogrowfinefruitunfamiliartotheplacetheylivedin,oraffordcooks trainedelsewhere to providevariety at mealtimes.Theundoubtedadvantagesofpresent-daylarge-scalemanufactureandorganization—wonderfulcleanliness,quickdistribution,pricesthatallowfarmorepeoplethan.everinthepas ttosatisfytheirhunger—havenotsofarcometoustogetherwithanexcellentqualityofflavor.Moreover,inaworldwher epossibilitiesareendless,businessseemstotolimitchoicebeyondacertainlevel.Ofthe300va rietiesofpearsthatlistedbyoneFrench17th-centurygardener-eventhoughhehadtoadmitthatonly30ofthemwerereallywortheating—onlyabouthalfaarenowproduced in Europe.33.Why do we knowso littleabout the food people ate in thepast?A.Eatinghabits usedtochange very quickly.B.There were no cookery books before the 19thcentury.C.Peopledid not changetheir tastesinfoodfast,D.There arevery few paintings of food.84.Changes in eating habits were caused byA.shortagesof basic foodB.the movement ofpopulationC.developments inagricultureD. the recipes of foreign cooks85.What is the problemwith our foodtoday?A.It's too expensive for many people.B.It'smanufactured tooquickly.C.It doesn't taste as it used to.D.It's exported in verylarge quantities.86.What point isthe authormaking in the example aboutthe pear inthe lastparagraph?A.There is not as much variety now as before.B.There wasmorevariety in the past but quality was not as high.C.Most pears that wereproduced were not fit toeat.D.There is more varietyoutside Europe than inEurope.(B)Everhadatimeinyourlifewherecashhasn'tbeenasplentifulasyou'dlikeandyou've needed to tightenthe budget?We all have,AccordingtotheWorldBank,globalfoodprices,driveninpartbyhigherfuelcosts,are36 %abovetheirlevelayearago.Foodanddrinkareamongourbiggestongoing expenses, which meansit's also one of thebiggestopportunities for savings.Therearechangesofthewayyoubuyandpreparefoodthatcanmakeabigimpactonyourb ankaccount.Makingsignificantsavingsonyourgrocerybillisn'tashard as you might think, and you canstill livewell. Hereis how.COOK FROMTHE BEGINNINGIngeneral,themorepreparationorprocessingstepsafoodhagundergone,thehighertheprice.Soit'sbesttocookthingsfromtheverybeginningwherepossible.Evensmallc hangacanmakeabigcostdifference.Trybuyingunwashedwholelettuceorsaladleavesandgi vingthemacleaningyourself,Theper-ki10ÿa mprice can be as much as ten timesmore for prepackaged salad leaves.STEWNGMEATSupplyanddemandplayabigroleinfoodpricing.Foodthatisn'tfashionablecanbeexcell entvalueformoney.Unfortunately,lambchopsarenotasinexpensiveastheyoncewere—buttherearestillplentyofotheroptions.Lookforoxtail,chicken wingtip'andfeet,or•other org anssuchasliversandlungs.Aquickinternetsearchwillgiveyouahostofideasonhowtoprepar eunfamiliaringredients.Agoodplaceto start is with slow cookedcasseroles or stews, BUYIN BULKAgoodrule,basedonexperience,isthatischeaper",butlargerpacksizesaren'talwaysbet tervalue.Itdoespaytocomparethepricetomakesureyoureallyaresaving-a—ifyoucan,checkthe''unitpricing",whichcc—patespricesbyvolumeandweight.Andmakesureyou'llnotendupthrowinghalfofitoutthanu singitbytheexpiry date.TAKEALISTPlanningaheadisgreatwaytoeconomizeandreducetheriskofexpensiveimpulsepurch ases,somakealistofwhatyouneedbeforeyouheadtothesupermarket.It'salsoagoodideanott oshopwhileyouarehungrytoresisttheurgeto pick up unnecessary food items, IGNORETHE LISTGiveyourselftheflexibilitytomakethemostofthesupermarketspecialsanddiscounts,a ndtousedifferentingredientsorchangeyourweeklyrecipeplansdepending onwhat's onsale orthe best value for money on the day.87.It can be learned from the passage thatA.t here maybe less opportunityfor you to save moneywhen food pricesareincreasinggreatly.B.checking the expiry date is important when buyingfoods.C.the way we prepare our food affects our expenses.D.changing your weekly recipe planswill help you stay healthy.88."Buy in bulk"most probably means"A.Buyfoods in largequantitiesB. Buy unpackaged goodsOnlyC.Bu y what's on sale D.Buy goods in smaller pack sizes89.W hich of the following isTRUE according tothe passage?A.Prepackaged food is usually of highquality.B.Wehadbettermakeaplaninadvanceaboutwhattobuybeforegoingtothesupermarket.C.Preparingunfamiliaringredientsusually costsyour moneyand time.D.We should ignorethe supermarket specials anddiscountstoavoid beingtricked.(C)RobertF.Kennedyoncesaidthata country’s GDPmeasures"everythingexceptthatwhic hmakeslifeworthwhile."WithBritainvotingtoleavetheEuropeanUnion,andGDPalreadypr edictedtoslowasaresult,itisnowatimelymomenttoassesswhat he was referring to.The question of GDPand its usefulness has annoyedpolicymakers for over half acentury.Manyarguethatitisaflawedconcept,Itmeasuresthingsthatdonotmatterandmisses thingsthatdo.Bymostrecentmeasures,theUK'sGDPhasbeentheenvyoftheWesternworld, withrecordlowunemploymentandhighgrowthfigures.Ifeverythingwasgoingsowell,then whydidover17millionpeoplevoteforBrexit,despite thewarnings about what it could do to theircountry'seconomicprospects?Arecentannualstudyofcountriesandtheirabilitytoconvertgrowthintowell-beingshedssomelightonthatquestion.Acrossthe163countriesmeasured,theUKisoneofthe poorestperformersinensuringthateconomicgrowthistranslatedintomeaningfulimprovem entsforitscitizens.RatherthanjustfocusingonGDP,over40differentsetsofcriteriafromhealt h,educationandcivilsocietyengagementhavebeen measured to geta more rounded of how countries areperforming.Whileallofthesecountriesfacetheirownchallenges,thereareanumberofconsistenttha nes.Yes,therehasbeenabuddingeconomicrecoverysincethe2008globalcrash,butinkeyindi catorsinareassuchashealthandeducation,majoreconomieshavecontinuedtodecline.Yetthi sisn'tthecasewithallcountries.SomerelativelypoorEuropeancountrieshaveseenhugeimpr ovementsacrossmeasuresincluding civil society,income equality andthe environment Thisisalessonthatrich countriescanlearn:WhenGDPisnolongerregardedasthe solemeasure of O country's success, the world looks verydifferent,So whatKennedy wasreferring to was thatwhileGDPhasbeenthemostcommonmethodformeasuringtheeconomicactivityofnati ons,asa•measure,itisnolongerenough.Itdoesnotincludeimportantfactorssuchasenvironm entalqualityor educationoutcomes—allthingsthat contribute to a person's sense of well-being.ThesharphittogrowthpredictedaroundtheworldandintheUKcouldleadtoadeclineint heeverydayserviceswedependonforourwell-beingandforgrowth.Butpolicymakerswhorefocuseffortsonimprovingwell-beingratherthansimplyworryingaboutGDPfigurescouldavoidtheforecasteddoomandma yevenseeprogress.90.Robert F. Kennedy is cited because heA.praised the UKfor its GDP.B.identified GDPwith happiness.C.misinterpreted the role ofGDP.D.had a low opinion ofGDP.9I. It can beinferred from Paragraph 2 thatA.the UKis reluctantto remold its economic pattern.B.GDPas the measure of successisWidely disapproved of in the UK.C.theUKwill contribute less tothe worldeconomy.D.policymakers in theUKare paying less attention to GDP. 92.Whichof the following is titleabout the recent annual study?A.It is sponsored by 163 countries.B.It excludes GDPas an indicator.C.Its criteria are questionable.D.Its results are enlightening.93.I n thelast twoparagraphs, the author suggests thatA.the UK is preparing for an economicboom.B.B. high GDPforeshadows an economic decline.C.it is essential to consider factors beyondGDP.D.it requires caution tohandle economic issues.(D)Inarareunanimousruling,theUSSupremeCourthasoverturnedthecorruptionconvicti onofaformerVirginiagovernor,RobertMcDonnell.putitdidsowhileholdingitsnoseattheeth içsofhisconduct,whichincludedacceptinggiftssuchasaRolex watch and aFerrariautomobile froma companyseeking accessto government.Thehighcourt'sdecisionsaidthejudgeinMr.McDonnell'strialfailedtotellajurythatitm ustlookonlyathis''officialacts,"ortheformergovernor'gdecisionson"specific"and'unsettle d" issues related to hisduties.Merelyhelpingagift-givergainaccesstootherofficials,unlessdonewithclearintent to pressure thoseofficials, is not corruption, the justices found.Thecourtdidsuggestthatacceptingfavorsinreturnforopeningdoorsis"distasteful"and "nasty,"Butunderanti-briberylaws,proofmustbemadeofconcretebenefits,suchasapprovalofacontractorregulati on.Simplyarrangingmeeting,makingaphonecall,or hostingan event isnot an "official act".Thecourt'srulingislegallysoundindefininga.kindoffavoritismthatisnotcriminal.Elec tedleadersmustbeallowedtohelpsupportersdealwithbureaucraticproblemswithoutfearof prosecutionforbribery."Thebasiccompactunderlyingrepresentativegovernment,"wroteC hiefJusticeJohnRobertsforthecourt,"assumesthat publicofficialswill hear fromtheir constituents and act on theirconcerns."Buttherulingreinforcestheneedforcitizensandtheirelectedrepresentatives,notthecou rts,toensureequalityofaccesstogovernment.Officialsmustnotbeallowedtoplayfavoritesin providinginformationorinarrangingmeetingssimplybecauseanindividualorgroupprovide sacampaigndonationorapersonalgift.Thistypeofintegrityrequireswell-enforcedlawsingovernmenttransparency,suchasrecordsofofficialmeetings,rulesonlobby ing,andinformationabouteachelectedleader's source of wealth.Favoritisminofficialaccesscanfanpublicperceptionsofcorruption.Butitisnotalwaysc orruption.Ratherofficialsmustavoiddoublestandards,ordifferenttypesofaccessforaverag epeopleandthewealthy.Ifconnectionscanbebought,abasicpremiseofdemocraticsociety—thatallareequalintreatmentbygovernment—isundermined.Goodgovernancerestsonanunderstandingoftheinherentworthofeach individual.Thecourt'srulingisastepforwardinthestruggleagainstbothcorruptionandofficialfavoritism.94.The underlined sentence (Para.l) most probably shows thatthecourtA.avoided defining theextent of McDonnell's duties.B.made no compromisein convicting McDonnell.C.wascontemptuous of McDonnell's conduct.D.refused to comment on McDonnell'sethics.95.According to Paragraph 4, an officialactis considered corruptive only if itinvolvesA.leakingsecrets intentionally.B.B. sizable gains inthe formofgifts.C.concrete returns for gift-givers.D.breakingcontractsofficially.96.T he court's is based on the assumption that publicofficials areA.justifiedin addressing the needs oftheir constituents.B.qualifiedto dealindependently withbureaucratic issues.C.allowed to focus on the concernsof theirsupporters.D.spare from conviction on the charge offavoritism.97.W ell-enforced laws ingovernmenttransparency are needed toA.awaken the conscience of officials.B.g uaranteefair playin officialaccess.C.allow for certain kinds oflobbying.D.inspire hopes in average people.98.The author's attitudetoward the court's rulingisA.sarcasticB. tolerantC.skepticalD. supportiveSectionA(10')Directions: Fillin the sentenceswith the proper formof the given verbs.1.H owhappyweare!TheNationalDayHolidaywehavebeenlookingforwardto(come) atlast.2.N ndhasbeenmarkedwilderness,buthowitisusedand managed is far from (settle).3.T hemanagersdiscussedtheplanthattheywouldliketosee(carry)outthe next year.4.When she said some people cheated in the exam, she (not refer) toyou.5.Therewasaknockatthedoor.Itwasthesecondtimesomeone (interrupt)me that evening.6.The watch he (mend)went wrong.7.The samething, happening in war time (amount) to a greatdisaster.8.If he follow the doctor's advice, he would be quite all rightnow.9.M ark oftenattempts to escape (fine) whenever he breaks trafficregulations10.HewillhavelearnedEnglishforeightyearsbythetimehefromthe(graduate) universitynextyear.Section B (10')Directions:Afterreadingthepassagebelow,fillintheblankstomakethepassagecoherentand grammaticallycorrect.Fortheblankswithagivenword,fillineachblankwiththeproperform ofthegivenword;fortheotherblanks,useonewordthatbest fits each blankHistory of PizzaPizzaisprobablythemostgloballyrecognizablefoodtoday.Originallyafoodforthepoo rhasnowbecomeadishforeveryone.Pizzaistheresultofatransformationdeveloped over the centuries.Theancestorofpizzaisthesimplebread.Itdominateshistoryasoneofthemostusedfoods forthepoorwhowereforced11_(eat)onlycerealsand12fewproductsthatnaturewasoffering.Theyusedtheirownimagina tionandfantasytocreatesomethingtastefulfortheirmeals,Barleywas 13 (common)ingredientusedtopreparebread.Throughthepizzatookitsowncourseanddevelop edintothedishthatweknowtoday.Thisdishhasinspiredmanywritersandartistsinourhistoryandit'sfromtheirdocu mentsandnotes 14_weknowthatpizzahastrulyoldorigins.Themodernpizza_15 (develop)aftertheintroductionofthetomatotoItaly in the16thcentury.Throughoutthe17thand18thcenturiesNaplesestablisheditselfasthepizzacapital ofItaly.However,creditfortheinventionofthetrulymodernpizzagoestoaman16(name)RafaelaEsposito.Forthe1889visitofKingUmbertoIandQueenMargherita,Esposi todesignedthe"pizzaMargherita",whichlookedliketheItalianflag 17 .the queenloved it, and today, pizzaMargherita is the world standard.PizzadidnotfinditswayintotheUnitedStates_18aftertheendoftheSecondWar.Atthatt ime,U.S.soldiersreturnedhome.Alongwiththeirreturn,theyalsobroughtbackanappetitefo rthepizzathey 19 (enjoy)inItaly,Pizzaparlorstospringupinandmorepiecesashungrypeoplestartedtoeatthiso ne-dishfoodon a regular basis.Today,20 (depend)onwhereyoulive,youcanusuallyfindapizza parlornearyourhome.Manyparlorswillevendelivertheirfoodrighttoyourfrontdoor!or,ifyo uprefer,mostgrocerystoressell,frozenpizzasandpizza-makingkitaswell asthe separate ingredientsformaking your own style of pizza too!Directions: Complete thefollowing sentences with the he p of the Chinese given-21.G raduatedfromoneofthetopfiveuniversitiesintheworld,thisaggressiveyoungman(声称自己已经确立了作为一名有名的经济学家的地位）22.Smokinganddrinking arealwayslisted asthe twomajorcause;of variousdiseasesbecause （它们会妨碍身体吸收氧气）(偏袒艺术23.TheGovernmentcameunderfireyesterdayfor领域的一些精英)24.He(保持浓厚的兴趣)the progress of the work.25.Engineersnoticedthatthepipes_ (没有如预期的那样在膨胀)26.WesternEuropeancountries (见证了移民的快速增长）afterWorldWarIlandmanyEuropeansnationstodayhavesizeableimmigrantpopulations,bo th of European and non-European origin.27.Itmaybethe21stCenturybuttheissueofvirginityinArabculturecanstillbe(生死攸关的大事，especiallyfor womenlikeSonia andNada. 28,Thereisastudyprogram (大学毕业生目前可以获得)in twenty-two countries inAsia,Africa and LatinAmerica.29.E mma Kelty (充分意识到危险）in theAmazon rainforest butkeptgoingregardlessofthewarnings,onlytobekilledbythedrugthetraffickersAmazon hidden in the forest.30.To (培养一个人的品位）Englishprose, the most effective way is to read English booksextensively.Section D (12')Directions:Translate the followingsentencesinto English, usingthe wordsgivenin the brackets.31.在这些细节上争来争去浪费了他们很多宝贵的时光。
上海交通大学附属中学2017-2018学年高二12月月考数学试题 Word版含答案
上海交通大学附属中学2017-2018学年度第一学期高二数学月考试卷一. 填空题 1. 124312⎛⎫⎛⎫=⎪⎪-⎝⎭⎝⎭2. △ABC 顶点(0,0)A 、(1,2)B 、(3,1)C -，则该三角形面积为3. 已知方程22146x y k k+=-+表示椭圆，则实数k 的取值范围是 4. 若关于,x y 的二元一次方程组12ax y a x ay a +=+⎧⎨+=⎩无解，则a =5. 已知点F 是抛物线24y x =的焦点，M 、N 是该抛物线上两点，||||6MF NF +=， 则MN 中点的横坐标为6. 过原点的直线l 与双曲线2222:1x y C a b-=(0,0)a b >>的左右两支分别相交于A 、B 两点，(F 是双曲线的左焦点，若||||4FA FB +=，0FA FB ⋅=，则双曲线的方程是7. 点(1,1)M 到抛物线2y ax =的准线的距离是2，则a =8. △ABC 外接圆半径为1，圆心为O ，3450OA OB OC ++= ，则OC AB ⋅=9. 已知圆22:(1)(1)4M x y -+-=，直线:60l x y +-=，A 为直线l 上一点，若圆M 上 存在两点B 、C ，使得60BAC ︒∠=，则点A 横坐标取值范围是10. 已知1F 、2F 分别是椭圆2214x y +=的两焦点，点P 是该椭圆上一动点，则12PF PF ⋅ 的取值范围是11. 若直线240ax by -+=(0,0)a b >>被圆222410x y x y ++-+=截得的弦长为4， 则ab 的最大值是12. 已知1F 、2F 分别为椭圆2214x y +=左右焦点，点P 在椭圆上，12||PF PF += ， 则12F PF ∠=13. 已知20a b ab +-=(0,0)a b >>，当ab 取得最小值时，曲线||||1x x y y a b-=上的点到直线y =的距离的取值范围是14. 在平面直角坐标系xOy 中，已知圆22:16O x y +=，点(2,2)P ，M 、N 是圆O 上相异两点，且PM PN ⊥，若PQ PM PN =+ ，则||PQ的取值范围是二. 选择题15. 若(2,3)a = ，(4,7)b =-，则a 在b 方向上的投影为（ ）5D. 16. 已知过定点(2,0)P 的直线l与曲线y =相交于A 、B 两点，O 为坐标原点，当△AOB 的面积取到最大值时，直线l 的倾斜角为（ ）A. 150︒B. 135︒C. 120︒D. 不存在17. 已知双曲线2222:1x y C a b -=(0,0)a b >>的左右焦点分别为1F 、2F ，点O 为双曲线的 中心，点P 在双曲线右支上，△12PF F 内切圆的圆心为Q ，圆Q 与x 轴相切于点A ，过2F 作直线PQ 的垂线，垂足为B ，则下列结论中成立的是（ ） A. ||||OA OB > B. ||||OA OB < C. ||||OA OB = D. ||OA 、||OB 大小关系不确定18. 若椭圆2212211:1x y C a b +=11(0)a b >>和椭圆2222222:1x y C a b +=22(0)a b >>的焦点相同，且12a a >，给出如下四个结论：① 椭圆1C 和椭圆2C 一定没有公共点；②1122a b a b >； ③ 22221212a ab b -=-；④ 1212a a b b -<-；其中，所有正确结论的序号是（ ） A. ①③ B. ①③④ C. ①②④ D. ②③④三. 解答题19. 已知,x y 满足约束条件10230x y x y --≤⎧⎨--≥⎩，当目标函数z ax by =+(0,0)a b >>在该约束条件下取到最小值22a b +最小值；20. 已知△ABC的三边长||AB ||4BC =，||1AC =，动点M 满足CM =CA CB λμ+ ，且14λμ=；（1）求cos ACB ∠；（2）求||CM最小值；21. 双曲线2222:1x y E a b-=(0,0)a b >>；（1）点1(,0)A a -、2(,0)A a ，动点P 在E 上，作11AQ A P ⊥，22A Q A P ⊥，求点Q 的 轨迹方程；（2）点00(,)M x y 、00(,)N x y --为E 上定点，点P 为E 上动点，作MP MQ ⊥，NP NQ ⊥，求Q 的轨迹方程；22. 两圆221111:0C x y D x E y F ++++=（圆心1C ，半径1r ），与2222:C x y D x +++ 220E y F +=（圆心2C ，半径2r ）不是同心圆，方程相减（消去二次项）得到的直线 121212:()()0l D D x E E y F F -+-+-=叫做圆1C 与圆2C 的根轴；（1）求证：当1C 与2C 相交于,A B 两点时，AB 所在直线为根轴l ；（2）对根轴上任意点P ，求证：22221122||||PC r PC r -=-；（3）设根轴l 与12C C 交于点H ，12||C C d =，求证：H 分12C C 的比2221222212d r r d r r λ+-=-+；23. 已知椭圆2222:1x y E a b+=(0)a b >>上动点P 、Q ，O 为原点；（1）若2222||||OP OQ a b +=+，求证：||OP OQ k k ⋅为定值；（2）点(0,)B b ，若BP BQ ⊥，求证：直线PQ 过定点； （3）若OP OQ ⊥，求证：直线PQ 为定圆的切线；参考答案一. 填空题 1. 810⎛⎫⎪⎝⎭2. 72 3. (6,1)(1,4)--- 4. 1- 5. 2 6.2212x y -= 7. 112-或14 8. 15- 9. [1,5] 10. [2,1]-11. 1 12. 2π 13. 14.二. 选择题15. C 16. A 17. C 18. B三. 解答题19. 4； 20.（1）12；（2 21.（1）22224a x b y a -=；（2）2222222200a xb y a x b y -=-； 22. 略； 23. 略；。
2017-2018年上海市交大附中高三下开学考
上海交大附中2017-2018学年第二学期高三数学摸底考试卷一、填空题（第1题至第6题，每题4分；第7题至12题，每题5分，共54分）1. 已知a 是实数，i 是虚数单位，若21(1)z a a i =-++是纯虚数，则a =____________．2. 已知二元一次方程组111222a x b y c a x b y c ì+=ïïíï+=ïî的增广矩阵是111113骣-÷ç÷ç÷ç÷桫，则此方程组的解是____________．3. 为了普及环保知识，增强环保意识，某大学随机抽取30名学生参加环保知识测试，得分（10分制）的频数分布统计图如图所示，如果得分值的中位数为a ，众数为b ，平均数为c ，则a 、b 、c 中的最大者是____________．4. 命题：“若a 、b 、c 成等比数列，则2b ac =”及其逆命题、否命题、逆否命题（这四个命题）中正确的个数是____________．5. 已知正数a 、b 满足430a b +=，使得14a b+取最小值的实数对(,)a b 是____________． 6. 不等式10x x ->的解集为____________．7. 已知2(1)(1)(1)f x x x +=-?，则1(1)f x -+=____________．8. 在平面直角坐标系xOy 中，以Ox 正半轴为始边的钝角a 的终边与圆22:4O x y +=交于点11(,)P x y ，点P 沿圆顺时针移动23p个单位弧长后到达点22(,)Q x y ，则12y y +的取值范围是____________．9. 已知三棱锥V ABC -，底面是边长为2的正三角形，VA ^底面ABC V ，2VA =，D 是VB 中点，则异面直线VC 、AD 所成角的大小为____________．（用反三角函数表示） 10. 如表给出一个“等差数阵”：其中每行、每列都是等差数列，ij a 表示位于第i 行第j 列的数，则112在这“等差数阵”中出现的次数为____________．11. 在平面直角坐标系xOy 中，设(1,1)A -，,B C 是函数1(0)y x x=>图像上的两点，且ABC V 为正三角形，则ABC V 的高为____________．12. 如图，平面上两点(0,1),(3,6)P Q ，在直线y x =上取两点,M N 使MN =，且使PM MN NQ ++的值取最小，则N 的坐标为____________．二、选择题（每题5分，共20分）13. 某电商设计了一种红色，打开每个红包都会获得三种福卡（“和谐”、“爱国”、“敬业”）中的一种，若集齐三种卡片可获得奖励，小明现在有4个此类红包，则它获奖的概率为（ ）A.38B.58C.49D.7914. 设,,l m n 表示三条直线，,,a b g 表示三个平面，给出下列四个命题： ①若,l m a a ^^，则//l m ；②若m b Ì，n 是l 在b 内的射影，m l ^，则m n ^； ③若,//m m n a Ì，则//n a ；④若,a g b g ^^，则//a b ，其中真命题为（ ） A. ①②B. ①②③C. ②③④D. ①③④15. 在平面内，定点,,,A B C O 满足2OA OB OC ===uu r uu u r uuu r ，AC AB BC BA OB AC AB BC BA骣骣鼢珑鼢珑鼢珑-=?鼢珑鼢珑鼢鼢珑桫桫uuu r uu u r uu u r uu r uu u r uuu r uu u r uu u r uu r 0=，动点P 满足1,AP PM MC ==uu u r uuu r uuu r ，则2BM uuu r 的最大值是（ ）A.434B.494C.374D.37216. 设等差数列{}n a 满足：22223535317cos cos sin sin cos 2sin()a a a a a a a --=+，4,2k a k Z p 刮且公差(1,0)d ?，若当且仅当8n =时，数列{}n a 的前n 项和n S 取得最大值，则首项1a 的取值范围是（ ）A. 3,22pp 轾犏犏臌B. 3,22pp 骣÷ç÷ç÷ç桫C. 7,24pp 轾犏犏臌D. 7,24pp 骣÷ç÷ç÷ç桫三、解答题17. （本题满分14分）已知函数221()cos sin ,(0,)2f x x x x p =-+?． （1）求()f x 的单调递增区间；（2）设ABC V 为钝角三角形，角A所对边a =角B 所对边5b =，若()0f A =，求ABCV 的面积．18. （本题满分14分，第1小题6分，第2 小题8分）如图1，在高为2的梯形ABCD 中，//,2,5AB CD AB CD ==，过A 、B 分别作,AE CD BF CD ^^，垂足分别为E 、F ，已知1DE =，将梯形ABCD 沿AE 、BF 同侧折起，使得,//AF BD DE CF ^，得空间几何体ADE BCF -，如图2． （1）证明：//BE 面ACD ；（2）求三棱锥B ACD -的体积．19. （本题满分14分）某小区有一块三角形空地，如图ABC V ，其中180AC =米，90BC =米，90C ?o ，开发商计划在这片空地上进行绿化和修建运动场所，在ABC V 内的点P 处有一服务站（其大小可忽略不计），开发商打算在AC 边上选一点D ，，然后过点P 和点D 画一分界线与边AB 相交于点E ，在ADE V 区域内绿化，在四边形BCDE 区域内修建运动场所，现已知点P 处的服务站与AC 距离为10米，与BC 距离为100米，设DC d =米，试问d 取何值时，运动场所面积最大？20. （本题满分16分，第1小题4分，第2小题6分，第3小题6分）已知O 为坐标原点，圆22:(1)16M x y ++=，定点(1,0)F ，点N 是圆M 上一动点，线段NF 的垂直平分线交圆M 的半径MN 于点O ，点Q 的轨迹为E ． （1）求曲线E 的方程；（2）已知点P 是曲线E 上但不在坐标轴上的任意一点，曲线E 与y 轴的交点分别为1B 、2B ，某直线1B P 和2B P 分别与x 轴交于C 、D 两点，请问线段之积OC OD ×是否为定值？如果是请求出定值，如果不是请说明理由；（3）在（2）的条件下，若点C 坐标为(1,0)-，过点C 的直线l 与曲线E 相交于A 、B 两点，求ABD V 面积的最大值．21. （本题满分18分，第1小题4分，第2小题6分，第3小题8分）设()f x 是定义在[,]a b 上的函数，若存在(,)x a b Î，使得()f x 在[,]a x 单调递增，在[,]x b 上单调递减，则称()f x 为[,]a b 上的单峰函数，x 为峰点，包含峰点的区间称为含峰区间，其含峰区间的长度为：b a -．（1）判断下列函数中，哪些是“[0,1]上的单峰函数”？若是，指出峰点；若不是，说出原因；2123241()2,()121,()log ,()sin 42f x x x f x x f x x f x x 骣÷ç=-=--=+=÷ç÷ç桫； （2）若函数3()(0)f x ax x a =+<是[1,2]上的单峰函数，求实数a 的取值范围； （3）若函数()f x 是区间[0,1]上的单峰函数，证明：对于任意的1212,(0,1),x x x x ?，若12()()f x f x ³，则2(0,)x 为含峰区间；若12()()f x f x £，则1(,1)x 为含峰区间；试问当12,x x 满足何种条件时，所确定的含峰区间的长度不大于0.6．参考答案一、填空题 1. 12. 21x y ì=ïïíï=ïî3. c4. 2个5. 15,154骣÷ç÷ç÷ç桫 6. (0,1)(1,)+?U7. )2[0,)x -??8.9. 1arccos410. 711. 212. 99,44骣÷ç÷ç÷ç桫二、选择题 13. C 14. A 15. B 16. D三、解答题17. （1）,2pp 轹÷ê÷÷êøë； （2）4 18. （1）证明略； （2）2319. 6020. （1）22143x y +=； （2）定值为4； （3）9221. （1）①21()2f x x x =-是[0,1]上的单峰函数，峰点为14； ②2()121f x x =--不是[0,1]上的单峰函数； ③321()log 2f x x 骣÷ç=+÷ç÷ç桫不是[0,1]上的单峰函数； ④4()sin 4f x x =是[0,1]上的单峰函数，峰点为8p（2）11,312a 骣÷ç?-÷ç÷ç桫 （3）证明略；120.40.6x x ì=ïïíï=ïî。
上海市交大附中2018~2019学年度 高二下期中数学考试卷(图片版)
上海交通大学附属中学2018-2019学年度第二学期高二数学期中考试试卷（满分150分，120分钟完成.答案一律写在答题纸上）命题：刘亚丽审核：杨逸峰一、填空题（本大题共12题，1-6题每题4分，7-12题每题5分，满分54分）1、如果一条直线与两条平行直线都相交，那么这三条直线共可确定个平面．答案：12、【2017高考上海，4】已知球的体积为36π，则该球主视图的面积等于.【答案】9π【解析】设球的半径为R ，则：34363R ππ=，解得：3R =，该球的主视图是一个半径为3的圆，其面积为：29S R ππ==.3、若正三棱柱的所有棱长均为a ，且其体积为a =．【答案】4【解析】236444a a a ⋅=⇒=⇒=4、【2017高考上海，7】如图，以长方体1111ABCD A B C D -的顶点D 为坐标原点，过D 的三条棱所在的直线为坐标轴，建立空间直角坐标系.若1DB 的坐标为()4,3,2，则1AC的坐标是.【答案】()4,3,2-【解析】将向量1AC的起点平移至点D ，则平移后的向量与向量1DB 关于平面11CDD C 对称，据此可得：()14,3,2AC =-.5、【2014上海,理6】若圆锥的侧面积是底面积的3倍，则其母线与底面所成的角的大小为（结果用反三角函数值表示）.【答案】1arccos3.6、【2013上海文10】已知圆柱Ω的母线长为l ，底面半径为r ，O 是上底面圆心，A 、B 是下底面圆周上两个不同的点，BC 是母线，如图．若直线OA 与BC 所成角的大小为6π，则l r＝______.【解析】由题知，tan63r l π==⇒l r =.7、已知ABC ∆三个顶点到平面α的距离分别是3，3，6，则其重心到平面α的距离为__________（写出所有可能值）答案：0，2，4。
8、正方体1111ABCD A B C D -的棱长为1，若动点P 在线段1BD 上运动,则DC AP ⋅的取值范围是______________．【答案】【解析】试题分析：以所在的直线为轴，以所在的直线为轴，以所在的直线为轴，建立空间直角坐标系．则、、、、．∴、．∵点在线段上运动，∴，且．，，故答案为．9、【2010上海理12，倒数第3题】如图所示，在边长为4的正方形纸片ABCD 中，AC 与BD 相交于O ，剪去AOB ∆，将剩余部分沿OC 、OD 折叠，使OA 、OB 重合，则以A （B ）、C 、D 、O 为顶点的四面体的体积为________；【答案】3【解析】在折叠过程中OC OB ⊥，OD OA ⊥始终没有改变，所以最后形成的四面体()A B CDO -中，OA ⊥底面CDO ，故其体积21182(22)22323V =⨯⨯⨯=，故答案为：823.10、某三棱锥的三视图如图所示，且三个三角形均为直角三角形，则34x y +的最大值为．【答案】55试题分析：由图可知，根据三视图得到三棱锥OABC 如图，OC=2，AC=y，BC=1，在OAC Rt ∆中，24y OA -=，2225y BC OA x -=+=，即522=+y x ，三角换元（或者称利用圆的参数方程）设5cos ,5sin x y θθ==，故3455cos()55x y θϕ+=+≤。
2017-2018学年交大附中高三下英语开学考
上海交通大学附属中学2017-2018学年第二学期高三英语摸底考(满分140分，考试时间120分钟，答案一律写在网上阅卷答题纸上)第I卷I. Listening ComprehensionPart A Short ConversationsDirections: In Part 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 conversations and the questions 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. In the kitchen. B. At home. C.In a fashion shop. D. In a net bar.2. A. 500 yen. B. 1500 yen. C.2500 yen. D. 5000 yen.3. A. He will pay for the lunch. B.Sue likes Spanish food very much.C. He doesn’t like Spanish food at all.D. Both of them should treat Sue wellat dinner.4. A. Boss and employee. B. Doctor and patient.C. Teacher and student.D.Interviewer and interviewee.5. A. She doesn’t have time to find a new flat.B. She has paid three months, rent in advance.C. She is unlikely to find such a satisfactory flat.D. She doesn’t like the idea of decorating an empty flat.6. A. Mike. B. Sandy. C. An unexpected friend. D. A handyman.7. A. Make a recovery plan. B. Find a full-time job.C. Drop out of school.D. Resign from her present job.8. A. The woman was fully absorbed in the music.B. The woman couldn’t understand the music very well.C. The woman lost her way to the conceit that evening.D. The concert was no better than what the woman imagined.9. A. Extreme sports. B. Travel insurance.C. Travel arrangements.D. Courage and safety.10. A. She is talkative. B. She is quiet. C. She is active. D. She is sociable.Part B PassagesDirections: In Section B, you will hear two short passages and one longer conversation. After each passage or conversation, you will be asked several questions. The passage and the conversations 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. 28 billion copies. B. 567 million copies.C. 1953 million copies.D. 2015 million copies.12. A. Interviewed the vice-president of the Commercial Press in China.B. Completed data investigation, collection and examination for the records.C. Compared the sales volumes of the Secret Garden and Harry Potter series.D. Delivered enough copies of Xinhua Dictionary to the presentation ceremony.13. A. It contains colorful cultures of many countries.B. It is the world’s most popular dictionary.C. It is a useful tool for learners of the Chinese language.D. It has influenced several generations of Chinese people.Questions 14 through 16 are based on the following passage.14. A. Over a billion people do not have enough safe drinking water.B. Diseases and 6acteria in the water kill millions of people every day.C. People in developed countries take their clean drinking water for granted.D. Clean,bottled water is expensive to transport to poor rural villages.15. A. It is a plastic straw.B. It uses batteries or electricity.C. It can be used for one year.D. It’s portable and cheap.16. A. Appeal to the government to supply clean water for everyone.B. Provide easy access to clean water to destroy the bacteria.C. Offer the WaterTube free to people in. developing countries.D. Sell the WaterTube to every person who needs one.Questions 17 through 20 are based on the following conversation.17. A. A little restaurant. B. A private organization.C. An ice cream shop.D. A snack bar.18. A. More flexible working time.B. Less conflicts with consumers.C. Closer relationship with customers.D. More freedom to make financial decisions.19. A. A program of the Small Business Administration.B. A grant from a private organization.C. Savings in his own bank account.D. Loans from the bank.20. A. He found owing his own business is rewarding.B. He didn’t recommend others to have their own business.C. He chose the location to compete with bigger businesses.D. He found managing a small business is high risky.II. Grammar and VocabularySection ADirections:After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Talking with—Not Just to—Kids Powers How They Learn Language Children from the poorer families begin life not only with material disadvantages but cognitive ones. Decades of research (21)____________(confirm) this, including a famous 1995 finding by psychologistsBetty Hart and Todd Risley: By age four children raised in poverty have heard 30 million fewer words, on average, than their peers from wealthier families. That gap has been linked to shakier language skills at the start of school, (22)____________, in turn, predicts weaker academic performance.But just the quantity of words a toddler hears is not the most significant influence on language acquisition. Growing evidence has led researchers to conclude quality matters more than quantity, and (23)____________(valuable) quality seems to be back-and-forth communication—what researchers call conversational turns.A paper (24)____________(publish) last week in Psychological Science brings a new kind of support to this idea, offering the first evidence that these exchanges play a vital role in the development of Broca’s area, the brain region most closely associated with producing speech. Further, the amount of conversational turns a child experiences daily outweighs socioeconomic status in predicting (25)____________ activity in Broca’s area and the child’s language skills.The researchers confirmed the classic 1995 finding that, overall, kids from wealthier families hear more words. And small (26)____________ their sample was, they even confirmed the 30-million-word gap between the poorest and richest children. But they found that “by far the biggest driver for brain development was not the number of words spoken (27)____________ the conversations,” Gabrieli sa ys.The researchers calculated that a child’s verbal ability score increased (28)____________ one point for every additional 11 conversational exchanges per hour.The study is a “very, very important” addition (29)____________ a growing body of work, says developmental psychologist Kathryn Hirsh-Pasek, director of the Infant Language Laboratory at Temple University. “We have known for quite a while that conversational turns—or (30)____________ in my work we call conversational duets—are very important for building a foundation for language and maybe for learning general ly. What hadn’t been done is to link it where we knew it had to be linked—to changes in the brain.”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.A. obedienceB. adaptedC. conflictD. receivedE. willF. strandedG. supervision H. harmony I. interpreted J. exception K. pursuitLord of the FliesLord of the Flies is a 1954 novel by Nobel Prize-winning British author William Golding. The book focuses on a group of British boys (31)__________ on an uninhabited islandand their disastrous attempt to govern themselves.The novel has been critically well (32)__________. It was named in theModern Library 100 Best Novels, reaching number 41 on the editor's list, and25 on the reader's list. In 2005 Time magazine named it as one of the 100 bestEnglish-language novels from 1923 to 2005.Published in 1954, Lord of the Flies was Golding's first novel. Although it was not a great success at the time, it soon went on to become a best-seller. It has been (33)__________ for film twice in English, in 1963 by Peter Brook and 1990 by Harry Hook.The book takes place in the middle of an unspecified nuclear war.A plane carrying a group of British schoolboys is shot down over the Pacific. The pilot of the plane is killed, but many of the boys survive the crash and find themselves deserted on an island, where they are alone without adult (34)__________. Some of them arrive as a musical choir(合唱队) under an established leader , while others are ordinary students. With the (35)__________ of the choirboys, some children like Sam and Eric, appear never to have encountered each other before. The book portrays their gradually becomingextremely cruel and violent. Left to themselves on a heavenly island, far from modern civilization, the well-educated children return to a primitive state.Golding wrote his book as a counterpoint(对照) to R.M. Ballantyne's youth novel The Coral Island (1858), and included specific references to it, such as the rescuing naval officer's description of thechildren's (36)__________ of Ralph, one of the kid leaders on island, as "a jolly(pleasant) good show, like the Coral Island". Golding's three central characters—Ralph, Piggy and Jack—have been (37)__________ as exaggerated versions of Ballantyne's Coral Island leading characters.At an allegorical(比喻的) level, the central theme is the (38)__________ ofhuman desirefor civilization and social organization—living by rules, peacefully—and for the (39)__________ to power. Several themes are revealed in the book, including the tensionbetween groupthink and individuality, between rational and emotional reactions, between (40)__________and aggression, and between morality and immorality. How these play out, and how different people feel the influences of these form a major subtext of Lord of the Flies.III. Reading Comprehension (15’+30’)Section 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.The picture of miseries and sufferings of the Black woman in America may appear, perhaps darker and more depressing than Dante’s(但丁) description of Hell because in the history of human race, the Black woman in America has suffered the most. A close and critical (41)_______ of the history of the Black woman’s life and condition since her arrival in a strange land will unfold very well the long story of her pain and sorrow (42)_______ her mutilated(被切断的) soul. Since her arrival on the foreign shore, the Black woman has been facing the worst kind of(43)_______ and oppression. As a Black she has had to endure all the horrors of slavery andhas been the (44)_______ of continually inhumane treatmentin all kinds of work, (45)_______ the lowest place on the wage scale and restricted to the lowest-status and the most uncreative jobs. Since times of slavery, Black womanhood has been destroyed, twisted and abused with racial and inhuman (46)_______ by black men and white men and women. In the process, they have lost their genuine "self", and have developed a(n)(47)_______ in themselves—though as black women, they see themselves with the eyes of white men and women and black men. This has ultimately been responsible for the destruction of their self-confidence and the feeling of being human. They (48)_______ themselves as master s’ belongings.As a mother she has seen her children sold into slavery. She has seen them left at home without attentionwhile she (49)_______ to the needs of the children of the ruling class. She has seen her children suffer fromdrug addiction, the (50)_______ of decent education and experience attacks by a racist society, and (51)_______ the prisons of this nation. In addition, besides suffering the common fate of all oppressed people, the African-American women continue to (52)_______ the oppression of woman by men, which existed for long. In the home she becomes “the slave of a slave”. Men may be cruelly treated and subjected to all sorts of dehumanizing treatmenton the part of the ruling class.(53)_______, at least they can take out their frustration on someone else, their women.Thus feminism in America means much more than what it stands for in other European countries, chiefly because it has different role and meaning(54)_______ the Black women. If a feminist is commonly defined as one who is involved in transforming and (55)_______ familiar realities, then Black women are innate(天生的) feminists. Black women writing exhibits and constructs a Black women’s literary tradition that is inherently feminist. Barbara Smith, an influential Black feminist critic, states that the ability of Black women to survive in the face of White America exhibits an innate feminist potential.41. A. respect B. reference C. account D. expression42. A. associated with B. completed with C. involved in D. joined by43. A. exploration B. exhaustion C. exploitation D. explosion44. A. article B. item C. substance D. object45. A. including B. occupying C. containing D. striking46. A. procedures B. processes C. descriptions D. practices47. A. complexity B. anxiety C. simplicity D. gravity48. A. look to B. look upon C. look over D. look through49. A. took B. kept C. attended D. appealed50. A. richness B. lack C. variety D. possibility51. A. populating B. emphasizing C. charging D. criticizing52. A. undertake B. underline C. undervalue D. undergo53. A. Therefore B. Anyway C. However D. Moreover54. A. with respect to B. in line with C. in addition to D. in parallel with55. A. reappointing B. representing C. reconsidering D. reinterpretingSection BDirections: Read the following four 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)I was at the funeral of my dearest friend--my mother. She finally had lost her long battle with cancer. The hurt was so intense; I found it hard to breathe at times. Always supportive, Mother clapped loudest at my school plays, held the box of tissues while listening to my first heartbreak, comforted me at my father's death, and prayed for me my entire life.When mother's illness was diagnosed, my sister had a new baby and my brother had recently married his childhood sweetheart, so it fell on me, the 27-year-old middle child, to take care of her. I counted it an honor.My place had been with our mother, preparing her meals, taking her to the doctor, reading the Bible together. Now she was in heaven.My work was finished, but I was alone.Deep in sorrow, suddenly, I heard a door open and slam shut at the back of the church. Quick footsteps hurried along the carpeted floor. Ayoung man looked around briefly and then sat next to me. He folded his hands and placed them on his lap. His eyes started to be filled with tears."I'm late," he explained, though no explanation was necessary.After several eulogies, he leaned over and commented, "Why do they keep calling Mary by the name of Margaret?''"Because, that was her name, Margaret. Never Mary." I whispered, wondering who the stranger was anyway."Isn't this the Lutheran church?""No, the Lutheran church is across the street.""Oh...""I believe you're at the wrong funeral, Sir."The solemness(庄重) of the occasion mixed with the realization of the man's mistake bubbled up inside me and came out as laughter. Sharp looks from other mourners(哀悼者) only made the situation seem more stupid. I peeked at the confused, misguided man seated beside me. He was laughing too, as he glanced around, deciding it was too late for an uneventful exit.I imagined Mother laughing.At the final 'Amen,' we rushed out a door and into the parking lot. "I do believe we'll be the talk of thetown. By the way, my name is Rick." he smiled.That afternoon began a lifelong journey for me with this man who attended the wrong funeral, but was in the right place. A year after our meeting, we were married at a country church. This time we both arrived at the same church, right on time.In my time of sorrow, God gave me laughter. In place of loneliness, God gave me love. This past June, we celebrated our twenty-second wedding anniversary. Whenever anyone asks us how we met, Rick tells them, 'Her mother and my Aunt Mary introduced us, and it's truly a match made in heaven.'56. Only author could take care of her mom mainly because ________.A. she was the only child in the familyB. a lovely baby came into her brother’s familyC. she was the only child without a new family’s burdenD. her mom loved her much more than other children57. What can we infer from the passage?A. The author and Rick met 22 years ago for the first time.B. The author was supposed to have been in Lutheran Church.C. Margaret should be the name of Rick’s aunt.D. The mourners considered the author’s joy improper.58. What could be the best title of the passage?A. Hope Remaining at the FuneralB. A Heavenly EncounterC. Two Funerals at One TimeD. Seeking God’s Everlasting Love(B)59. You and your friend just watched the eruption of Old Faithful at 12:26 p.m., at what time is it possible for you to enjoy the next one?A. 13:10B. 15:06C. 14:06D. 13:1660. Which one of the following statement about Old Faithful is true?A. The geyser’s name gives people an indication that it always erupts regularly, about 20 times each day, once every 74 minutes.B. When it is erupting, people should keep a safe distance from that due to the huge amount of water it expels as well as its freezing coldness.C. Old Faithful is a well-known geyser which can expel at least 3700 gallons water each time and it’s located in the world’s largest national park.D. If visitors want to check the eruption time, they may refer to the posted timetables, on which the predictions are calculated by the naturalists.61. Where does the article most probably appear?A. The Yellowstone official website.B. Local travel pamphlets introducing Yellowstone.C. A recently-issued guide book on Yellowstone.D. A travel magazine column about Yellowstone.(C)Science advances fastest when data and conclusions are shared as quickly as possible. Yet it is common practice for medical researchers to conceal results for months or years until research is published in an academic journal. Even then, the data supporting a study are often not revealed.The motive to withhold findings is powerful. Journal papers are the key measure of a scientist’s productivity. To win research money and get promoted, scientists need to increase an impressive list ofpublications. Yet the delays in spreading knowledge have the capacity to do real harm. Nor are elite journals the guardians of quality that they often claim to be. The number of papers so flawed that they need to be retracted has risen sharply in the past two decades. Studies in elite journals (such as Nature and Science) are no statistically stronger than those in lesser journals.Science should not, and need not, be restrained by journal publication. Three sensible reforms would ensure that researchers’ results could be commu nicated to more people more quickly, without any decline in quality. Step one is for the organizations that finance research to demand that scientists put their academic papers, along with their experimental data, in publicly accessible “databases” before they are sent to a journal. That would allow other researchers to make use of the findings without delay. Those opposing such “preprints” argue that they allow work of poor quality to grow wildly because it has not yet been peer-reviewed. That may surprise physicists and mathematicians, who have been posting work to arXiv, a preprint database, for more than 25 years with no ill effects.Step two is to improve the process of peer review itself. Journals currently administer a system of organizing anonymous peer reviewers to pass judgment on new research—a fact they use, in part, to justify their sky-high subscription prices. But this process tends to be abused. At its worst, groups of researchers are suspected of guaranteeing favorable reviews for each other’s work. Better that reviewers are named and that the reviews are published. The Gates foundation has announced its support for an online database where such open peer review of papers takes place. The database was launched last year by the Wellcome Trust, meaning that the world’s two largest medical charities have thrown their weight behind it. Others should follow.Finally, science needs to stop relying so much on journal publication as the only recognized credential for researchers and the only path to career progression. Tools exist that report how often a preprint has been viewed, for example, or whether a clinical data set has been cited in guidelines for doctors. A handful of firms are using artificial intelligence to assess the scientific importance of research, disregarding how it has been spread. Such approaches need encouragement. Journals may lose out, but science itself will benefit.62. The passage primarily argues ________.A. why scientists should timely make public their information with other peersB. how a reliable system that promotes fast sharing of academic outcomes can be establishedC. how the scientific circle ensures the quality of those research papers published in journalsD. what the difference is between the studies posted to preprint databases and the elite journals63. Which of the following is NOT the reason why scientists hold back their findings?A. They want to climb up the career ladders.B. They want to show their capacity of yielding great researches.C. They want to ensure the scientific validity before releasing them.D. They want to get funded for their study.64. What does the underlined word “anonymous” probably mean?A. showing strong abilityB. identified by many peopleC. unknown by nameD. having great authority65. What can help reduce academics’dependence on journal publication?A. Using tools to report the frequency of a preprint being viewed.B. Firms using advanced technology to produce a research.C. Citing doctors’ guideline in processing a clinical data.D. Finding methods to assess the way those studies are shared or spread.66. Which one of the following statements will the author most probably agree with?A. Full trust can be put in elite journals like Science and Nature.B. The prospect of scientific journals undoubtedly outweighs that of science.C. The popularity of preprint will inevitably bring about a careless academic world.D. Organizations ought to employ an open measure to conduct the peer review.Section CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there is one sentence more than you need.Gun laws in America are already so weak, it must be tough to find ways to make them even looser. But you have to give gun lobbyists(游说者) credit. They've come up with some new ideas and found friends in Congress to promote them, even as the nation is in deep sadnessdue to the 58 people killed at a music festival in Las Vegas.At the top of the wish list: a push to make it easier to buy gun silencers. (67)__________. At worst, they would put innocent people and police officers in graver danger.The changes are tucked(塞) into a measure called the Sportsmen's Heritage and Recreational Enhancement (SHARE) Act. The seemingly harmless title tries to hide its most controversial content. The proposal getting the most attention is a move to make it easier to buy silencers. (68)__________. But when opponents argue that silencers will make it harder to hear gunshots, those same supporters insist it isn't so. Sorry, but they can't have it both ways.Silencers have been available to hunters and others under the current law since 1934. But the National Rifle Association thinks it takes too much trouble to buy one. Buyers must pass a criminal background check, turn over their fingerprints to the government, and pay a $200 fee. Silencers are registered, and transfers are tracked. (69)__________.Two days after the Las Vegas massacre(屠杀), a group of House Democrats called on House Speaker Paul Ryanto ensure that the SHARE measure won’t be taken up by the House. But the best Ryan could know was, “I don’t know when it’s going to be scheduled.”Ryan and other Republicans who have opposed every previous effort to pass gun measures said this week they'd consider banning "bump stocks," anextra piece of equipment the Las Vegas shooter used to make his semi-automatic weapon even more dangerous. Also, the NRA talked favorably about "additional regulations" but not a new law. (70)__________.Controlling gun violence should be at the top of Congress' agenda. Instead, according to the gun lobby's demand, it is looking at weakening existing restrictions. That is pathetic.Section DDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Virginia is killing its U.S. history testsAs a history exam supporter, I have seen so many people appealing to the government to cancel it. Many students scored poorly in it and complained about hard work. In 2014, the Virginia legislature(立法) ordered a cut in the number of tests taken by the state’s children, and specifically eliminated the fifth grade and middle school U.S. history tests. The U.S. and Virginia history exam in high school is also about to disappear. Politicians seem to think the best way to reduce testing pressure is to abandon tests, no matter how important.I met Steve Constantino, now chief academic officer of the Virginia Education Department, and he told me, “even without tests, the course, requirements, standards, etc., are all still very much required and schools are accountable to ensure that students master the content. I believe what this does is move the course toward more of a deeper learning experience, like IB.”But when I met him the first time in the late 1990s, he was then the brilliant principal of Stonewall Jackson High School. He turned that school into a model for the demanding International Baccalaureate(IB) program.One of the reasons his IB program at Stonewall Jackson High School worked so well was that his IB students had to take the nation’s most difficult final exams, some of them five hours long.My suggestion is that Virginia replace its state history tests with IB exams. You can learn a great deal preparing for one of those even if you fail.But that’s not going to happen. Constantino’s dream of deeper learning without challenging tests has never been achieved in American schools. So we will muddle along, not learning much history, since despite what we say, we really don’t think we need it.第II卷V. Translation(3’+3’+4’+5’)Directions: Translate the following sentences into English, using the words given in the brackets.1. 没有什么比网游更容易浪费你的时间了。
2017-2018学年上海市交大附中高三(下)开学数学试卷(解析版)
2017-2018学年上海市交大附中高三（下）开学数学试卷一、填空题：第1题至第6题，每题4分第7题至12题，每小题4分，共54分）.1．（4分）已知a是实数，i是虚数单位，若z＝a2﹣1+（a+1）i是纯虚数，则a＝．2．（4分）已知二元一次方程组的增广矩阵是，则此方程组的解是．3．（4分）为了普及环保知识，增强环保意识，某大学随机抽取30名学生参加环保知识测试，得分（10分制）的频数分布统计图如图所示，如果得分值的中位数为a，众数为b，平均数为c，则a、b、c中的最大者是．4．（4分）命题：“若a，b，c成等比数列，则b2＝ac”及其逆命题、否命题、逆否命题中正确的个数是．5．（4分）已知正数a，b满足4a+b＝30，使得取最小值的实数对（a，b）是．6．（4分）不等式x|x﹣1|＞0的解集为．7．（5分）已知f（x+1）＝（x﹣1）2（x≤1），则f﹣1（x+1）＝．8．（5分）在平面直角坐标系xOy中，设钝角α的终边与圆O：x2+y2＝4交于点P（x1，y1），点P沿圆顺时针移动个单位弧长后到达点Q（x2，y2），则y1+y2的取值范围是；若x2＝，则x1＝．9．（5分）已知三棱锥V﹣ABC，底面是边长为2的正三角形，VA⊥底面△ABC，VA＝2，D 是VB中点，则异面直线VC、AD所成角的大小为（用反三角函数表示）．10．（5分）如表给出一个“等差数阵”：其中每行、每列都是等差数列，a ij表示位于第i行第j列的数．则112在这“等差数阵”中出现的次数为．11．（5分）在平面直角坐标系xOy中，设A（﹣1，1），B，C是函数图象上的两点，且△ABC为正三角形，则△ABC的高为．12．（5分）如图：平面上两点P（0，1），Q（3，6），在直线y＝x上取两点M，N，使且使|PM|+|MN|+|NQ|的值取最小，则N的坐标为．二、选择题：本大题共4小题，每小题5分，共60分．在每小题给出的四个选项中，只有一个是符合题目要求的．13．（5分）某电商设计了一种红包，打开每个红包都会获得三种福卡（“和谐”、“爱国”、“敬业”）中的一种，若集齐三种卡片可获得奖励，小明现在有4个此类红包，则他获奖的概率为（）A．B．C．D．14．（5分）设l，m，n表示三条直线，α，β，γ表示三个平面，给出下列四个命题：①若l⊥α，m⊥α，则l∥m；②若m⊂β，n是l在β内的射影，m⊥l，则m⊥n；③若m⊂α，m∥n，则n∥α；④若α⊥γ，β⊥γ，则α∥β．其中真命题为（）A．①②B．①②③C．②③④D．①③④15．（5分）在平面内，定点A，B，C，O满足|＝2，＝，动点P，M满足的最大值是（）A．B．C．D．16．（5分）设等差数列{a n}满足：cos2a3cos2a5﹣sin2a3sin2a5﹣cos2a3＝sin（a1+a7），a4≠，k∈Z且公差d∈（﹣1，0），若当且仅当n＝8时，数列{a n}的前n项和S n取得最大值，则首项a1的取值范围是（）A．[，2π]B．（，2π）C．[，2π]D．（，2π）三、解答题：本大题共5小题，共48分．解答写出文字说明、证明过程或演算过程．17．（14分）已知函数f（x）＝cos2x﹣sin2x+，x∈（0，π）．（1）求f（x）的单调递增区间；（2）设△ABC为锐角三角形，角A所对边a＝，角B所对边b＝5，若f（A）＝0，求△ABC的面积．18．（14分）如图1，在高为2的梯形ABCD中，AB∥CD，AB＝2，CD＝5，过A、B分别作AE⊥CD，BF⊥CD，垂足分别为E、F．已知DE＝1，将梯形ABCD沿AE、BF同侧折起，使得AF⊥BD，DE∥CF，得空间几何体ADE﹣BCF，如图2．（Ⅰ）证明：BE∥面ACD；（Ⅱ）求三棱锥B﹣ACD的体积．19．（14分）某小区有一块三角形空地，如图△ABC，其中AC＝180米，BC＝90米，∠C ＝90°，开发商计划在这片空地上进行绿化和修建运动场所，在△ABC内的P点处有一服务站（其大小可忽略不计），开发商打算在AC边上选一点D，然后过点P和点D画一分界线与边AB相交于点E，在△ADE区域内绿化，在四边形BCDE区域内修建运动场所．现已知点P处的服务站与AC距离为10米，与BC距离为100米．设DC＝d米，试问d取何值时，运动场所面积最大？20．（16分）已知O为坐标原点，圆M：（x+1）2+y2＝16，定点F（1，0），点N是圆M上一动点，线段NF的垂直平分线交圆M的半径MN于点Q，点Q的轨迹为E．（1）求曲线E的方程；（2）已知点P是曲线E上但不在坐标轴上的任意一点，曲线E与y轴的交点分别为B1、B2，直线B1P和B2P分别与x轴相交于C、D两点，请问线段长之积|OC|•|OD|是否为定值？如果是请求出定值，如果不是请说明理由；（3）在（2）的条件下，若点C坐标为（﹣1，0），过点C的直线l与E相交于A、B两点，求△ABD面积的最大值．21．（18分）设f（x）是定义在[a，b]上的函数，若存在x∈（a，b），使得f（x）在[a，x]单调递增，在[x，b]上单调递减，则称f（x）为[a，b]上的单峰函数，x为峰点，包含峰点的区间称为含峰区间，其含峰区间的长度为：b﹣a．（1）判断下列函数中，哪些是“[0，1]上的单峰函数”？若是，指出峰点；若不是，说出原因：f1（x）＝x﹣2x2，f2（x）＝1﹣|2x﹣1|，f3（x）＝|log2（x+）|，f4（x）＝sin4x；（2）若函数f（x）＝ax3+x（a＜0）是[1，2]上的单峰函数，求实数a的取值范围；（ 3 ）若函数f（x）区间[0，1]上的单峰函数，证明：对于任意的x1，x2∈（0，1），x1＜x2，若f（x1）≥f（x2），则（0，x2）为含峰区间；若f（x1）≤f（x2），则（x1，1）为含峰区间；试问当x1，x2满足何种条件时，所确定的含峰区间的长度不大于0.6．2017-2018学年上海市交大附中高三（下）开学数学试卷参考答案与试题解析一、填空题：第1题至第6题，每题4分第7题至12题，每小题4分，共54分）. 1．【解答】解：∵z＝a2﹣1+（a+1）i是纯虚数，∴，解得a＝1．故答案为：1．2．【解答】解：由题意，方程组解之得故答案为3．【解答】解：由频率分布直方图知，众数为b＝5；由中位数的定义知是第15个数与第16个数的平均值，将数据从小到大排第15 个数是5，第16个数是6，∴中位数为a＝＝5.5；平均数是c＝×（2×3+3×4+5×10+6×6+3×7+2×9+2×10）≈6.0，∴b＜a＜c，即a、b、c中最大者是c．故答案为：c．4．【解答】解：因为原命题为真，故逆否命题：“若b2≠ac，则a，b，c不成等比数列”为真，又逆命题：“若b2＝ac，则a，b，c成等比数列”，当a＝b＝c＝0时，结论不成立，故命题为假，所以否命题也未假，故答案为2．5．【解答】解：∵正数a，b满足4a+b＝30，∴＝＝≥＝，当且仅当b＝4a＝15时取等号．∴使得取最小值的实数对（a，b）是．故答案为：．6．【解答】解：∵x|x﹣1|＞0，∴x＞0，|x﹣1|＞0，故x﹣1＞0或x﹣1＜0，解得：x＞1或0＜x＜1，故不等式的解集是（0，1）∪（1，+∞），故答案为：（0，1）∪（1，+∞）．7．【解答】解：∵f（x+1）＝（x﹣1）2（x≤1），∴f（x）＝（x﹣2）2（x≤2），∴f﹣1（x）＝2﹣，（x≥0）∴f﹣1（x+1）＝（x≥﹣1）故答案为：（x≥﹣1）8．【解答】解：圆的半径r＝2，点P沿圆顺时针移动个单位弧长后到达点Q，则移动的弧度为＝，由三角函数定义知，x1＝2cosα，y1＝2sinα，＜α＜π，x2＝2cos（α﹣），y2＝2sin（α﹣），则y1+y2＝2sinα+2sin（α﹣）＝2sinα+2（sinαcos﹣cosαsin）＝2sinα+sinα﹣cosα＝3sinα﹣cosα＝2（sinα﹣cosα）＝2sin（α﹣），∵＜α＜π，∴＜α﹣＜，∴＜sin（α﹣）≤1，＜2sin（α﹣）≤2，即y1+y2的取值范围是（，2]，∵x2＝2cos（α﹣）＝，∴cos（α﹣）＝，∵＜α＜π，∴＜α﹣＜，∵cos（α﹣）＝＞0，∴＜α﹣＜，则sin（α﹣）＝＝＝，则x1＝2cosα＝2cos（α﹣+）＝2[cos（α﹣）cos﹣sin（α﹣）sin]＝2（﹣×）＝，故答案为：（3，2]，9．【解答】解：取BC的中点E，连接AE，DE，则DE∥VC，故∠ADE是异面直线VC、AD所成角，在△ADE中，AD＝．DE＝VC＝，AE＝，由余弦定理得：cos∠ADE＝＝∴∠ADE＝，则异面直线VC、AD所成角的大小为，故答案为：（等）．10．【解答】解：根据图象和每行、每列都是等差数列，该等差数阵的第一行是首项为4，公差为3的等差数列：a1j＝4+3（j﹣1），第二行是首项为7，公差为5的等差数列：a2j＝7+5（j﹣1）第i行是首项为4+3（i﹣1），公差为2i+1的等差数列，因此a ij＝4+3（i﹣1）+（2i+1）（j﹣1）＝2ij+i+j，要找112在该等差数阵中的位置，也就是要找正整数i，j，使得2ij+i+j＝112，所以j＝，当i＝1时，j＝37，当i＝2时，j＝22，当i＝4时，j＝12，当i＝7时，j＝7，当i＝12时，j＝4，当i＝22时，j＝2，当i＝37时，j＝1．∴112在这“等差数阵”中出现的次数为7．故答案为：7．11．【解答】解：设B、C为直线y＝kx+b（k＜0，b＞0）与y＝的交点，由得kx2+bx﹣1＝0．设B（x1，y1），C（x2，y2），则x1+x2＝﹣，y1+y2＝+＝＝b，设BC的中点为D，则D（﹣，）．因为A（﹣1，1），依题意，k AD•k BC＝﹣1，即•k＝﹣1，由于k＜0，故1﹣k≠0，∴b＝（b＞0）．∵|BC|＝|x1﹣x2|＝•＝•＝•∴d A﹣BC＝|BC|，即＝×|BC|＝×2•，即＝ו，解得：k＝．∵b＝＞0，∴k＝，k2＝，∴d A﹣BC＝＝＝＝＝＝2．故△ABC的高为2．故答案为：2．12．【解答】解：P点关于直线y＝x的对称点为P′（1，0），则有|PM|+|MN|+|NQ|＝|P′M|+|MN|+|NQ|，又过Q（3，6）作平行于y＝x的直线为y＝x+b；由6＝3+b得b＝3，即此时直线为y＝x+3，过M点作MQ′∥NQ，则|MQ′|＝|NQ|，|QQ′|＝|MN|＝；则|P′M|+|MN|+|NQ|＝|P′M|+|MN|+|MQ′|，∵|MN|是常数，∴要使|PM|+|MN|+|NQ|的值取最小，则|P′M|+|MQ′|的值取最小，即P′，M，Q′三点共线时最小，则设Q′（a，a+3），（a＜3），由|QQ′|＝|MN|＝；得，即2（a﹣3）2＝2，即（a﹣3）2＝1，∴a＝4（舍）或a＝2，即Q′（2，5），设M（x，x），则x，解得x＝，即M（，），设N（b，b），b＞，由|MN|＝；得，得2（b﹣）2＝2，即（b﹣）＝1，则b﹣＝1或b﹣＝﹣1，即b＝或b＝（舍），故N（，），故答案为：（，）二、选择题：本大题共4小题，每小题5分，共60分．在每小题给出的四个选项中，只有一个是符合题目要求的．13．【解答】解：小明现在有4个此类红包，基本事件总数n＝3×3×3×3＝81，他获奖包含的基本事件个数m＝＝36，∴他获奖的概率p＝＝．故选：C．14．【解答】解：选项①，可以根据直线与平面垂直的性质定理得出的，故其正确；选项②，根据由三垂线定理的逆定理可证可知正确；选项③，n在平面α内时不正确；选项④，若α⊥β，α⊥γ，则γ⊥β，不正确，如正方体共顶点的三个平面；故选：A．15．【解答】解：由|＝2知，O是△ABC的外心；＝，∴﹣＝﹣＝0，当﹣＝0时，＝，即＝，∴cos∠DAC＝cos∠DAB∴∠DAC＝∠DAB，∴O点在三角形的角A平分线上；同理，O点在三角形的角B，角C平分线上；∴点定O的一定是△ABC的内心，如图1所示；∴△ABC是正三角形，且边长为＝2；如图2所示，建立平面直角坐标系；则B（0，0），C（2，0），A（，3）；∵M满足||＝1，∴点P的轨迹方程为：+（y﹣3）2＝1；令x＝+cosθ，y＝3+sinθ，θ∈[0，2π），由＝，得M（+cosθ，+sinθ），∴||2＝+＝+3sin（θ+）≤；∴||2的最大值是．故选：B．16．【解答】解：∵cos2a3cos2a5﹣sin2a3sin2a5﹣cos2a3＝sin（a1+a7），∴cos2a3cos2a5﹣sin2a3sin2a5﹣cos2a3+sin2a3＝sin（a1+a7），即cos2a3（cos2a5﹣1）﹣sin2a3（sin2a5﹣1）＝sin2a4，即﹣cos2a3sin2a5+sin2a3cos2a5＝sin2a4，即（sin a3cos a5﹣cos a3sin a5）（sin a3cos a5+cos a3sin a5）＝sin2a4，即sin（a3﹣a5）sin（a3+a5）＝sin2a4，即﹣sin2d sin（2a4）＝sin2a4，∵a4≠，∴sin2a4≠0，∴sin（2d）＝﹣1．∵d∈（﹣1，0），∴2d∈（﹣2，0），则2d＝，d＝﹣．由S n＝na1+＝na1+×（﹣）＝﹣n2+（a1+）n．对称轴方程为n＝（a1+），由题意当且仅当n＝8时，数列{a n}的前n项和S n取得最大值，∴＜（a1+）＜，解得：＜a1＜2π．∴首项a1的取值范围是（，2π），故选：D．三、解答题：本大题共5小题，共48分．解答写出文字说明、证明过程或演算过程．17．【解答】解：（1）函数f（x）＝cos2x﹣sin2x+＝cos2x+，x∈（0，π），由2kπ﹣π≤2x≤2kπ，解得kπ﹣π≤x≤kπ，k∈Z，k＝1时，π≤x≤π，可得f（x）的增区间为[，π）；（2）设△ABC为锐角三角形，角A所对边a＝，角B所对边b＝5，若f（A）＝0，即有cos2A+＝0，解得2A＝π，即A＝π，由余弦定理可得a2＝b2+c2﹣2bc cos A，化为c2﹣5c+6＝0，解得c＝2或3，若c＝2，则cos B＝＜0，即有B为钝角，c＝2不成立，则c＝3，△ABC的面积为S＝bc sin A＝×5×3×＝．18．【解答】（Ⅰ）证明：证法一、连接BE交AF于O，取AC的中点H，连接OH，则OH是△AFC的中位线，∴OH∥CF，OH＝．由已知得DE∥CF，DE＝，∴DE∥OH，DE＝OH，连接DH，则四边形DHOE是平行四边形，∴EO∥DH，又∵EO⊄面ADC，DH⊂面ADC，∴EO∥面ACD，即BE∥面ACD；证法二、延长FE，CD交于点K，连接AK，则面CKA∩面ABFE＝KA，由已知得DE∥CF，DE＝，∴DE是△KFC的中位线，则KE＝EF．∴KE∥AB，KE＝AB，则四边形ABEK是平行四边形，得AK∥BE．又∵BE⊄面ADC，KA⊂面ADC，∴BE∥面ACD；证法三、取CF的中点G，连接BG，EG，得DE∥CG，DE＝CG，即四边形CDEG是平行四边形，则EG∥DC，又GE⊄面ADC，DC⊂面ADC，∴GE∥面ADC，又∵DE∥GF，DE＝GF，∴四边形DGFE是平行四边形，得DG∥EF，DG＝EF，又ABFE是平行四边形，∴AB∥EF，AB＝EF，得AB∥DG，AB＝DG，∴四边形ABGD是平行四边形，则BG∥AD，又GB⊄面ADC，DA⊂面ADC，∴GB∥面ADC，又GB∩GE＝G，∴面GBE∥面ADC，又BE⊂面GBE，∴BE∥面ACD；（Ⅱ）解：∵GB∥面ADC，∴V B﹣ACD＝V E﹣ACD，由已知得，四边形ABFE为正方形，且边长为2，则在图2中，AF⊥BE，由已知AF⊥BD，且BE∩BD＝B，可得AF⊥平面BDE，又DE⊂平面BDE，∴AF⊥DE，又AE⊥DE，AF∩AE＝A，∴DE⊥平面ABFE，且AE⊥EF，∴AE⊥面CDE，∴AE是三棱锥A﹣DEC的高，∵四边形DEFC是直角梯形．且AE＝2，DE＝1，EF＝2，∴．19．【解答】解：法一：以C为坐标原点，CB所在直线为x轴，CA所在直线为y轴建立直角坐标系，则C（0，0），A（0，180），B（90，0），P（10，100），D（0，d）．DE直线方程：，①AB所在直线方程为2x+y＝180，②解①、②组成的方程组得，，∵直线DE经过点B时，∴，设，＝，∵（当且仅当t＝60，即k＝4时取等号），此时d＝120﹣t＝60，∴当d＝60时，绿化面积最小，从而运动区域面积最大．法二：如图，分别过点P，E作AC的垂线，垂足为Q，F，设EF＝h，若如图1所示，则PQ＝10，CQ＝100，DQ＝100﹣d，由△AFE～△ACB得，即AF＝2h，从而CF＝180﹣2h，DF＝180﹣2h﹣d，由△DPQ～△DEF得，解得若如图2所示，则PQ＝10，CQ＝100，DQ＝d﹣100，AF＝2h，CF＝180﹣2h，DF＝2h+d ﹣180，由△DPQ～△DEF得，解得；由0＜h＜90得，由，设，＝，∵（当且仅当t＝60，即k＝4时取等号），此时d＝120﹣t＝60，∴当d＝60时，绿化面积最小，从而运动区域面积最大．20．【解答】（1）解：连结FQ，则FQ＝NQ，∵MQ+FQ＝MQ+QN＝MN＝4＞ME，椭圆的定义即得点Q的轨迹为以点M、F为焦点，长轴为4的椭圆∴2a＝4，即a＝2，又∵焦点为（1，0），即c＝1，∴b2＝a2﹣c2＝4﹣1＝3，故点Q的轨迹C的方程为：（2）证明：设P（x0，y0），直线B1P的方程为：y＝．令y＝0，得，|OC|•|OD|＝|x C|•|x D|＝||…①∵点P是曲线E上但不在坐标轴上的任意一点，∴．即3x02＝4（3﹣y02），代入①得|OC|•|OD|为定值4．（3）当点C的坐标为（﹣1，0）时，点D（﹣4，0），|CD|＝3，设直线l的方程为：x＝my﹣1，A（x1，y1），B（x2，y2）联立得（3m2+4）y2﹣6my﹣9＝0解得：．|y1﹣y2|＝，△ABD面积s＝×|y1﹣y2|•|CD＝•＝＝；∵，根据∵在[1，+∞）递增可得3．∴∴m＝0，即直线AB：x＝﹣1时，△ABD面积的最大为．21．【解答】解：（1）画出f1（x）＝x﹣2x2，其对称轴为x＝，如图所示，根据图象可知，函数f（x）在[0，1]上为单峰函数，峰点为x＝，如图所述；画出f2（x）＝1﹣|2x﹣1|，其对称轴为x＝，如图所示，根据图象可知，函数f（x）在[0，1]上为单峰函数，峰点为x＝，画出f3（x）＝|log2（x+）|，如图所示，根据图象可知，函数f（x）在[0，1]上不是单峰函数，画出f4（x）＝sin4x，其一条对称轴为x＝，如图所示，根据图象可知，函数f（x）在[0，1]上为单峰函数，峰点为x＝，（2）∵f（x）＝ax3+x，x∈[1，2]，a＜0，函数f（x）＝ax3+x（a＜0）是[1，2]上的单峰函数∴f′（x）＝3ax2+1，令f′（x）＝0得x＝，当1≤x＜时，f′（x）＞0，函数单调递增，当＜x≤2时，f′（x）＞0，函数单调递减，∴1＜＜2，解得﹣＜a＜﹣故a的范围为（﹣，﹣）（3）证明：设x0为f（x）的峰点，则由单峰函数定义可知，f（x）在[0，x0]上单调递增，在[x0，1]上单调递减．当f（x1）≥f（x2）时，假设x0∉（0，x2），则x1＜x2＜x0，从而f（x0）≥f（x2）＞f（x1），这与f（x1）≥f（x2）矛盾，所以x0∈（0，x2），即（0，x2）是含峰区间．当f（x1）≤f（x2）时，假设x0∉（x1，1），则x0≤x1＜x2，从而f（x0）≥f（x1）＞f（x2），这与f（x1）≤f（x2）矛盾，所以x0∈（x1，1），即（x1，1）是含峰区间：在所得的含峰区间内选取x3，由x3与x1或x3与x2类似地可确定是一个新的含峰区间，对先选择的x1；x2，x1＜x2，x1+x2＝l，①在第一次确定的含峰区间为（0，x2）的情况下，x3的取值应满足x3+x1＝x2，②由①与②可得，当x1＞x3时，含峰区间的长度为x1．由条件x1﹣x3≥0.2，得x1﹣（1﹣2x1）≥0.2，从而x1≥0.4．因此确定的含峰区间的长度不大于0.6，只要取x1＝0.4，x2＝0.6。
2017-2018学年上海交通大学附属中学高二下学期期末考试数学试题(解析版)
2017-2018学年上海交通大学附属中学高二下学期期末考试数学试题一、单选题1．设地球的半径为，地球上，两地都在北纬的纬度线上去，且其经度差为，则，两地的球面距离是（）A. B. C. D.【答案】C【解析】分析：设在北纬纬圆的圆心为，球心为，连结，根据地球纬度的定义，算出小圆半径，由两地经度差为，在中算出，从而得到，利用球面距离的公式即可得到两地球面的距离.详解：设在北纬纬圆的圆心为，球心为，连结，则平面，在中，，同理，两地经度差为，，在中，，由此可得是边长为的等边三角形，得，两地球面的距离是，故选C.点睛：本题考查地球上北纬圆上两点球的距离，着重考查了球面距离及相关计算，经纬度等基础知识，考查运算求解能力，考查空间想象能力，属于中档题. 2．对于不重合的两个平面与，给定下列条件： ①存在平面，使得、都垂直于； ②存在平面，使得、都平行于； ③内有不共线的三点到的距离相等； ④存在异面直线，，使得，，，其中，可以判定与平行的条件有（ ）A. 个B. 个C. 个D. 个 【答案】B【解析】试题分析：直线与平面的位置关系，平面与平面的位置关系，对选项进行逐一判断，确定正确选项即可．：①与平行．此时能够判断①存在平面γ，使得都平行于γ；两个平面平行，所以正确． ②存在平面γ，使得都垂直于γ；可以判定与β平行，如正方体的底面与相对的侧面．也可能与不平行．②不正确．③不能判定与平行．如面内不共线的三点不在面的同一侧时，此时与相交；④可以判定与平行．∵可在面内作，则与必相交．又．故选B ．【考点】平面与平面平行的性质；平面与平面平行的判定；平面与平面垂直的判定． 3．一个正方体的展开如图所示，点，，为原正方体的顶点，点为原正方体一条棱的中点，那么在原来的正方体中，直线与所成角的余弦值为（ ）A. B. C. D.【答案】D【解析】分析：先还原正方体，将对应的字母标出，与所成角等于与所成角，在三角形中，再利用余弦定理求出此角的余弦值即可.详解：还原正方体，如图所示，设，则，与所成角等于与所成角，余弦值为，故选D.点睛：本题主要考查异面直线所成的角以及空间想象能力，属于中档题题.求异面直线所成的角的角先要利用三角形中位线定理以及平行四边形找到，异面直线所成的角，然后利用直角三角形的性质及余弦定理求解，如果利用余弦定理求余弦，因为异面直线所成的角是直角或锐角，所以最后结果一定要取绝对值.4．已知函数的图像是一条连续不断的曲线，若，，那么下列四个命题中①必存在，使得；②必存在，使得；③必存在，使得；④必存在，使得.真命题的个数是（）A. 个B. 个C. 个D. 个【答案】A【解析】分析：函数是连续的，故在闭区间上，的值域也是连续的，令，根据不等式的性质可得①正确；利用特值法可得②③④错误，从而可得结果.详解：函数是连续的，故在闭区间上，的值域也是连续的，令，对于①，，故①正确.对于②，若，则，无意义，故②错误.对于③，时，不存在，使得，故③错误.对于④，可能为，则无意义，故④错误，故选A.点睛：本题主要通过对多个命题真假的判断，主要综合考查函不等式的性质及连续函数的性质，属于难题.这种题型综合性较强，也是高考的命题热点，同学们往往因为某一处知识点掌握不好而导致“全盘皆输”，因此做这类题目更要细心、多读题，尽量挖掘出题目中的隐含条件，利用定理、公理、结论以及特值判断，另外，要注意从简单的自己已经掌握的知识点入手，然后集中精力突破较难的命题.二、填空题5．函数的定义域为__________．【答案】【解析】分析：解不等式组即可得结果.详解：要使函数有意义，则有，故答案为.点睛：定义域的三种类型及求法：(1)已知函数的解析式，则构造使解析式有意义的不等式(组)求解；(2) 对实际问题：由实际意义及使解析式有意义构成的不等式(组)求解；(3) 若已知函数的定义域为，则函数的定义域由不等式求出.6．表面积为的球的体积为__________．【答案】【解析】分析：先根据球的表面积公式，列方程得到球半径，再利用球的体积公式求解该球的体积即可.详解：，，故答案为.点睛：本题主要考查球的体积公式和表面积公式，意在考查学生对基础知识的掌握情况，属于基础题.7．的二项展开式中，项的系数是__________．（用数字作答）【答案】【解析】分析：先求出二项式的展开式的通项公式，令的指数等于，求出的值，即可求得展开式中项的系数.详解：的二项展开式的通项为，，展开式项的系数为故答案为.点睛：本题主要考查二项展开式定理的通项与系数，属于简单题. 二项展开式定理的问题也是高考命题热点之一，关于二项式定理的命题方向比较明确，主要从以下几个方面命题：（1）考查二项展开式的通项公式；（可以考查某一项，也可考查某一项的系数）（2）考查各项系数和和各项的二项式系数和；（3）二项展开式定理的应用. 8．高一（10）班有男生人，女生人，若用分层抽样的方法从该班的全体同学中抽取一个容量为的样本，则抽取男生的人数为__________人．【答案】6【解析】分析：根据分层抽样的定义直接计算即可.详解：设抽取男生的人数为，因为男生人，女生人，从该班的全体同学中抽取一个容量为的样本，所以，取男生的人数为，故答案为.点睛：本题主要考查分层抽样的应用以及古典概型概率公式的应用，属于中档题.分层抽样适合总体中个体差异明显，层次清晰的抽样，其主要性质是，每个层次，抽取的比例相同.9．人并排站成一行，其中甲、乙两人必须相邻，那么不同的排法有__________种.（用数学作答）【答案】240【解析】分析：甲、乙两人必须相邻，利用捆绑法与其余的人全排即可.详解：甲乙相邻全排列种排法，利用捆绑法与其余的人全排有种排法，共有，故答案为.点睛：常见排列数的求法为：（1）相邻问题采取“捆绑法”；（2）不相邻问题采取“插空法”；（3）有限制元素采取“优先法”；（4）特殊顺序问题，先让所有元素全排列，然后除以有限制元素的全排列数.10．若交大附中共有名教职工，那么其中至少有两人生日在同一天的概率为__________．【答案】1【解析】分析：根据每年有天，可判断名教职工，中至少有两人生日在同一天为必然事件,从而可得结果.详解：假设每一天只有一个人生日，则还有人，所以至少两个人同日生为必然事件，所以至少有两人生日在同一天的概率为，故答案为.点睛：本题考查必然事件的定义以及必然事件的概率，属于简单题.11．设函数，则使得成立的的取值范围是__________．【答案】【解析】分析：根据函数的奇偶性和单调性之间的关系，将不等式转化为，两边平方利用一元二次不等式的解法求解即可.详解：且在时，，导数为，即有函数在单调递增，函数为偶函数，等价为，即，平方得，解得，所求的取值范围是，故答案为.点睛：本题主要考查抽象函数的奇偶性与单调性的应用，属于难题.将奇偶性与单调性综合考查是，一直是命题的热点，解这种题型往往是根据函数在所给区间上的单调性，根据奇偶性判断出函数在对称区间上的单调性(偶函数在对称区间上单调性相反，奇函数在对称区间单调性相同)，然后再根据单调性列不等式求解.12．在长方体中，，，则直线与平面所成角的正弦值为__________．【答案】【解析】分析：过作，垂足为，则平面，则即为所求平面角，从而可得结果.详解：依题意，画出图形，如图，过作，垂足为，由平面，可得，所以平面，则即为所求平面角，因为，，所以，故答案为.点睛：本题考查长方体的性质，以及直线与平面所成的角，属于中档题.求直线与平面所成的角由两种方法：一是传统法，证明线面垂直找到直线与平面所成的角，利用平面几何知识解答；二是利用空间向量，求出直线的方向向量以及平面的方向向量，利用空间向量夹角余弦公式求解即可.13．一个正方体的个顶点可以组成__________个非等边三角形.【答案】48【解析】分析：从正方体的个顶点中人取三个点共有种取法，其中等边三角形共有个，作差即可得结果.详解：从正方体的个顶点中人取三个点共有种取法，其中等边三角形共有个，所以非等边三角形共有个，故答案为.点睛：本题主要考查组合数的应用，属于简单题.14．将集合的元素分成互不相交的三个子集：，其中,，，且，，则满足条件的集合有__________个．【答案】3【解析】分析：由可得，令，则，，，然后列举出的值，从而可得结果.详解：，所以，令，根据合理安排性，集合的最大一个元素，必定为：，则，又，，①当时，同理可得.②当时，同理可得或，综上，一共有种，故答案为.点睛：本题考查主要考查集合与元素的关系，意在考查抽象思维能力，转化与划归思想，分类讨论思想应用，属于难题.解得本题的关键是首项确定，从而得到，由此打开突破点.15．设非空集合为实数集的子集，若满足下列两个条件：（1），；（2）对任意，都有，，，则称为一个数域，那么命题：①有理数集是一个数域；②若为一个数域，则；③若，都是数域，那么也是一个数域；④若，都是数域，那么也是一个数域.其中真命题的序号为__________．【答案】①②③④【解析】分析：根据“数域”的定义，对四个结论逐一验证即可，验证过程一定注意“照章办事”，不能“偷工减料”.详解：，则①正确；对于②，若是一个数域，则，于是任何一个分数，都可以构造出来，即，②正确；对于③，，③正确；定义④，④正确，故答案为①②③④.点睛：本题考查集合与元素的关系，以及新定义问题，属于难题. 新定义题型的特点是：通过给出一个新概念，或约定一种新运算，或给出几个新模型来创设全新的问题情景，要求考生在阅读理解的基础上，依据题目提供的信息，联系所学的知识和方法，实现信息的迁移，达到灵活解题的目的.遇到新定义问题，应耐心读题，分析新定义的特点，弄清新定义的性质，按新定义的要求，“照章办事”，逐条分析、验证、运算，使问题得以解决.16．已知函数在时有最大值，，并且时，的取值范围为，则__________．【答案】【解析】分析：由函数在时有最大值，可得，先判断在上单调递减，可得，解高次方程即可得结果.详解：函数在时有最大值，则可得，，，在上单调递减，则满足，，，解得，又，故答案为.点睛：本题考查求二次函数闭区间上的最值，二次函数的应用，体现了分类讨论的数学思想以及转化与划归思想，属于难题.解答本题的关键是判断出函数的单调性，求出解析式，将问题转化为解高次方程.三、解答题17．某公司生产一种产品，每年投入固定成本万元.此外，每生产件这种产品还需要增加投入万元.经测算，市场对该产品的年需求量为件，且当出售的这种产品的数量为（单位：百件）时，销售所得的收入约为（万元）.（1）若该公司这种产品的年产量为（单位：百件），试把该公司生产并销售这种产品所得的年利润表示为年产量的函数；（2）当该公司的年产量为多少时，当年所得利润最大？最大为多少？【答案】(1) ;(2) 当年产量为件时，所得利润最大.【解析】分析：（1）利用销售额减去成本即可得到年利润关于年产量的函数解析式；（2）分别利用二次函数的性质以及函数的单调性，求得两段函数值的取值范围，从而可得结果.详解：（1）由题意得：；（2）当时，函数对称轴为，故当时，；当时，函数单调递减，故，所以当年产量为件时，所得利润最大.点睛：本题主要考查阅读能力及建模能力、分段函数的解析式，属于难题.与实际应用相结合的题型也是高考命题的动向，这类问题的特点是通过现实生活的事例考查书本知识，解决这类问题的关键是耐心读题、仔细理解题，只有吃透题意，才能将实际问题转化为数学模型进行解答.理解本题题意的关键是构造分段函数，构造分段函数时，做到分段合理、不重不漏，分段函数的最值是各段的最大(最小)者的最大者(最小者). 18．解关于的不等式.（）【答案】见解析.【解析】分析：对分五种情况讨论，分别利用一元一次不等式与一元二次不等式的解法求解即可.详解：①当时，；②当时：，，因为，故等式左边因式分解得：；当时，；当时，，此时解集为空集；当时，；点睛：本题主要考查一元二次不等式的解法、分类讨论思想的应用.属于中档题.分类讨论思想解决高中数学问题的一种重要思想方法，是中学数学四种重要的数学思想之一，尤其在解决含参数问题发挥着奇特功效，大大提高了解题能力与速度.运用这种方法的关键是将题设条件研究透，这样才能快速找准突破点. 充分利用分类讨论思想方法能够使问题条理清晰，进而顺利解答，希望同学们能够熟练掌握并应用与解题当中.19．如图，二面角的大小为，四边形是边长为的正方形，，为上的点，且平面.（1）求证：；（2）求二面角的大小；（3）求点到平面的距离.【答案】(1)见解析；（2）；（3）.【解析】试题分析：（1）由平面可证，由二面角为直二面角及是正方形可证，再由线面垂直判定定理得平面，即可得证；（2）取的中点，连接，，由四边形为正方形可证，，即可得为二面角的平面角，根据题设条件求出及，即可得二面角的余弦值；（3）利用等体积法，由即可得点到平面的距离.试题解析：（1）∵平面，∴.又∵二面角为直二面角，且，∴平面，∴，∴平面，∴.（2）取的中点，连接，.∵四边形为正方形，∴，∴，即为二面角的平面角，又，∴，由（1）知，且，∴，∴，由，解得，∴，即∴，即二面角的余弦值为.（3）取的中点，连接，∵，二面角为直二面角，∴平面，且.∵，，∴平面，∴，∴，又，由，得，∴.点睛：立体几何的证明需要对证明的逻辑关系清楚，证明线线垂直，先由线面垂直得到线线垂直，再由线线垂直证明线面垂直；用普通法求二面角，讲究“一作、二证、三求”，通过辅助线先把二面角的平面角及计算所需线段作出来，再证明所作角是二面角的平面角；点到面的距离还原到体积问题，则利用等体积法解题.20．设全体空间向量组成的集合为，为中的一个单位向量，建立一个“自变量”为向量，“应变量”也是向量的“向量函数”.（1）设，，若，求向量；（2）对于中的任意两个向量，，证明：；（3）对于中的任意单位向量，求的最大值.【答案】（1）或；（2）见解析；（3）最大值为.【解析】分析：（1），设，代入运算得：，从而可得结果；（2）设，，，则利用“向量函数”的解析式化简，从而可得结果；（3）设与的夹角为，则，则，即最大值为.详解：（1）依题意得：，设，代入运算得：或；（2）设，，，则从而得证；（3）设与的夹角为，则，则，故最大值为.点睛：新定义问题一般先考察对定义的理解，这时只需一一验证定义中各个条件即可.二是考查满足新定义的函数的简单应用，如在某些条件下，满足新定义的函数有某些新的性质，这也是在新环境下研究“旧”性质，此时需结合新函数的新性质，探究“旧”性质.三是考查综合分析能力，主要将新性质有机应用在“旧”性质，创造性证明更新的性质.21．对于函数，若关系式中变量是变量的函数，则称函数为可变换函数.例如：对于函数，若，则，所以变量是变量的函数，所以是可变换函数.（1）求证：反比例函数不是可变换函数；（2）试判断函数是否是可变换函数并说明理由；（3）若函数为可变换函数，求实数的取值范围.【答案】（1）见解析；（2）见解析；（3）见解析.【解析】分析：（1）利用反证法，假设是可变换函数，，利用关变量的一元二次方程无解但导出矛盾，从而可得结论；（2）利用必须有交点，而连续且单调递减，值域为，连续且单调递增，值域为，进而可得结论；.（3），则恒大于，即无交点，不满足题意；若，则必定有交点，即方程有解，从而可得结果.详解：（1）假设是可变换函数，则，因为变量是任意的，故当时，此时有关变量的一元二次方程无解，则与假设矛盾，故原结论正确，得证；（2）若是可变换函数，则，则有关的两个函数：必须有交点，而连续且单调递减，值域为，连续且单调递增，值域为，所以这两个函数与必定有交点，即：变量是变量的函数，所以是可变换函数；（3）函数为可变换函数，则，若，则恒大于，即无交点，不满足题意；若，则必定有交点，即方程有解，从而满足题意.点睛：本题主要考查函数的性质、新定义问题，属于难题.新定义题型的特点是：通过给出一个新概念，或约定一种新运算，或给出几个新模型来创设全新的问题情景，要求考生在阅读理解的基础上，依据题目提供的信息，联系所学的知识和方法，实现信息的迁移，达到灵活解题的目的.遇到新定义问题，应耐心读题，分析新定义的特点，弄清新定义的性质，按新定义的要求，“照章办事”，逐条分析、验证、运算，使问题得以解决.本题定义“可变换函数”达到考查函数性质的目的.。
上海市交大附中2017-2018学年下学期高二开学考试数学试题
2017-2018年上海市交大附中高二下开学考一、填空题1、复数i 32+（i 是虚数单位）的模是;2、在如图所示的正方体1111D C B A ABCD -中,异面直线B A 1与C B 1所成角的大小为3、已知点()3,1A 、()1,4-B ,则与→AB 方向相同的单位向量的坐标为4、已知双曲线15422=-y x ,则以双曲线的焦点为顶点,以双曲线顶点为焦点的椭圆方程 为5、已知两圆1022=+y x 和()()203122=-+-y x 相交于B A ,两点,则直线AB 的方程是6、将参数方程⎩⎨⎧=+=θθsin 2cos 21y x （θ为参数）化为普通方程,所得方程是7、已知椭圆15222=+ty t x 的焦距为62,则实数=t 8、已知ai +2,i b +是实系数一元二次方程02=++q px x 的两根,则q p +的值为 9、若b a ,为非零实数,则下列四个命题都成立：①01≠+aa ；②()2222b ab a b a ++=+；③若b a =,则b a ±=；④若ab a =2,则b a =则对于任意非零复数b a ,,上述命题仍然成立的序号是10、如图,S 是三角形 ABC 所在平面外的一点,SC SB SA ==,且2π=∠=∠=∠CSA BSC ASB ,NM 、分别是AB 和SC 的中点,则异面直线SM 与BN 所成角的大小为（用反三角函数表示）11、已知直线n m ,及平面α,其中n m //,那么在平面α内到两条直线n m ,距离相等的点的集合可能是：①一条直线；②一个平面；③一个点；④空集。
其中正确的是12、动点()y x P ,在直角坐标系平面上能完成下列动作,先从原点O 沿正偏北⎪⎭⎫⎝⎛≤≤20παα方向行走一段时间后,再向正北方向行走,但何时改变方向不定,假定()y x P ,速度为10米/分钟,则当α变化时()y x P ,行走2分钟内的可能落点的区域面积是 二、选择题13、在下列命题中,不是公理的是（ ）A 、平行于同一个平面的两个平面相互平行B 、过不在同一条直线上的三点,有且只有一个平面C 、如果同一条直线上的两点在同一个平面内,那么这条直线上所有的点都在此平面内D 、如果两个不重合的平面有一个公共点,那么他们有且只有一条过该点的公共直线14、若空间三条直线c b a ,,满足b a ⊥,c b //,则直线a 与c （ ）A 、一定平行B 、一定相交C 、一定是异面直线D 、一定垂直15、在四边形ABCD 中,()2,1=→AC ,()2,4-=→BD ,则四边形的面积为（ ）A 、5B 、52C 、5D 、1016、已知动点P 的横坐标x 、纵坐标y 满足：①1sin cos =+ααy x （R ∈α）；②422≤+y x ,那么当α变化时,点P 形成的图形的面积为（ ）A 、πB 、π3C 、π4D 、π-4三、解答题17、如图,ABCD 是正方形,直线⊥PD 底面ABCD ,PC PD =,E 是PC 的中点。
2017-2018学年交大附中高二下学期期末考试化学试题(试卷版)
C.钙元素的化合价在反应前后发生了改变D.单质碳的状态发生了变化
13.兴奋剂是体育比赛中禁用的药物，有一种兴奋剂的结构如图。关于它的说法中正确的是
A. 该有机物能发生银镜反应
B. 分子中有4个碳碳双键
C. 既能被还原，又能被氧化
D.1mol该物质最多跟6mol氢气发生反应
B. 常温下，Br2为液体，而Cl2为气体
C. 向溴化亚铁溶液中滴入少量氯水，溶液颜色变黄
D.BrCl+H2O=HBrO+HCl是非氧化还原反应
16.对于常温下0.01mol/L的氨水，下列叙述正确的是
A.c(H+)+c(NH4+)>c(OH－)
B. 由H2O电离出的c(H+)=1.0×10－12mol·L－1
14.洁厕灵与84消毒液混合会产生氯气：2HCl + NaClO → NaCl + Cl2↑ + H2O，下列说法错误的是
A NaClO作氧化剂B.n(氧化剂)：n(还原剂)=1：2
C. 氧化性：NaClO > Cl2D.Cl2既 氧化产物又是还原产物
15.下列能说明氯的非金属性强于溴的事实是
A.HClO酸性强于HBrO
C. 与等体积等浓度的盐酸混合后所得溶液显酸性
D. 与等体积等浓度的氯化铵溶液混合c(NH4+)+c(NH3·H2O)=0.02mol﹒L-1
17.已知25℃时，向0.1mo/L的碳酸氢钠溶液滴几滴酚酞，呈微红色，则该溶液中
A.若滴加甲基橙溶液呈红色
B. c（HCO3-）>c（CO32-）>c（H2CO3）
上海交通附属中学2018-2019学年高二下学期开学考试英语试题(详解版)
上海交通附属中学 2018-2019 学年度第二学期高二英语摸底考试试卷II. Grammar and VocabularySection A(A)There is a tendency to think of each of the arts as ___1___ separate area of activity. Many artists, however, would prove ___2___ there has always been a warm relationship between the a warm relationship areas of human activity. For example, in the late nineteenth century the connections between music and painting were especially close. Artists ___3___ (invite) to design clothes and settings for operas and ballets, ___4___ sometimes it was the musicians who were inspired by the work of contemporary painters. Of the musical compositions as responses to the visual arts, perhaps the most famous is Mussorgsky’s Pictures at an Exhibition.Mussorgsky composed the piece in 1874 after the death, at the age of 39, of the artist Victor Hartman. ___5___ their friendship had not been a particularly long-lasting one, Mussorgsky was shocked by Hartmann’s unexpected death. The following year the critic, Vladimir Stasov, ___6___ decided to told a n exhibition of Hartmann’s work, suggested that Mussorgsky ___7___ (try) to relieve his grief by writing something in memory of Hartmann.The exhibition served as Mussorgsky’s inspiration. The ten pieces that make up Pictures at an Exhibition ___8___ (intend) as symbols rather than representation of the paintings in the exhibition. Between each is a promenade (舞曲中的行进), as the composer walks from one painting to ___9___ The music is sometimes witty and playful, sometimes almost alarming and frightening. Thought a range of surprising contrasts, Mussorgsky manages ____10____ (convey) the spirit of the artist and his work.(B)Today, home-ownership has reached extremely high levels. Modern generations tend to believe there is something wrong with them ___11___ they rent. However, is high home-ownership really as people imagine ？___12___ (stare) at data first, we realize that the most successful, stable, attractive country in the Western world is Switzerland. It has tiny unemployment; wealth; high happiness and mental-health scores. Does it have high home-ownership rates? Absolutely not. In Switzerland, about seven in ten of the population are renters. Yet, with Europe’s ___13___ (low) home-ownership rate, the nation thrives. Now go to the other end of the miserydistribution. Spain has approximately the highest home-ownership rate in Europe （at more than 80%）. But one-quarter of its population are unemployed.A likely reason is that high levels of home-ownership mess up the labour market. In a sensibly functioning economy it is easy for people to move around to drop into the vibrant job slots ___14___ (throw) up by technological change. With a high degree of owner-occupation, everything slows. Folk get stuck. Renters can go to new jobs. In that way they do the economy a favours. ___15___ Friedman said, the rate of unemployment depends on the flexibility of the housing market.Next we come to economic breakdown. Most analysts accept that at heart it was the housing market-obsessive pursuit of homes, the engendered mortgage（房贷）lending and an unavoidable house-price crash--- ___16___ sank the Western world. Germany, say, with its more efficient rental market, had a far smoother ride through trouble.As for the monetary system, in the past few decades, in the hope of getting untaxed capital gains way above their true labour earnings, many people threw their spare cash into buying larger houses or building extra bedrooms. TV programmes about how to make easy money, beautiful rising house prices, and most importantly, our faulty tax system encouraged that. When ___17___ some point market broke down, everyone suffered. Our countries ought, instead, to design tax systems that encourage people to invest in productive real activities and in innovation. Renting leaves money free for better purposes. That also points to the role of sensible budgeting over a person’s lifetime. Why should we think that when we die it is necessary ___18___ (pay) off an entire house？Our children do not deserve it. Let them pay for themselves. We ___19___ rent-and enjoy our lives with the money saved.Finally, moderation usually pays off. Our scientific understanding of how economies function is horribly limited. This suggests that the golden rule should be to avoid extremes. A50-50mix of home-ownership and renting, not the 70-30split that is now observed in so many Western nations, ______20______ (make) sense.Section BRapidly advancing technology and its impact on education has been a subject of debate. How can schools equip students with the skills they need to succeed in a challenging job market?Since technology is driving these changes, there is a theory that governments should keep focusing on STEM subjects. These are often referred to as “hard skills,” which are ___21___ in primary school and right through to university level. In the meantime, ‘soft skills’ are being ___22___.This is a mistake. Much evidence suggests that soft skills are far more ___23___ to graduates in the long term.Research from Harvard University on the global job market has shown that STEM- related careers grew strongly between 1989 and 2000 but have slowed down since. In contrast, jobs in the creative industries----the sector probably most ___24___ with the need for soft skills---are growing rapidly. Soft skills are, in fact, increasingly in demand in the workplace: Google cites creativity, leadership ___25___ and communication skills as top requirements for both potential and current employees.So why are soft skills so highly ___26___ for?With the rapid evolution of technology, a focus on hard skills leaves students ___27___ to change, as these often have a short shelf life. According to research by World Economic Forum, more than one in four adults reports a mismatch between their skills and those needed for their job role. If soft skills are taught well, these skills should enable students to adapt to change more easily and progress further in their ___28___ career.Of course, technical skills are important. But without the curriculum placing equal--- if not greater ---___29___ on soft skills, our education systems are missing a huge trick. Hard skills may help a student get a job in a particular industry, but soft skills will help them disrupt it, achieving a wider ___30___ in their chosen field. III. Reading ComprehensionSection ASaddleworth Moor in the north of England is a bare place. It seemed almost wired to me, then, that anyone should ___31___ the building of seven wind turbines( 风力发电机) to produce clean, renewable energy. Surely this was the perfect place to situate them--- basically dull, unattractive to tourists and ----- ___32___ ---windy. Yet Saddleworth is becoming another battleground in an increasingly confusing ___33___ over wind farming and the future of the planet.Typical of this confusion is hearing Professor David Bellamy ___34___ the fight against wind farms. I had always thought of Professor Bellamy as an environmentalist had made the ___35___ assumption that he would be a natural supporter of wind power. However, on reflection, Bellamy would be better described as a conservationist, whose main aim is to preserve natural habitats of plants and animals from destruction, rather than a(n) ___36___ on climate change. He has fought against other renewable energies that ___37___ wildlife and wildness, and has described the wind turbines as weapons of mass destruction killing birds and bats.Bellamy, along with other opponents, has argued the wind farms are in fact ___38___, and are only commercially successfully because they are so heavily funded. This argument has been put forward by several newspaper commentators recently , who have then gone to ___39___ nuclear power. This doesn’t take into account years of ___40___ from Greens who claim that nuclear power is both expensive and dangerous. And yet nuclear energy has recently been ___41___ by a leading green scientist , Professor James Lovelock, who was one of the firstto draw attention to the problem of climate change. He argues that renewable energy such as wind simply cannot provide sufficient electricity for our energy needs.And so it goes on. There are so many ___42___ claims, each apparently fronted by some outstanding scientists and backed up by a lot of statics. So who’s actually right? What’s the right solution? What ___43___ me is that we will take so long in deciding that it will be too late. The damage will have been done. Yet what I also ___44___ is how convenient these conflicting arguments are. We can avoid making any changes to our personal lifestyles by just doing nothing. Global warming isn’t down to me going to Barcelona by air for the weekend or having a dishwasher or driving everywhere; no, it’s because those people in Saddleworth won’t let us build our ___45___!31. A. take over B. call for C. look into D. object to32. A. vice versa B. or rather C. above all D. to date33. A. debate B. concern C. advantage D. control34. A. leading B. reporting C. watching D. abandoning35. A. cautious B. basic C. common D. false36. A. expert B. campaigner C. commentator D. columnist37. A. exploited B. threatened C. restored D. attracted38. A. unaided B. unfriendly C. uneconomic D. unbalanced39. A. produce B. praise C. eliminate D. research40. A. intentions B. passions C. opinions D. protest41. A. advocated B. rejected C. proposed D. overlooked42. A. expected B. successful C. conflicting D. personal43. A. relieves B. amazes C. disappoints D. worries44. A. predict B. recognize C. question D. ski45. A. wind farms B. power plants C. animal habitats D. nuclear engines SectionBAEvery April I am troubled by the same concern -- that spring might not occur this year. The landscape looks dull, with hills, sky and forest appearing gray. My spirits ebb, as they did during an April snowfall when I first came to Maine 15 years ago. "Just wait," a neighbor advised. "You’ll wake up one morning and spring will just be here."And look, on May 3 that year I awoke to a green so amazing as to be almost electric, as if spring were simply a matter of flipping a switch. Hills, sky and forest revealed their purples, blues and green. Leaves had unfolded and daffodils were fighting their way heavenward.Then there was the old apple tree. It sits on an undeveloped lot in my neighborhood. It belongs to no one and therefore to everyone. The tree’s dark twisted branches str etch out in unpruned（未经修剪的）abandon. Each spring it blossoms so freely that the air becomes filled with the scent of apple.Until last year, I thought I was the only one aware of this tree. And then one day, in a bit of spring madness, I set out with pruner to remove a few disorderly branches. No sooner had I arrived under the tree than neighbors opened their windows and stepped onto their porches(门廊). These were people I barely knew and seldom spoke to, but it was as if I had come uninvited into their personal gardens.My mobile-home neighbor was the first to speak. "You’re not cutting it down, are you?" she asked anxiously. Another neighbor frowned as I cut off a branch. "Don’t kill it, now," he warned. Soon half the neighborhood had joined me under the apple tree. It struck me that I had lived there for five years and only now was learning these people’s names, what they did for a living and how they passed the winter. It was as if the old apple tree was gathering us under its branches for the purpose of b oth acquaintanceship and shared wonder. I couldn’t help recalling Robert Frost’s words:The trees that have it in their pent-up buds To darken nature and be summer woods One thaw led to another. Just the other day I saw one of my neighbors at the local store. He remarked how this recent winter had been especially long and complained of not having seen or spoken at length to anyone in our neighborhood. And then, he looked at me and said, "We need to prune that apple tree again."46. By saying that “my spirits ebb” (Para. 1), the author means that _____.A. he feels relievedB. he feels blueC. he is surprisedD. he is tired47. The apple tree mentioned in the passage is most likely to _____.A. be regarded as a delight in the neighborhoodB. have been abandoned by its original ownerC. have been neglected by everyone in the communityD. be appealing only to the author48. According to Para. 4, why did the neighbors open their windows and step onto their porches?A. They were surprised that someone unknown was pruning the tree.B. They wanted to prevent the author from pruning the treeC. They were concerned about the safety of the treeD. They wanted to get to know the author49. It can be inferred that the author’s neighbor mentioned in the last paragraph most cared about _____.A. when spring would arriveB. how to pass the long winterC. the neighborhood gatheringD. the pruning of the apple treeBUniversity Shotokan Karate ClubLearn Karate—self-defense and fitnessThe university karate club was founded in 1962. All grades from beginners to advanced are welcome. The classes are suitable for both men and women and several women have obtained their black belt.TUESDAY AND THURSDAY6.00—8.00 p.m.WEA VER HOUSE GYMFIRST LESSON FREEHardly any of us have experienced real violence, but, over the last few years, attacks on innocent people have increased. Each of us has some ability to defend ourselves, and by learning a form of self-defence, we are not only increasing that ability, but also doing something to build our own sense of respect. Karate will show you a lot of simple and effective techniques to protect yourself, giving you increase self-confidence.Far too many people think martial arts (武术) are about violence. Martial arts training is based on a lot of respect, self-discipline, self-control and non-violence. We learn basic etiquette, courtesy and tolerance. Good manners and consideration for others are expected at all times.Karate is the practice of blocking and striking techniques for the purpose of self-defence, health andself-development. Karate exercises the entire body. Techniques are practised on both sides of the body, therefore muscle imbalances do not occur and the strength, coordination, flexibility and agility of both sides of the body are impro ved. Regular training in Karate improves the body’s physical endurance and flexibility. It also helps concentration and produces the mental calm and assurance that come from knowing we can defend ourselves.Karate has many benefits but they do not come easily or overnight. Training requires ongoing commitment and hard work. Some of you will give up, but a few of you will get your black belt.50. The passage is mainly intended to_________.A. tell readers the benefits of KarateB. attract readers to join the clubC. encourage people to get their black beltD. correct people’s misunderstand of Karate51. What can be learned about University Shotokan Karate Club?A. More men than women have been its members.B. It tests members to see what level they are at.C. Members don’t need to pay for the lessons.D. It offers 4 hours of lessons every week.52. According to the passage, which of the following statements about Karate is true?A. It cultivates people’s respect for themselves as well as for others.B. It invo lves people’s arm muscles more than body musclesC. It makes people healthy physically instead of mentally.D. It focuses as much on self-defence as on attack.CSpeaking in ClicksClick sounds, such as those found in some languages in Africa, make perfectly good consonants. So why do they appear so rarely in most human speech? One culprit may be anatomy(骨骼).Previous studies have suggested that in some speakers of click languages, the alveolar ridge(齿龈) the rounded bump between the upper teeth and the roof of the mouth-is small or even absent. In recent research, Scott Moisik of Nanyang Technological University in Singapore and Dan Dediu of the Max Planck Institute for Psycholinguistics in Nijmegen, the Netherlands, built biomechanical models that simulated clicks in vocal tracts(声道) with alveolar ridges of varying sizes. Their results, published in January in the Journal of Language Evolution, showed a clear disadvantage for tracts with large edges. These allowed less air to be trapped in the mouth, requiring more muscular force to produce a click.The authors interpret this finding as support for an anatomical bias against clicks. They believe the bias is probably weak at the individual level; people with large alveolar ridges can still learn click consonants or that their pronunciations may be wrong. Amplified over generations, this bias might explain why such consonants are so rarely found in language worldwide.These results are not the first to challenge the traditional premise among linguists that language evolution is largely immune to external factors. Several other researchers have recently argued that geographical context, environmental conditions and genetics could all play a role. But Moisik and Dediu’s work goes a step further by singling out a single feature of human anatomy and quantifying its contribution to a particular type of speech sound.Susanne Fuchs, senior researcher at the Leibniz Center of General Linguistics in Berlin, who was not involved in the work, says the study’s conclusions are valid. But she cautions that they may present a chicken-and-egg problem: “The palate( 味蕾) shape of an individual matures from early childhood to puberty and , may be affectedby frequent productions of clicks,” Fuchs says, “Therefore, over the course of h istory, it may well be that vocal tract properties and click productions developed in parallel.53. The underlined word “one culprit” in Paragraph 1 means _____.A. something that must cause sufferingB. something that may be the causeC. something that could be concludedD. something that never happened before.54. According to the findings by Moisik and Dediu, who can make click sounds easily?A.People with a small or absent alveolar ridge.B. People with strong muscles inside the mouthC. People with a normal alveolar rid.D. People with a large alveolar ridge. 55. Which of the following statements is true? A. People with large alveolar ridges cannot learn click language. B. Having less air trapped in the mouth makes it easier to produce a click. C. Both Moisik and Dediu believe that language evolution is largely independent of external factors. D. Language evolution may be subject to geographical context, environmental conditions and genetics.56. What is Susanne Fuchs most likely to agree with?A. The conclusions of the study by Moisik and Dediu are unreliable.B. V ocal tract properties and click productions might evolve at the same time.C. Frequent productions of clicks decide the palate shape of an individual.D. The palate shape of an individual decides whether one can produce click sounds.DVisit the grocery store on an empty stomach, and you will probably come home with a few things you had not planned to buy. But hunger is not the only culprit behind such purchases. The location of store displays also influences our shopping and may make or break some healthy eating habits.The checkout area is a particular hotspot for junk food. Studies have found that the products most commonly found there are sugary and salty snacks-and a few studies have suggested that simply swapping in healthier options can shift customer behavior. A 2012 study in the Netherlands found that hospital workers were more likely to give up junk food for healthy snacks when the latter were more readily available on canteen shelves for example, In2014 Norwegain and Icelandic researcher likewise found that replacing unhealthy items with healthy ones in the checkout area significantly increased last-minutes sales of healthier foods.These findings caught the attention of the New York City Department of Health and Mental Hygiene, which has been working with more than 1,000 store owners to encourage them to stock and promote consumption,” says Tamar Adjoian, a research scientist at the department. “Making healthy food more convenient or appealing can lead to increased sales of those products.”Adjoian and her colleagues wondered if such findings would apply to their city’s dense urban checkout areas, so they recruited three Bronx supermarkets for their own study. They gave one checkout line in each store a healthy makeover, replacing candy, cookies and other processed snacks with fruit , nuts and similar items containing 200 or fewer calories per serving. Then they recorded purchases over six three-hour periods in each store for two weeks.Of the more than 2,100 shoppers they observed, just 4 precent bought anything from the checkout area. Among those who did, however, customers in the healthy lines purchased nutritious items more than twice as often as those in the standard lines –and they bought unhealthy items 40 percent less often. The findings were reported in September in the Journal of Nutrition Education and Behavior.The potential impact may seem small, but Adjoian believes that converting more checkout lines would open customers’ e yes to nutritious, lower-calories foods. Health department officials are now exploring ways to expand healthy options at checkout areas throughout New York City.57. The word “ culprit” in Para. 1 is closest in meaning to _____in the passage.A. something to blameB. something related to cultureC. something that resultsD. something concealed58. The findings of the 2012 study and the 2014 study proved that _____.A. shops put great emphasis on increase in last-minute sales of food.B. healthier options were rarely seen in company canteens or in supermarket.C. customers’ shopping behavior could be changed by what is easily available.D. sugary and salty snacks were among the most common items in checkout areas.59. Why does the writer say “ the potential impact may seem small” in the last paragraph?A. Only a little attention has been drawn to Adjoian’s researchB. Only several healthy items are displayed and then purchased.C Only three Bronx supermarkets have been involved in the study.D. Only a small proportion of shoppers buy things from checkout areas.60. Which of the following might be the best title of the passage?A. Behavior that MattersB. Healthy See, Healthy DoC. Changing Checkout AreasD. Shoppers and Their NeedsSection CHow to protect children Web fans from unsuitable material on-line while encouraging them to use the Internet has long been discussed in the U.S.___61___. But jungles contain wonders as well as dangers and with good guides, some education, and a few precautions , the wilds of the Internet can be safely navigated . “Kids have to be on-line. If we tell our kids they can’t have access(机会) to the Internet, we’re cutting them off from their future,” said an expert.Most kids have started to use search engines. ___62___. There are search engines designed just for kids. A certain software contains only sites that have been selected as safe. The most popular way to limit access would be to use what is known as a “content screener(过滤器)”. But this can’t be wholly reliable , and the bes t thing parents can do is to talk to their kids and let them know what is OK or not OK to see or do on the Internet.___63___A few other tips●Don’t put the PC in a child’s room but keep it in an area where mum or dad can keep an eye on things. That also makes the Internet more of a family activity.●___64___●Tell your child not to give on-line strangers personal information, especially like address and phone number.●And tell your children never to talk to anyone they meet on-line over the phone, send them anything, accept anything from them or agree to meet with them unless you go along.A. Many of them are great for finding tons of interesting Internet sites, and they can also locate places where you might not want your kids to go.B. Another way is that mum or dad is nearby when the child is surfing the Internet.C. Ask your child what he or she has been doing and about any friends they make on-line.D. For some parents, the Internet can seem like a jungle, filled with danger for their children.E. Most kids are taught to have started to use search engines.F. Don’t give online strangers personal information.第II 卷IV. Section ATranslation65. Starbucks says ___________ .(这个举措旨在使各式各样的群体凝聚在一起)，thus creating a surprisinglycalm atmosphere.66. And even then they would not be able to ___________.(提供足够的光照，来消除对地面照明的需求). Space mirrors strikes me as a solution in search of a problem.67. For ___________ (那些比半个世纪以前聘用更多拥有大学学历的工作岗位中), real wage, adjusted for the effects of inflation, have actually fallen.68. Not only are more people taking to running, but they are also using wearable devices and software on their phones to record ___________.(他们跑过的路程距离和燃烧的卡路里数量)69. Many institution ___________, (在某些学术领域有特别大的优势) making them a good choice for students interested in those fields.Section BTranslation70. 最低工资的不断提高可能会使雇主们不愿意雇佣缺乏经验的年轻人，他在阿里巴巴工作过五年， 所以在经验方面， 他较其他求职者而言有着明显的优势。
交通大学附属中学学高二下学期期末历考试(化学)
上海交通大学附属中学 2018—2018学年度第二学期高二相关化学期终试卷（满分150分，120分钟完成。
答案一律写在答题纸上 >命题：尤颖欣 审核：陈晓芬第I 卷（共66分>原子量：H-1,O-16,C-12,Na-23, P-31, CI-35.5,Ca-40,Fe-56, Mn-55 ,V-51 ,Cr-52 , Cu-64,一、选择题（本题共10分，每小题2分，只有一个正确选项。
>1 •下列有机物不会发生水解反应的是 A. 脂肪 B 蛋白质 C 葡萄糖 D 纤维素2 •在陆地生态系统研究中， 2H 、13C 、15N 、180、34S 等常用作环境分析指示物。
下列说法正确的是A.34S 原子核内中子数为16 B.在相同条件下1H 216O 比1H 218O 更易挥发C. 13C 和15N 原子核内的质子数相差 2D. 2H +的酸性比1H +的酸性更强3 .禽流感是一种由甲型流感病毒的一种亚型引起的传染性疾病综合症，被国际兽疫局定为A 类传染病，又称真性鸡瘟或欧洲鸡瘟。
禽流感病毒对常用消毒剂十分敏感，容易将其杀灭。
下列不能杀灭禽流 感病毒的方法或物质的是 A.高温蒸煮B.用纯净水多次洗涤C.双氧水D.过氧乙酸4 .易燃易爆有毒的化学物质在其包装上应贴上危险警告标签。
下面所列物质贴错了标签的是AB C D 物质的化学式 CC 4 NaOH CH s OH NH 4NO 3 危险警告标签易燃品腐蚀品有毒品爆炸品5 .下列各化合物的命名中正确的是、选择题（本题共36分，每小题3分，只有一个正确选项。
>6 .下列化学式能代表其分子组成的是 A. SQ B. SiO 2C NH 4CI D. Mg7 .在一定条件下，将钠与氧气反应的生成物1.5g 溶于水，所得溶液恰好能被 HCl 溶液中和，则该生成物的成分是 A. Na 2O B. Na 2O 2C. Na 2O 和 Na 2O 2 D. NaOH &若N A 表示阿佛加德罗常数，下列说法正确的是 A. 1 mol 氦气分子中含 2N A 个氦原子2，4 —二硝基甲苯80mL 浓度为 0.50mol/L 的B . (CH 3CH2>2CHCf 3—甲基戊烷B. 25C 时，I L pH=13的Ba （OH>2溶液中含有Ba 2+的数目为0.1N AC.在标准状况下，22.4L 空气中约有 N A 个气体分子 D. 在0 C, 101kPa 时，22.4L 氢气中含有 N A 个氢原子9 .把500mL 有BaC 2和KCI 的混合溶液分成 5等份，取一份加入含 amol 硫酸钠的溶液，恰好使钡离子 全沉淀；另取一份加入含 bmol 硝酸银的溶液，恰好使氯离子完全沉淀。
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2017-2018年上海市交大附中高二下开学考
一、填空题
1、复数（是虚数单位）的模是 ;
i 32+i 2、在如图所示的正方体中，异面直线与所成角的大小为
1111D C B A ABCD -B A 1C B 1
3、已知点、，则与方向相同的单位向量的坐标为 (
)3,1A ()1,4-B →
AB 4、已知双曲线，则以双曲线的焦点为顶点，以双曲线顶点为焦点的椭圆方程15
42
2=-y x 为
5、已知两圆和相交于两点，则直线的方程1022=+y x ()()203122=-+-y x B A ,AB 是
6、将参数方程（为参数）化为普通方程，所得方程是 ⎩⎨⎧=+=θ
θsin 2cos 21y x θ7、已知椭圆的焦距为，则实数 152
22=+t
y t x 62=t 8、已知，是实系数一元二次方程的两根，则的值为 ai +2i b +02
=++q px x q p +9、若为非零实数，则下列四个命题都成立：①；②；③若b a ,01≠+a
a ()2222
b ab a b a ++=+，则；④若，则b a =b a ±=ab a =2b
a =则对于任意非零复数，上述命题仍然成立的序号是
b a ,10、如图，是三角形 所在平面外的一点，，且，
S ABC SC SB SA ==2π
=∠=∠=∠CSA BSC ASB 分别是和的中点，则异面直线与所成角的大小为
（用反三角函数
N M 、AB SC SM BN 表示）
11、已知直线及平面，其中，那么在平面内到两条直线距离相等的点的集合可能是：n m ,αn m //αn m ,①一条直线；②一个平面；③一个点；④空集。
其中正确的是
12、动点在直角坐标系平面上能完成下列动作，先从原点沿正偏北方向行走一()y x P ,O ⎪⎭
⎫ ⎝⎛
≤≤20παα段时间后，再向正北方向行走，但何时改变方向不定，假定速度为10米/分钟，则当变化时()y x P ,α行走2分钟内的可能落点的区域面积是
()y x P ,二、选择题
13、在下列命题中，不是公理的是（ ）
、平行于同一个平面的两个平面相互平行
A 、过不在同一条直线上的三点，有且只有一个平面
B 、如果同一条直线上的两点在同一个平面内，那么这条直线上所有的点都在此平面内
C 、如果两个不重合的平面有一个公共点，那么他们有且只有一条过该点的公共直线
D 14、若空间三条直线满足，，则直线与（ ）
c b a ,,b a ⊥c b //a c 、一定平行 、一定相交
、一定是异面直线 、一定垂直A B C D 15、在四边形中，，，则四边形的面积为（ ）ABCD (
)2,1=→AC ()2,4-=→BD 、 、 、 、
A 5
B 52
C 5
D 1016、已知动点的横坐标、纵坐标满足：①（）；②，那么
P x y 1sin cos =+ααy x R ∈α422≤+y x 当变化时，点形成的图形的面积为（ ）
αP 、 、 、 、A πB π3C π4D π
-4三、解答题
17、如图，是正方形，直线底面，，是的中点。
ABCD ⊥PD ABCD PC PD =E PC （1）证明：直线平面；
//PA EDB （2）求直线与平面所成角的正切值。
PB ABCD
18、已知椭圆的焦点为，，（），为椭圆上一点，且是，的等差()0,1t F -()0,2t F 0>t P 21F F 1PF 2PF 中项。
（1）求椭圆方程；
（2）如果点在第二象限且，求的值。
P
12021=∠F PF 21tan PF F ∠19、已知平面与平面的交线为直线，为平面内一条直线；为平面一条直线，且直线αβl m αn β互不重合。
n m l ,,（1）若直线与直线交于点，判断点与直线的位置关系并证明；
m n P P l （2）若，判断直线与直线的位置关系并证明。
n m //l m
20、现代城市大多是棋盘式布局（如北京道路几乎都是东西和南北走向）。
在这样的城市中，
我们说的两点间的距离往往不是指两点间的直线距离（位移），而是实际路程（如图）。
在直
角坐标平面内，我们定义两点间的“直角距离”为：
1122(,),(,)A x y B x y ()1212||||.
AB D x x y y =-+-（1）在平面直角坐标系中，写出所有满足到原点的“直角距离”为2的“格点”的坐标。
（格点指横、纵坐标均为整数的点）
（2）求到两定点F 1、F 2的“直角距离”和为定值的动点轨迹方程，并在直角坐
2(0)a a >标系内作出该动点的轨迹。
（在以下三个条件中任选一个做答，多做不计分，基保选择条件
①，满分3分；条件②满分4分；条件③，满分6分）
①；②；③12(1,0),(1,0),2F F a -=12(1,1),(1,1),2F F a --=12(1,1),(1,1), 4.
F F a --=（3）写出同时满足以下两个条件的“格点”的坐标，并说明理由（格点指横、纵坐标均为
整数的点）。
①到A （-1，-1），B （1，1）两点“直角距离”相等；
②到C （-2，-2），D （2，2）两点“直角距离”和最小。
21、过抛物线的一条弦的中点作平行于抛物线对称轴的平行线（或与对称轴重合），交抛物线于一点，称以该点及弦的端点为顶点的三角形为这条弦的阿基米德三角形（简称阿氏三角形）.
现有抛物线，直线（其中是常数，且），直线交抛物线于，2
:ax y M =c bx y l +=:c b a ,,0>a l M A 两点，设弦的阿氏三角形是.
B AB AB
C ∆（1）指出抛物线的焦点坐标和准线方程；
M （2）求的面积（用表示）；
ABC ∆c b a ,,（3）称的阿氏为一阶的；、的阿氏、为二阶的；
AB ABC ∆AC BC ACD ∆BCE ∆、、、的阿氏三角形为三阶的；……，由此进行下去，记所有的阶阿氏三角AD DC CE EB ()
*∈N k k 形的面积之和为，探索与之间的关系，并求.k S k S 1+k S ()n n S S S +++∞→ 21lim 参考答案
一、填空题：
1、；
2、；
3、；
4、；
5、；
6、；133π()4,3-=→
AB 15922=+y x 03=+y x ()4122=+-y x 7、；8、1；9、②④；10、
；11、（1）（2）（4）；12、；6,3,2=t 510200100-π二、选择题：
13、A ；14、D ；15、C ；16、B ；
三、解答题：
17、（1）证明略；（2）；2
218、（1）；（2）；1342222=+t y t x 11
3519、（1），证明略；（2），证明略；
l p ∈l m //20、（1），，，，，，，，，()0,2()1,1()2,0()1,1-()0,2-()1,1--()0,2-()1,1--()2,0-()1,1-（2）①；②；③
1124x x y ++-+=11114x y x y ++++-+-=，图略
11118x y x y ++++-+-=（3），，，，，，，，()2,2-()1,2-()2,1-()1,1-()0,0()1,1-()2,1-()1,2-()
2,2-21、（1），；（2）；（3），⎪⎭
⎫ ⎝⎛a F 41,0a y 41-=ac b a c S 422+=k k S S 411=+；
()ac b a c S S S n n 432lim 221+⋅=+++∞→。