华师大二附中2015高一上期末考试试卷(无答案PDF版)

上海华师大二附中高一物理联考试题含解析一、选择题：本题共5小题，每小题3分，共计15分．每小题只有一个选项符合题意1. （单选）水平匀速转动的圆盘上的物体相对于圆盘静止，则圆盘对物体的摩擦力方向是（）A．沿圆盘平面指向转轴B．沿圆盘平面背离转轴C.沿物体做圆周运动的轨迹的切线方向D．无法确定参考答案：A2. 某物体运动的速度图象如图，根据图象可知（）A．0－5s内的位移为10mB．0－2s内的加速度为C．第1s末与4.5s末的速度方向相反D．第1s末与第5s末加速度方向相同参考答案：B3. 农民在精选谷种时，常用一种叫“风车”的农具进行分选。

在同一风力作用下，谷种和瘪谷都从洞口水平飞出，结果谷种和瘪谷落地点不同，自然分开，如图所示。

若不计空气阻力，对这一现象，下列分析正确的是（）A．谷种飞出洞口时的速度比瘪谷飞出洞口时的速度大些B．谷种和瘪谷飞出洞口后都做匀变速曲线运动C．谷种和瘪谷从飞出洞口到落地的时间不相同D．M处是瘪谷，N处为谷种参考答案：BD4. （2014秋?苍山县期中）为了使公路交通安全有序，路旁立了许多交通标志，如图所示，甲图是限速标志，表示允许行驶的最大速度是100km/h；乙图是路线指示标志，表示此处到青岛还有170km．上述两个数据表达的物理意义是（）A．100km/h是平均速度，170km是位移B．100km/h是瞬时速度，170km是路程C．100km/h是瞬时速度，170km是位移D．100km/h是平均速度，170km是路程参考答案：B解：允许行驶的最大速度表示在某一位置的速度，是瞬时速度，所以100 km/h是指瞬时速度；到诸城还有170km，170km是运动轨迹的长度，是路程．故B正确，A、C、D错误．故选：B5. （单选）在倾角为30°的光滑固定斜面上，用两根轻绳跨过两个固定的定滑轮接在小车上，两端分别悬挂质量为2m和m的物体A、B，当小车静止时两绳分别平行、垂直于斜面，如图所示。

2015-2016学年上海市华师大二附中高一（上）期中数学试卷一、填空题（每小题4分，共40分）1．（4分）已知集合A={x|x≤a}，B={x|﹣2≤x＜1}，若A∪B=A，则实数a的取值范围是．2．（4分）命题“若实数a，b满足a+b＜7，则a=2且b=3”的否命题是．3．（4分）已知函数，则f（x）•g（x）=．4．（4分）已知四组函数：①与；②f（x）=x0与；③与；④f（x）=2x，D={0，1，2，3}与．表示同一函数的是．（写出所有符合要求的函数组的序号）5．（4分）设集合M={x|5﹣|2x﹣3|∈N*}，则M的所有真子集的个数是．6．（4分）不等式（1+x）（1﹣|x|）＞0的解为．7．（4分）已知命题，命题q：x2﹣2x+1﹣m2＜0（m＞0），若p 是q的充分不必要条件，则实数m的范围是．8．（4分）已知集合，且2∈A，3∉A，则实数a的取值范围是．9．（4分）若函数，则不等式的解集为．10．（4分）我们用记号“|”表示两个正整数间的整除关系，如3|12表示3整除12．试类比课本中不等关系的基本性质，写出整除关系的两个性质．①；②．11．（4分）设a+b=2，b＞0，则当a=时，取得最小值．二、选择题（每小题3分，共12分）12．（3分）若，有下面四个不等式：①|a|＞|b|；②a＜b；③a+b＜ab，④a3＞b3，正确的不等式的个数是（）A．0 B．1 C．2 D．313．（3分）函数f（x）=x|x+a|+b是奇函数的充要条件是（）A．ab=0 B．a+b=0 C．a2+b2=0 D．a=b14．（3分）若a和b均为非零实数，则下列不等式中恒成立的是（）A．B．C．D．15．（3分）设整数n≥4，集合X={1，2，3，…，n}．令集合S={（x，y，z）|x，y，z∈X，且三条件x＜y＜z，y＜z＜x，z＜x＜y恰有一个成立}．若（x，y，z）和（z，w，x）都在S中，则下列选项正确的是（）A．（y，z，w）∈S，（x，y，w）∉S B．（y，z，w）∈S，（x，y，w）∈S C．（y，z，w）∉S，（x，y，w）∈S D．（y，z，w）∉S，（x，y，w）∉S三、解答题（共48分，写出必要的解答过程）16．（8分）已知命题P：|1﹣a|＜6，命题Q：{x|x2+（a+2）x+1=0}∩R+=∅．命题P真Q假，求实数a的取值范围．17．（8分）甲厂以x千克/小时的速度匀速生产某种产品（生产条件要求1≤x≤10），每小时可获得的利润是100（5x+1﹣）元．（1）要使生产该产品2小时获得的利润不低于3000元，求x的取值范围；（2）要使生产900千克该产品获得的利润最大，问：甲厂应该选取何种生产速度？并求此最大利润．18．（10分）解关于x的不等式x2+a（a+1）x+a3＞0．19．（10分）已知f（x）=|x2﹣1|+x2+kx．（Ⅰ）若k=2，求方程f（x）=0的解；（Ⅱ）若关于x的方程f（x）=0在（0，2）上有两个解x1，x2，求k的取值范围，并证明．20．（12分）已知函数（m为实常数）．（1）若函数y=f（x）图象上动点P到定点Q（0，2）的距离的最小值为，求实数m的值；（2）若函数y=f（x）在区间[2，+∞）上是增函数，试用函数单调性的定义求实数m的取值范围；（3）设m＜0，若不等式f（x）≤kx在有解，求k的取值范围．2015-2016学年上海市华师大二附中高一（上）期中数学试卷参考答案与试题解析一、填空题（每小题4分，共40分）1．（4分）已知集合A={x|x≤a}，B={x|﹣2≤x＜1}，若A∪B=A，则实数a的取值范围是a≥1．【解答】解：∵A∪B=A，∴B⊆A∵A={x|x≤a}，B={x|﹣2≤x＜1}，∴a≥1故答案为：a≥1．2．（4分）命题“若实数a，b满足a+b＜7，则a=2且b=3”的否命题是若实数a，b满足a+b≥7，则a≠2或b≠3．【解答】解：命题“若实数a，b满足a+b＜7，则a=2且b=3”的否命题是“若实数a，b满足a+b≥7，则a≠2或b≠3”，故答案为：若实数a，b满足a+b≥7，则a≠2或b≠33．（4分）已知函数，则f（x）•g（x）=﹣，x∈（﹣1，3）．【解答】解：由3+2x﹣x2＞0得：x∈（﹣1，3），∵函数，∴f（x）•g（x）=﹣（3+2x﹣x2）•=﹣，x∈（﹣1，3），故答案为：﹣，x∈（﹣1，3）4．（4分）已知四组函数：①与；②f（x）=x0与；③与；④f（x）=2x，D={0，1，2，3}与．表示同一函数的是②③．（写出所有符合要求的函数组的序号）【解答】解：①中，=|x|﹣1，=x﹣1，解析式不一致，不是同一函数；②中，f（x）=x0=1（x≠0），=1（x≠0），定义域，解析式均一致，是同一函数；③中，=|x|，（x≠0），，定义域，解析式均一致，是同一函数；④中，f（x）=2x，D={0，1，2，3}与，解析式不一致，不是同一函数．故表示同一函数的是：②③，故答案为：②③5．（4分）设集合M={x|5﹣|2x﹣3|∈N*}，则M的所有真子集的个数是15．【解答】解：M={x|5﹣|2x﹣3|∈N*}，当x≥时，M={x|8﹣2x∈N*}，x=2，3当x＜时，M={x|2+2x∈N*}，x=1，0∴M={0，1，2，3}真子集的个数为24﹣1=15，故答案为：15．6．（4分）不等式（1+x）（1﹣|x|）＞0的解为（﹣∞，﹣1）∪（﹣1，1）．【解答】解：当x≥0时，|x|=x，原不等式变形为：（1+x）（1﹣x）＞0，可化为或，解得：﹣1＜x＜1，不等式的解集为[0，1）；当x＜0时，|x|=﹣x，原不等式变形为：（1+x）（1+x）＞0，解得x≠﹣1，不等式的解集为（﹣∞，﹣1）∪（﹣1，+∞），综上，原不等式的解集为（﹣∞，﹣1）∪（﹣1，1）．故答案为：（﹣∞，﹣1）∪（﹣1，1）7．（4分）已知命题，命题q：x2﹣2x+1﹣m2＜0（m＞0），若p 是q的充分不必要条件，则实数m的范围是（2，+∞）．【解答】解：由命题得|x﹣1|≤2，解得﹣1≤x≤3，即p：﹣1≤x≤3．由x2﹣2x+1﹣m2＜0（m＞0），得1﹣m＜x＜1+m，即q：1﹣m＜x＜1+m，m＞0，∵p是q的充分不必要条件，∴，即，∴m＞2，故答案为：（2，+∞）．8．（4分）已知集合，且2∈A，3∉A，则实数a的取值范围是．【解答】解：∵，且2∈A，3∉A，∴，解得：≤a或2＜a≤3故答案为．9．（4分）若函数，则不等式的解集为．【解答】解：由题意可得x＞0，由基本不等式可得f（x）≥2，再根据函数f（x）在（0，1]上是减函数，在[1，+∞）上是增函数，令x+=，求得x=，或x=2，故不等式的解集为（，2），故答案为：（，2）．10．（4分）我们用记号“|”表示两个正整数间的整除关系，如3|12表示3整除12．试类比课本中不等关系的基本性质，写出整除关系的两个性质．①若a|b，b|c，则a|c；②若a|b，c|d，则ac|bd．【解答】解：类比课本中不等关系的基本性质，①传递性a＜b，b＜c，则a＜c；②0＜a＜b，0＜c＜d，则ac＜bd．写出整除关系的两个性质：①若a|b，b|c，则a|c；②若a|b，c|d，则ac|bd；故答案为：①若a|b，b|c，则a|c；②若a|b，c|d，则ac|bd．11．（4分）设a+b=2，b＞0，则当a=﹣2时，取得最小值．【解答】解：∵a+b=2，b＞0，∴=，（a＜2）设f（a）=，（a＜2），画出此函数的图象，如图所示．利用导数研究其单调性得，当a＜0时，f（a）=﹣+，f′（a）==，当a＜﹣2时，f′（a）＜0，当﹣2＜a＜0时，f′（a）＞0，故函数在（﹣∞，﹣2）上是减函数，在（﹣2，0）上是增函数，∴当a=﹣2时，取得最小值．同样地，当0＜a＜2时，得到当a=时，取得最小值．综合，则当a=﹣2时，取得最小值．故答案为：﹣2．二、选择题（每小题3分，共12分）12．（3分）若，有下面四个不等式：①|a|＞|b|；②a＜b；③a+b＜ab，④a3＞b3，正确的不等式的个数是（）A．0 B．1 C．2 D．3【解答】解：由，可得0＞a＞b，∴|a|＜|b|，故①②不成立；∴a+b＜0＜ab，a3＞b3都成立，故③④一定正确，故选：C．13．（3分）函数f（x）=x|x+a|+b是奇函数的充要条件是（）A．ab=0 B．a+b=0 C．a2+b2=0 D．a=b【解答】解：由奇函数的性质可得：f（0）=b=0，∴f（x）=x|x+a|，则f（﹣x）+f（x）=0，∴﹣x|﹣x+a|+x|x+a|=0，x≠0时，|x﹣a|=|x+a|恒成立，则a=0．∴函数f（x）=x|x+a|+b是奇函数的充要条件是a=b=0，即a2+b2=0．故选：C．14．（3分）若a和b均为非零实数，则下列不等式中恒成立的是（）A．B．C．D．【解答】解：当a=1，b=﹣1时，选项A、B、C中的不等式都不成立，只有D成立，故选：D．15．（3分）设整数n≥4，集合X={1，2，3，…，n}．令集合S={（x，y，z）|x，y，z∈X，且三条件x＜y＜z，y＜z＜x，z＜x＜y恰有一个成立}．若（x，y，z）和（z，w，x）都在S中，则下列选项正确的是（）A．（y，z，w）∈S，（x，y，w）∉S B．（y，z，w）∈S，（x，y，w）∈S C．（y，z，w）∉S，（x，y，w）∈S D．（y，z，w）∉S，（x，y，w）∉S【解答】解：方法一：特殊值排除法，取x=2，y=3，z=4，w=1，显然满足（x，y，z）和（z，w，x）都在S中，此时（y，z，w）=（3，4，1）∈S，（x，y，w）=（2，3，1）∈S，故A、C、D 均错误；只有B成立，故选B．直接法：根据题意知，只要y＜z＜w，z＜w＜y，w＜y＜z中或x＜y＜w，y＜w＜x，w＜x ＜y中恰有一个成立则可判断（y，z，w）∈S，（x，y，w）∈S．∵（x，y，z）∈S，（z，w，x）∈S，∴x＜y＜z…①，y＜z＜x…②，z＜x＜y…③三个式子中恰有一个成立；z＜w＜x…④，w＜x＜z…⑤，x＜z＜w…⑥三个式子中恰有一个成立．配对后有四种情况成立，第一种：①⑤成立，此时w＜x＜y＜z，于是（y，z，w）∈S，（x，y，w）∈S；第二种：①⑥成立，此时x＜y＜z＜w，于是（y，z，w）∈S，（x，y，w）∈S；第三种：②④成立，此时y＜z＜w＜x，于是（y，z，w）∈S，（x，y，w）∈S；第四种：③④成立，此时z＜w＜x＜y，于是（y，z，w）∈S，（x，y，w）∈S．综合上述四种情况，可得（y，z，w）∈S，（x，y，w）∈S．故选：B．三、解答题（共48分，写出必要的解答过程）16．（8分）已知命题P：|1﹣a|＜6，命题Q：{x|x2+（a+2）x+1=0}∩R+=∅．命题P真Q假，求实数a的取值范围．【解答】解：根据命题Q知，方程x2+（a+2）x+1=0无解，或实根小于0；∴△=（a+2）2﹣4＜0，或；解得﹣4＜a＜0，或a≥0；∴a＞﹣4；解|1﹣a|＜6得﹣5＜a＜7；∵P真Q假；∴；∴﹣5＜a≤﹣4；∴实数a的取值范围为（﹣5，﹣4]．17．（8分）甲厂以x千克/小时的速度匀速生产某种产品（生产条件要求1≤x≤10），每小时可获得的利润是100（5x+1﹣）元．（1）要使生产该产品2小时获得的利润不低于3000元，求x的取值范围；（2）要使生产900千克该产品获得的利润最大，问：甲厂应该选取何种生产速度？并求此最大利润．【解答】解：（1）生产该产品2小时获得的利润为100（5x+1﹣）×2=200（5x+1﹣）根据题意，200（5x+1﹣）≥3000，即5x2﹣14x﹣3≥0∴x≥3或x≤﹣∵1≤x≤10，∴3≤x≤10；（2）设利润为y元，则生产900千克该产品获得的利润为y=100（5x+1﹣）×=90000（）=9×104[+]∵1≤x≤10，∴x=6时，取得最大利润为=457500元故甲厂应以6千克/小时的速度生产，可获得最大利润为457500元．18．（10分）解关于x的不等式x2+a（a+1）x+a3＞0．【解答】解：x2+（a2+a）x+a3＞0，即（x+a）（x+a2）＞0．当a=0时，原不等式化为x2＞0，不等式的解集为{x|x≠0}；当a=1时，原不等式化为（x+1）2＞0，不等式的解集为{x|x≠﹣1}；当0＜a＜1时，a2＜a，解得：﹣a＜x＜a2．所以，原不等式的解集为{x|﹣a＜x ＜﹣a2}；当a＜0或a＞1时，a＜a2，解得：﹣a2＜x＜﹣a，所以，原不等式的解集为{x|﹣a2＜x＜a}．综上：当a=0原不等式的解集为{x|x≠0}；a=1时，不等式的解集为{x|x≠﹣1}；当0＜a＜1时，原不等式的解集为{x|﹣a＜x＜﹣a2}；当a＜0或a＞1时，原不等式的解集为{x|﹣a2＜x＜a}．19．（10分）已知f（x）=|x2﹣1|+x2+kx．（Ⅰ）若k=2，求方程f（x）=0的解；（Ⅱ）若关于x的方程f（x）=0在（0，2）上有两个解x1，x2，求k的取值范围，并证明．【解答】解：（Ⅰ）解：（1）当k=2时，f（x）=|x2﹣1|+x2+kx①当x2﹣1≥0时，即x≥1或x≤﹣1时，方程化为2x2+2x﹣1=0解得，因为，故舍去，所以．②当x2﹣1＜0时，﹣1＜x＜1时，方程化为2x+1=0解得（II）解：不妨设0＜x1＜x2＜2，因为所以f（x）在（0，1]是单调函数，故f（x）=0在（0，1]上至多一个解，若1＜x1＜x2＜2，则x1x2=＜0，故不符题意，因此0＜x1≤1＜x2＜2．由f（x1）=0得，所以k≤﹣1；由f（x2）=0得，所以；故当时，方程f（x）=0在（0，2）上有两个解．当0＜x1≤1＜x2＜2时，，2x22+kx2﹣1=0消去k得2x1x22﹣x1﹣x2=0即，因为x2＜2，所以．20．（12分）已知函数（m为实常数）．（1）若函数y=f（x）图象上动点P到定点Q（0，2）的距离的最小值为，求实数m的值；（2）若函数y=f（x）在区间[2，+∞）上是增函数，试用函数单调性的定义求实数m的取值范围；（3）设m＜0，若不等式f（x）≤kx在有解，求k的取值范围．【解答】解：（1）设P（x，y），则，=，当m＞0时，解得；当m＜0时，解得，∴或．（2）由题意，任取x1、x2∈[2，+∞），且x1＜x2，则=＞0，∵x2﹣x1＞0，x1x2＞0，所以x1x2﹣m＞0，即m＜x1x2，（3）由f（x）≤kx，得，∵，∴，令，则t∈[1，2]，所以k≥mt2+2t+1，令g（t）=mt2+2t+1，t∈[1，2]，于是，要使原不等式在有解，当且仅当k≥g（t）min（t∈[1，2]）．∵m＜0，∴图象开口向下，对称轴为直线，∵t∈[1，2]，∴当，即时，g（t）min=g（2）=4m+5；当，即时，g（t）min=g（1）=m+3，综上，当时，k∈[4m+5，+∞）；当时，k∈[m+3，+∞）．。

华师大二附中2014学年第一学期期末考试卷高一 数学时间：90分钟 满分：100分一、填空题（每题4分，10小题，共40分）1．已知幂函数()f x 的图像经过点2⎛ ⎝⎭，则()16f = ． 2．函数()253x f x x -=-的值域是： ． 3．若函数()24f x x x a =--的零点个数为3，则a = ．4．命题“单调函数不是周期函数”的逆否命题为： ．5．对于集合A 和B ，定义运算：{}|A B x x A x B -=∈∉，且，又()()*A B A B B A =-- ．设{}1357X =，，，，{}|4Y x x x =<∈Z ，且，则*X Y = ．6．已知关于x 的方程122x a a+=-只有正实根，则实数a 的取值范围为： ． 7．若a b +∈R ，，1a b +=，则1ab ab +的最小值为： ． 8．函数()12log 1f x x =-的单调递增区间是： ．9．不等式11x x m +--<的解集是R 是非真子集，则实数m 的取值范围是： ．10．定义在R 上的奇函数()f x 满足()()4f x f x =+，且()02x ∈，时，()241xx f x =+．则()f x 在[]22-，上的解析式为： ．二、选择题（每题4分，4小题，共16分）11．下列各组函数中，表示同一函数的是（ ）A ．()f x x =，()g t =B ．()f x =()3g x =C ．()211x f x x -=-，()1g x x =+D ．()f x =()g x =12．要使函数12x y m +=+的图像不经过第二象限，则实数m 的取值范围是（ ）A ．1m -≤B ．1m <-C ．2m -≤D ．2m -≥13．已知()f x 是单调减函数，若将方程()f x x =与()()1f x f x -=的解分别称为函数()f x 的不动点与稳定点，则“x 是()f x 的不动点”是“x 是()f x 的稳定点”的（ ）A ．充要条件B ．充分不必要条件C ．必要不充分条件D ．既不充分也不必要条件14．定义域是一切实数的函数()y f x =，其图像是连续不断的，且存在常数()λλ∈R 使得()()0f x f x λλ++=对任意实数x 都成立，则称()f x 是一个“λ附属函数”．有下列关于“λ附属函数”的结论：①()0f x =是常数函数中唯一一个“λ附属函数”；②“12附属函数”至少有一个零点；③()2f x x =是一个“λ附属函数”；其中正确结论的个数是（ ）A ．0个B ．1个C ．2个D ．3个三、解答题（4小题，共44分，请写出必要的解答步骤）15．（10分）解下列方程：⑴ 943450x x -⋅-=⑵ ()()222log 2log 5x x x -=--16．（10分）已知()()23f x x a x a =+-+⑴ 对于x ∈R ，()0f x >总成立，求a 的取值范围；⑵ 当[]12x ∈-，时，()0f x >，求a 的取值范围．17．（12分）设()()1011x xa f x a a a +=>≠-， ⑴ 求()f x 的反函数()1f x -；⑵ 判断并证明()1f x -在()1+∞，上的单调性；⑶ 令()1log a g x x =+，当[]()()1m n m n ⊆+∞<，，时，()1f x -在[]m n ，上的值域是()()g n g m ⎡⎤⎣⎦，，求a 的取值范围．18．（12分）已知函数()2211x b f x a x ⎛⎫⎛⎫=-+- ⎪ ⎪⎝⎭⎝⎭，()0x ∈+∞，，其中0a b <<． ⑴ 当12a b ==，时，求()f x 的最小值；⑵ 若()21m f a -≥对任意0a b <<恒成立，求实数m 的取值范围；⑶ 设k 、0c >，当()22a k b k c ==+，时，记()()1f x f x =；当()2a k c =+，()22b k c =+时，记()()2f x f x =．求证：()()()2124c f x f x k k c +>+．。

2015-2016学年上海市华师大二附中高一（下）期中数学试卷一、填空题（共10小题）.1．求值arctan（cot）=．2．函数f（x）=的定义域是．3．若tanθ=﹣3，则sinθ（sinθ﹣2cosθ）=．4．若x∈（0，2π），则使=sinx﹣cosx成立的x的取值范围是．5．若arcsinx﹣arccosx=，则x=．6．函数f（x）=log cos1（sinx）的单调递增区间是．7．若0＜θ＜，则cosθ，cos（sinθ），sin（cosθ）的大小顺序为．8．若关于x的函数y=sinωx在[﹣，]上的最大值为1，则ω的取值范围是．9．已知，且，则cos（x+2y）=．10．设函数f（x）=，关于f（x）的性质，下列说法正确的是．①定义域是{x|x≠kπ+，k∈Z}；②值域是R；③最小正周期是π；④f（x）是奇函数；⑤f（x）在定义域上单调递增．二、选择题（4*4=16分）11．为了得到y=3sin（2x+）的图象，只需将y=3cos2x的图象（）A．向左平移B．向右平移C．向右平移D．向左平移12．α，β∈（，π），且tanα＜cotβ，则必有（）A．α＜β B．α＞β C．α+β＜D．α+β＞13．下列函数中以π为周期，在（0，）上单调递减的是（）A．y=（cot1）tanx B．y=|sinx|C．y=﹣cos2x D．y=﹣tan|x|14．下列命题中错误的是（）A．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（cosy）=cos2y成立B．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（siny）=sin2y成立C．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（cosy）=cos3y成立D．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（siny）=sin3y成立三、解答题（8+10+12+14=44分）15．已知α，β∈（0，π），并且sin（5π﹣α）=cos（π+β），cos（﹣α）=﹣cos （π+β），求α，β的值．16．若关于x的方程sinx+cosx+a=0在（0，2π）内有两个不同的实数根α，β，求实数a 的取值范围及相应的α+β的值．17．已知函数y=．（1）设变量t=sinθ+cosθ，试用t表示y=f（t），并写出t的范围；（2）求函数y=f（t）的值域．18．用a，b，c分别表示△ABC的三个内角A，B，C所对边的边长，R表示△ABC的外接圆半径．（1）R=2，a=2，B=45°，求AB的长；（2）在△ABC中，若∠C是钝角，求证：a2+b2＜4R2；（3）给定三个正实数a，b，R，其中b≤a，问a，b，R满足怎样的关系时，以a，b为边长，R为外接圆半径的△ABC不存在，存在一个或存在两个（全等的三角形算作同一个）？在△ABC存在的情况下，用a，b，R表示c．2015-2016学年上海市华师大二附中高一（下）期中数学试卷参考答案与试题解析一、填空题（4*10=40分）1．求值arctan（cot）=．【考点】反三角函数的运用．【分析】利用特殊角的三角函数，反正切函数的定义和性质，求得arctan（cot）的值．解：arctan（cot）=arctan（）=，故答案为：．2．函数f（x）=的定义域是{x|x=2kπ，k∈z} ．【考点】函数的定义域及其求法．【分析】根据二次根式的性质得到cosx=1，解出即可．解：由题意得：cosx﹣1≥0，cosx≥1，∴cosx=1，∴x=2kπ，k∈Z，故答案为：{x|x=2kπ，k∈z}．3．若tanθ=﹣3，则sinθ（sinθ﹣2cosθ）=．【考点】同角三角函数基本关系的运用．【分析】利用同角三角函数的基本关系，求得要求式子的值．解：∵tanθ=﹣3，∴sinθ（sinθ﹣2cosθ）====，故答案为：．4．若x∈（0，2π），则使=sinx﹣cosx成立的x的取值范围是[]．【考点】三角函数的化简求值．【分析】把根式内部的代数式化为完全平方式的形式，由已知等式可得sinx≥cosx，再由已知x的范围求得x的具体范围．解：∵===sinx﹣cosx，∴sinx≥cosx，又x∈（0，2π），∴x∈[]．故答案为：∈[]．5．若arcsinx﹣arccosx=，则x=．【考点】反三角函数的运用．【分析】由题意可得arcsinx与arccosx=均为锐角，x＞0，求得cos（arcsinx﹣arccosx）的值，可得x的值．解：∵arcsinx∈（﹣，），arccosx∈（0，π），arcsinx﹣arccosx=，∴arcsinx与arccosx 均为锐角，x＞0．又cos（arcsinx﹣arccosx）=cos=，即cos（arcsinx）•cos（arccosx）+sin（arcsinx）sin（arccosx）=•x+x•=，∴•x=，∴x2（1﹣x2）=，∴x2=，或x2=，∴x=，或x=．经检验，x=不满足条件，故舍去．故答案为：．6．函数f（x）=log cos1（sinx）的单调递增区间是[）（k∈Z）．【考点】复合函数的单调性．【分析】由0＜cos1＜1，得外函数y=log cos1t在定义域内单调递减，再求出内函数t=sinx的减区间，取使t大于0的部分得答案．解：令t=sinx，∵0＜cos1＜1，∴外函数y=log cos1t在定义域内单调递减，又sinx＞0，∴当x∈[）（k∈Z）时，内函数t=sinx大于0且单调递减，∴函数f（x）=log cos1（sinx）的单调递增区间是[）（k∈Z），故答案为：[）（k∈Z）．7．若0＜θ＜，则cosθ，cos（sinθ），sin（cosθ）的大小顺序为cos（sinθ）＞cosθ＞sin（cosθ）；．【考点】三角函数线．【分析】观察知道，利用x＞0时，sinx＜x，结合余弦函数的单调性解答．解：因为sinx＜x，所以0＜θ＜，sinθ＜θ，所以cos（sinθ）＞cosθ，令x=cosθ，所以cosθ＞sin（cosθ），故答案为：cos（sinθ）＞cosθ＞sin（cosθ）；8．若关于x的函数y=sinωx在[﹣，]上的最大值为1，则ω的取值范围是{ω|ω≥1或ω≤﹣}．【考点】正弦函数的图象．【分析】利用正弦函数的图象特征，正弦函数的最大值，分类讨论求得ω的取值范围．解：∵关于x的函数y=sinωx在[﹣，]上的最大值为1，∴当ω＞0时，由ω•≥，ω≥1，当ω＜0时，由ω•（﹣）≥，求得ω≤﹣，故答案为：{ω|ω≥1或ω≤﹣}．9．已知，且，则cos（x+2y）=1．【考点】三角函数的恒等变换及化简求值；两角和与差的余弦函数．【分析】设f（u）=u3+sinu．根据题设等式可知f（x）=2a，f（2y）=﹣2a，进而根据函数的奇偶性，求得f（x）=﹣f（2y）=f（﹣2y）．进而推断出x+2y=0．进而求得cos（x+2y）=1．解：设f（u）=u3+sinu．由①式得f（x）=2a，由②式得f（2y）=﹣2a．因为f（u）在区间上是单调增函数，并且是奇函数，∴f（x）=﹣f（2y）=f（﹣2y）．∴x=﹣2y，即x+2y=0．∴cos（x+2y）=1．故答案为：1．10．设函数f（x）=，关于f（x）的性质，下列说法正确的是②④．①定义域是{x|x≠kπ+，k∈Z}；②值域是R；③最小正周期是π；④f（x）是奇函数；⑤f（x）在定义域上单调递增．【考点】三角函数的化简求值．【分析】利用二倍角公式化简函数解析式，根据正切函数的图象和性质逐一分析各个选项即可得解．解：f（x）===tanx（cosx），对于①，函数f（x）的定义域是{x|x≠2kπ+，x≠kπ+，x≠2kπ+，k∈Z}，故错误；对于②，函数f（x）的值域是R，故正确；对于③，由于f（x+π）===tanx（其中cosx ≠），故错误；对于④，由于f（﹣x）==﹣=﹣f（x），故正确；对于⑤，由正切函数的图象可知函数在整个定义域上不单调，有无数个单调增区间，故错误．故答案为：②④．二、选择题（4*4=16分）11．为了得到y=3sin（2x+）的图象，只需将y=3cos2x的图象（）A．向左平移B．向右平移C．向右平移D．向左平移【考点】函数y=Asin（ωx+φ）的图象变换．【分析】把函数y=3sin（2x+）变形为y=3sin[2（x+）]即可得到答案．解：∵y=3sin（2x+）=3sin[2（x+）]．∴要得到y=3sin（2x+）的图象，只需将y=3cos2x的图象向左平移个单位．故选：D．12．α，β∈（，π），且tanα＜cotβ，则必有（）A．α＜β B．α＞β C．α+β＜D．α+β＞【考点】正切函数的图象．【分析】由题意可得α+β∈（π，2π），再根据tan（α+β）=＞0，可得α+β∈（π，），从而得出结论．解：α，β∈（，π），且tanα＜cotβ=＜0，∴tanα•tanβ＞1，α+β∈（π，2π），∴tan（α+β）=＞0，∴α+β∈（π，），故选：C．13．下列函数中以π为周期，在（0，）上单调递减的是（）A．y=（cot1）tanx B．y=|sinx|C．y=﹣cos2x D．y=﹣tan|x|【考点】正弦函数的图象．【分析】利用三角函数的周期性和单调性，逐一判断各个选项是否正确，从而得出结论．解：由于y=tanx的周期为π，0＜cot1＜1，故y=（cot1）tanx的周期为π，且在（0，）上单调递减，故A满足条件．由于y=|sinx|在（0，）上单调递增，故排除B．由于在（0，）上，2x∈（0，π），函数y=﹣cos2x在（0，）上单调递增，故排除C．由于函数y=﹣tan|x|不是周期函数，故排除D，故选：A．14．下列命题中错误的是（）A．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（cosy）=cos2y成立B．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（siny）=sin2y成立C．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（cosy）=cos3y成立D．存在定义在[﹣1，1]上的函数f（x）使得对任意实数y有等式f（siny）=sin3y成立【考点】二倍角的余弦；二倍角的正弦．【分析】利用二倍角公式、三倍角公式，函数的定义，判断各个选项是否正确，从而得出结论．解：令x=cosy∈[﹣1，1]，则对任意实数y，有等式f（cosy）=cos2y成立，即f（x）=2x2﹣1成立，故A成立．对任意实数y有等式f（cosy）=cos3y=4cos3y﹣3cosy 成立，即f（x）=4x3﹣3x成立，故B 正确．令t=siny∈[﹣1，1]，则对任意实数y，有等式f（siny）=sin2y=2sinycosy=2t•（±）成立，即f（x）=2•（±）成立，故B错误．则对任意实数y，有等式f（sin3y）=sin3y=3siny﹣4sin3y 成立，即f（t）=3t﹣4t3成立，故D成立，故选：B．三、解答题（8+10+12+14=44分）15．已知α，β∈（0，π），并且sin（5π﹣α）=cos（π+β），cos（﹣α）=﹣cos （π+β），求α，β的值．【考点】三角函数的化简求值．【分析】利用诱导公式化简已知可得sinα=sinβ，cosα=cosβ，将两式平方后利用同角三角函数基本关系式解得或，结合角的范围即可得解α，β的值．解：∵由sin（5π﹣α）=cos（π+β），可得：sinα=sinβ，两边平方可得：sin2α=2sin2β，①由cos（﹣α）=﹣cos（π+β），可得：cosα=cosβ，两边平方可得：3cos2α=2cos2β，②∴①+②可得：sin2α+3cos2α=2sin2β+2cos2β=2，又∵sin2α+cos2α=1，∴解得：cos2α=，即：或，∵α，β∈（0，π），∴解得或．16．若关于x的方程sinx+cosx+a=0在（0，2π）内有两个不同的实数根α，β，求实数a 的取值范围及相应的α+β的值．【考点】三角函数中的恒等变换应用．【分析】由sinx+cosx+a=0，得sinx+cosx=﹣a，画出函数y=sinx+cosx=的图象，数形结合得答案．解：由sinx+cosx+a=0，得sinx+cosx=﹣a，令y=sinx+cosx=，∵x∈（0，2π），∴x+∈（，），作出函数的图象如图：若关于x的方程sinx+cosx+a=0在（0，2π）内有两个不同的实数根α，β，则﹣2，或，即或．当a∈（﹣2，﹣）时，；当a∈（﹣，2）时，．17．已知函数y=．（1）设变量t=sinθ+cosθ，试用t表示y=f（t），并写出t的范围；（2）求函数y=f（t）的值域．【考点】三角函数中的恒等变换应用．【分析】（1）由t=sin（t+）利用正弦函数的性质可求t的范围，平方后利用同角三角函数基本关系式可求sinθcosθ=，进而即可用t表示y=f（t）．（2）由y== [（t+2）+﹣4]，利用基本不等式即可求其最小值，进而求得最大值即可得解函数y=f（t）的值域．解：（1）∵t=sinθ+cosθ，∴t=sinθ+cosθ=sin（θ+）∈[﹣，]，∴t2=sin2θ+cos2θ+2sinθcosθ=1+2sinθcosθ，∴sinθcosθ=，∴y===，t∈[﹣，]．（2）∵y==（）= [（t+2）+﹣4]，∵t∈[﹣，]．∴t+2∈[2﹣，2+]．∴（t+2）+=2，当且仅当（t+2）=，即t+2=时取等号．∵t+2∈[2﹣，2+]．∴函数的最小值为 [2﹣4]=．当t=﹣时，f（﹣）=，t=时，f（）=，∴函数的最大值为，故函数y=f（t）的值域为：[，]．18．用a，b，c分别表示△ABC的三个内角A，B，C所对边的边长，R表示△ABC的外接圆半径．（1）R=2，a=2，B=45°，求AB的长；（2）在△ABC中，若∠C是钝角，求证：a2+b2＜4R2；（3）给定三个正实数a，b，R，其中b≤a，问a，b，R满足怎样的关系时，以a，b为边长，R为外接圆半径的△ABC不存在，存在一个或存在两个（全等的三角形算作同一个）？在△ABC存在的情况下，用a，b，R表示c．【考点】正弦定理．【分析】（1）由已知及正弦定理可sinA，b，利用大边对大角可得A为锐角，利用同角三角函数基本关系式可求cosA，利用三角形内角和定理，两角和的正弦函数公式可求sinC的值，利用正弦定理即可得解AB的值．（2）利用余弦定理推出a2+b2＜c2，利用正弦定理推出a2+b2＜4R2．（3）分类讨论判断三角形的形状与两边a，b的关系，以及与直径的大小的比较，分类讨论即可．解：（1）∵R=2，a=2，B=45°，∴由正弦定理可得：，解得：sinA=，b=2，又∵a＜b，可得：A＜B，可得cosA==，∴sinC=sin（A+B）=sinAcosB+cosAsinB==，∴AB=c=4sinC=4×=．证明：（2）由余弦定理得cosC=，∵C为钝角，可得cosC＜0，∴a2+b2＜c2又∵由正弦定理得c=2RsinC＜2R，∴c2＜4R2，∴a2+b2＜4R2．解：（3）①a＞2R≥b或a≥b≥2R时，不存在；②当a=2R且b＜2R时，A=90°，存在一个，c=；③当a=b＜2R，∠A=∠B且都是锐角sinA=sinB=时，△ABC存在且只有一个，c=2RsinC=；④当b＜a＜2R，存在两个，c=．。

【全国名校】2014-2015学年湖北省武汉华中师大附中高一上学期期末考试数学试卷（带解析）副标题一、选择题（本大题共8小题，共40.0分） 1. 设全集U 是实数集R ，集合，，则为A.B.C.D.2. 若且，则A.B.C.D.3. 下列函数中，对于任意R ，同时满足条件和的函数是A.B.C.D.4. 设，，，则 A.B. C.D.5. 函数，有零点，则m 的取值范围是A.B.C.D.6. 已知，，则等于A.B.C.D.7. 若函数，分别是R 上的奇函数，偶函数，且满足，则有A.B.C.D.8. 在△ABC 中，内角A ，B ，C 所对边的长分别为a , b , c ，且，，满足，若，则的最大值为A.B. 3C.D. 9二、填空题（本大题共8小题，共40.0分）9.若函数且在上既是奇函数又是增函数，则的图象是10.设满足，则A. 2B.C. 1D.11.已知，且，则的值用a表示为__________．12.在平面直角坐标系中，已知，，点C在第一象限内，，且，若，则的值是__________．13.已知△ABC的三个内角A，B，C的对边依次为a, b, c，外接圆半径为1，且满足，则△ABC面积的最大值为__________．14.已知A是半径为5的圆O上的一个定点，单位向量在A点处与圆O相切，点P是圆O上的一个动点，且点P与点A不重合，则的取值范围是__________．15.已知函数，给出下列五个说法：①；②若，则Z)；③在区间上单调递增；④函数的周期为；⑤的图象关于点成中心对称。

其中正确说法的序号是__________．16.（14分）已知函数．（1）若，求的值域；（2）若存在实数t，当，恒成立，求实数m的取值范围．三、解答题（本大题共5小题，共60.0分）17.A，B，C为△ABC的三内角，其对边分别为a, b, c，若．（1）求；（2）若，，求△ABC的面积．18.设集合，集合，集合C为不等式的解集．（1）求；（2）若，求a的取值范围．19.已知向量，设函数．（1）求的单调增区间；（2）若，求的值．20.已知向量，，，.（1）当时，求向量与的夹角；（2）当时，求的最大值；（3）设函数，将函数的图像向右平移s个长度单位，向上平移t个长度单位后得到函数的图像，且，令，求的最小值．21.（1）利用已学知识证明：．（2）已知△ABC的外接圆的半径为1，内角A，B，C满足，求△ABC的面积．答案和解析1.【答案】C【解析】由e卷通组卷系统，得e卷通组卷系统，又e卷通组卷系统，故e卷通组卷系统=e卷通组卷系统考点：1、解不等式；2、集合的运算.2.【答案】B【解析】由e卷通组卷系统，得e卷通组卷系统，又e卷通组卷系统，得e卷通组卷系统又e卷通组卷系统，所以e卷通组卷系统e卷通组卷系统.考点:三角函数的诱导公式.3.【答案】D【解析】若函数满足条件e卷通组卷系统，则函数为偶函数，若函数满足条件e卷通组卷系统，则函数为周期为e卷通组卷系统的周期函数，e卷通组卷系统，不满足条件e卷通组卷系统，故A不对；e卷通组卷系统也不满足条件e卷通组卷系统，故B不对；e卷通组卷系统满足条件e卷通组卷系统，但其最小正周期为e卷通组卷系统，故选D考点：1、二倍角的正弦、余弦公式；2、函数的奇偶性、周期性.4.【答案】C【解析】分析可知e卷通组卷系统e卷通组卷系统,由e卷通组卷系统，e卷通组卷系统即e卷通组卷系统e卷通组卷系统,e卷通组卷系统故e卷通组卷系统.考点：对数、指数、三角函数的综合考察.5.【答案】D【解析】若函数在e卷通组卷系统有零点，则应满足e卷通组卷系统，又e卷通组卷系统e卷通组卷系统则e卷通组卷系统解得e卷通组卷系统考点：1、三角函数求值；2、函数的零点.6.【答案】D【解析】令e卷通组卷系统，得e卷通组卷系统e卷通组卷系统分析得e卷通组卷系统因而e卷通组卷系统e卷通组卷系统.考点：1、两角和（差）的正、余弦公式；2、构造法.7.【答案】D【解析】用-x代换x得：f（-x）-g（-x）=e-x，即f（x）+g（x）=-e-x，又∵f（x）-g（x）=e x∴解得：f(x)=，g(x)=-，故f（x）单调递增，又f（0）=0，g（0）=-1，有g（0）＜f（2）＜f（3）故选D．考点：函数奇偶性的性质．8.【答案】C【解析】由正弦定理得e卷通组卷系统，由二倍角公式及两角和的正弦公式得，e卷通组卷系统，所以e卷通组卷系统，由余弦定理得e卷通组卷系统即e卷通组卷系统，解得e卷通组卷系统e卷通组卷系统e卷通组卷系统 .考点：1、正弦定理、余弦定理；2、基本不等式.9.【答案】C【解析】若函数e卷通组卷系统且e卷通组卷系统在e卷通组卷系统上是奇函数,则有e卷通组卷系统，即e卷通组卷系统e卷通组卷系统又函数是增函数，则有e卷通组卷系统，所以e卷通组卷系统，e卷通组卷系统图像是将e卷通组卷系统向右平移一个单位得到的，故选C.考点：对数函数的图像和性质.10.【答案】B【解析】由题意分析，当时，，解得，不符合条件，当时，，解得，即，则考点：指数函数、对数函数求值11.【答案】2【解析】由两角差的余弦公式，由e卷通组卷系统，得e卷通组卷系统由e卷通组卷系统，则e卷通组卷系统=2e卷通组卷系统.考点：三角函数的诱导公式及三角恒等变换.12.【答案】【解析】在三角形e卷通组卷系统中e卷通组卷系统，e卷通组卷系统，由余弦定理得e卷通组卷系统，所以三角形e卷通组卷系统为直角三角形，即e卷通组卷系统，由e卷通组卷系统，得e卷通组卷系统，即e卷通组卷系统，e卷通组卷系统，所以e卷通组卷系统=e卷通组卷系统考点：1、余弦定理；2、向量的坐标表示.13.【答案】【解析】由e卷通组卷系统，得e卷通组卷系统，又e卷通组卷系统，得e卷通组卷系统e卷通组卷系统e卷通组卷系统 又e卷通组卷系统，得e卷通组卷系统即e卷通组卷系统，由三角形面积公式e卷通组卷系统，由e卷通组卷系统，e卷通组卷系统展开即可得当三角形为等边三角形时面积最大，为e卷通组卷系统.考点：1、正弦定理；2、三角恒等变化.14.【答案】【解析】如图所示：e卷通组卷系统e卷通组卷系统e卷通组卷系统e卷通组卷系统e卷通组卷所以e卷通组卷系统考点：三角函数.15.【答案】①③【解析】①e卷通组卷系统正确；②若e卷通组卷系统，即e卷通组卷系统，则e卷通组卷系统时也成立，不正确；③在区间e卷通组卷系统上，e卷通组卷系统单调递增，正确；④e卷通组卷系统，故函数e卷通组卷系统的周期为e卷通组卷系统不正确；⑤函数e卷通组卷系统是奇函数，关于原点（0,0）对称，所以点e卷通组卷系统不是函数的对称中心，不正确考点：1、函数的单调性；2、函数的奇、偶性；3、三角函数求值.16.【答案】（1）;（2）【解析】（1）函数的对称轴为e卷通组卷系统，研究函数的值域可分三种情况讨论对称轴的位置：对称轴在e卷通组卷系统的左侧，内部，右侧；\（2）将e卷通组卷系统在e卷通组卷系统恒成立，转化为e卷通组卷系统恒成立，即e卷通组卷系统在e卷通组卷系统上的最大值e卷通组卷系统恒成立，由e卷通组卷系统恒成立知e卷通组卷系统，化简得e卷通组卷系统, 令e卷通组卷系统，则原题可转化为：存在e卷通组卷系统，使得e卷通组卷系统。

华东师大二附中2024学年第一学期9月英语试卷高三英语考试时间：120分钟满分：140分I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The 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. 145 minutes. B. 120 minutes. C 130 minutes. D. 160 minutes2. A. Teacher and student B. Eye doctor and patient.C. Salesman and customer D Interviewer and applicant.3. A. On Saturday. B. On Monday C. On Thursday. D. On Friday.4. A. Neither of them knows the composer of the music.B. The style of the music is not familiar to the man.C. The woman is as good a composer as the man.D. They share the same opinion of the odd music.5. A. They should talk about the apartment later.B. The apartment is still available to customers.C. The apartment had already been sold.D. It is not a suitable time to buy the apartment6. A. The customer's feedback. B. The responsibilities of her jobC. The prospects of her job.D. The manager's opinion of her7. A. The woman should think of giving up the subject.B. The woman should seek help from the tutoring service.C. The woman should work as a tutor to help others.D. The woman should major in accounting8. A. He is rejected for lack of experience. B. He quit his job not long agoC. He doesn't care about his appearanceD. He shaves himself every day.9. A. The woman had violated traffic regulationsB. The woman had been fined many times beforeC. The woman knows how to deal with the situation.D. The woman crossed the traffic light for poor eyesight.10. A. He is too busy to attend the lecture on Friday.B. Professor Simpson's lecture is not interestingC. He might miss the lecture if he was not reminded.D. The lecture has an opposite effect on himSection BDirections: In Section B, you will hear two short passages and one longer conversation, and you will be asked several questions on each of the passages and the conversation. The passages and the conversation 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. Aluminum (铝) cans. B. Plastic bags.C. Glass bottles.D. Cigarette-related litter.12. A. By 60 million. B. By 500 per cent. C. By 500 million. D. By 120 per cent.13. A. Simply leaving rubbish where it belongs is all that we can do.B. Littering is a more pressing problem than people might think.C. Only measuring the harm of rubbish by its lifetime is not enough.D. A large sum of money has been spent in order to keep streets clean.Questions 14 through 16 are based on the following passage.14. A. The rise of sea level. B. Flooding. C. High temperature. D. Bad light.15. 40% English football league grounds will be flooded every year.B. Many more matches will be shortened because of bad weather.C. Ticket prices of football matches will continue to rise.D. The revenue from ticket sales will be reduced.16. A. Spectators should be banned from watching sporting matches.B. Players, teams and sponsors promote carbon neutralization.C. Sports leaders keep the venue's address secret from the public.D. The government may cancel all the matches to be carbon-neutral.Questions 17 through 20 are based on the following conversation.17. A. Computer programmer. B. General manager. C. Salesman. D. 6ales manager.18. A. Two years. B. Three years. C. Five years. D. Six years.19. A. Achieving the assigned sales revenue target.B. Managing 50 employees in the department.C. Cooperating with her colleagues efficiently.D. Dealing with angry customers' complaints.20. A. Because she saw no chance for further advancement.B. Because she couldn't stand the pressure of the job.C. Because she was not satisfied with the low pay.D. Because she didn't want to work extra hours.II. Grammar and vocabularySection ADirections: After reading the passage below, fil 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 otherblanks, use one word that best fits each blank.A Day in the Life of a Curator (馆长)What are some of the most enjoyable aspects of being senior curator at the National Gallery?Among the joys of being a curator are getting (21) _________ (know) thoroughly great works of art; working alongside and learning from expert colleagues in different departments; and feeling that one's work, (22)_________ that concerns new acquisitions, displays and exhibitions, lectures or publications, can help shed important new light on our paintings, (23) _________, in turn, offers new ways for visitors to engage with them. What are some of the challenges of your role?One particular challenge is having to accept the frustrating reality (24) _________ although the gallery is there to connect people with pictures, it sadly doesn't have the capacity or resources to reach everyone all the time. (25)_________ is simply finding the time, amidst a busy workload of daily museum tasks and an intensive shorter-term exhibition schedule, to undertake longer-term research projects, involving thorough investigation of the pictures themselves, secondary research (26) _________ (conduct) in libraries, and conversations with peers worldwide. Have you personally had any unusual experiences during your work for the National Gallery?I have had plenty of memorable and exciting experiences, (27) _________ _________ ascending scaffolding (脚手架) to see Bridget Riley's Messengers in progress or looking at technical images to detect fascinating under drawing lying beneath the visible painted surface of a painting. A particularly happy moment for me (28) _________ (occur) when an album of 200 drawings came to light, the work by Elizabeth, Lady Eastlake (1809-1893) , wife of the gallery's first director. They included her sketches of places she visited abroad with her husband, as well as her pencil copies of paintings they inspected during his search to find qualified masterpieces for the national collection.I (29) _________ (imagine) that Lady Eastlake's sketches were lost or destroyed, so it was an extraordinary moment when I realised they were staring me in the face-and were so well preserved! The generous owner soon donated the precious album to the National Gallery so that her drawings (30) _________ be reunited with her husband's working notebooks and accessible for others to enjoy.Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can be used only once. Note that there is one word more than you need.A. analyzedB. boundaryC. compoundD. detectE. orbitF. potentialG. primitive H. stretching I. subsequently J. tubular K. unquestionablyVisitors are hereIn 2017, when astronomers discovered the asteroid (小行星) 11/2017 U1, it soon dawned on them that they had a strange object on their hands. The calculated 31 showed this long and thin rock to be simply passing through the solar system, and therefore its origin not of this world, as they like to say in the movies. The object was 32 named 'Oumuamua (from the Hawaiian word for “a person sent ahead to get information about the enemy's position, strength, etc.”) , raising memories of an old Star Trek episode, “For the World is Hollow and I Have Touched the Sky,” in which a long, 33 asteroid proved to be an alien ship in disguise (伪装) .There's nothing surprising about interstellar objects passing through our neighborhood. Or there shouldn't be, at least. There's no magical barrier at the 34 of our solar system. Although we see a sky full of stars a and inor telescopes 35 hundreds of clusters a and nebulae (星云) , most of the space in a galaxy's disk is practically empty, save for the thin interstellar medium.This month, science journalist David Chandler delivers a fascinating look at the 36 for spacecraft missions to interstellar intruders. Catching up to Oumuamua now would be virtually impossible. This thin, cigar-shaped rock, 37 about 1, 300 feet long, is rushing along at about 16 miles per second and is already as far away as the average distance to Pluto.But there's no doubt that other visitors from other stars will come by again. This has 38 happened countless times in the 4. 6-billion-year history of our star and its planets- and now, astrophysics is in an advanced state. Every day we learn about 39 conditions long ago in the solar system by studying pieces of rock or metal from space that have landed on Earth and other objects. With the chemistry of materials from the origin days of other stars 40 , who knows what could be found from such priceless relics.That's another comforting thought to keep in mind as you read David's story and then gaze up into a dark sky full of wonder.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B. C and D. Fill in each blank with the word or phrase that best fits the context.Should period dramas reflect modern sensibilities?Simon Jenkins The most popular films of my youth were war films. They were about how Britain won a war—and that could be any war you cared to mention. In my opinion, they were great fun and, mostly, patriotic 41 .I'm a journalist and occasional historian, and something the two professions share is a respect for the 42of truth. Both are in the business of bringing events to life through the power of fact, not falsification. They may sometimes be 43 of distortion (歪曲) and thoughtless analysis, but to be plain wrong is unethical and unprofessional. The gap between fact and fiction is one that should not be crossed—or if crossed, should stand corrected.Many playwrights, filmmakers and novelists 44 . To them, history is a stimulus to artistic licence, material to be exploited and 45 for dramatic effect. Their considerations are audience appeal, profit and, often, politics. They leave it to historians to worry about 46 . This, to me, is lying.I have always found ‘ 47 ’ (in which facts are the basis for fiction) hard to stomach. I can appreciate‘docudrama’, which dramatises the events, or the novels of Hilary Mantel, as attempts to deepen our understanding of the past. Mantel insisted that her goal was always to be as 48 as the facts allowed. She did not 49 create false events. The same was not true of The Crown and its much-documented faking of stories. The fact the team behind the show took such pains to cast actors that 50 their real-life counterparts simply added a touch of reality to the made-up story. The result was an audience 51 of what was true or false.I appreciate that history—as with 52 —involves selection, and that selection itself can be motivated by a desire to twist the truth. Each age puts pressure on historians to select material in a manner that respects the 53 or bias (偏向) of nations, groups or individuals. The duty of the historian is to see behind such bias. The task is toreveal what happened, why and how.In an age of artificial intelligence and online ‘deep fakery’, the truth has never been more 54 . The world of fiction has no need to be a parasite (寄生虫) on history: it has all of human imagination to supply it with plots. Every work that claims to be ‘based on real events’ should, in my view, be identified as lies, and should display a large ‘T’ or ‘NT’ —true or not true. Artistic licence should not be a (n) 55 to deceive.41. A. honour B. instinct C. rubbish D. masterpiece42. A. victory B sacredness C. suspicion D. degree43. A. proud B typical C. desperate D. guilty44. A. disagree B. persist C. hesitate D. echo45. A. recorded B. publicized C. abused D. corrected46. A. plot B. artistry C. edition D. truth47. A faction B. profile C nonfiction D. social-drama48. A. awesome B. imaginative C. accurate D. comprehensive49. A. necessarily B deliberately C. merely D. duly50. A. adored B. falsified C. documented D. resembled51. A. conscious B ignorant C. clear D. insightful52. A. journalism B. literature C. politics D. patriotism53. A. literacy B. reason C. justice D. sensitivity54. A. immoral B. precious C. sufficient D. revolutionary55. A. instance B. tendency C. licence D submissionSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)“Who says it's Father's Day?” my son says to me, with the questioning angry look of someone who's been told they have extra tax to pay. “Well, the world does,” I tell him, suddenly self-conscious. “It's a special day for daddies.”Something about this—I can't think what—comes out sounding quite desperate and he looks at me as if I've just suggested he prove his love for me with a face tattoo (文身) . It's a look of suspicion, but also of tender concern for my mental state.This is his fifth Father's Day, so I can't help feeling slightly wounded that the concept hasn't stuck with him. I also can't help noting that he has never had any such issue with Mother's Day, which has always seemed to him like common sense.The event's nearness to his own birthday two weeks from now—is making things more difficult for him to tolerate. It would seem he finds it impolite that the run-up to his special day should be interrupted so close to the finish line by a day that celebrates me, the lesser of his two parents. In any case, if he's planning to make or gift me something, this conversation has been a masterstroke of expectation management“So, will all daddies get a Father's Day?” he asks. “Yes,” I reply, “and this isn't new-it's every year!” I attempt to regulate my voice/offence, but also make it very clear I haven't made this idea up on the spot by myself. “You've been doing it since you were born. And it's been around longer than that. I get things for Grandad every year, too.”At this he stirs himself up. He has never quite stopped being fascinated by the idea that his grandad is my dad, in the same way that I am his. I suppose it's the same thrill I feel when I see pictures of massive cranes (起重机) being built by other, even bigger cranes.“What do you get him?” he asks. “Well,” I say, “things like CDs or socks-and always a card.” At this he seems inspired. “I'll do a card!” he says, brightening.“You could buy me something, too. . .” I begin, but he is no longer listening, running to grab coloured paper and glittery pens. Not wishing to see this tribute to myself a whole week early, I smile and tell him I really shouldn't be watching and get up to leave him to it.“Yes,” he says, just in time for me to see he's actually writing “Dear Grandad” on the page. “Don't tell him!”56. How does the writer feel when explaining Father's Day?A. Suspicious.B. Embarrassed.C. Proud.D. Ridiculous.57. It can be inferred from the son's response that ________.A. the son feels hurt because the concept escapes himB. the writer looks relieved due to his son's tender heartC. a face tattoo is the way to prove a son's love for his fatherD. the son identifies with Mother's Day more than Father's Day58. The writer mentions Grandad in order to ________.A. justify the annual celebration of Father's DayB. practise skills of expectation managementC. narrow the gap between the three generationsD. link Father's Day to his son's birthday59. Which of the following best summarises the passage?A. When celebrating Father's Day, you should also send a DIY card to your grandpa.B. It is a universally acknowledged fact that Father's Day is a special day for daddies.C. Father's Day is a special time to celebrate Dad, but for my son that's a bit of a stretch.D. My son and father have agreed to keep the Father's Day greetings card secret from me.(B)611907Elizabeth Maconchy is born on 19 March in Broxbourne,Hertfordshire. Her parents are both Irish, and the family later move toHowth, close to Dublin on the east coast of Ireland.Edward VII opens the new Old Bailey criminal court building inLondon, its dome decorated by Lady Justice, a bronze sword-holdingsculpture.1930In the same year that her PianoConcerto receives its world premiere(首次公演) in Prague, her orchestralsuite (组曲) The Land enjoys greatpraise when Sir Henry Wood conductsit at the BBC Concerts.At London's Queen's Hall, Adrian Boult conducts the recently founded BBC Symphony Orchestra in its first ever concert, featuring works by Wagner, Brahms and Ravel.1947Married since 1930 toWilliam LeFanu, a librarian at theRoyal College of Surgeons, shegives birth to their seconddaughter, Nicola LeFanu, who will also go on to enjoy a career as a composer.An exceptionally harsh winter results firstly in power cuts due to difficulties in transporting coal and then, as the snow melts in March, the most damaging flooding of the River Thames for more than 100 years.1968Her Aristophanes-inspired operaThe Birds, one of a number of piecesthat she composes for children, isperformed for the first time atBishop's Stortford College for Boys.After his controversial ‘Riversof Blood’ speech about immigration,MP Enoch Powell is removed fromthe Shadow Cabinet by Conservativeleader Edward Heath.1994Seven years after receiving a Damehood for services to music, she dies in Norwich, aged 87. During the week of May 13-17, 2024, she is featured as Composer of the Week on BBC RadioAt a ceremony in Calais on 6 May, Queen Elizabeth II and French president Fran gois Mitterrand officially open the Channel Tunnel, six years after tunnelling began on Dec. 1st , 1987. 1983She composes ‘Quartetto Corto’, the 13th and last of her string quartets (弦乐四重奏) , a series that, begun some half-a-century earlier, she describes as‘my best and most deeply felt works’.Jenny Pitman becomes the first woman to train a winner of the Grand National when Corbiere, ridden by Ben De Haan, finishes three-quarters of a length ahead of Greasepaint at Aintree.60. Elizabeth Maconchy is probably ________.A. Lady JusticeB. a conductorC. a librarianD. a composer61. The best title (numbered 61 ) of the passage is probably ________.A MACONCHY Life &Times B. MACONCHY AchievementsC. Irish Lady's ContemporariesD. Uphill Battle for Recognition62. What happened in the 1930s?A. Adrian Boult conducted Maconchy's Piano Concerto.B. William LeFanu, a surgeon, got married to Maconchy.C. Maconchy began composing a series of string quartets.D. Sir Henry Wood composed an orchestral suite The Land.(C)A theme at this year's World Economic Forum (WEF) meeting in Switzerland was the perceived need to “speed up breakthroughs in research and technology.” Some of this framing was motivated by the climate emergency, some by the opportunities and challenges presented by generative artificial intelligence. Yet in various conversations, it seemed to be taken for granted that to address the world's problems, scientific research needs to move faster. The WEF mindset is similar to the Silicon Valley dictate—to move fast and break things. But what if the thing being broken is science? Or public trust?The WEF meeting took place just two weeks after Harvard University President Claudine Gay stepped down after complaints were made about her political science scholarship. In response, Gay requested corrections to several of her papers. Although it may be impossible to determine just how widespread such problems really are,it's hard to imagine that the scene of high-profile scholars correcting and retracting papers has not had a negative impact on public trust in science and perhaps in experts broadly.In recent years we've seen important papers, written by outstanding scientists and published in celebrated journals, retracted because of questionable data or methods, hence a question: Are scholars at supercompetitive places such as Harvard and Stanford rushing to publish rather than taking the time to do their work right?It's impossible to answer this question scientifically because there's no scientific definition of what constitutes "rushing. "But there's little doubt that we live in a culture where academics at leading universities are under tremendous pressure to produce results—and a lot of them—quickly.The problem is not unique to the U. S. In Europe, formal research assessments—which are used to allocate (分配) future funding—have for years judged academic departments largely on the quantity of their output. A recent reform urging an emphasis on quality over quantity allowed that the existing system had created “counterincentives. ”Good science takes time. More than 50 years elapsed between the 1543 publication of Copernicus's On the Revolutions of the Heavenly Spheres. And it took just about half a century for geologists and geophysicists to accept geophysicist Alfred Wegener's idea of continental drift.There's plenty of circumstantial evidence that scientists and other scholars are pushing results out far faster than they used to. Consider the sheer volume of academic papers being published these days. One recent study put the number at more than seven million a year, compared with fewer than a million as recently as 1980. Some of this growth is driven by more scientists and more co-authorship of papers, but the numbers also suggest that the research world has prioritized quantity over quality. Researchers may need to slow down if we are to produceknowledge worthy of trust.63. WEF meeting in Switzerland advocated that ________.A. researchers need to achieve breakthroughs more rapidlyB. public trust in science is not supposed to be easily brokenC. WEF and Silicon Valley reach an agreement to move fastD. climate emergency and AI push scientific research hard64. Which of the following examples fails to prove that good science takes time?A. Gay's correction and retraction of papers.B. Publication of Copernicus's theory.C. High-profile scholars' tremendous output.D. Acceptance of the idea of continental drift.65. The underlined word “counterincentives” in para. 5 probably means ________.A. measures to increase quantities of outputB. discouragements of high quality papersC. rewards for leading universities' researchD. contradictory motives for future funding66. Which of the following is best title of the passage?A. WEF Coincides with Silicon ValleyB. Collapse of Public TrustC. Dilemma between Quantity and QualityD. Trouble in the Fast LaneSection CDirections: Read the passage carefully. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A. Brentford FC has taken a different approach.B. Statistics have helped the team win on the pitch, too.C. He applied his talent to identifying the underlying strength of football teams.D. Analytics underlay and supported a remarkably profitable buy-low-sell-high transfer strategy.E. They were told to focus not on how many goals a team was scoring, which was subject to too much randomness.F. Like “Moneyball” , a hit book about the use of statistics in baseball, “Smart Money” is both informative and entertaining.Football and dataA numbers gameAt most football clubs, the equation is simple: you put in vast amounts of money, and you get out star players and win victories. Take Manchester City, the Premier League's reigning champions. Before its takeover in 2008 by a Middle East plutocrats (财阀) , the club often struggled in the bottom half of the table; it has won English football's championship seven times since 2011.67. _________ The club was promoted into the Premier League in 2021 after striving for decades in the lower reaches of football. What makes its success surprising is not how much money its owner, Matthew Benham, has put into the team, but how little. In a new book “Smart Money”. Alex Duff, a lifelong Brentford fan, explains how a money-saving plan made profits.Mr Benham studied physics at Oxford University and then went to work in banking. In his early 30s, sensing correctly that bookmakers (赌注登记人) were inaccurate when setting odds for football matches, Mr Benham leftbanking to become a full-time gambler.He set up his own company, Smartodds, and competed with financial institutions to hire the best mathematicians. 68. _________ Instead, their focus should be the "goal-scoring opportunities it was creating. In time, he reckoned, the goals would comeThe approach was so efficient that when Brentford, going through one of its periodic financial crises, put out a general appeal for help in 2005, Mr Benham offered his services. Within a decade he owned the club and was applying his ideas to how the team was constructed. 69. _________There were plenty of doubters within football about Brentford's philosophy. But in time it worked. In their first season in the Premier League, Mr Benham's investment of around f100m in the club—a tiny sum compared with competitors—realised its first profit of £25m.70. _________ Mr Benham identifed ser pieces (定位球) as an important part of creating scoring opportunities, and hired Gianni Vio, an Italian coach with 4, 000 such plays in his career. Players are instructed to press the opposition and tackle players within ten seconds of them receiving the ball. Though Brentford is not competing for the title, in recent years the team has beaten several of football's plutocrats—including Manchester City. What were the odds?IV. Summary WritingDirections: 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.Build better boundariesDoing a good deed like helping your friend with their homework or sharing a snack can make you feel happy. Studies show that kindness is good for your wellbeing. However, if you often agree to things you don't want to do, or feel guilty saying no, you could be falling into a people-pleasing trap. It's not always easy to tell when this happens but one clue is that it's difficult to stop. Pleasing other people may feel good for a short while but the feeling doesn't last. This is why it's a good idea to set limits on what you'll do for others. These are called “boundaries”.Spending too much energy on someone else can stop you doing things you want or need to do. Research has found that trying to please others can leave us feeling stressed and uncomfortable. We can also feel angry and frustrated with ourselves and our friends. “There's nothing wrong with being kind to other people,” says Dr Toru Sato, an author and expert in thoughts and feelings, but we need to be sure we're doing it out of kindness, not because we're worried about what other people think.Taking on so much that you end up letting others down doesn't make you a better friend. The youth mental health charity YoungMinds says boundaries include taking time alone when you need it and being able to explain your needs. If saying no feels hard, practise with small things, like if someone offers a straw in a restaurant. This can help you feel more confident. Thinking about how to say no also helps, YoungMinds says, and allows you to communicate what you want clearly and calmly. Remember, you don't need an excuse to say no; you don't owe anyone an explanation. The good people in your life will respect this boundary.71. _________________________________________________________________________________________ V. Translation。

上海华师大二附中2015届高一数学上册《集合与命题、不等式》单元测试题 沪教版一、 填空题：（每题4分，共40分）1.已知集合},02{2R x x x x A ∈=--=，集合}31|{≤≤=x x B ，则A ∩B = .2.集合{}52<<-=x x A ，集合{}121-≤≤+=m x m x B ，若A B ⊆，且B 为非空集合，则m 的取值范围为 .3.命题“若实数b a ,满足,7<+b a 则2=a 且3=b ”的否命题是 .4. “y x >”是“y x >”的 条件.5. 不等式1312>+-x x 的解是 6. 已知不等式052>+-b x ax 的解集是}23|{-<<-x x ，则不等式052>+-a x bx 的解是___________ .7. 不等式（1＋x ）（1－｜x ｜）＞0的解集是⎽⎽⎽⎽⎽⎽⎽⎽⎽⎽⎽⎽⎽8.设集合(){}()(){}521,,31,2+-==-==x m y y x B x y y x A ，其中R m R y x ∈∈,,. 若∅=⋂B A ，则实数m 的取值范围是 .9.集合A 中有10个元素，B 中有6个元素，全集U 有18个元素，B A ≠φ.设集合)(B A C U 有x 个元素，则x 的取值集合为______________.10.已知非空集合{},6,5,4,3,2,1⊆S 满足：若S a ∈，则必有S a ∈-7.问这样的集合S 有 个将该问题推广到一般情况： . 二、选择题（每题5分，共20分）11.设{}{}为质数，为合数x x B x x A ==，N 表示自然数集，若E 满足N E B A =⋃⋃，则这样的集合E （ ）Ａ.只有一个； Ｂ.只有两个 Ｃ.至多３个 Ｄ.有无数个 1２.定义集合运算：A ⊙B =｛z ︳z = xy (x+y )，x ∈A ，y ∈B ｝，设集合A=｛0，1｝，B=｛2，3｝，则集合A ⊙B 的所有元素之和为 （ ）A.0B.6C.12D.181３.四个条件：a b >>0；b a >>0；b a >>0；0>>b a 中，能使ba 11<成立的充分条件的个数是( )A.1B.2C.3D.4 14. 设a 、b 、c 是互不相等的正数，则下列不等式中不恒成立的是 （ ）A ．c b c a b a -+-≤-B ．a a a a 1122+≥+C ．a a a a -+<+-+213D ．21≥-+-ba b a 三、解答题：（8+10++10+12=40分）１5. 若集合{}{}2230,,0,A x x mx x R B x x x n x R =+-=∈=-+=∈， 且{}3,0,1A B =-，求实数,m n 的值。

2014-2015学年第一学期高一化学期末考试试卷可能用到的相对原子质量：H 1- C 12- N 14- O 16- Na 23- Mg 24- Al 27- Si 28-S 32- Cl 35.5- K 39- Ca 40- Mn 55- Fe 56- Cu 64-Zn 65- Ag 108- I 127- Ba 137- 一、选择题(每小题只有一个正确选项,每小题2分，共40分） 1．下列叙述中正确的是（ ）A ．分子晶体中一定含有共价键，不含有离子键B ．仅含共价键的物质不一定是共价化合物C ．原子晶体中只存在非极性共价键D ．不同原子形成的纯净物一定是化合物 2．下列化学用语错误的是( ）A ．过氧化钡的电子式22Ba:O :O:-+⎡⎤⎣⎦B ．钠23-的原子符号2311NaC ．次氯酸的结构式H Cl O --D ．锰酸根：24MnO -3．下列说法正确的是（ )A ．氧化还原反应中,一定有一种元素被氧化,另一种元素被还原B ．物质所含元素的化合价越高，该物质的氧化性就越强C ．HClO 既有氧化性,又有还原性D ．21 mol Cl 参加反应转移2 mol 电子4．下列化学式中,真实表示物质组成的是( )A ．2MgCl B ．22Na O C ．SiC D ．2H O5．某金属分别与足量的硫和氯气化合时，1 mol 金属推动的电子分别为236.0210⨯和241.20410⨯，该金属是（ )A ．NaB ．CaC ．FeD ．Cu6．在通常条件下,下列各组物质的性质排列正确的是( )A ．熔点：22COKCl SiO >> B ．水溶性：22HCl H S SO >>C ．沸点：22NaH H O H S >> D ．硬度：SiC >石墨>3CaCO7．纯锌和稀硫酸反应速率很小,为了加快锌的溶解和放出2H 的速率，可以加入少量( ）A ．硫酸铜晶体B ．稀盐酸C ．镁带D ．碘化钾溶液8．下列过程中,既有离子键被破坏，又有共价键被破坏的是（ )A ．溴蒸汽被木炭吸附B ．烧碱熔化C ．碳酸钙高温分解D ．HCl 溶于水 9．下列物质中,不能与硫化氢反应的是（ ）A ．2K S 溶液 B ．2SO C ．双氧水 D ．4FeSO 溶液10．已知:①()()()2222H g O g 2H O g +→+kJ Q ,0Q >；②键能是指将1 mol 气体分子拆开为气态原子所需要的能量。

华师大二附中2014-2015学年高一下学期期中考试数学试题（时间90分钟 满分100分 命题人：郑同 审核人：龚杰曾）一、填空题（每小题3分，共36分）１、扇形的半径为cm 1，圆心角为2弧度，则扇形的面积为________2cm ．2、已知角α的终边过点()12,5--P ，则=αcos ______．3、已知),2(,41)sin(ππααπ∈=-，则=α2sin _________. 4、已知α是锐角，则=+)tan 1(log 2cos αα ．5、化简：=--⋅+-⋅+-)2sin()cos()2sin()2cot()tan()sin(απααπαπαπαπ ． 6、若α是第三象限角，且1312)cos(sin cos )sin(-=+-+βαβββα，则=2tan α ． 7、在ABC ∆中，32,3,1π=∠==C c b ，则=∆ABC S ． 8、隔河测算B A ,两目标的距离，在岸边取D C ,两点，测得m CD 200=，︒=∠105ADC ，︒=∠15BDC ，︒=∠120BCD ，︒=∠30ACD ，则B A ,间的距离 m ．9、定义bc ad d c b a -=，则函数)(sin 1cos 4sin )(R x xx x x f ∈-=的值域为 ． 10、定义在区间⎪⎭⎫ ⎝⎛20π,上的函数x y cos 6=的图像与x y tan 5=的图像的交点为P ，过点P 作x PP ⊥1轴于点1P ，直线1PP 与x y sin =的图像交于点2P ,则线段21P P 的长为____ ．11、已知函数12)(2+-=ax x x f ，存在)2,4(ππϕ∈，使得)(cos )(sin ϕϕf f =，则实数a 的取值范围是 ．12、设函数]),[(42cos 322sin 3cos 1224)(4234ππ-∈+++-+-=x x x x x x x x f 的最大值为M ，最小值为m ，则=+m M __ __．二、选择题（每小题4分，共16分）13、已知k Z ∈，下列各组角的集合中，终边相同的角是 （ ）A ．2k π 与 2k ππ± B ．2k ππ+与4k ππ± C ．6k ππ+ 与26k ππ± D ．3k π 与 3k ππ+14、在ABC ∆中，若A B B A sin sin cos cos >，则此三角形一定是 （ ）A ．钝角三角形B ．直角三角形C ．锐角三角形D ．形状不确定15、给出下列三个等式：()()()()()()f xy f x f y f x y f x f y =++=，，()()()1()()f x f y f x y f x f y ++=-,下列函数中不满足其中任何一个等式的是 （ ）A ．()3x f x =B ．()sin f x x =C ．2()log f x x =D ．()tan f x x =16、定义在R 上的偶函数)(x f 满足)()2(x f x f =-，且在]2,3[--上是减函数，βα,是钝角三角形的两个锐角，且βα<，则下列不等式关系中正确的是 （ ）A ．(sin )(cos )f f αβ>B ．(cos )(cos )f f αβ<C ．(cos )(cos )f f αβ>D ．(sin )(cos )f f αβ<三、解答题（本大题共48分）17、（本题6分）若2tan 1tan 1=+-A A ，求)4cot(A +π的值．18、（本题8分）设ABC ∆的内角C B A 、、所对的边分别为c b a 、、.已知1=a ，2=b ，41cos =C . （1）求ABC ∆的周长；（2）求()C A -cos 的值.19、（本题10分）已知函数()f x =223sin cos 2cos 1()x x x x R +-∈．(1)求函数()f x 的最小正周期及在0,2π⎡⎤⎢⎥⎣⎦上的单调递增区间； (2)若06()5f x =，0,42x ππ⎡⎤∈⎢⎥⎣⎦，求0cos 2x 的值．20、（本题10分）如图, 单位圆（半径为1的圆）的圆心O 为坐标原点，单位圆与y 轴的正半轴交与点A ,与钝角α的终边OB 交于点),(B B y x B ，设BAO β∠=.（1）用β表示；（2）如果4sin 5β=,求点),(B B y x B 的坐标； （3）求B B y x -的最小值.21、（本题14分）已知函数)1,0(112log )(≠>+--=a a x mx m x f a 是奇函数，定义域为区间D （使表达式有意义的实数x 的集合.（1）求实数m 的值，并写出区间D ；（2）当1>a ，试判断函数)(x f y =的定义域D 内的单调性，并说明理由；（3）当),[b a A x =∈（A ⊂≠B ，a 是底数）时，函数)(x f 为),1[+∞，求实数b a ,的值．参考答案一、填空题（每小题3分，共42分）１、扇形的半径为cm 1，圆心角为2弧度，则扇形的面积为____１____2cm ．2、已知角α的终边过点()12,5--P ，则=αcos __135-____． 3、已知),2(,41)sin(ππααπ∈=-，则=α2sin ___815-______. 4、已知α是锐角，则=+)tan 1(log 2cos αα 2- ．5、化简：=--⋅+-⋅+-)2sin()cos()2sin()2cot()tan()sin(απααπαπαπαπ 1- ． 6、若α是第三象限角，且1312)cos(sin cos )sin(-=+-+βαβββα，则=2tan α 23- ． 7、在ABC ∆中，32,3,1π=∠==C c b ，则=∆ABC S 43 ． 8、隔河测算B A ,两目标的距离，在岸边取D C ,两点，测得m CD 200=，︒=∠105ADC ，︒=∠15BDC ，︒=∠120BCD ，︒=∠30ACD ，则B A ,间的距离 2200 m ．9、定义bc ad d c b a -=，则函数)(sin 1cos 4sin )(R x xx x x f ∈-=的值域为 ]4,4[- ． 10、定义在区间⎪⎭⎫ ⎝⎛20π,上的函数x y cos 6=的图像与x y tan 5=的图像的交点为P ，过点P 作x PP ⊥1轴于点1P ，直线1PP 与x y sin =的图像交于点2P ,则线段21P P 的长为___32_ ． 11、已知函数12)(2+-=ax x x f ，存在)2,4(ππϕ∈，使得)(cos )(sin ϕϕf f =，则实数a 的取值范围是 )22,2( ．12、设函数]),[(42cos 322sin 3cos 1224)(4234ππ-∈+++-+-=x x x x x x x x f 的最大值为M ，最小值为m ，则=+m M 4 ．二、选择题（每小题4分，共16分） 13、已知k Z ∈，下列各组角的集合中，终边相同的角是 （ Ｂ ）A ．2k π 与 2k ππ± B ．2k ππ+与4k ππ±C ．6k ππ+ 与26k ππ±D ．3k π 与 3k ππ+ 14、在ABC ∆中，若A B B A sin sin cos cos >，则此三角形一定是 （ A ） A ．钝角三角形 B ．直角三角形 C ．锐角三角形 D ．形状不确定15、给出下列三个等式：()()()()()()f xy f x f y f x y f x f y =++=，，()()()1()()f x f y f x y f x f y ++=-,下列函数中不满足其中任何一个等式的是 （ Ｂ ）A ．()3x f x =B ．()sin f x x =C ．2()log f x x =D ．()tan f x x =16、定义在R 上的偶函数)(x f 满足)()2(x f x f =-，且在]2,3[--上是减函数，βα,是钝角三角形的两个锐角，且βα<，则下列不等式关系中正确的是 （ Ｂ ）A ．(sin )(cos )f f αβ>B ． (cos )(cos )f f αβ>C ．(cos )(cos )f f αβ<D ．(sin )(cos )f f αβ<三、解答题（本大题共48分）与钝角 的终边OB 交于点),(B B y x B .21（本题14分）已知函数)1,0(112log )(≠>+--=a a x mx m x f a 是奇函数，定义域为区间D （使表达式有意义的实数x 的集合.（1）求实数m 的值，并写出区间D ；（2）当1>a ，试判断函数)(x f y =的定义域D 内的单调性，并说明理由；（3）当),[b a A x =∈（A ⊂≠D ，a 是底数）时，函数)(x f 为),1[+∞，求实数b a ,的值．。