高一数学秋季学期期末考试2

高一数学秋季学期期末考试高一数学（示范高中卷）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．满分150分．考试时间120分钟．说明：可以使用计算器，但未注明精确度的计算问题不得采取近似计算，建议根据题型特点把握好使用计算器的时机．相信你一定会有出色的表现！第Ⅰ卷本卷共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．请把符合题目要求的选项的字母填入答题卷的答题卡中．一、选择题：1．不等式｜2－x｜≥1的解集是（A）｛x｜1≤x≤3｝（B）｛x｜x≤1或x≥3｝（C）｛x｜x≤1｝（D）｛x｜x≥3｝2．设A＝｛x｜x2－5x＋4≤0｝，B＝｛x｜x2－5x＋6≥0｝，则A∩B（A）［1，2］［3，4］（B）［1，2］［3，4］（C）｛1，2，3，4｝（D）［－4，－1］［2，3］3．命题“若x2＋y2＝0，则x、y全为0”的逆否命题是（A）若x、y全为0，则x2＋y2≠0（B）若x、y不全为0，则x2＋y2＝0（C）若x、y全不为0，则x2＋y2≠0（D）若x、y不全为0，则x2＋y2≠04．如果a，b，c都是实数，那么P：ac＜0，是q：关于x的方程ax2＋bx＋c＝0有一个正根和一个负根的（A）充分不必要条件（B）必要不充分条件（C）充要条件（D）既不充分也不必要条件5．f（x）是一次函数且2f（1）＋3f（2）＝3，2f（－1）－f（0）＝－1，则f（x）等于（A）499xy（B）36x－9（C）4199x（D）9－36x6．已知三个命题：①方程x2－x＋2＝0的判别式小于或等于零；②若｜x｜≥0，则x≥0；③5＞2且3＜7．其中真命题是（A）①和②（B）①和③（C）②和③（D）只有①7．已知函数f（x）＝23,(0),log,(0),xxxx≤那么14ff的值为（A）9（B）19（C）－9（D）－198．若定义在（－1，0）内的函数f（x）＝log2a（x＋1）＞0，则a的取值范围是（A）10,2（B）10,2（C）1,2（D）(0,)9．设｛an｝为递增等差数列，前三项的和为12，前三项的积为48，则它的首项为（A）1（B）2（C）4（D）610．若｛an｝是等比数列，且an＞0，a2a4＋2a3a5＋a4a6＝25，则a3＋a5的值为（A）5（B）10（C）15（D）2011．若数列｛an｝是公差为12的等差数列，它的前100项和为145，则a1＋a3＋a5＋…＋a99的值是（A）60（B）72.5（C）85（D）12012．若2x－3－x≥2－y－3y，则（A）x－y≥0（B）x－y≤0（C）x＋y≥0（D）x＋y≤0第Ⅱ卷（本卷共10小题，共90分）二、填空题：本大题共4小题；每小题4分，共16分．请将答案填写在答．题卷．．中的横线上．13．（x，y）在映射f下的象是（xy，x＋y），则点（2，3）在f下的象是▲．14．若f（10x）＝x，则f（5）＝▲．15．若数列｛an｝满足an＋1＝323na且a1＝0，则a7＝▲．16．若函数y＝21x（x≤－1），则f－1（2）＝▲．高一数学（示范高中卷）题号一二三总分1～1213～16171819202122得分一、选择题答题卡：（每小题5分，共60分）题号123456789101112答案二、填空题：（每小题4分，共16分）13．；14．；15．；16．．三、解答题：本大题共6小题；共74分．解答应写出文字说明、证明过程或演算步骤．17．（本小题满分12分）求（lg2）2＋lg2·lg50＋lg25的值．得分评卷人得分评卷人18．（本小题满分12分）已知集合A＝｛2，－1，x2－x＋1｝，B＝｛2y，－4，x＋4｝，C＝｛－1，7｝，且A∩B＝C，求实数x，y的值．19．（本小题满分12分）判断函数f（x）＝211x在区间（1，＋∞）上的单调性，并用单调性定义证明．得分评卷人得分评卷人20．（本小题满分12分）数列｛an｝，Sn为它的前n项的和，已知a1＝－2，an＋1＝Sn，当n≥2时，求：an和Sn．21．（本小题满分12分）已知一扇形的周长为c（c＞0），当扇形的弧长为何值时，它有最大面积？并求出面积的最大值．得分评卷人得分评卷人22．（本小题满分14分）已知函数f（x）＝log2（x＋m），且f（0）、f（2）、f（6）成等差数列．（1）求实数m的值；（2）若a、b、c是两两不相等的正数，且a、b、c成等比数列，试判断f（a）＋f（c）与2f（b）的大小关系，并证明你的结论．得分评卷人高一数学（示高期末）答案第1页（共3页）高一数学（示范高中卷）参考答案及评分标准一、选择题：（每小题5分，共60分）题号123456789101112答案BADCCBBABAAC二、填空题：（每小题4分，共16分）13．（6，5）14．lg515．416．－3三、解答题：17．解：原式＝（lg2）2＋lg2·（lg2＋2lg5）＋2lg5···············································2分＝2（lg2）2＋2lg2·lg5＋2lg5·························································4分＝2lg2（lg2＋lg5）＋2lg5······························································6分＝2lg2＋2lg5···············································································8分＝2（lg2＋lg5）··········································································10分＝2．·······················································································12分18．解：∵A∩B＝C，且C＝｛－1，7｝，∴7∈A，－1∈B，7∈B．························2分∵A＝｛2，－1，x2－x＋1｝，∴x2－x＋1＝7，·········································4分∴x＝3或x＝－2．·············································································6分当x＝－2时，B＝｛2y，－4，2｝，与－1∈B，7∈B矛盾．······································································8分当x＝3时，B＝｛2y，－4，7｝，∴2y＝－1．∴y＝－12．····································································10分∴3,1.2xy·····················································································12分高一数学（示高期末）答案第2页（共3页）19．解：f（x）在区间（1，＋∞）上是减函数．证明如下：····································2分取任意的x1，x2∈（1，＋∞），且x1＜x2，则··············································3分f（x1）－f（x2）＝2111x－2211x＝22212212(1)(1)xxxx＝21212212()()(1)(1)xxxxxx．··5分∵x1＜x2，∴x2－x1＞0．·······································································6分又∵x1，x2∈（1，＋∞），∴x2＋x1＞0，21x－1＞0，22x－1＞0，················8分∴（21x－1）（22x－1）＞0．（x2＋x1）（x2－x1）＞0·································10分∴f（x1）－f（x2）＞0．·····································································11分根据定义知：f（x）在区间（1，＋∞）上是减函数．·······························12分20．解：∵an＋1＝Sn，又∵an＋1＝Sn＋1Sn，∴Sn＋1＝2Sn．·······································2分∴｛Sn｝是以2为公比，首项为S1＝a1＝－2的等比数列．·························6分∴Sn＝a1×2n1＝－2n．·······································································10分∵当n≥2时，an＝Sn－Sn1＝－2n1．····················································12分21．解：设扇形的半径为R，弧长为l，面积为S∵c＝2R＋l，∴R＝2cl（l＜c）．·························································3分则S＝12Rl＝12×2cl·l＝14（cl－l2）·················································5分＝－14（l2－cl）＝－14（l－2c）2＋216c．···············································7分∴当l＝2c时，Smax＝216c．··································································10分答：当扇形的弧长为2c时，扇形有最大面积，扇形面积的最大值是216c．······12分高一数学（示高期末）答案第3页（共3页）22．解：（1）由f（0）、f（2）、f（6）成等差数列，可得2log2（2＋m）＝log2m＋log2（6＋m），·····································3分即（m＋2）2＝m（m＋6），且m＞0，解得m＝2．·····························5分（2）由f（x）＝log2（x＋2），可得2f（b）＝2log2（b＋2）＝log2（b＋2）2，··································6分f（a）＋f（c）＝log2（a＋2）＋log2（c＋2）＝log2［（a＋2）（c＋2）］，·········································7分∵a、b、c成等比数列，∴b2＝ac．·················································8分又a、b、c是两两不相等的正数，故（a＋2）（c＋2）－（b＋2）2＝ac＋2（a＋c）＋4－（b2＋4b＋4）·········································10分＝2（a＋c－2ac）＝22ac＞0，····································12分∴log2［（a＋