四川省成都市龙泉实验中学2017届高三英语9月月考试题
2017-2018学年四川成都实验高级中学高三上学期9月月考英语试题

成都实验高级中学高三上学期月考试题英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。
用2B铅笔将答题卡上试卷类型A后的方框涂黑。
2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后,请将本试题卷和答题卡一并上交。
第I卷(100分)第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题。
从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What will the man do after psychology class?A. Study in the library.B. Have a leisure afternoon.C. Practice running.2. What does the woman suggest the man do?A. See a doctor.B. Take one more pill.C. Buy a different kind of medicine.3. What problem does the man have?A. Have no other choices.B. Get to class on time.C. Make a quick decision.4. Who comes from Australia?A. The woman.B. The man.C. Darcy.5. What time is it now?A.5:45.B.5:30.C.5:15.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。
2017届四川省成都市高新区高三9月月考英语试题及答案

成都高新区高2017届统一测试试题卷英语全卷分第I卷(选择题)和第II卷(非选择题)两部分。
考试时间120分钟,满分150分。
( 命题:中和中学高三英语备课组;审题:李祖诚 )注意事项:1.答第I卷前,考生务必将自己的姓名、考籍号、考试科目涂写在答题卡上。
2.每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其他答案标号。
不能答在试卷上。
3.答非选择题时,必须将0.5毫米黑色签字笔,将答案书写在答题卷规定的位置上。
4.考试结束时,监考人只将答题卡和答题卷收回。
第I卷选择题(共90分)第一部分英语知识运用(共两节,满分40分)第一节语法和词汇知识(共10小题;每小题1分,满分10分)从A、B、C、D四个选项中,选出可填入空白处的最佳选项,并在答题卡上将该项涂黑。
1.A few stars from the entertainment industry, such as Ke Zhendong and Fang zuming, have recently been arrested, which shows that drug taking is becoming _____ increasingly serious threat to society.A. theB. anC. aD. 不填2.Peter, though a green hand, has earned not only the trust of his fellow workers but also ________ of his boss.A.one B.onesC.that D.those3.Life is like riding a bicycle. _____ your balance, you must keep moving.A. To keepB. KeepingC. Being keptD. Having kept 4.Only after one loses something ________ how precious it is.A.he will realize B.will he realizeC.he has realized D.has he realized 5.No student ________ go out of the school campus after 10:00 at night without permission according to the schoolregulations.A. shallB. mustC. shouldD. would6.In my opinion, ________ matters is whether we can win together as a team instead of individuals.A.how B.itC.that D.what7.A big earthquake hit Ludian in Yunnan, ________ many deaths and injuries and a big loss of property.A. to causeB. causedC. causingD. having caused 8.If you find yourself ______ no one will help you, pleasecall me.A. whichB. whenC. whereD. what9.A new product must be _________ before it comes into use.A. given outB. tried outC. put outD. made out10. --- Could I ask you a rather personal question?--- Of course, _________.A. go aheadB. good ideaC. pardon meD. don’t mention it第二节完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C和D)中,选出可填入空白处的最佳选项,并在答题卡上将该项涂黑。
四川省成都市龙泉中学2018届高三9月月考英语试卷(含答案)

成都龙泉中学2015级高三上学期9月月考试题英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。
用2B铅笔将答题卡上试卷类型A后的方框涂黑。
2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后,请将本试题卷和答题卡一并上交。
第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题l.5分,满分7.5分) 听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后;你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What is the woman’s nationality?A. Australian.B. American.C. Indian.2. Why was the woman able to get an A on the test?A. She read the whole textbook.B. She reviewed all the notes.C. She attended every lecture.3. What does the man plan to do on the weekend?A. Host a dance party.B. Visit Bill with the woman.C. Help the woman with a party.4. When will the woman meet the man tomorrow?A. At 10:00 am.B. At 10:30 am.C. At 11:00 am.5. What do we know about the man?A. He worked abroad for a few years.B. He is planning to go to Canada.C. He wants to meet the woman again.第二节(共15小题;每小题1.5分,满分22.5分)听第6段材料,回答第6至7题。
四川省成都市龙泉驿区高三英语9月月考试题

2015级高三上学期9月月考试题英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。
用2B铅笔将答题卡上试卷类型A后的方框涂黑。
2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后,请将本试题卷和答题卡一并上交。
第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题l.5分,满分7.5分) 听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后;你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What is the woman’s nationality?A. Australian.B. American.C. Indian.2. Why was the woman able to get an A on the test?A. She read the whole textbook.B. She reviewed all the notes.C. She attended every lecture.3. What does the man plan to do on the weekend?A. Host a dance party.B. Visit Bill with the woman.C. Help the woman with a party.4. When will the woman meet the man tomorrow?A. At 10:00 am.B. At 10:30 am.C. At 11:00 am.5. What do we know about the man?A. He worked abroad for a few years.B. He is planning to go to Canada.C. He wants to meet the woman again.第二节(共15小题;每小题1.5分,满分22.5分)听第6段材料,回答第6至7题。
四川省成都市龙泉实验中学2017届高三上学期月考物理试卷(9月份) 含解析

2016-2017学年四川省成都市龙泉实验中学高三(上)月考物理试卷(9月份)二、选择题(本题共8小题,每小题6分,共48分.在每小题给出的四个选项中,第14~18题只有一项符合题目要求,第19~21题有多项符合题目要求.全部选对的得6分,选对但不全的得3分,有选错的得0分.)1.在法拉第时代,下列验证“由磁产生电”设想的实验中,能观察到感应电流的是()A.将绕在磁铁上的线圈与电流表组成一闭合回路,然后观察电流表的变化B.在一通电线圈旁放置一连有电流表的闭合线圈,然后观察电流表的变化C.将一房间内的线圈两端与相邻房间的电流表连接,往线圈中插入条形磁铁后,再到相邻房间去观察电流的变化D.绕在同一铁环上的两个线圈,分别接电源和电流表,在给线圈通电或断电的瞬间,观察电流表的变化2.如图,一均匀金属圆盘绕通过其圆心且与盘面垂直的轴逆时针匀速转动.现施加一垂直穿过圆盘的有界匀强磁场,圆盘开始减速.在圆盘减速过程中,以下说法不正确的是()A.处于磁场中的圆盘部分,靠近圆心处电势高B.所加磁场越强越易使圆盘停止转动C.若所加磁场反向,圆盘将加速转动D.若所加磁场穿过整个圆盘,圆盘将匀速转动3.如图所示,在匀强磁场中,AB为长度为L粗细均匀的金属丝,输出电压恒定的电源接A、B两端时,金属丝受到的安培力为F;若将金属丝截取一半再弯成一个半圆形,仍然接在刚才的电源两端,则金属丝受到的安培力为( )A.B. C. F D.F4.如图所示,固定在竖直平面内的光滑圆环的最高点有一光滑的小孔.质量为m的小球套在圆环上,一根细线的下端系着小球,上端穿过小孔用手拉住,现拉动细线,使小球沿圆环缓慢上移.在移动过程中手对线的拉力F和圆环对小球的弹力N的大小变化情况是()A.F不变,N增大B.F减小,N不变C.F减小,N减小D.F增大,N不变5.两电荷量分别为q1和q2的点电荷放在x轴上的O、M两点,两电荷连线上各点电势φ随x变化的关系如图所示,其中A、N两点的电势均为零,ND段中的C 点电势最高,则()A.N点的电场强度大小为零B.q1小于q2C.NC间场强方向向x轴正方向D.将负点电荷从N点移到D点,电场力先做正功后做负功6.如图所示理想变压器原、副线圈的匝数比为5:1,原线圈接交流电源和交流电压表,副线圈接有“220V,440W"的热水器、“220V,220W"的抽油烟机.如果副线圈电压按图乙所示规律变化,则下列说法正确的是()A.副线圈两端电压的瞬时值为u=220 sinVB.交流电压表的示数为1100VC.1min内变压器输出的电能为3.96×104D.热水器的法人热功率是抽油烟机发热功率的2倍7.如图所示,轻质弹簧一端固定在水平面上O点的转轴上,另一端与一质量为m、套在粗糙固定直杆A 处的小球(可视为质点)相连,直杆的倾角为30°,OA=OC,B为AC的中点,OB等于弹簧原长.小球从A处由静止开始下滑,初始加速度大小为a A,第一次经过B处的速度大小为v,运动到C处速度为0,后又以大小为a C的初始加速度由静止开始向上滑行.设最大静摩擦力等于滑动摩擦力,重力加速度为g.下列说法正确的是()A.小球可以返回到出发点A处B.撤去弹簧,小球可以在直杆上处于静止C.弹簧具有的最大弹性势能为D.a A﹣a C=g8.如图所示,水平抛出的物体,抵达斜面上端P处时速度恰好沿着斜面方向,紧贴斜面PQ无摩擦滑下;如下四图图为物体沿x方向和y方向运动的位移﹣时间图象及速度﹣时间图象,其中可能正确的是()A.B.C.D.三、非选择题:包括必考题和选考题两部分.第22题~第32题为必考题,每个试题考生都必须作答.第33题~第40题为选考题,考生根据要求作答.(一)必考题(共129分)9.某同学用如图甲所示装置研究匀变速直线运动,得到如图乙所示的纸带,纸带上相邻计数点间还有4个点未画出.打点计时器交流电的频率为50Hz.(1)通过分析纸带数据,可判断物块在两相邻计数点和之间某时刻开始减速.(2)计数点5对应的速度大小为m/s,计数点6对应的速度大小为m/s(均保留三位有效数字).(3)物块减速运动过程中加速度的大小为a= m/s2(保留三位有效数字).10.要测量某种合金的电阻率.(1)若合金丝长度为L,直径为D,阻值为R,则其电阻率ρ=.用螺旋测微器测合金丝的直径如图甲所示,读数为mm.(2)图乙是测量合金丝阻值的原理图,S2是单刀双掷开关.根据原理图在图丙中将实物连线补充完整.(3)闭合S1,当S2处于位置a时,电压表和电流表的示数分别为U1=1.35V,I1=0.30A;当S2处于位置b时,电压表和电流表的示数分别为U2=0。
高三英语9月月考试题(无答案)

四川省成都市石室佳兴外国语学校2017届高三英语9月月考试题(无答案)第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. When will the two speakers attend the meeting?A. At 7:30 a.m.B. At 8:00 a.m.C. At 8:30 p.m.2. Where is Mary now?A. She is back home.B. She is coming here now.C. She is at the bus station.3. How many ships of China sailed through the waters of Okinawa?A. Six.B. Seven.C. Eight.4. What is the probable relationship between the two speakers?A. Salesman and customer.B. Husband and wife.C. Doctor and patient.5. Why were there so many people at Bill’s party?A. He has always been popular.B. He is popular with children.C. He had a surprising party.听第6段材料,回答第6至7题。
6. Where will the man spend his summer vacation?A. In the seaside.B. In the countryside.C. In his home.7. What will they do in the man’s hometown?A. To get around his village.B. To do some farming.C. To study in the lab.听第7段材料,回答第8至10题。
【月考试卷】四川省成都市龙泉驿区2017届高三12月月考英语试题 Word版含答案
成都龙泉中学高三(上)12月月考试题英语注意事项:1.答卷前,考生务必用2B铅笔在―考生号‖处填涂考生号。
用黑色字迹的钢笔或签字笔将自己所在的学校以及自己的姓名和考生号、试室号、座位号填写在答题卡上。
用2B铅笔将试卷类型(A)填涂在答题卡相应位置上。
2.选择题每小题选出答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
答案不能答在试卷上。
3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内的相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液。
不按以上要求作答的答案无效。
4.作答选做题时,请先用2B铅笔填涂选做题题组号对应的信息点,再作答。
漏涂、错涂、多涂的,答案无效。
5.考生必须保持答题卡的整洁。
考试结束后,收卷时只交答题卷。
第Ⅰ卷第一部分:英语听力(共两节,共20小题;每小题1.5分,满分30分。
)做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5 小题;每小题1.5 分,满分7.5分)听下面5 段对话。
每段对话后有一个小题,从题中所给的A、B、C 三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10 秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. Where does this conversation probably take place?A. In a bookstore.B. In a classroom.C. In a library.2. At what time will the film begin?A. 7:20.B. 7:15.C. 7:00.3. What are the two speakers mainly talking about?A. Their friend Jane.B. A weekend trip.C. A radio programme.4. What will the woman probably do?A. Catch a train.B. See the man off.C. Go shopping.5. Why did the woman apologize?A. She made a late delivery.B. She went to the wrong place.C. She couldn’t take the cake back.第二节(共15 小题;每小题1.5 分,满分22.5 分)听下面5 段对话。
四川省成都市龙泉实验中学2017-2018学年高三上学期9月月考数学试卷(文科) Word版含解析
2017-2018学年四川省成都市龙泉实验中学高三(上)9月月考数学试卷(文科)一、选择题:本大题共12小题,每小题5分,满分60分.1.设集合A={x∈N|≤2x≤16},B={x|y=ln(x2﹣3x)},则A∩B中元素的个数是()A.1 B.2 C.3 D.42.已知i为虚数单位,复数z满足iz=1+i,则=()A.1+i B.1﹣i C.﹣1+i D.﹣1﹣i3.有3个兴趣小组,甲、乙两位同学各自参加其中一个小组,每位同学参加各个小组的可能性相同,则这两位同学参加同一个兴趣小组的概率为()A.B.C.D.4.若偶函数f(x)在(﹣∞,0]上单调递减,a=f(log23),b=f(log45),c=f(2),则a,b,c满足()A.a<b<c B.b<a<c C.c<a<b D.c<b<a5.已知f(x)=3sinx﹣πx,命题p:∀x∈(0,),f(x)<0,则()A.p是假命题,¬p:∀x∈(0,),f(x)≥0B.p是假命题,¬p:∃x0∈(0,),f(x0)≥0C.p是真命题,¬p:∀x∈(0,),f(x)>0D.p是真命题,¬p:∃x0∈(0,),f(x0)≥06.一个几何体的三视图如图所示,则此几何体的体积是()A.112 B.80 C.72 D.647.已知S n是公差不为0的等差数列{a n}的前项和,且S1,S2,S4成等比数列,则=()A.4 B.6 C.8 D.108.已知函数f(x)=log2,若f(a)=,则f(﹣a)=()A.2 B.﹣2 C.D.﹣9.若某程序框图如图所示,则输出的p的值是()A.21 B.26 C.30 D.5510.已知双曲线C:﹣=1(a>0,b>0)的右焦点与抛物线y2=20x的焦点重合,且其渐近线方程为y=±x,则双曲线C的方程为()A.﹣=1 B.﹣=1C.﹣=1 D.﹣=111.在同一平面直角坐标系中,函数y=f(x)和y=g(x)的图象关于直线y=x对称.现将y=g(x)的图象沿x轴向左平移2个单位,再沿y轴向上平移1个单位,所得的图象是由两条线段组成的折线(如图所示),则函数f(x)的表达式为()A .f (x )=B .f (x )=C .f (x )=D .f (x )=12.某学校餐厅每天供应500名学生用餐,每星期一有A ,B 两种菜可供选择.调查资料表明,凡是在星期一选A 种菜的学生,下星期一会有20%改选B 种菜;而选B 种菜的学生,下星期一会有30%改选A 种菜.用a n ,b n 分别表示在第n 个星期的星期一选A 种菜和选B 种菜的学生人数,若a 1=300,则a n +1与a n 的关系可以表示为( )A .a n +1=+150B .a n +1=+200C .a n +1=+300D .a n +1=+180二.填空题:本大题共4小题,每小题5分.13.若存在b ∈[1,2],使得2b (b +a )≥4,则实数a 的取值范围是 .14.已知一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是 .15.已知平面直角坐标系内的两个向量, =(1,2),=(m ,3m ﹣2),且平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则m 的取值范围是 . 16.对于函数f (x )定义域中任意的x 1,x 2(x 1≠x 2),有如下结论: ①f (x 1+x 2)=f (x 1)•f (x 2); ②f (x 1•x 2)=f (x 1)+f (x 2); ③>0;④.当f (x )=lgx 时,上述结论中正确结论的序号是 .三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.已知函数f(x)和g(x)的图象关于原点对称,且f(x)=x2+2x.(Ⅰ)求函数g(x)的解析式;(Ⅱ)解不等式g(x)≥f(x)﹣|x﹣1|.18.袋中有大小、形状相同的红、黑球各一个,现一次有放回地随机摸取3次,每次摸取一个球(Ⅰ)试问:一共有多少种不同的结果?请列出所有可能的结果;(Ⅱ)若摸到红球时得2分,摸到黑球时得1分,求3次摸球所得总分为5的概率.19.已知函数f(x)=sinx﹣ax,g(x)=bxcosx(a∈R,b∈R).(1)讨论函数f(x)在区间(0,π)上的单调性;(2)若a=2b且a≥,当x>0时,证明f(x)<g(x).20.如图1,正方形ABCD的边长为,E、F分别是DC和BC的中点,H是正方形的对角线AC与EF的交点,N是正方形两对角线的交点,现沿EF将△CEF折起到△PEF的位置,使得PH⊥AH,连结PA,PB,PD(如图2).(Ⅰ)求证:BD⊥AP;(Ⅱ)求三棱锥A﹣BDP的高.21.已知函数f(x)=,g(x)=alnx﹣x(a≠0).(Ⅰ)求函数f(x)的单调区间;(Ⅱ)求证:当a>0时,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立.请考生在第22、23、24题中任选一题作答,如果多做,则按所做的第一题记分.作答时,用2B铅笔在答题卡上把所选题目对应的题号涂黑.[选修4-1:几何证明选讲]22.如图,已知AB是圆O的直径,C、D是圆O上的两个点,CE⊥AB于E,BD交AC 于G,交CE于F,CF=FG.(Ⅰ)求证:C是劣弧的中点;(Ⅱ)求证:BF=FG.[选修4-4:坐标系与参数方程选讲]23.在直角坐标系xOy中,直线l的参数方程为(t为参数),以原点为极点,x轴正半轴为极轴建立极坐标系,⊙C的极坐标方程为ρ=2sinθ.(Ⅰ)写出⊙C的直角坐标方程;(Ⅱ)P为直线l上一动点,当P到圆心C的距离最小时,求P的直角坐标.[选修4-5:不等式选讲]24.选修4﹣5:不等式选讲设函数f(x)=|2x﹣4|+|x+2|(Ⅰ)求函数y=f(x)的最小值;(Ⅱ)若不等式f(x)≥|a+4|﹣|a﹣3|恒成立,求a的取值范围.2016-2017学年四川省成都市龙泉实验中学高三(上)9月月考数学试卷(文科)参考答案与试题解析一、选择题:本大题共12小题,每小题5分,满分60分.1.设集合A={x∈N|≤2x≤16},B={x|y=ln(x2﹣3x)},则A∩B中元素的个数是()A.1 B.2 C.3 D.4【考点】交集及其运算.【分析】求出A中不等式的解集,确定出解集的自然数解确定A,求出B中x的范围确定出B,找出两集合的交集,即可作出判断.【解答】解:由A中不等式变形得:2﹣2≤2x≤24,即﹣2≤x≤4,x∈N,∴A={0,1,2,3,4},由B中y=ln(x2﹣3x),得到x2﹣3x>0,解得:x<0或x>3,即B={x|x<0或x>3},则A∩B={4},即A∩B中元素个数为1,故选:A.2.已知i为虚数单位,复数z满足iz=1+i,则=()A.1+i B.1﹣i C.﹣1+i D.﹣1﹣i【考点】复数代数形式的乘除运算.【分析】利用复数的运算法则与共轭复数的定义即可得出.【解答】解:∵iz=1+i,∴﹣i•iz=﹣i(1+i),化为z=1﹣i,∴=1+i.故选:A.3.有3个兴趣小组,甲、乙两位同学各自参加其中一个小组,每位同学参加各个小组的可能性相同,则这两位同学参加同一个兴趣小组的概率为()A.B.C.D.【考点】古典概型及其概率计算公式.【分析】本题是一个古典概型,试验发生包含的事件数是3×3种结果,满足条件的事件是这两位同学参加同一个兴趣小组有3种结果,根据古典概型概率公式得到结果.【解答】解:由题意知本题是一个古典概型,试验发生包含的事件数是3×3=9种结果,满足条件的事件是这两位同学参加同一个兴趣小组,由于共有三个小组,则有3种结果,根据古典概型概率公式得到P=,故选A.4.若偶函数f(x)在(﹣∞,0]上单调递减,a=f(log23),b=f(log45),c=f(2),则a,b,c满足()A.a<b<c B.b<a<c C.c<a<b D.c<b<a【考点】函数单调性的性质;对数值大小的比较.【分析】由偶函数f(x)在(﹣∞,0]上单调递减,可得f(x)在{0,+∞)上单调递增,比较三个自变量的大小,可得答案.【解答】解:∵偶函数f(x)在(﹣∞,0]上单调递减,∴f(x)在{0,+∞)上单调递增,∵2>log23=log49>log45,2>2,∴f(log45)<f(log23)<f(2),∴b<a<c,故选:B.5.已知f(x)=3sinx﹣πx,命题p:∀x∈(0,),f(x)<0,则()A.p是假命题,¬p:∀x∈(0,),f(x)≥0B.p是假命题,¬p:∃x0∈(0,),f(x0)≥0C.p是真命题,¬p:∀x∈(0,),f(x)>0D.p是真命题,¬p:∃x0∈(0,),f(x0)≥0【考点】复合命题的真假;命题的否定.【分析】由三角函数线的性质可知,当x∈(0,)时,sinx<x可判断p的真假,根据全称命题的否定为特称命题可知¬p.【解答】解:由三角函数线的性质可知,当x∈(0,)时,sinx<x∴3sinx<3x<πx∴f(x)=3sinx﹣πx<0即命题p:∀x∈(0,),f(x)<0为真命题根据全称命题的否定为特称命题可知¬p:∃x0∈(0,),f(x0)≥0故选D6.一个几何体的三视图如图所示,则此几何体的体积是()A.112 B.80 C.72 D.64【考点】由三视图求面积、体积.【分析】由三视图可知此几何体是由一个棱柱和一个棱锥构成的组合体,代入数据分别求棱柱与棱锥的体积即可.【解答】解:由三视图可知,此几何体是由一个棱柱和一个棱锥构成的组合体,棱柱的体积为4×4×4=64;棱锥的体积为×4×4×3=16;则此几何体的体积为80;故选B.7.已知S n是公差不为0的等差数列{a n}的前项和,且S1,S2,S4成等比数列,则=()A.4 B.6 C.8 D.10【考点】等比数列的性质;等差数列的前n项和.【分析】由等比中项的性质列出,再代入等差数列的通项公式和前n项和公式,用a1和d表示出来,求出a1和d的关系,进而求出式子的比值.【解答】解:设等差数列{a n}的公差为d,且d≠0,∵S1,S2,S4成等比数列,∴,∴=a1×,∴=2a1(2a1+3d),∴d2=2a1d,解得d=2a1或d=0(舍去),∴===8,故选C.8.已知函数f(x)=log2,若f(a)=,则f(﹣a)=()A.2 B.﹣2 C.D.﹣【考点】对数的运算性质;函数奇偶性的性质.【分析】先证明函数f(x)是奇函数,从而得到f(﹣a)=f(a),结合条件求得结果.【解答】解:∵已知函数f(x)=log2,∴f(﹣x)=log2=﹣=﹣f(x),故函数f(x)是奇函数,则f(﹣a)=﹣f(a)=﹣,故选D.9.若某程序框图如图所示,则输出的p的值是()A.21 B.26 C.30 D.55【考点】循环结构.【分析】先根据已知循环条件和循环体判定循环的次数,然后根据运行的后P的值找出规律,从而得出所求.【解答】解:根据题意可知该循环体运行3次第1次:n=2,p=1+22=5第2次:n=3,p=5+32=14,第3次:n=4,p=14+42=30因为P=30>20,结束循环,输出结果p=30.故选C.10.已知双曲线C:﹣=1(a>0,b>0)的右焦点与抛物线y2=20x的焦点重合,且其渐近线方程为y=±x,则双曲线C的方程为()A.﹣=1 B.﹣=1C.﹣=1 D.﹣=1【考点】双曲线的简单性质.【分析】求出抛物线的焦点坐标,根据双曲线的焦点坐标和抛物线的焦点关系,得到c=5,根据双曲线的渐近线方程得到=,联立方程组求出a,b即可.【解答】解:抛物线的焦点坐标为(5,0),双曲线焦点在x轴上,且c=5,∵又渐近线方程为y=±x,可得=,即b=a,则b2=a2=c2﹣a2=25﹣a2,则a2=9,b2=16,则双曲线C的方程为﹣=1,故选A11.在同一平面直角坐标系中,函数y=f(x)和y=g(x)的图象关于直线y=x对称.现将y=g(x)的图象沿x轴向左平移2个单位,再沿y轴向上平移1个单位,所得的图象是由两条线段组成的折线(如图所示),则函数f(x)的表达式为()A.f(x)=B.f(x)=C.f(x)=D.f(x)=【考点】函数的图象;分段函数的解析式求法及其图象的作法.【分析】要求f(x)的解析式,可先求g(x)的解析式,通过逆向平移画出g(x)的图象,写出g(x)解析式,根据对称求出f(x)解析式.【解答】解:由图可知g(x)=,则f(x)=,故选A.12.某学校餐厅每天供应500名学生用餐,每星期一有A ,B 两种菜可供选择.调查资料表明,凡是在星期一选A 种菜的学生,下星期一会有20%改选B 种菜;而选B 种菜的学生,下星期一会有30%改选A 种菜.用a n ,b n 分别表示在第n 个星期的星期一选A 种菜和选B 种菜的学生人数,若a 1=300,则a n +1与a n 的关系可以表示为( )A .a n +1=+150B .a n +1=+200C .a n +1=+300D .a n +1=+180【考点】数列递推式.【分析】由题意可得数列递推式,结合a n +b n =500,两式联立消去b n 得数列{a n }的递推公式.【解答】解:依题意得,消去b n 得:a n +1=a n +150.故选:A .二.填空题:本大题共4小题,每小题5分.13.若存在b ∈[1,2],使得2b (b +a )≥4,则实数a 的取值范围是 [﹣1,+∞) . 【考点】指数函数的定义、解析式、定义域和值域.【分析】由b ∈[1,2],知2b ∈[2,4],,由2b (b +a )≥4,能求出实数a的取值范围.【解答】解:∵b ∈[1,2],∴2b ∈[2,4],∴,∵2b (b +a )≥4,∴a ≥≥﹣1.设函数f (b )=﹣b ,b ∈[1,2].函数f (b )是区间[1,2]上的减函数.故f (b )∈[f (2),f (1)]=[﹣1,1].原题目可转化为实数a 不小于函数f (b )的最小值﹣1即可. ∴实数a 的取值范围是[﹣1,+∞). 故答案为:[﹣1,+∞).14.已知一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是 12 .【考点】棱柱、棱锥、棱台的侧面积和表面积.【分析】求出球的半径,然后求解棱柱的底面边长与高,即可求解侧面积.【解答】解:球的体积为:,可得=,r=1,棱柱的高为:2,底面正三角形的内切圆的半径为:1,底面边长为:2=2,一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是:6×2=12.15.已知平面直角坐标系内的两个向量,=(1,2),=(m,3m﹣2),且平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则m的取值范围是(﹣∞,2)∪(2,+∞).【考点】平面向量的坐标运算.【分析】平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则=(1,2),=(m,3m﹣2)为基底,由基底的条件即可解出m.【解答】解:∵平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则=(1,2),=(m,3m﹣2)为基底,即基底不共线.∴1×(3m﹣2)﹣2×m≠0,∴m≠2.故答案为:(﹣∞,2)∪(2,+∞).16.对于函数f(x)定义域中任意的x1,x2(x1≠x2),有如下结论:①f(x1+x2)=f(x1)•f(x2);②f(x1•x2)=f(x1)+f(x2);③>0;④.当f(x)=lgx时,上述结论中正确结论的序号是②③.【考点】函数单调性的判断与证明;函数奇偶性的判断.【分析】利用对数的基本运算性质进行检验:①f(x1+x2)=lg(x1+x2)≠f(x1)f(x2)=lgx1•lgx2,②f(x1•x2)=lgx1x2=lgx1+lgx2=f(x1)+f(x2)③f(x)=lgx在(0,+∞)单调递增,可得④,=,由基本不等式可得从而可得【解答】解:①f(x1+x2)=lg(x1+x2)≠f(x1)f(x2)=lgx1•lgx2②f(x1•x2)=lgx1x2=lgx1+lgx2=f(x1)+f(x2)③f(x)=lgx在(0,+∞)单调递增,则对任意的0<x1<x2,d都有f(x1)<f(x2)即④,=∵∴故答案为:②③三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.已知函数f(x)和g(x)的图象关于原点对称,且f(x)=x2+2x.(Ⅰ)求函数g(x)的解析式;(Ⅱ)解不等式g(x)≥f(x)﹣|x﹣1|.【考点】绝对值不等式的解法;函数解析式的求解及常用方法.【分析】(Ⅰ)设函数y=f(x)的图象上任意一点Q(x0,y0)关于原点的对称点为P(x,y),则P在g(x)的图象上,由线段的中点公式解出x0和y0 的解析式,代入函数y=f(x)可得g(x)的解析式.(Ⅱ)不等式可化为2x2﹣|x﹣1|≤0,分类讨论,去掉绝对值,求出不等式的解集.【解答】解:(Ⅰ)设函数y=f(x)的图象上任意一点Q(x0,y0)关于原点的对称点为P (x,y),则P在g(x)的图象上,且,即∵点Q(x0,y0)在函数y=f(x)的图象上,∴﹣y=x2﹣2x,即y=﹣x2+2x,故,g(x)=﹣x2+2x.(Ⅱ)由g(x)≥f(x)﹣|x﹣1|,可得2x2﹣|x﹣1|≤0当x≥1时,2x2﹣x+1≤0,此时不等式无解.当x<1时,2x2+x﹣1≤0,解得﹣1≤x≤.因此,原不等式的解集为[﹣1,].18.袋中有大小、形状相同的红、黑球各一个,现一次有放回地随机摸取3次,每次摸取一个球(Ⅰ)试问:一共有多少种不同的结果?请列出所有可能的结果;(Ⅱ)若摸到红球时得2分,摸到黑球时得1分,求3次摸球所得总分为5的概率.【考点】等可能事件的概率;随机事件.【分析】(1)由分步计数原理知这个过程一共有8个结果,按照一定的顺序列举出所有的事件,顺序可以是按照红球的个数由多变少变化,这样可以做到不重不漏.(2)本题是一个等可能事件的概率,由前面可知试验发生的所有事件数,而满足条件的事件包含的基本事件为:(红、红、黑)、(红、黑、红)、(黑、红、红),根据古典概型公式得到结果.【解答】解:(I)一共有8种不同的结果,列举如下:(红、红、红、)、(红、红、黑)、(红、黑、红)、(红、黑、黑)、(黑、红、红)、(黑、红、黑)、(黑、黑、红)、(黑、黑、黑)(Ⅱ)本题是一个等可能事件的概率记“3次摸球所得总分为5”为事件A事件A包含的基本事件为:(红、红、黑)、(红、黑、红)、(黑、红、红)事件A包含的基本事件数为3由(I)可知,基本事件总数为8,∴事件A的概率为19.已知函数f(x)=sinx﹣ax,g(x)=bxcosx(a∈R,b∈R).(1)讨论函数f(x)在区间(0,π)上的单调性;(2)若a=2b且a≥,当x>0时,证明f(x)<g(x).【考点】导数在最大值、最小值问题中的应用;三角函数的最值.【分析】(1)求出函数的导数f'(x)=cosx﹣a通过余弦函数的值域,讨论a与[﹣1,1]的范围,判断导数的符号,然后得到函数的单调性.(2)用分析法证明f(x)<g(x),转化为证明,构造函数M(x)=,通过求解函数的导数,求出函数的最值,然后证明即可.【解答】(本小题13分)解:(1)f(x)=sinx﹣ax,则f'(x)=cosx﹣a…当a≥1时,f'(x)<0,所以函数f(x)在区间(0,π)上单调递减…当a≤﹣1时,f'(x)>0,所以函数f(x)在区间(0,π)上单调递增…当﹣1<a<1时,存在ϕ∈(0,π),使得cosϕ=a,即f'(ϕ)=0,x∈(0,ϕ)时,f'(x)>0,所以函数f(x)在区间(0,ϕ)上单调递增,x∈(ϕ,π)时,f'(x)<0,所以函数f(x)在区间(ϕ,π)上单调递减…(2)要证明f(x)<g(x),只须证明f(x)﹣g(x)<0当a=2b时,…等价于…记M(x)=,则…M'(x)==…当,即时,M'(x)≤0,M(x)在区间上(0,+∞)单调递减,M(x)<M(0)=0所以,当x>0,f(x)<g(x)恒成立.…20.如图1,正方形ABCD的边长为,E、F分别是DC和BC的中点,H是正方形的对角线AC与EF的交点,N是正方形两对角线的交点,现沿EF将△CEF折起到△PEF的位置,使得PH⊥AH,连结PA,PB,PD(如图2).(Ⅰ)求证:BD⊥AP;(Ⅱ)求三棱锥A ﹣BDP 的高.【考点】棱柱、棱锥、棱台的体积;空间中直线与直线之间的位置关系. 【分析】(1)由PH ⊥AH ,PH ⊥EF 可得PH ⊥平面ABCD ,故PH ⊥BD ,又AC ⊥BD ,得出BD ⊥平面PAH ,得出BD ;(2)分别把△ABD 和△BDP 当做底面求出棱锥的体积,列出方程解出. 【解答】(Ⅰ)证明:∵E 、F 分别是CD 和BC 的中点,∴EF ∥BD . 又∵AC ⊥BD ,∴AC ⊥EF ,故折起后有PH ⊥EF .又∵PH ⊥AH ,∴PH ⊥平面ABFED . 又∵BD ⊂平面ABFED ,∴PH ⊥BD , ∵AH ∩PH=H ,AH ,PH ⊂平面APH ,∴BD ⊥平面APH ,又∵AP ⊂平面APH ,∴BD ⊥AP (Ⅱ)解:∵正方形ABCD 的边长为, ∴AC=BD=4,AN=2,NH=PH=1,PE=PF∴△PBD 是等腰三角形,连结PN ,则PN ⊥BD ,∴△PBD 的面积设三棱锥A ﹣BDP 的高为h ,则三棱锥A ﹣BDP 的体积为由(Ⅰ)可知PH 是三棱锥P ﹣ABD 的高,∴三棱锥P ﹣ABD 的体积:∵V A ﹣BDP =V P ﹣ABD ,即,解得,即三棱锥A ﹣BDP 的高为.21.已知函数f (x )=,g (x )=alnx ﹣x (a ≠0).(Ⅰ)求函数f (x )的单调区间;(Ⅱ)求证:当a >0时,对于任意x 1,x 2∈(0,e ],总有g (x 1)<f (x 2)成立. 【考点】利用导数研究函数的单调性;函数单调性的性质.【分析】(I)先求函数f(x)的导数,再对字母a进行分类讨论,根据导数大于0函数单调递增,导数小于0时函数单调递减可得答案.(Ⅱ)欲证当a>0时,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立,只须证明对于任意x1,x2∈(0,e],总有g(x)max<f(x)min.由(Ⅰ)可知,当a>0时,f(x)在(0,1)上单调递增,f(x)在(1,e]上单调递减,从而有f(x)min=a,同样地利用导数可得,当a>0时,g(x)在(0,a)上单调递增,g(x)在(a,e]上单调递减,从而g (x)max=g(a)=alna﹣a,最后利用作差法即可得到g(x)max<f(x)min.【解答】解:(Ⅰ)函数f(x)的定义域为R,.当a>0时,x f'x f xx f'x f x当a>0时,f(x)的单调递增区间为(﹣1,1),单调递减区间为(﹣∞,﹣1),(1,+∞);当a<0时,f(x)的单调递增区间为(﹣∞,﹣1),(1,+∞),单调递减区间为(﹣1,1).…(Ⅱ)由(Ⅰ)可知,当a>0时,f(x)在(0,1)上单调递增,f(x)在(1,e]上单调递减,又f(0)=a,f(e)=所以f(x)min=a,同样地,当a>0时,g(x)在(0,a)上单调递增,g(x)在(a,e]上单调递减,所以g(x)max=g(a)=alna﹣a,因为a﹣(alna﹣a)=a(2﹣lna)>a(2﹣lne)=a>0,所以对于任意x1,x2∈(0,e],总有g(x)max=g(e)=alna﹣a<a=f(x)min.所以对于任意x1,x2∈(0,e],仍有x1,x2∈(0,e].综上所述,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立.…请考生在第22、23、24题中任选一题作答,如果多做,则按所做的第一题记分.作答时,用2B铅笔在答题卡上把所选题目对应的题号涂黑.[选修4-1:几何证明选讲]22.如图,已知AB是圆O的直径,C、D是圆O上的两个点,CE⊥AB于E,BD交AC 于G,交CE于F,CF=FG.(Ⅰ)求证:C是劣弧的中点;(Ⅱ)求证:BF=FG.【考点】与圆有关的比例线段.【分析】(I)要证明C是劣弧BD的中点,即证明弧BC与弧CD相等,即证明∠CAB=∠DAC,根据已知中CF=FG,AB是圆O的直径,CE⊥AB于E,我们易根据同角的余角相等,得到结论.(II)由已知及(I)的结论,我们易证明△BFC及△GFC均为等腰三角形,即CF=BF,CF=GF,进而得到结论.【解答】解:(I)∵CF=FG∴∠CGF=∠FCG∵AB圆O的直径∴∵CE⊥AB∴∵∴∠CBA=∠ACE∵∠CGF=∠DGA∴∴∠CAB=∠DAC∴C为劣弧BD的中点(II)∵∴∠GBC=∠FCB∴CF=FB又因为CF=GF∴BF=FG[选修4-4:坐标系与参数方程选讲]23.在直角坐标系xOy中,直线l的参数方程为(t为参数),以原点为极点,x轴正半轴为极轴建立极坐标系,⊙C的极坐标方程为ρ=2sinθ.(Ⅰ)写出⊙C的直角坐标方程;(Ⅱ)P为直线l上一动点,当P到圆心C的距离最小时,求P的直角坐标.【考点】点的极坐标和直角坐标的互化.【分析】(I)由⊙C的极坐标方程为ρ=2sinθ.化为ρ2=2,把代入即可得出;.(II)设P,又C.利用两点之间的距离公式可得|PC|=,再利用二次函数的性质即可得出.【解答】解:(I)由⊙C的极坐标方程为ρ=2sinθ.∴ρ2=2,化为x2+y2=,配方为=3.(II)设P,又C.∴|PC|==≥2,因此当t=0时,|PC|取得最小值2.此时P(3,0).[选修4-5:不等式选讲]24.选修4﹣5:不等式选讲设函数f(x)=|2x﹣4|+|x+2|(Ⅰ)求函数y=f (x )的最小值;(Ⅱ)若不等式f (x )≥|a +4|﹣|a ﹣3|恒成立,求a 的取值范围. 【考点】绝对值不等式的解法;函数恒成立问题.【分析】(Ⅰ)去绝对值可得f (x )=,分段求最值可得;(Ⅱ)问题等价于|a +4|﹣|a ﹣3|≤f (x )min =4,解之可得.【解答】解:(Ⅰ)由于f (x )=|2x ﹣4|+|x +2|=可得当x <﹣2时,﹣3x +2>8,当﹣2≤x <2时,4<6﹣x ≤8, 当x ≥2时,3x ﹣2≥4,所以函数的最小值为f (2)=4.(Ⅱ)若不等式f (x )≥|a +4|﹣|a ﹣3|恒成立,则|a +4|﹣|a ﹣3|≤f (x )min =4,又解不等式|a +4|﹣|a ﹣3|≤4可解得a ≤.所以a 的取值范围为a ≤2016年12月10日。
四川省成都龙泉实验中学2017届高三上学期9月月考理科综合试题(含解答) 含答案
成都龙泉实验中学2016—2017学年度高三9月月考理科综合能力测试试题1、本试卷分第I卷和第II卷两部分.满分300分。
考试时间150分钟。
2、以下数据可供解题时参考:可能用到的相对原子质量:H 1 C 12 O 16 Mg 24 Al 27 K 39 Ca 40第I卷选择题本题1-18题为单项选择题,19—21为多项选择题。
在每小题列出的A、B、C和D四个选项中.选出符合题目要求的所有项。
一、选择题:本大题共13小题,每小题6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。
1.下列有关实验或调查的叙述,不正确的是A.研究遗传病发病率需在人群中随机抽样调查,研究遗传方式需分析患者家系系谱图B.统计显微镜下各期细胞数占计数细胞总数的比例,能比较细胞周期各期时间的长短C.选取经低温诱导的洋葱根尖制成的临时装片,在显微镜下观察不到联会现象D.加入无水乙醇越多,叶绿体色素提取液的绿色越深2.生物学中的实验部分是学好生物这门学科必不可少的。
下列叙述正确的是A.恩格尔曼的水绵实验中好氧细菌的作用是确定水绵光合作用释放氧气的部位B.模拟细胞大小与物质运输的关系实验表明细胞体积越大,物质运输效率越高C.绿叶中色素的分离原理是色素能溶解在无水乙醇中,且不同色素溶解度不同D.探究培养液中酵母菌种群数量随时间变化的实验中需要另设置对照实验3。
下图甲表示温度对淀粉酶活性的影响;下图乙是将一定量的淀粉酶和足量的淀粉混合后麦芽糖积累量随温度变化的情况。
下列说法不正确的是A.To表示淀粉酶催化该反应的最适温度B.图甲中,Ta、Tb时淀粉酶催化效率都很低,但对酶活性的影响却有区别C.图乙中Tb麦芽糖的积累量不再上升,是因为淀粉已完全水解D.图乙中温度To对应曲线上的的点最可能是4.下列对有关鉴定性实验、观察类实验、探究性实验以及调查类实验的叙述,正确的是A.鉴定组织中还原糖实验时,可依次向待测样液中加入甲液和乙液B.在“低温诱导染色体加倍”实验中,可以设置不同梯度的低温进行实验,确定染色体加倍的适宜温度C.设计探究实验时,自变量的设置应考虑周全,而无关变量可忽略不计D.调查某荒地内蒲公英的种群密度时,所选择的样方数不影响调查结果5.某二倍体生物(染色体数为2N)染色体上的DNA用3H充分标记,置于不含3H的培养基中培养,其有丝分裂过程中细胞局部化过程如图所示,有关叙述错误的是A.图A中DNA复制的酶一定是从1运进的,与3直接相连的膜性细胞器是内质网B.图B中DNA数:染色单体数:染色体数:同源染色体对数=4:4:2:1C.图C表示细胞处于有丝分裂的末期,其判断依据是核膜重新构建D.经过连续两次细胞有丝分裂,子细胞含3H的染色体数一定为N6.如图表示淋巴细胞起源和分化过程(其中a、b、c、d表示不同种类的细胞,①、②表示有关过程),下列有关叙述正确的是A。
四川省成都市龙泉实验中学2017届高三上学期9月月考数学试卷(文科)Word版含解析
2016-2017学年四川省成都市龙泉实验中学高三(上)9月月考数学试卷(文科)一、选择题:本大题共12小题,每小题5分,满分60分.1.设集合A={x∈N|≤2x≤16},B={x|y=ln(x2﹣3x)},则A∩B中元素的个数是()A.1 B.2 C.3 D.42.已知i为虚数单位,复数z满足iz=1+i,则=()A.1+i B.1﹣i C.﹣1+i D.﹣1﹣i3.有3个兴趣小组,甲、乙两位同学各自参加其中一个小组,每位同学参加各个小组的可能性相同,则这两位同学参加同一个兴趣小组的概率为()A.B.C.D.4.若偶函数f(x)在(﹣∞,0]上单调递减,a=f(log23),b=f(log45),c=f(2),则a,b,c满足()A.a<b<c B.b<a<c C.c<a<b D.c<b<a5.已知f(x)=3sinx﹣πx,命题p:∀x∈(0,),f(x)<0,则()A.p是假命题,¬p:∀x∈(0,),f(x)≥0B.p是假命题,¬p:∃x0∈(0,),f(x0)≥0C.p是真命题,¬p:∀x∈(0,),f(x)>0D.p是真命题,¬p:∃x0∈(0,),f(x0)≥06.一个几何体的三视图如图所示,则此几何体的体积是()A.112 B.80 C.72 D.647.已知S n是公差不为0的等差数列{a n}的前项和,且S1,S2,S4成等比数列,则=()A.4 B.6 C.8 D.108.已知函数f(x)=log2,若f(a)=,则f(﹣a)=()A.2 B.﹣2 C.D.﹣9.若某程序框图如图所示,则输出的p的值是()A.21 B.26 C.30 D.5510.已知双曲线C:﹣=1(a>0,b>0)的右焦点与抛物线y2=20x的焦点重合,且其渐近线方程为y=±x,则双曲线C的方程为()A.﹣=1 B.﹣=1C.﹣=1 D.﹣=111.在同一平面直角坐标系中,函数y=f(x)和y=g(x)的图象关于直线y=x对称.现将y=g(x)的图象沿x轴向左平移2个单位,再沿y轴向上平移1个单位,所得的图象是由两条线段组成的折线(如图所示),则函数f(x)的表达式为()A .f (x )=B .f (x )=C .f (x )=D .f (x )=12.某学校餐厅每天供应500名学生用餐,每星期一有A ,B 两种菜可供选择.调查资料表明,凡是在星期一选A 种菜的学生,下星期一会有20%改选B 种菜;而选B 种菜的学生,下星期一会有30%改选A 种菜.用a n ,b n 分别表示在第n 个星期的星期一选A 种菜和选B 种菜的学生人数,若a 1=300,则a n +1与a n 的关系可以表示为( )A .a n +1=+150 B .a n +1=+200C .a n +1=+300D .a n +1=+180二.填空题:本大题共4小题,每小题5分.13.若存在b ∈[1,2],使得2b (b +a )≥4,则实数a 的取值范围是 .14.已知一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是 .15.已知平面直角坐标系内的两个向量, =(1,2),=(m ,3m ﹣2),且平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则m 的取值范围是 . 16.对于函数f (x )定义域中任意的x 1,x 2(x 1≠x 2),有如下结论:①f (x 1+x 2)=f (x 1)•f (x 2);②f (x 1•x 2)=f (x 1)+f (x 2);③>0;④.当f (x )=lgx 时,上述结论中正确结论的序号是 .三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.已知函数f(x)和g(x)的图象关于原点对称,且f(x)=x2+2x.(Ⅰ)求函数g(x)的解析式;(Ⅱ)解不等式g(x)≥f(x)﹣|x﹣1|.18.袋中有大小、形状相同的红、黑球各一个,现一次有放回地随机摸取3次,每次摸取一个球(Ⅰ)试问:一共有多少种不同的结果?请列出所有可能的结果;(Ⅱ)若摸到红球时得2分,摸到黑球时得1分,求3次摸球所得总分为5的概率.19.已知函数f(x)=sinx﹣ax,g(x)=bxcosx(a∈R,b∈R).(1)讨论函数f(x)在区间(0,π)上的单调性;(2)若a=2b且a≥,当x>0时,证明f(x)<g(x).20.如图1,正方形ABCD的边长为,E、F分别是DC和BC的中点,H是正方形的对角线AC与EF的交点,N是正方形两对角线的交点,现沿EF将△CEF折起到△PEF的位置,使得PH⊥AH,连结PA,PB,PD(如图2).(Ⅰ)求证:BD⊥AP;(Ⅱ)求三棱锥A﹣BDP的高.21.已知函数f(x)=,g(x)=alnx﹣x(a≠0).(Ⅰ)求函数f(x)的单调区间;(Ⅱ)求证:当a>0时,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立.请考生在第22、23、24题中任选一题作答,如果多做,则按所做的第一题记分.作答时,用2B铅笔在答题卡上把所选题目对应的题号涂黑.[选修4-1:几何证明选讲]22.如图,已知AB是圆O的直径,C、D是圆O上的两个点,CE⊥AB于E,BD交AC 于G,交CE于F,CF=FG.(Ⅰ)求证:C是劣弧的中点;(Ⅱ)求证:BF=FG.[选修4-4:坐标系与参数方程选讲]23.在直角坐标系xOy中,直线l的参数方程为(t为参数),以原点为极点,x轴正半轴为极轴建立极坐标系,⊙C的极坐标方程为ρ=2sinθ.(Ⅰ)写出⊙C的直角坐标方程;(Ⅱ)P为直线l上一动点,当P到圆心C的距离最小时,求P的直角坐标.[选修4-5:不等式选讲]24.选修4﹣5:不等式选讲设函数f(x)=|2x﹣4|+|x+2|(Ⅰ)求函数y=f(x)的最小值;(Ⅱ)若不等式f(x)≥|a+4|﹣|a﹣3|恒成立,求a的取值范围.2016-2017学年四川省成都市龙泉实验中学高三(上)9月月考数学试卷(文科)参考答案与试题解析一、选择题:本大题共12小题,每小题5分,满分60分.1.设集合A={x∈N|≤2x≤16},B={x|y=ln(x2﹣3x)},则A∩B中元素的个数是()A.1 B.2 C.3 D.4【考点】交集及其运算.【分析】求出A中不等式的解集,确定出解集的自然数解确定A,求出B中x的范围确定出B,找出两集合的交集,即可作出判断.【解答】解:由A中不等式变形得:2﹣2≤2x≤24,即﹣2≤x≤4,x∈N,∴A={0,1,2,3,4},由B中y=ln(x2﹣3x),得到x2﹣3x>0,解得:x<0或x>3,即B={x|x<0或x>3},则A∩B={4},即A∩B中元素个数为1,故选:A.2.已知i为虚数单位,复数z满足iz=1+i,则=()A.1+i B.1﹣i C.﹣1+i D.﹣1﹣i【考点】复数代数形式的乘除运算.【分析】利用复数的运算法则与共轭复数的定义即可得出.【解答】解:∵iz=1+i,∴﹣i•iz=﹣i(1+i),化为z=1﹣i,∴=1+i.故选:A.3.有3个兴趣小组,甲、乙两位同学各自参加其中一个小组,每位同学参加各个小组的可能性相同,则这两位同学参加同一个兴趣小组的概率为()A.B.C.D.【考点】古典概型及其概率计算公式.【分析】本题是一个古典概型,试验发生包含的事件数是3×3种结果,满足条件的事件是这两位同学参加同一个兴趣小组有3种结果,根据古典概型概率公式得到结果.【解答】解:由题意知本题是一个古典概型,试验发生包含的事件数是3×3=9种结果,满足条件的事件是这两位同学参加同一个兴趣小组,由于共有三个小组,则有3种结果,根据古典概型概率公式得到P=,故选A.4.若偶函数f(x)在(﹣∞,0]上单调递减,a=f(log23),b=f(log45),c=f(2),则a,b,c满足()A.a<b<c B.b<a<c C.c<a<b D.c<b<a【考点】函数单调性的性质;对数值大小的比较.【分析】由偶函数f(x)在(﹣∞,0]上单调递减,可得f(x)在{0,+∞)上单调递增,比较三个自变量的大小,可得答案.【解答】解:∵偶函数f(x)在(﹣∞,0]上单调递减,∴f(x)在{0,+∞)上单调递增,∵2>log23=log49>log45,2>2,∴f(log45)<f(log23)<f(2),∴b<a<c,故选:B.5.已知f(x)=3sinx﹣πx,命题p:∀x∈(0,),f(x)<0,则()A.p是假命题,¬p:∀x∈(0,),f(x)≥0B.p是假命题,¬p:∃x0∈(0,),f(x0)≥0C.p是真命题,¬p:∀x∈(0,),f(x)>0D.p是真命题,¬p:∃x0∈(0,),f(x0)≥0【考点】复合命题的真假;命题的否定.【分析】由三角函数线的性质可知,当x∈(0,)时,sinx<x可判断p的真假,根据全称命题的否定为特称命题可知¬p.【解答】解:由三角函数线的性质可知,当x∈(0,)时,sinx<x∴3sinx<3x<πx∴f(x)=3sinx﹣πx<0即命题p:∀x∈(0,),f(x)<0为真命题根据全称命题的否定为特称命题可知¬p:∃x0∈(0,),f(x0)≥0故选D6.一个几何体的三视图如图所示,则此几何体的体积是()A.112 B.80 C.72 D.64【考点】由三视图求面积、体积.【分析】由三视图可知此几何体是由一个棱柱和一个棱锥构成的组合体,代入数据分别求棱柱与棱锥的体积即可.【解答】解:由三视图可知,此几何体是由一个棱柱和一个棱锥构成的组合体,棱柱的体积为4×4×4=64;棱锥的体积为×4×4×3=16;则此几何体的体积为80;故选B.7.已知S n是公差不为0的等差数列{a n}的前项和,且S1,S2,S4成等比数列,则=()A.4 B.6 C.8 D.10【考点】等比数列的性质;等差数列的前n项和.【分析】由等比中项的性质列出,再代入等差数列的通项公式和前n项和公式,用a1和d表示出来,求出a1和d的关系,进而求出式子的比值.【解答】解:设等差数列{a n}的公差为d,且d≠0,∵S1,S2,S4成等比数列,∴,∴=a1×,∴=2a1(2a1+3d),∴d2=2a1d,解得d=2a1或d=0(舍去),∴===8,故选C.8.已知函数f(x)=log2,若f(a)=,则f(﹣a)=()A.2 B.﹣2 C.D.﹣【考点】对数的运算性质;函数奇偶性的性质.【分析】先证明函数f(x)是奇函数,从而得到f(﹣a)=f(a),结合条件求得结果.【解答】解:∵已知函数f(x)=log2,∴f(﹣x)=log2=﹣=﹣f(x),故函数f(x)是奇函数,则f(﹣a)=﹣f(a)=﹣,故选D.9.若某程序框图如图所示,则输出的p的值是()A.21 B.26 C.30 D.55【考点】循环结构.【分析】先根据已知循环条件和循环体判定循环的次数,然后根据运行的后P的值找出规律,从而得出所求.【解答】解:根据题意可知该循环体运行3次第1次:n=2,p=1+22=5第2次:n=3,p=5+32=14,第3次:n=4,p=14+42=30因为P=30>20,结束循环,输出结果p=30.故选C.10.已知双曲线C:﹣=1(a>0,b>0)的右焦点与抛物线y2=20x的焦点重合,且其渐近线方程为y=±x,则双曲线C的方程为()A.﹣=1 B.﹣=1C.﹣=1 D.﹣=1【考点】双曲线的简单性质.【分析】求出抛物线的焦点坐标,根据双曲线的焦点坐标和抛物线的焦点关系,得到c=5,根据双曲线的渐近线方程得到=,联立方程组求出a,b即可.【解答】解:抛物线的焦点坐标为(5,0),双曲线焦点在x轴上,且c=5,∵又渐近线方程为y=±x,可得=,即b=a,则b2=a2=c2﹣a2=25﹣a2,则a2=9,b2=16,则双曲线C的方程为﹣=1,故选A11.在同一平面直角坐标系中,函数y=f(x)和y=g(x)的图象关于直线y=x对称.现将y=g(x)的图象沿x轴向左平移2个单位,再沿y轴向上平移1个单位,所得的图象是由两条线段组成的折线(如图所示),则函数f(x)的表达式为()A.f(x)=B.f(x)=C.f(x)=D.f(x)=【考点】函数的图象;分段函数的解析式求法及其图象的作法.【分析】要求f(x)的解析式,可先求g(x)的解析式,通过逆向平移画出g(x)的图象,写出g(x)解析式,根据对称求出f(x)解析式.【解答】解:由图可知g(x)=,则f(x)=,故选A.12.某学校餐厅每天供应500名学生用餐,每星期一有A ,B 两种菜可供选择.调查资料表明,凡是在星期一选A 种菜的学生,下星期一会有20%改选B 种菜;而选B 种菜的学生,下星期一会有30%改选A 种菜.用a n ,b n 分别表示在第n 个星期的星期一选A 种菜和选B 种菜的学生人数,若a 1=300,则a n +1与a n 的关系可以表示为( )A .a n +1=+150B .a n +1=+200C .a n +1=+300D .a n +1=+180【考点】数列递推式.【分析】由题意可得数列递推式,结合a n +b n =500,两式联立消去b n 得数列{a n }的递推公式.【解答】解:依题意得,消去b n 得:a n +1=a n +150.故选:A .二.填空题:本大题共4小题,每小题5分.13.若存在b ∈[1,2],使得2b (b +a )≥4,则实数a 的取值范围是 [﹣1,+∞) . 【考点】指数函数的定义、解析式、定义域和值域.【分析】由b ∈[1,2],知2b ∈[2,4],,由2b (b +a )≥4,能求出实数a的取值范围.【解答】解:∵b ∈[1,2],∴2b ∈[2,4],∴,∵2b (b +a )≥4,∴a ≥≥﹣1.设函数f (b )=﹣b ,b ∈[1,2].函数f (b )是区间[1,2]上的减函数.故f (b )∈[f (2),f (1)]=[﹣1,1].原题目可转化为实数a 不小于函数f (b )的最小值﹣1即可. ∴实数a 的取值范围是[﹣1,+∞). 故答案为:[﹣1,+∞).14.已知一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是 12 .【考点】棱柱、棱锥、棱台的侧面积和表面积.【分析】求出球的半径,然后求解棱柱的底面边长与高,即可求解侧面积.【解答】解:球的体积为:,可得=,r=1,棱柱的高为:2,底面正三角形的内切圆的半径为:1,底面边长为:2=2,一个三棱柱,其底面是正三角形,且侧棱与底面垂直,一个体积为的球与该棱柱的所有面均相切,那么这个三棱柱的侧面积是:6×2=12.15.已知平面直角坐标系内的两个向量,=(1,2),=(m,3m﹣2),且平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则m的取值范围是(﹣∞,2)∪(2,+∞).【考点】平面向量的坐标运算.【分析】平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则=(1,2),=(m,3m﹣2)为基底,由基底的条件即可解出m.【解答】解:∵平面内的任一向量都可以唯一的表示成=λ+(λ,μ为实数),则=(1,2),=(m,3m﹣2)为基底,即基底不共线.∴1×(3m﹣2)﹣2×m≠0,∴m≠2.故答案为:(﹣∞,2)∪(2,+∞).16.对于函数f(x)定义域中任意的x1,x2(x1≠x2),有如下结论:①f(x1+x2)=f(x1)•f(x2);②f(x1•x2)=f(x1)+f(x2);③>0;④.当f(x)=lgx时,上述结论中正确结论的序号是②③.【考点】函数单调性的判断与证明;函数奇偶性的判断.【分析】利用对数的基本运算性质进行检验:①f(x1+x2)=lg(x1+x2)≠f(x1)f(x2)=lgx1•lgx2,②f(x1•x2)=lgx1x2=lgx1+lgx2=f(x1)+f(x2)③f(x)=lgx在(0,+∞)单调递增,可得④,=,由基本不等式可得从而可得【解答】解:①f(x1+x2)=lg(x1+x2)≠f(x1)f(x2)=lgx1•lgx2②f(x1•x2)=lgx1x2=lgx1+lgx2=f(x1)+f(x2)③f(x)=lgx在(0,+∞)单调递增,则对任意的0<x1<x2,d都有f(x1)<f(x2)即④,=∵∴故答案为:②③三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤. 17.已知函数f (x )和g (x )的图象关于原点对称,且f (x )=x 2+2x . (Ⅰ)求函数g (x )的解析式;(Ⅱ)解不等式g (x )≥f (x )﹣|x ﹣1|.【考点】绝对值不等式的解法;函数解析式的求解及常用方法. 【分析】(Ⅰ)设函数y=f (x )的图象上任意一点Q (x 0,y 0)关于原点的对称点为P (x ,y ),则P 在g (x )的图象上,由线段的中点公式解出 x 0和y 0 的解析式,代入函数y=f (x )可得g (x )的解析式.(Ⅱ)不等式可化为 2x 2﹣|x ﹣1|≤0,分类讨论,去掉绝对值,求出不等式的解集. 【解答】解:(Ⅰ)设函数y=f (x )的图象上任意一点Q (x 0,y 0)关于原点的对称点为P (x ,y ),则P 在g (x )的图象上,且,即∵点Q (x 0,y 0)在函数y=f (x )的图象上,∴﹣y=x 2﹣2x ,即y=﹣x 2+2x ,故,g (x )=﹣x 2+2x . (Ⅱ)由g (x )≥f (x )﹣|x ﹣1|,可得2x 2﹣|x ﹣1|≤0 当x ≥1时,2x 2﹣x +1≤0,此时不等式无解.当x <1时,2x 2+x ﹣1≤0,解得﹣1≤x ≤.因此,原不等式的解集为[﹣1,].18.袋中有大小、形状相同的红、黑球各一个,现一次有放回地随机摸取3次,每次摸取一个球(Ⅰ)试问:一共有多少种不同的结果?请列出所有可能的结果;(Ⅱ)若摸到红球时得2分,摸到黑球时得1分,求3次摸球所得总分为5的概率. 【考点】等可能事件的概率;随机事件. 【分析】(1)由分步计数原理知这个过程一共有8个结果,按照一定的顺序列举出所有的事件,顺序可以是按照红球的个数由多变少变化,这样可以做到不重不漏.(2)本题是一个等可能事件的概率,由前面可知试验发生的所有事件数,而满足条件的事件包含的基本事件为:(红、红、黑)、(红、黑、红)、(黑、红、红),根据古典概型公式得到结果. 【解答】解:(I )一共有8种不同的结果,列举如下: (红、红、红、)、(红、红、黑)、(红、黑、红)、(红、黑、黑)、(黑、红、红)、(黑、红、黑)、(黑、黑、红)、(黑、黑、黑) (Ⅱ)本题是一个等可能事件的概率 记“3次摸球所得总分为5”为事件A事件A包含的基本事件为:(红、红、黑)、(红、黑、红)、(黑、红、红)事件A包含的基本事件数为3由(I)可知,基本事件总数为8,∴事件A的概率为19.已知函数f(x)=sinx﹣ax,g(x)=bxcosx(a∈R,b∈R).(1)讨论函数f(x)在区间(0,π)上的单调性;(2)若a=2b且a≥,当x>0时,证明f(x)<g(x).【考点】导数在最大值、最小值问题中的应用;三角函数的最值.【分析】(1)求出函数的导数f'(x)=cosx﹣a通过余弦函数的值域,讨论a与[﹣1,1]的范围,判断导数的符号,然后得到函数的单调性.(2)用分析法证明f(x)<g(x),转化为证明,构造函数M(x)=,通过求解函数的导数,求出函数的最值,然后证明即可.【解答】(本小题13分)解:(1)f(x)=sinx﹣ax,则f'(x)=cosx﹣a…当a≥1时,f'(x)<0,所以函数f(x)在区间(0,π)上单调递减…当a≤﹣1时,f'(x)>0,所以函数f(x)在区间(0,π)上单调递增…当﹣1<a<1时,存在ϕ∈(0,π),使得cosϕ=a,即f'(ϕ)=0,x∈(0,ϕ)时,f'(x)>0,所以函数f(x)在区间(0,ϕ)上单调递增,x∈(ϕ,π)时,f'(x)<0,所以函数f(x)在区间(ϕ,π)上单调递减…(2)要证明f(x)<g(x),只须证明f(x)﹣g(x)<0当a=2b时,…等价于…记M(x)=,则…M'(x)==…当,即时,M'(x)≤0,M(x)在区间上(0,+∞)单调递减,M(x)<M(0)=0所以,当x>0,f(x)<g(x)恒成立.…20.如图1,正方形ABCD的边长为,E、F分别是DC和BC的中点,H是正方形的对角线AC与EF的交点,N是正方形两对角线的交点,现沿EF将△CEF折起到△PEF的位置,使得PH⊥AH,连结PA,PB,PD(如图2).(Ⅰ)求证:BD⊥AP;(Ⅱ)求三棱锥A ﹣BDP 的高.【考点】棱柱、棱锥、棱台的体积;空间中直线与直线之间的位置关系. 【分析】(1)由PH ⊥AH ,PH ⊥EF 可得PH ⊥平面ABCD ,故PH ⊥BD ,又AC ⊥BD ,得出BD ⊥平面PAH ,得出BD ;(2)分别把△ABD 和△BDP 当做底面求出棱锥的体积,列出方程解出. 【解答】(Ⅰ)证明:∵E 、F 分别是CD 和BC 的中点,∴EF ∥BD . 又∵AC ⊥BD ,∴AC ⊥EF ,故折起后有PH ⊥EF .又∵PH ⊥AH ,∴PH ⊥平面ABFED . 又∵BD ⊂平面ABFED ,∴PH ⊥BD , ∵AH ∩PH=H ,AH ,PH ⊂平面APH ,∴BD ⊥平面APH ,又∵AP ⊂平面APH ,∴BD ⊥AP (Ⅱ)解:∵正方形ABCD 的边长为, ∴AC=BD=4,AN=2,NH=PH=1,PE=PF∴△PBD 是等腰三角形,连结PN ,则PN ⊥BD ,∴△PBD 的面积设三棱锥A ﹣BDP 的高为h ,则三棱锥A ﹣BDP 的体积为由(Ⅰ)可知PH 是三棱锥P ﹣ABD 的高,∴三棱锥P ﹣ABD 的体积:∵V A ﹣BDP =V P ﹣ABD ,即,解得,即三棱锥A ﹣BDP 的高为.21.已知函数f (x )=,g (x )=alnx ﹣x (a ≠0).(Ⅰ)求函数f (x )的单调区间;(Ⅱ)求证:当a >0时,对于任意x 1,x 2∈(0,e ],总有g (x 1)<f (x 2)成立. 【考点】利用导数研究函数的单调性;函数单调性的性质.【分析】(I)先求函数f(x)的导数,再对字母a进行分类讨论,根据导数大于0函数单调递增,导数小于0时函数单调递减可得答案.(Ⅱ)欲证当a>0时,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立,只须证明对于任意x1,x2∈(0,e],总有g(x)max<f(x)min.由(Ⅰ)可知,当a>0时,f(x)在(0,1)上单调递增,f(x)在(1,e]上单调递减,从而有f(x)min=a,同样地利用导数可得,当a>0时,g(x)在(0,a)上单调递增,g(x)在(a,e]上单调递减,从而g (x)max=g(a)=alna﹣a,最后利用作差法即可得到g(x)max<f(x)min.【解答】解:(Ⅰ)函数f(x)的定义域为R,.当a>0时,x f'x f x当a>0时,f(x)的单调递增区间为(﹣1,1),单调递减区间为(﹣∞,﹣1),(1,+∞);当a<0时,f(x)的单调递增区间为(﹣∞,﹣1),(1,+∞),单调递减区间为(﹣1,1).…(Ⅱ)由(Ⅰ)可知,当a>0时,f(x)在(0,1)上单调递增,f(x)在(1,e]上单调递减,又f(0)=a,f(e)=所以f(x)min=a,同样地,当a>0时,g(x)在(0,a)上单调递增,g(x)在(a,e]上单调递减,所以g(x)max=g(a)=alna﹣a,因为a﹣(alna﹣a)=a(2﹣lna)>a(2﹣lne)=a>0,所以对于任意x1,x2∈(0,e],总有g(x)max=g(e)=alna﹣a<a=f(x)min.所以对于任意x1,x2∈(0,e],仍有x1,x2∈(0,e].综上所述,对于任意x1,x2∈(0,e],总有g(x1)<f(x2)成立.…请考生在第22、23、24题中任选一题作答,如果多做,则按所做的第一题记分.作答时,用2B铅笔在答题卡上把所选题目对应的题号涂黑.[选修4-1:几何证明选讲]22.如图,已知AB是圆O的直径,C、D是圆O上的两个点,CE⊥AB于E,BD交AC 于G,交CE于F,CF=FG.(Ⅰ)求证:C是劣弧的中点;(Ⅱ)求证:BF=FG.【考点】与圆有关的比例线段.【分析】(I)要证明C是劣弧BD的中点,即证明弧BC与弧CD相等,即证明∠CAB=∠DAC,根据已知中CF=FG,AB是圆O的直径,CE⊥AB于E,我们易根据同角的余角相等,得到结论.(II)由已知及(I)的结论,我们易证明△BFC及△GFC均为等腰三角形,即CF=BF,CF=GF,进而得到结论.【解答】解:(I)∵CF=FG∴∠CGF=∠FCG∵AB圆O的直径∴∵CE⊥AB∴∵∴∠CBA=∠ACE∵∠CGF=∠DGA∴∴∠CAB=∠DAC∴C为劣弧BD的中点(II)∵∴∠GBC=∠FCB∴CF=FB又因为CF=GF∴BF=FG[选修4-4:坐标系与参数方程选讲]23.在直角坐标系xOy中,直线l的参数方程为(t为参数),以原点为极点,x轴正半轴为极轴建立极坐标系,⊙C的极坐标方程为ρ=2sinθ.(Ⅰ)写出⊙C的直角坐标方程;(Ⅱ)P为直线l上一动点,当P到圆心C的距离最小时,求P的直角坐标.【考点】点的极坐标和直角坐标的互化.【分析】(I)由⊙C的极坐标方程为ρ=2sinθ.化为ρ2=2,把代入即可得出;.(II)设P,又C.利用两点之间的距离公式可得|PC|=,再利用二次函数的性质即可得出.【解答】解:(I)由⊙C的极坐标方程为ρ=2sinθ.∴ρ2=2,化为x2+y2=,配方为=3.(II)设P,又C.∴|PC|==≥2,因此当t=0时,|PC|取得最小值2.此时P(3,0).[选修4-5:不等式选讲]24.选修4﹣5:不等式选讲设函数f(x)=|2x﹣4|+|x+2|(Ⅰ)求函数y=f(x)的最小值;(Ⅱ)若不等式f(x)≥|a+4|﹣|a﹣3|恒成立,求a的取值范围.【考点】绝对值不等式的解法;函数恒成立问题.【分析】(Ⅰ)去绝对值可得f(x)=,分段求最值可得;(Ⅱ)问题等价于|a+4|﹣|a﹣3|≤f(x)min=4,解之可得.【解答】解:(Ⅰ)由于f(x)=|2x﹣4|+|x+2|=可得当x<﹣2时,﹣3x+2>8,当﹣2≤x<2时,4<6﹣x≤8,当x≥2时,3x﹣2≥4,所以函数的最小值为f(2)=4.(Ⅱ)若不等式f(x)≥|a+4|﹣|a﹣3|恒成立,则|a+4|﹣|a﹣3|≤f(x)min=4,又解不等式|a+4|﹣|a﹣3|≤4可解得a≤.所以a的取值范围为a≤2016年12月10日。
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成都龙泉实验中学高2014级高三上学期9月月考英语试题本试卷分第I卷(选择题)和第 II 卷(非选择题)。
满分120分,考试用时120分钟。
注意事项:答题前,考生务必将自己的学校、姓名、考试号写在答题纸上。
考试结束后,将答题纸交回。
第 I 卷第一部分:听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Where did the woman go?A.To downtown Berkeley.B.To the man’s house. C.To Amherst.2.When will the man leave for home?A.On Thursday.B.On Friday. C.On Sunday.3.What does the man mean?A.The diamond is not beautiful.B.The design is not fashionable.C.The price is too high for them now.4.What will the woman do?A.Go to the movies.B.Do some shopping. C.Attend a party.5.Who is the man most probably?A.The woman’s colleague. B.The woman’s brother. C.The woman’s husband.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6至7题。
6.When did the man have his first class of Civil Engineering?A.Last Monday. B.Last Tuesday. C.Last Friday.7.What is the woman’s advice?A.Sharing the cost of the textbook and using it together.B.Thinking about Prof.Qian’s course more se riously.C.Sitting together in class from now on.听第7段材料,回答第8至9题。
8.Where does the man ask the woman to hand in her research report at first?A.At his meeting room. B.At his office.C.At his house.9.What will the man be doing at 5:00 this afternoon?A.Having a meeting.B.Having a class.C.Talking to his secretary.听第8段材料,回答第10至12题。
10.Who wants to go to Egypt?A.The man and his wife.B.The man’s wife and son.C.The man and his son 11.Why does the woman advise the man to go to Egypt by ship?A.He can have a safe and happy journey.B.He can have a better rest during the voyage.C.He can save a lot and enjoy the ocean scenery.12.What does the woman think of the man?A.Hesitant. B.Mean. C.Careless.请听第9段材料,回答第13至16题。
13.When will the woman leave her home?A.At 6:30. B.At 7:30. C.At 8:30.14.What does the woman plan to take with her?A.One large bag and two small bags. B.Two large bags and one small bag.C.One large bag and a handbag.15.Where will the man say goodbye to the woman?A.In the waiting room. B.Outside the train. C.At the entrance.16.What does the man ask the woman to do?A.Wait for him at the airport.B.Come back a week later.C.Give him a call.请听第10段材料,回答第17至20题。
17.When was Greenpeace founded?A.Twenty years ago. B.Fifty years ago. C.Seventy years ago.18.What role does the man play in the organization?A.He is the action organizer. B.He is a leading nuclear scientist.C.He is one of the plan makers.19.What has the man been involved in recently?A.Protesting the use of nuclear weapons.B.Protecting animals and plants.C.Environmental campaign.20.How did people react to the man’s appeal last week?A.They showed little interest in it.B.They were against the demonstration.C.They expressed their support in different ways.第二部分:阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
AParents usually go to malls to find new school supplies for their children. One of the most important things to get during this time is the backpack. Most students choose to bring a backpack because it allows them to have both hands free. This makes the bag less likely to hinder (妨碍)other activities that the person carrying it may be doing at the same time. Here are the ways parents can choose the perfect backpack for school.How many books will children be carrying to school? How many notebooks should be inside the bag? Parents should know these details before buying a bag for their children. Each situation requires a different bag so it pays to ask children about the expected load. Do children need special storage for the bag? If they will be carrying a laptop computer inside the new backpac k, make sure that parents get a model that’s designed specificallyfor laptop use. This will protect the computer from bumps and falls that might damage the device if it had less protection.Are children going to use the bag for a long time? If they want to use a bag for a long time,parents will be better off investing in a good brand that is known for quality. Parents may have to pay extra cash to get the top of the line and the best backpack in the market.Count the pockets. Backpacks usually come with a lot of pockets, and generally, the more pockets, the better. This allows children to organize their things, like pens and pencils inside small pockets.Picking the perfect school backpack is all about getting the sweet compromise (折中)between form, function and price. Remember to choose the best fit for children’s needs.21. What does the second paragraph mainly talk about?A. Parents should find out how many books their children have.[B. Parents must ask teachers for help before buying a backpack.C. Parents should know the weight of things that will go inside the backpack.D. Parents must ask their children’s requirements when buying a backpack.22. The underlined word “bumps” in the third paragraph is closest in meaning to“”.A. beatsB. jumpsC. standsD. knocks23. Which of the following can we know from the text?A. Picking a backpack is very easy to do.B. The backpacks that have a good brand are more popular with children.C. It’s better for children to have a backpack that has more pockets.D. Parents always feel worried when buying a backpack for their children.BThousands of taxi drivers in Shenyang, Liaoning Province, reportedly blocked streets with their vehicles on Sunday in protest against unlicensed vehicles using taxi-hiring apps (打车软件) and apps-based car rental companies providing passenger services, including high-end cars. Although the drivers also complained about the withdrawal of the fuel subsidy(补贴) by the government, their main complaint was the loss of business because ofthe rising number of Internet-based car services companies.On Wednesday, news reports came that Beijing transport authorities will take measures to stop the illegal “taxi business” of private cars through the newly rising Internet apps, following the footsteps of Shenyang and Nanjing.It is not yet clear how the Shenyang city government will handle the issue and whether it will declare the services offered by market leaders such as Didi Dache, a taxi-hiring app provider backed by Tencent Holdings, and Kuaidi Dache illegal. But Shanghai transport regulators have set a rule, by banning Didi Zhuanche, or car services offered by Didi Dache in December.Such regulations will cause a setback to the car-hiring companies and investors that are waiting to cash in on the potentially booming business. Just last month, Didi Dache got $700 million in funding from global investors, including Singapore state investment company Temasek Holdings, Russian investment company DST Global and Tencent. Besides, the market is uncertain that Kuaidi Dache is about to finalize its latest round of funding after getting $800 million from global investors.Regulatory uncertainties, however, could cast a shadow on the future of the Internet-based car-hiring services, whi ch have become popular in most of China’s big cities. To be fair, these companies’ business model is anything but bad. For example, Didi Zhuanche works side by side with established car rental companies to provide high-end car service mainly for businesspeople through the Internet and mobile phone apps.Every link in this business model chain has legal companies and services. Therefore, it is hard to define it as illegal and ban it.24. Why did taxi drivers in Shenyang block the streets with their vehicles?A. Because they wanted the authority to increase their driving allowances.B. Because they wanted to be taught how to use the taxi-hiring apps.C. Because they wanted to appeal to passengers not to hire the private cars.D. Because they wanted to make their main complaints known to the author ity.25. The author’s attitude to banning internet car-hiring service is__________.A. positiveB. negativeC. neutralD. unclear26. Which of the following statements is True?A. All the taxi drivers in China benefit a lot from taxi-hiring apps.B. Taxi drivers mainly complained about the withdrawal of the fuel subsidy by thegovernment.C. The Internet-based car-hiring services have become popular all over China.D. The Internet-based car-hiring services are challenging the traditional taxi industry.27. We can learn from the passage that __________.A. Shenyang forbade apps-based car rental companiesB. Shanghai is the second city banning Didi ZhuancheC. it is not difficult to picture the apps-based car rental companies illegalD. some international investment companies have strong faith in the future of apps-basedcar rental companiesCMany years ago, my dad was facing a serious heart condition. He was unable to do a steady job. He fell suddenly ill and had to be admitted to the hospital.He wanted to do something to keep himself busy, so he decided to volunteer at the local children’s hospital. My dad loved kids. It was the perfect job for him. He ended up working with the seriously ill children. He would talk, play, and do arts with them.One of his kids was a girl with a rare disease that paralyzed (瘫痪) her from the neck down. She couldn’t do anything, and she was very depressed. My dad decided to try to help her. He started visiting her in her room, bringing paints, brushes and paper. He stood the paper up, put the paintbrush in his mouth and began t o paint. He didn’t use his hands at all. All the while he would tell her, “See, you can do anything you set your mind to.”At the end of the day, she began to paint using her mouth, and she and my dad became friends. Soon after, the little girl was sent home because the doctors felt there was nothing else they could do for her. My dad also left the children’s hospital for a little whilebecause he became ill. Some time later after my dad had recovered and returned to work, in came the little girl who had been paralyzed and only this time she was walking. She ran straight over to my dad and hugged him really tight. She gave him a picture she had done using her hands. At the bottom it read: “Thank you for helping me walk.”My dad would cry every time he told us this story and so would we. He would say sometimes love is more powerful than doctor, and my dad—who died just a few months after the little girl gave the picture—loved every single child in that hospital.28.The author’s father worked at the local hosp ital to .A. realize his childhood dreamB. ease his serious heart conditionC. keep himself occupied and happyD. earn money to pay for treatment29. How did the author’s father help the paralyzed little girl?()A. He showed her she could still do things.B. He visited her and made a toy for her.C. He helped her practice walking.D. He painted special pictures for her.30.According to Paragraph 4,the paralyzed girl .A. sent him a picture painted with her mouthB. eventually became a unique painterC. was sent home and never seen againD. gradually recovered and walked31. What message does the passage mainly convey?A. It’s better to give than to receive.B. Love can sometimes bring great results.C. Volunteering is a worthwhile thing to do.D.A sick person should not focus on his disease.DU.S. Bank ScholarshipAre you a United States citizen and a high school senior or college undergraduate student?Then you can win $1,000. No other qualifications. Just be what you already are. Each year, U.S.Bank gives away 40$1,000 scholarships for those simply being an American citizen who will be or is attending college. All you have to do is fill out some basic information about yourself. It takes hardly any time at all.“No Essay Scholarship”Most scholarships come but once a year and some once in a lifetime. The beauty of the $2,000 “No Essay” Scholarship is that it comes once a month and you can apply every single month. And the application is very easy. In fact. You could have filled it out probably 20 times since starting to read this article. You can only apply once a month, but if you are a high school senior or college student, then you can win $2,000. Winners are drawn at random. $1,000 Weekly ScholarshipEven better than once-a-month is once-a-week. The Zinch. Com weekly scholarship is just as simple as filling out basic information about yourself. All U.S.high schoolers and college students are fit for this $1,000 scholarship. All it requires is filling out a short form about yourself and answering an unusual question in 280 characters or less. The question changes weekly. You can apply every week, once a week. Have fun with it. Scholarship Zone ScholarshipIt seems nearly every website wants you to register. But not many sites offer as great a possible reward as Scholarship Zone does. By simply registering, you are automatically entered for the next scholarship drawing of $10,000.You are not really required to do anything after that. Are you in college? Then why not take minutes to apply?32. What do we know about the “No Essay” Scholarship?A. It chooses its winners in a strict way.B. It is only for high school student.C. It can be won several times a year.D. It takes a long time to apply for.33. To apply for the $1,000 Weekly Scholarship, one needs to answer a question which ________.A. is common but interestingB. can be different every weekC. deals with personal experiencesD. should be answered in at least 280 words34. Which scholarship is only intended for college students?A. U.S. Bank ScholarshipB. “No Essay” ScholarshipC.$1,000. Weekly ScholarshipD. Scholarship Zone Scholarship35. What do the four scholarships have in common?A. They are easy to apply for.B. They are for American citizensC. They must be applied for online.D. They can be applied for again and again. 第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,并在答题卡上将该项涂黑。