1、2014-2015重庆南开高2017级高一(上)期末考试数学试卷(2015.1)本试卷分第卷(选择题)和第卷(非选择题),满分150分,考试时间120分钟一、选择题(本大题10个小题,每小题5分,共50分,每小题只有一个选项符合要求)1、计算( )A、 B、 C、 D、2、若设a=20.5,b=log0.5e,c=ln2,则下列结论正确的是( ) A、bac B、cab C、cba D、bcsinB”是“AB”的什么条件( ) A、充分不必要 B、必要不充分 C、充要 D、既不充分也不必要6、求函数y= 的单调区间( ) A、 B、 C、 D、7、已知函数f(x)Acos()(A0,)为奇函数
2、该函数的部分图像如图所示,EFG是边长为2的等边三角形,则f(1)的值为( ) A、 B、 C、 D、8、定义在R上的函数f(x)满足f(x+2)=f(x),当x1,3,f(x)=2-|x-2|,则下列结论中正确的是( ) A、f(sin)f(cos) B、f(sin1)f(cos1) C、f(sin)0sin(2x+) 2k+2x+2k+ k-xk+设u=2sin(2x+)-1在k-xk+上是减函数所以y=的单调递减区间为(k-,k+)u=2sin(2x+)-1在k+xk+上是增函数所以y=的单调递增区间为(k+,k+),选C7、解:f(x)=Acos(x+)为奇函数,f(0)=Acos=
3、00,=, f(x)=Acos(x+)=-Asinx,EFG是边长为2的等边三角形,则yE=A,又函数的周期 T=2FG=4,根据周期公式可得,=,f(x)=-Asinx=-sinx,则f(1)=-, 故答案为:- 选D8、解:设x-1,1,则x+21,3,由题意知此时f(x)=f(x+2)=2-|x+2-2|=2-|x|,这是一个偶函数,图象关于y轴对称,且当0x1时,f(x)=2-x是减函数,所以当-1x0时f(x)是增函数,在-1,1上,自变量的绝对值越小,函数值越大,因为0|sin|cos|1,|cos|sin|,所以排除A、C;又因为0cos1=sin(1)sin11,所以排除B,
4、故选D9、解:方程cos2x+(4t+2)sinx=2t2+2t+1化为1-2sin2x+(4t+2)sinx=2t2+2t+1,化为sin2x-(2t+1)sinx+t2+t=0,即(sinx-t-1)(sinx-t)=0,sinx=t+1或sinx=t画出函数y=sinx,y=t,y=t+1的图象,由图象可以看出:当且仅当-1t0时,函数y=sinx的图象分别与函数y=t,y=t+1的图象有一个交点、两个交点故所求的t的取值范围是-1t0故选A10、(2)设COE=,设OE交AD于E,交BC于F,显然矩形ABCD关于OE对称,而E,F均为AD,BC的中点,在RtOFC中,CF=2sin,O
5、F=2cos则BC=AD=4sin,OE=OF-EF=2cos-CD,DE=OEtan60= OE=2sin,即有CD=2cos- sin,则矩形ABCD的面积S=CDBC=(2cos-sin)4sin=8sincos-sin2=4sin2-(1cos2)=(sin2+cos2)-=sin(2+)-,当sin(2+)=1即时,面积取得最大值,且为,选C11、612、f(-)=sin(-)=sin(-2)=-sin()=-f()=f(-1)-1=f()-1=f(-1)-1-1=f(-)-2=sin(-)-2=-2=-f(-)+f()=-=-313、14、解:设原不等式的解集为A,当k=0时,则x
6、4,不合题意,当k0且k2时,原不等式化为x-( k+)(x-4)0,k+4,A(4,k+),要使不存在整数x使不等式(kx-k2-4)(x-4)0成立,须k+5,解得:1k4;当k=2时,A=,合题意,当k0时,原不等式化为x-( k+)(x-4)0,A=(-,k+)(4,+),不合题意,故答案为:1k415、解:f(x)=+1=+10x,由原式得到xsinx(0,1)(1,+)f(x)(2,+)函数没有最小值没有最大值, 故选D16、解:(1)当a=-1时,A=x|-2x2,B=x|2x或log5x1=x|x-1或x5则AB=x|x5或x2;(RA)B=x|x5或x-2(2)当A=时,2a
7、a+3,解得a3当A,若AB=,则,解得a2综上所述,a的取值范围a|a3或a217、解:(1)因为x(,), 所以x-(,),sin(x-)=sinx=sin(x-)+=sin(x-)cos+cos(x-)sin=+=(2)因为x(,),故cosx=-sin2x=2sinxcosx=-,cos2x=2cos2x-1=-所以sin(2x+)=sin2xcos+cos2xsin=-18、(1)证明:由acosA+bcosB=c,利用余弦定理化简得:a +b =c,整理得:a2(b2+c2-a2)+b2(a2+c2-b2)=2abc2,即(a-b)2c2-(a+b)2=0,ca+b,c2(a+b)
8、2,a=b,则ABC为等腰三角形;(2)解:设ABC的外接圆半径为R,由R2=,得到R=1,由(1)得:A=B,由正弦定理得:= = =6sinB+1+8cosB=10sin(B+)+1,记为f(B),其中sin= ,cos= ,且(,),ABC为锐角三角形,结合A=B,得到B,B+(,),f(B)在B上单调递减,当B=时,f(B)=10sin(+)+1=10cos+1=7;当B=时,f(B)=10sin(+)+1=10(sin+cos)+1=7+1,f(B)(7,7+1),即(7,7+1)19、解:(1)由f(x)=0,得cos(sin+cos)=0,由cos=0,得=k+,x=2k+(kZ
9、由sin+cos=0,得tan=-,=k-,x=2k-(kZ)所以方程f(x)=0的解集为x|x=2k+或x=2k-,kZ;(2)f(x)=sinx+(cosx+1)=sin(x+)+,x0,x+,sin(x+),1,f(x),1+20、解:(1)2,=2,f(x)=sin(2x+)-b又g(x)sin2(x)+b+为奇函数,且0,则,b,故f(x)sin(2x+)(2)令 2k-2x+2k+,kz,求得 +kx+k,(kZ),故函数的增区间为+k,+k(kZ)令2k+2x+2k+,kz,求得+kx+k,(kZ),故函数的减区间为+k,+k(kZ)(3)f2(x)-(2+m)f(x)+2+m
10、0恒成立,f(x)0,f(x)-1mf2(x)-2f(x)+2=f(x)-12+1,整理可得m,即 m+f(x)1x0,0sin(2x+)1,-f(x)1-,故1f(x)1则有+f(x)1,故+f(x)1的最小值为 ,故 m,即m取值范围是(,21、解:(1)由,得,解得g(x)=,h(x)=(2)因为h(x)在R上时单调递增的奇函数,所以h(x2+2x)+h(x-4)0h(x2+2x)h(4-x),所以x2+3x-40,解得x1或x-4,所以不等式的解集为:x|x1或x-4(3)g(2x)-ah(x)0,即得a0,参数分离得a=ex-e-x+,令t=ex-e-x,则ex-e-x+=t+=F(
11、t),于是F(t)=t+,t,因为F(t)min=F()=,所以a Shang, and action Shang with to XI comrade for General Secretary of Central keep height consistent; any when are put discipline rules is in front, enhanced organization concept, and obey Organization decided; to set right of worldview, and Outlook on life And to lea
12、d a clean slim family, always adhere to the Communists spiritual home; to consciously practice the partys purpose, keeping qualities. CPC requirements in section-level cadres above to be a benchmark, play an exemplary role, adhere to the loyal clean play, Jiao Yulu-a good cadre of activities carried
13、 out in the ordinary party members practicing pioneer standard post set up activities. We all party members to further enhance political par core, situation awareness, awareness, awareness, belief and faith, according to positions, downtown River regulation, diversion of hydraulic Process, protectio
14、n of the mother River, rural safety water drinking project, construction of flood control and drought relief, water in ecological civilization construction and other key, think, often, more active, hands-on, disciplined and strict rules, play, courage, give full play to the exemplary role, the succe
15、ssful completion of tasks. Three reform as the core of real change, completely change, and change image. This fundamental aim of education in order to solve the problem, for this reason, we insist on problem-oriented, enhance the awareness of problems, problems of reform in place, continuous improve
16、ment in style, to better serve the people, to better promote the water conservancy work. First of all, to find the problem. Im learning education programme, in view of the current construction of the party, proposed to focus on solution of fuzzy ideals and beliefs waver, consciousness, purpose of th
17、e party weak sense, honesty and self-discipline awareness is not strong, depressed, six issues of ethical misconduct. Vast numbers of party members and cadres should closely real life personal and ideological work, reflections on the standard control, check the pendulum problem, pinpoint problems, get real, deep, identification, find the main target, laying the Foundation for further corrective action implementation.11