江西省“三新”协同教研共同体2023-2024学年高一上学期12月联考数学答案

2023年“三新”协同教研共同体高一联考数学试卷参考答案选择题123456789CBBDDACDABACACDAC1.C∵A={1,2,3,4},B=[3,+∞),∁RB=(-∞,3),∴A∩(∁RB)={1,2}.3.B由命题p可得0<3a-1<1,解得<a<,所以p可以推出q,q推不出p,故p是q的充分不必要条件.4.DA选项,当c=0时不符合;B选项,当a=-2,b=-1时不符合;C选项,当a=3,b=1,c=-1,d=-2时不符合;D选项,,由a>0,可得.5.Df(x)图象的对称轴为直线x=1,且最大值为f(1)=4.令f(x)=-5,可得x=-2或x=4.由f(x)∈[-5,4],可得a∈[1,4].6.A由2a+b-ab=0,得=1,则a+2b=(a+2b)()=5++≥5+2··=9,当且仅当,即a=b=3时,等号成立.7.C由题知f(x)的周期为4.又函数f(x)为奇函数,所以f(20)+f()+f()=f(0)+f(-8.D由题意可设0≤x1<x2≤m,由x--2)<3,得x1f(x则x1f(x1)-x2f(x2)>3x1-3x2,所以x1f(x1)-3x1>x2f(x2)-3x2.则y=h(x)在[0,m]上单调递减.由y=h(x)的图象(图略),可得m∈(0,3].9.ABA选项,8-=(23)-=2-2;B选项,由题可得a<0,=-a;C选项,6a6=|a|;D选项,12(-2)4>0,3-2<0.10.AC由f(x)的图象(图略),可知f(m)=|4m-1|=1-4m,f(n)=|4n-1|=4n-1.由1-4m=4n-1,得4m+4n=2,故A正确,B错误;由4m+4n=2≥24m·4n=24m+n,可得m+n<0,故C正确,D错误.11.ACD当x=0时,可得g(0)>0,故A,D均不可能;设g(x)=a+bx+c(a≠0),由g(x)>g(2x),可得a+bx+c>(4a+2bx+c),化简可得2a-c<0在R上恒成立,所以故C选项不可能.12.AC在R上任取x1,x2,且x1>x2,则f(x1)-f(x2)=f[(x1-x2)+x2]-f(x2)=f(x1-x2)+f(x2)+1-f(x2)=f(x1-x2)+1.因为x1-x2>0,所以f(x1-x2)>-1,即f(x1-x2)+1>0,则f(x1)>f(x2),所以f(x)在R上为增函数.令x=1,y=1,可得f(2)=3,令x=1,y=2,可得f(3)=5,同理可得f(-1)=-3,故f(x)在[-1,3]上的值域为[-3,5].令y=1-x,可得f(x)+f(1-x)=0,故f(x)的图象关于点(,0)对称.由f(a+2x)+f(x+1)<6,可得f(a+3x+1)<7=f(4),所以a+3x+1<4在R上恒成立,则解得a∈(-∞,).填空题13.30≤1等价于≤0,所以(-≤0,解得x∈(0,5],则A={1,2,3,4,5},故集合A的非空真子集的个数为30.14.原式=33×(-)+log22(-6)×=-3=.15.(0,2],则g(x)=()x2-4x2+3=()x2-4|x|+3∈(0,2].16.[-1,1]由|+ax+2|≤4,可得x2+ax-2≤0,当不等式---,0,解得-1≤a≤1,此时+ax+6≥>0,所以a的取值范围是[-1,1].解答题17.解:∵log2x>2,∴x>4...................................................................................................................2分(1)∵“x∈A”是“x∈B”的充分条件,∴A=B,∴a≤4...............................................................................................................................................5分(2)∵A∪C=A,∴C=A.........................................................................................................................7分当C=⌀时,b+1≥2b+1,∴b≤0.........................................................................................................8分当C≠⌀时,+1,∴b≥3..............................................................................................9分综上,b≤0或b≥3..........................................................................................................................10分18.解:(1)当x∈[0,2]时,可设y=xα.由表知2=2α,∴α,............................................................2分当x∈(2,5]时,可设y=a+bx+c(a≠0),解得b=6,......................................................................................................................................5分∴y=+6x-4.1综上,y=x2,x∈[0,2],...........................................................................................................6分-x2+6x-4,x∈(2,5].(2)当x∈[0,2]时,函数单调递增,ymax=2.........................................................................................8分当x∈(2,5]时,函数在(2,3)上单调递增,在[3,5]上单调递减,∴当x=3时,ymax=5...........................................................................................................................10分又5>2,∴当广告宣传费投入3万元时,超市的利润最大........................................................12分19.(1)证明:令2a=3b=5c=m>1,∴a=log2m,b=log3m,c=log5m,..............................................................2分∴=logm2+logm5=logm10,=2logm3=logm9...................................................................................4分∴logm10>logm9,∴>..................................................................................................................6分x1x2=m=-6,t=6m.(2)解:∵∴结合图象(图略)可知,方程3m=2-m的根是y=3x的图象与y=2-x交点的横坐标,方程log3n=2-n的根是y=log3x的图象与y=2-x交点的横坐标.又y=3x的图象与y=log3x的图象关于直线y=x对称, ∴,-x,解得..................................................................................................................10分∴点(m,3m)与点(n,log3n)关于点(1,1)对称,所以m+n=2.................................................................12分20.解:(1)由题知m<0,方程m-nx-t=0的两根为-3,2,∴-tm解得, 由n+mx+t≤0,可得-m+mx+6m≤0,即-x-6≤0, ∴-2≤x≤3,∴不等式的解集为{x|-2≤x≤3} (2)-n=+m-n-m................................................................................................................8分∵f(1)>0,∴m-n>0,∴+m-n≥2,当且仅当m-n=1,即m=n+1时,等号成立................................10分又y-m在[1,3]上单调递减,当m=3时,ymin=0,∴当m=3,n=2时,-n有最小值2.........................................................................................12分21.解:(1)f(x)+2f(-x)=2x+2-x.①令x=-x,得f(-x)+2f(x)=2-x+2x.②联立①②,解得f(x)=...............................................................................................................4分(2)∵g(x)在[22-5,]上单调递增,∴g(n)∈[22,].........................................................................5分又f(x)=,∴f(am)=.对∀m1,m2∈[1,2],不妨设m1<m2,则f(am1)-f(am2)=-=[(2a)m1+m2-1]·[(2a)m1-(2a)m2].∵a>0,∴2a>1,∴y=(2a)x单调递增.又1≤m1<m2≤2,∴(2a)m1+m2-1>0,(2a)m1<(2a)m2,∴f(am1)<f(am2),即y=f(am)在[1,2]上单调递增,...............................................................................8分∴ymin=f(a)=,ymax=f(2a)=.又由题意可知f(am)的值域=g(n)的值域....................................................................................10分,4a4-a≤,可得≤a≤1...........................................................................................12分22.解:(1)∵f(x)为奇函数,∴f(x)的定义域关于原点对称.又b->0,∴>0的解集关于原点对称,∴b+b-2=0,即b=1,.............................................................................................................................2分∴f(x)=loga(1-)>0=loga1.当a>1时,1->1,∴x<-1;当0<a<1时,0<1-<1,∴x>1.综上,当a>1时,不等式的解集为{x|x<-1};当0<a<1时,不等式的解集为{x|x>1}......................5分(2)f(x)的定义域为{x|x<-1或x>1},由题意可知m<n,且loga(2n)<loga(2m),∴0<a<1,............