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湖北省孝感市重点高中教科研协作体*数学答案(共4页)第1页2022年湖北省孝感市重点高中教科研协作体高一期中考试高一数学参考答案及评分标准123456789101112BCDADABCADABBCDBD13.[0,4]14.[0,12]1
5.a(x²−1)(a>0)(答案不唯一,a>0即可)16.428.【解析】对于C,当x=y=0.5,[x+y]=1,[x]+[y]=0,[x+y]>[x]+[y],C错误.12.【解析】对于A,当0k时,121,22211kkk,所以fx在R上不是减函数,A错误.对于
B,当0k时,fx在1,上是减函数,无最小值,又2fxx在,1上是减函数,也无最小值,因此fx无最小值;当102k时,2kfxkx在1,上是增函数,122fk,
但221k,所以fx无最小值.综上,当12k时,fx无最小值,B正确.对于C,当1x时,21,fxx,当1x时,由1k,得11121fxxx是增函数,所以110,1fxx
,所以fx的值域是0,11,U,C错误.对于D,当1x时,由3k,得314,1fxx,所以1,4fx.而当1x时,1,fx,1,41,,因此11x,21x,使
得12fxfx,即120fxfx,D正确.16.【解析】由不等式ax²+4x+b≥0对于一切实数x恒成立可得���>016−4������≤0,解得������≥4,又存在实数���0
,使得a���0²+4���0+b=0成立,则∆=16−4������≥0,得������≤4,所以������=4.∴b=4���∵a>b∴a−b=a−4���>0湖北省孝感市重点高中教科研协作体*数学答案(共4页)第2页∴���²+���²���−���=���²+(4���
)²���−4���=(���−4���)²+8���−4���=a−4���+8���−4���≥42(当(���−4���)²=8时取等)17.【答案】(1)由x(x+2)>x(3−x)+6得:2x²−x−6>0,即(2x+3)(x−2)>0∴���<−32或���>2,即不等式
的解集为{x|���<−32或���>2}················5分(2)由题意可得:���+1≥0−���²−���+6>0���−1≠0,解得:−1≤x<2且���≠1所以函数的定义域为:
−1,1∪1,2·········································10分18.【答案】(1)当a=2时,A={x|1≤x≤5}····································1分∴A∩B={x|1≤x≤2},A∪B
={x|−1≤x≤5}·····································5分(2)若选①,由A∪B=B可知:A⊆B··············································6分当A=Ø时,则
a−1>2a+1,即a<−2,满足A⊆B·····························8分当A≠Ø时,由A⊆B得:���−1≤2���+1���−1≥−12���+1≤2,解得0≤���≤12··················10
分综上所述:实数a的取值范围为−∞,−2∪0,12························12分若选②,由题意可知:A⫋B························································
····6分当A=Ø时,则a−1>2a+1,即a<−2,满足A⫋B······························8分当A≠Ø时,由A⊆B得:���−1≤2���+1���−1>−12���+1≤2或���−1≤2���+1���−1≥−
12���+1<2···············10分解得0<���≤12或0≤���<12·····························································11分综上所述:实数a的取值范围为−∞,−2∪0,12···············
·········12分若选③,当A=Ø时,则a−1>2a+1,即a<−2,满足A∩B=Ø··············7分当A≠Ø时,由A∩B=Ø得:���−1≤2���+12���+1<−1或���−1≤2���+1���−1>2··········
··10分解得−2≤���<−1或���>3·····························································11分综上所述:实数a的取值范围为−∞,−1∪3,+
∞···················12分19.【答案】(1)设f(x)=ax²+bx+c(a≠0),则f(x+1)=a(x+1)²+b(x+1)+c∴f(x+1)−f(x)=2ax+a+b=2x+2·············································
············2分∴2���=2���+���=2,解得���=1,���=1···················································4分由f(x)的图象
经过点A(3,2)得f(3)=9+3+c=2,∴���=−10····················5分湖北省孝感市重点高中教科研协作体*数学答案(共4页)第3页∴f(x)=x²+x−10···································
········································6分(2)设g(x)=f(x)−mx=x²+(1−���)x−10.因为当x∈[−2,2]时,不等式f(x)≤mx恒成立,∴���(−2)≤0���(2)≤0············8分即4−2(1−���)−10≤
04+2(1−���)−10≤0·····························································10分解得−2≤���≤4.故实数m的取值范围是−2,4.······················
·····12分另解:由题意可知:x²+x−10≤mx对∀x∈[−2,2]恒成立①当x=0时,−10≤0恒成立,所以m∈R.········································
7分②当x∈−2,0时,m≤(���+1−10���)���������即可.由x∈−2,0得(���+1−10���)���������=4,故m≤4···································9分③当x∈0,2时,m≥(���+1−10���)����
�����即可.由x∈0,2得(���+1−10���)���������=−2,故m≥−2·····························11分综上所述:实数m的取值范围是−2,4.························
···············12分20.【答案】(1)由题意可知y=100n−(4n²+20n)−144=−4n²+80n−144(n∈N+),·······2分令y>0,得−4n²+80n−144>0,解得2<n<18,··············
···················4分所以从第3年起开始盈利;·····························································5分(2)若选择方案①,设年平均利润为1y万元,则13636804804232yynnnnn
,当且仅当36nn,即6n时等号成立,所以当6n时,1y取得最大值32,此时该项目共获利32672264(万元).········································8分若选择方案②,纯利润22480144410256y
nnn,所以当10n时,y取得最大值256,此时该项目共获利2568264(万元).11分以上两种方案获利均为264万元,但方案①只需6年,而方案②需10年,所以仅考虑该项目的获利情况时,选择方案①更有利于该公司的发展.··········12分21.【答案】(1)证明
:fabfafb,令0ab==,(0)2(0)ff,则(0)0f.令ax,bx,()()()fxxfxfx,即()()(0)fxfxf,而(0)0f,湖北省孝感市重点高中教科研协作体*数学答案
(共4页)第4页∴f(−���)=−f(���),即函数y=f(���)是奇函数;········································4分(2)任取1211xx,则
210xx,∵当0x时,fx>0恒成立∴f(���2−���1)>0,∴f(���2)-f(���1)=f(���2−���1+���1)-f(���1)=f(���2−���1)+f(���1)-f(���1)=f(���2−���1)>0即f(x1)<f(x
2)∴函数y=f(x)是−1,1上的增函数;···············································8分(3)由2110fafa,可得211fafa,又函数()yf
x是奇函数,∴211fafa,∵()fx在定义域上单调递增∴−1<1−���<1−1<1−���2<11−���<���2−1,得0<���<2−2<���<0,0<���<2���<−2或���>1,∴1<���<2,故a的取值范围为1,2.··············
·······················12分22.【答案】(1)∵y=x²+2x≥−1且y=x²+2x在−1,+∞上单调递增,由x²+2x=x得���=−1或0∴函数y=x²+2x(x∈R)存在“优美区间”,是[−1,0]··························2分∵y=
1−2���在0,+∞上是增函数,若存在“优美区间”[m,n],则有1−2���=���1−2���=���,方程组无实数解,则函数y=1−2���(x>0)不存在“优美区间”·················
·4分(2)f(x)=(���+1)���−1���������=1+1���−1���²���在−∞,0和0,+∞上都是增函数,因此“优美区间”[m,n]⊆−∞,0或[���,���]⊆0,+∞································6分由题意可
知:���(���)=������(���)=���,所以f(x)=x有两个同号的不等实根即���²���²−(���²+���)���+1=0有两个同号的不等实根∴∆=(���²+���)²−4���²>0,即���²(���+3)(���−1)>0,解得���<−3或�
��>1……8分∵���1���2=1���²>0(���1、���2同号,满足题意),���1+���2=���²+������²,∴n−m=���1−���2=(���1+���2)²−4���1���2=(���²+������²)²−4��
�²=−3���²+2���+1=−3(1���−13)²+43································································10分∵���<−3或���>1∴当1���=13,即���=3时,(���−���)
������x=233····································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com