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高一数学参考答案第1页共4页2020～2021学年第一学期期中考试高一数学参考答案及评分标准一、单项选择题：共8小题，每小题5分，共40分．1．A2．B3．D4．C5．B6．A7．C8．B二、多项选择题：共4小题，每小题5分，共20分．在

每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，部分选对的得3分，有选错的得0分．9．AC10．AC11．BD12．ABD三、填空题：共4小题，每小题5分，共20分．13．314．815．3016．36四、解

答题：共6小题，共70分．解答时写出必要的文字说明、证明过程或演算步骤．17．（本小题满分10分）解：若选择①因为()(1)0Axxaxa，故(,1)Aaa；·························3分ABB，则AB，·

········································································5分所以1,11,aa≥≤·········································

·············································8分解得10a≤≤所以选择①，实数a的取值范围是[1,0]．···············································10分

若选择②因为()(1)0Axxaxa，故(,1)Aaa；·······························3分因为AB，所以11a≤或1a≥，·····················

·····················································6分解得2a≤或1a≥，··········································

···································8分所以选择②，实数a的取值范围是(,2][1,)．·································10分

若选择③因为()(1)0Axxaxa，故(,1)Aaa；·······························3分因为ABB，则BA·························································

·············5分所以1,11,aa≤≥·····················································································8分所以a所以选择③，实数a不存在．············

·····················································10分高一数学参考答案第2页共4页18．（本小题满分12分）解：（1）由30,10xx≥≥得13x≤≤，所以

1,3A；·········································3分又22()23122gxxxx≥，所以2,B．···················

···········6分（2）由（1）知1,32,1,AB；······································9分因为,2UB，············································

·································10分所以1,3,21,2UAB．··················································12分19．（本小题满分

12分）解：（1）由17aa知0a，因为122()7aa，即224)29aa，所以2247aa；·······························································

················3分又112122()29aaaa，且11220aa，所以11223aa．················································

································6分（2）因为lg3a，lg5b，所以5lg3log3lg5ab；································

·····9分36lg3.6lglg36lg102lg61102lg22lg312(1lg5)2lg31221ab．··························12分20．（本小题满分12分）解：（

1）因为不等式()0fx的解集是(0,3)，所以0和3是方程()0fx的两个根，即22000,330,bcbc解得3,0,bc··································

····························3分所以函数()fx的解析式为：2()3fxxx．··············································4分（2）不等式2()3

0fxxx的解集为：(,0)(3,)，不等式2()()3()0fxtxtxt的解集为：(,3)tt，··························6分当3t≥时，不等式组()0,()0fxfxt的解集为(,3)tt，高一数学参考答案第3页共

4页(,3)tt中至少有2个整数，不满足题意，舍去；······································8分当03t时，不等式组()0,()0fxfxt的解集为(,0)t，因为满足不等式组()0,()0fxfxt的整数解有且只有

一个，所以1(,0)t，2(,0)t，即1,2,tt≤解得12t≤；···························11分综上，正实数t的取值范围是1,2．················

········································12分21．（本小题满分12分）解：（1）此人两次投资的总收益率为bcxax；··········································

·········3分（2）设此人第*()nnN次投资后的总收益率为()fn，则(1)()(1)bnxfnanx*()nN，所以第1n次投资后的总收益率为(1)bnxfnanx*()nN，······················5分有(1)()(1)

()(1)()[(1)]bnxbnxabxfnfnanxanxanxanx，·····················7分因为0a，0b，0x，1n≥，所以()[(1)]0anxanx，因此，当ab时，(1)()0fnfn

，即(1)()fnfn；当ab时，(1)()0fnfn，即(1)()fnfn；当ab时，(1)()0fnfn，即(1)()fnfn．·································10分所以，当ab时，每次投资后的总

收益率不变；当ab时，每次投资后的总收益率减少；当ab时，每次投资后的总收益率增加．················································12分22．（本小题满分12分）（1）证明：()fx的定

义域为R，对xR，()()fxxxxxfx，··········································2分所以()fx为奇函数．·····················································

······················3分高一数学参考答案第4页共4页（2）解：22,()(),20xkxkxgxfxkxkxkxkx0≤≤2，≤≤，①当0k≥时，因为()gx为2,0和0,2上增函数，所以()gx为2,

2上增函数，所以()gx在2,2上的最大值为(2)4gk；·········································5分②当4k≤时，因为()gx为2,0和0,2上减函数，所以()

gx为2,2上减函数，所以()gx在2,2上的最大值为(2)43gk；····································7分③当40k时，因为2yxkxk在2,2k上是增函数，在,02k上是减函数，

因为2yxkxk在0,2k上是减函数，,22k上是增函数，所以()gx为2,2k上增函数，为,22kk上减函数，,22k上增函数，因此()gx最大值为()2kg和(2)g中较大者，由2816()

(2)024kkkgg≥，得442k≤或442k≥，所以当4442k≤时，()(2)2kgg≥，()gx的最大值为2()24kkgk，所以当4420k时，()(2)2kgg，()gx的最大值为(2)4gk，·······11分综上，当4k≤时，(

)gx的最大值为(2)43gk；当4442k≤时，()gx的最大值为2()24kkgk；当442k时，()gx的最大值为(2)4gk．····································12分