河南省實驗中學20232024學年上期期中試卷答案（高一）一、單選題（共8小題）14BCCA58AABD二、多選題（共4小題）9.ABC10.ABC11.AC12.ABD三、填空題（共4小題）13.314.﹣ex+2x+115.[-3，3]16.四．解答題（共6小題）17．解：（1）原式=(=327+27-21=3+27﹣21＝9（2）原式＝lg52+lg2?lg50+（lg2）2﹣eln8＝2lg5+lg2（1+lg5）+（lg2）2﹣8＝2lg5+lg2+lg2?lg5+（lg2）2﹣8＝2lg5+lg2+lg2（lg5+lg2）﹣8＝2（lg5+lg2）﹣8＝﹣6．················································（10分）18．解：（1）當m＝﹣1時，A＝{x|﹣3≤x≤1}，集合B＝{x|﹣1＜x≤2}，所以?UB＝{x|x＞2或x≤﹣1}，···········································（2分）所以①A∪B＝{x|﹣3≤x≤2}；············································（4分）②A∩（?UB）＝{x|﹣3≤x≤﹣1}；········································（6分）（2）若A∩B＝?，當A＝?時，2m﹣1＞m+2，即m＞3，······································（8分）當A≠?時，2m解得m≤﹣3或32＜m≤3，··········································綜上，m的范圍為{m|m≤﹣3或m＞32}．··································（19.解：（1）∵f(x)=(3m2-2m)解得m＝1或m=-13，又∵冪函數在（0，+∞）上單調遞增，∴m-12＞0，得m＝1（2）由第一問得f(x)=x12，在[1所以f（x）的值域為[1，2]，即集合A＝{x|1≤x≤2}，·······················（6分）而g（x）＝﹣3x+t在[1，4]上遞減，所以g（x）的值域為[t﹣81，t﹣3]，即B＝{x|t﹣81≤x≤t﹣3}，··············································（8分）由命題q是命題p的必要不充分條件可得A?B，···························（10分）所以t-3≥2t-81≤1，解得5即t的取值范圍為[5，82]．··············································（12分）20．解：（1）由f（x+y）＝f（x）+f（y），令x＝y＝0得f（0）＝f（0）+f（0），∴f（0）＝0．·························································（2分）（2）f（x）是奇函數，證明：f（x）定義為R，關于原點對稱·································（3分）由f（x+y）＝f（x）+f（y），令y＝﹣x，得f（x﹣x）＝f（x）+f（﹣x），即f（x）+f（﹣x）＝f（0）＝0，f（﹣x）＝﹣f（x），所以f（x）是奇函數．······························（6分）（3）任取x1，x2∈R，x1＜x2，x2﹣x1＞0，·······························（7分）由f（x+y）＝f（x）+f（y）知f（x+y）﹣f（x）=f（y）f（x1）﹣f（x2）＝f（x1﹣x2）＝﹣f（x2﹣x1），···························（8分）由于x2﹣x1＞0，所以f（x2﹣x1）＜0，所以f（x1）﹣f（x2）＝﹣f（x2﹣x1）＞0，即f（x1）＞f（x2），所以f（x）是減函數，·················································（9分）f（6）＝f（3+3）＝f（3）+f（3）＝﹣8，································（10分）所以不等式f（t﹣1）+f（t）＜﹣8即f（t﹣1+t）＜f（6），所以2t﹣1＞6，t＞7所以不等式f（t﹣1）+f（t）＜﹣8的解集為（72，+∞）．····················（1221．解：（1）由題意得W（x）＝800x﹣R（x）﹣250，∵R（x）=10∴當0＜x＜40時，R（x）＝10x2+200x+1000，則W（x）＝800x﹣（10x2+200x+1000）﹣250＝﹣10x2+600x﹣1250，·········（2分）當x≥40時，R（x）＝701x+10000x則W（x）＝800x﹣（801x+10000x-8450）﹣250＝﹣x-10000綜上所述，W（x）=-10x2+600x-1250，（2）由（1）得W（x）=-10則當0＜x＜40時，W（x）＝﹣10x2+600x﹣1250＝﹣10（x﹣30）2+7750，二次函數W（x）的圖象開口向下，且對稱軸為直線x＝30，∴W（x）max＝W（30）＝7750，·········································（8分）當x≥40時，W（x）＝﹣x-10000x又x+10000x≥2x?10000x=200，當且僅當∴W（x）＝﹣x-10000x+8200≤﹣200+8200＝8000，···················∵8000＞7750，∴2023年產量為100（千部）時，企業所獲利潤最大，最大利潤是8000萬元．··（12分）22．解：（1）∵f（x）為R上的奇函數，∴f（0）＝0，可得b＝1················································（1分）又∵f（﹣1）＝﹣f（1）∴1-2-12-1+a=-1-22+a經檢驗當a＝1且b＝1時，f（x）=1-滿足f（﹣x）＝﹣f（x）是奇函數．·······································（3分）故a＝1，b＝1·························································（4分）（2）由（1）得f（x）=1-2x任取實數x1、x2，且x1＜x2················································（5分）則f（x1）﹣f（x2）=22x1∵x1＜x2，可得2x1∴f（x1）﹣f（x2）＞0，即f（x1）＞f（x2），······························（7分）∴函數f（x）在（﹣∞，+∞）上為減函數；·······························（8分）（3）根據（1）（2）知，函數f（x）是奇函數且在（﹣∞，+∞）上為減函數．∴不等式f（t2﹣2t）+f（2t2﹣k）＜0恒成立，即