2025福建高三3月适应性考试数学参考答案

2025届高中毕业班适应性练习卷数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、选择题：本题考查基础知识和基本技能。每小题5分，满分40分。1．C2．D3．A4．A5．B6．C7．A8．B二、选择题：本题考查基础知识和基本技能。每小题6分，满分18分。全部选对的得6分，部分选对的得部分分，有选错的得0分。9．AD10．ABD11．ACD三、填空题：本题考查基础知识和基本技能。每小题5分，满分15分。2112．yx=413．gx()=2x（写出一个满足条件的函数即可）14．16四、解答题：本题考查基础知识、基本技能、基本思想和基本经验，考查发现问题、提出问题、分析问题和解决问题的能力。共5小题，满分77分。15.本小题主要考查数列的基本概念、等比数列等基础知识，考查运算求解能力、逻辑推理能力等，考查函数与方程思想、分类与整合思想等，考查数学运算、逻辑推理等核心素养，体现基础性和综合性．满分13分．解法一：（1）因为aSnn+1−=1，所以当n2时，aSnn−=−11，··················································1分所以an+−11−−+aSSnnn=0，························································································2分（）所以ann+1=22an．·································································································4分当n=1时，aS21=+=12，满足aa21=2，也符合aann+1=2．·············································5分an+1=因为a1≠0，所以an≠0，2，an所以{an}是以1为首项，2为公比的等比数列，································································6分n−1所以an=2．··········································································································7分数学参考答案及评分细则第1页（共19页）2−n（2）由（1）知a=2n1，所以b=，·······································································8分nn2n−1(n+1)2bn(n+1)2所以n+1=2=．······················································································9分22bnnn22n−1b令n+1>1，解得−21+1，又因为bn>0，所以bbnn+1>；·······································11分bn*同理，当n3且n∈N时，bbnn+1<．··········································································12分9所以b的最大值为b=．························································································13分n34解法二：（1）同解法一．···································································································7分2−n（2）由（1）知a=2n1，所以b=，·······································································8分nn2n−1(nn+1)221所以bb−=−=(−+nn22+1)．·······························································9分nn+12n22nn−1令，解得，分bbnn+1−>0−21+；··············································································11分*同理，当n3且n∈N时，bbnn+1<．··········································································12分9所以b的最大值为b=．························································································13分n34解法三：（1）同解法一．···································································································7分2−n（2）由（1）得a=2n1，所以b=，·······································································8分nn2n−1bbnn−1,当n2时，令·······························································································9分bbnn+1,nn22(−1),nn−−12则22解得21++n22，······································································11分22nn(+1)nn−1,22又n∈N*，所以n=3．·······························································································12分数学参考答案及评分细则第2页（共19页）9又因为b=1，b=，所以bb<，134139所以b的最大值为b=．·························································································13分n34解法四：（1）同解法一．···································································································7分2−n（2）由（1）知a=2n1，所以b=．·······································································8分nn2n−1x2设fx()=(x>0)，································································································9分2x−1xx(2−ln2)则fx′()=，·····························································································10分2x−122所以当x∈(0,)时，fx′()>0；当x∈(,)+∞时，fx′()<0．ln2ln222所以fx()在(0,)单调递增，在(,)+∞单调递减．····················································11分ln2ln229又23<<，b=fn()，b=2，b=，ln2n234所以bbbbb12345<<>>>，·····················································································12分9所以b的最大值为b=．·························································································13分n3416．本小题主要考查直线与平面垂直的判定与性质、平面与平面垂直的判定与性质、直线与平面所成的角、二面角、点到直线的距离、空间向量和解三角形等基础知识，考查直观想象能力、逻辑推理能力、运算求解能力等，考查函数与方程思想、数形结合思想、化归与转化思想等，考查直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性．满分15分．解法一：（1）由题可知，∠=°ADC45，CD=22，AC=AP=2，CDAC在△ACD中，由正弦定理，得=.sin∠∠CADsinADC222所以=，sin∠=CAD1，··········································································2分sin∠CAD22所以∠=°CAD90，即AD⊥AC，·················································································3分所以AD=2.又AP=2，PD=22，所以AP222+=ADPD，即AD⊥AP.···········································5分又因为APAC=A，AP,AC⊂平面PAC，所以AD⊥平面PAC.······································6分又因为AD⊂平面ACD，所以平面PAC⊥平面ACD.·······················································7分数学参考答案及评分细则第3页（共19页）（2）过M作ME⊥AC于点E，过E作EF⊥CD于点F，连接MF.因为平面PAC⊥平面ACD，平面PAC∩平面ACD=AC，ME⊂平面PAC，所以ME⊥平面ACD.··················································