【山东卷】.山东省青岛市四区县（胶州、平度、黄岛、城阳）2024-2025学年高三上学期期中学业水平检测考试（11.6-11.8）数学试卷答案

一、选择题：本题共8小题，每小题5分，共40分。1-8：ADBBCCDA二、多项选择题：本大题共3小题，每小题6分，共18分。9．AD10．BCD11．ACD三、填空题：本大题共3小题，每小题5分，共15分。12．g(x)=x（答案不唯一）13．202514．5，乙（第一空3分，第二空2分）四、解答题：本大题共5小题，共77分。解答应写出文字说明、证明过程或演算步骤。 解:（1）依题意， 即x=kπ+,k∈Z时，f(x)max=，··························································4分由g(x)在区间(0,)上单调递增，解1）设等差数列{an}的公差为d（d≠0a2是a1和a4的等比中项，+d)2=9，····························································6分所以{bn+1-bn-10}是首项为-2，公比为2的等比数列，·········································9分n+1-bn-10=-2n，所以bn+1-bn=10-2n 13分24 n取最大值时n的值为4 即sinAcosC+3sinAsinCsinAcosCcosAsinCsinC=0··························3分因为A是ΔABC的内角，所以A=60o··································································6分所以·······························································8分所以·····································································10分)，sin2θ··················································································12分所以f(—)=0，且f(x)=aeax—1为增函数，······················································4分当x<时，f(x)<0，f(x)在(∞,)单调递减；当x>时，f(x)>0，f(x)在(,+∞)单调递增；af(x)在(—∞,)单调递减，在(,+∞)单调递增；所以f(0)=0>f(—)···················································································7分所以存在唯一x0∈(lna,)，使得f(x0)=综上，当0<a<1时，函数f(x)有两个零点·····························································10分设曲线y=g(x)+kx+b与直线y=kx+b切点(x0,y0)，则{，0l2exex0x00l2exex0x0=0易证ex≥x+1，等号当且仅当x=0时成立；即···········································································6分定义f=数列记Q表示1,2,…,中任取m项构成的单调递增数列的全体，对于Q中的任意数列b1,b2,…,bm，因为bk+1所以a1,a2,…,am单调递增数列······································································1