数学试题卷答案山东省九五高中协作体2026届高三年级12月质量检测（九五联考）(12.17-12.18)

2026高三年级12月质量检测(九五联考)数学试题参考答案及评分标准命题及审核：北京时代凤凰教育研究院说明：一、本解答只给出一种解法供参考，如考生的解法与本解答不同，可根据试题的主要考查内容参照评分标准酌情赋分．二、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与难度，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答案应得分数的一半；如果后继部分的解答有较严重的错误或又出现错误，就不再给分．三、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．四、只给整数分数，选择题和填空题不给中间分．一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的。题号12345678答案CBCABABD二、选择题：本题共3小题，每小题6分，共18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6分，部分选对的得部分分，有选错的得0分。题号9答案ACACDACD三、填空题：本题共3小题，每小题5分，共15分。四、解答题：共77分.解答应写出文字说明、证明过程或演算步骤。因为f(x)的图象关于y轴对称，所以f=2sin，······························································4分所以．····················································································6分（2）因为f=2sin所以cos········································7分所以sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα,（1）因为PA丄平面ABCD，CD平面ABCD，所以PA丄CD；······················2分因为AE平面PAD，所以CD丄AE；··················································4分因为PDCD=D，所以AE丄平面PCD；·············································6分（2）以点A为坐标原点，AD为y轴，AP为z轴建立如图所示的空间直角坐标系，·································································································7分因为点G在平面AEF内，所以n.=0，所以2λ-λ-2+2λ=0，所以λ=，因为AB=2AG，所以上GAB························································2分延长BG交AC于点D，所以DG丄AG；因为，所以G为△ABC的重心，····································4分所以BG=2DG；所以tan上GAC==.=；·························因为CE为△ABC的中线，所以4CE2+AB2=2CA2+2CB2，在△ABC中，cos上ACB··························14分当且仅当a=b时取等号，（1）因为f所以g(x)<g(0)=0，··········································································3分22设r(x)=ln(x+1)xax，注意到r(0)=0，设s(x)=a—xx2lna(1+lna)x1，注意到s(0)=0，s,(x)=axlna2xlna(1+lna)，注意到s,(0)=12lna，····················13分s,,(x)=axln2a2lna>0，所以s,(x)在(1，+∞)上单调递增．则s(x)<s(0)=0，故r,(x)<0，r(x)单调递减，所以s故存在x，使得s,(x2)=0，2)时，s,(x)<0，s(x)单调递减，则s(x)<s(0)=0，所以当x∈(一1，0)时，s,(x)<s,(0)=0，所以s(x)单调递减，则s(x)>s(0)=0，故r,(x)>0，r(x)单调递增，r(x)<r(0)=0，符合题意；当x∈(0，+∞)时，s,(x)>s,(0)=0，所以s(x)单调递增，则s(x)>s(0)=0，综上ae．·················································································17分S5的含有5但不含1的“膨胀子集”···············································4分（2）由题意可知，集合Sn一1的“膨胀子集”均为集合Sn的“膨胀子集”；故集合Sn不含元素n的“膨胀子集”的个数为cn一1；·····································5分下面分析集合Sn含有元素n的“膨胀子集”A：将A中n去掉，剩余的所有元素减去1，得到新集合A*，所以A*为Sn一2的“膨胀子集”，共有cn一2个，由于任意一个满足条件的A都对应一个满足条件的A*，且任意一个满足条件的A*都对应一个满足条件的A，所以满足条件的A共有cn一2个．·················9分所以cnn22025c20262025=(c3-c2)+(c5-c4)++(c2025-c2024)+(c2025-c2026)+(c3+23)++(c2023+22023)-c2024-22024=(c2+c3+c5