2026高三年級(jí)12月質(zhì)量檢測(cè)(九五聯(lián)考)數(shù)學(xué)試題參考答案及評(píng)分標(biāo)準(zhǔn)命題及審核：北京時(shí)代鳳凰教育研究院說明：一、本解答只給出一種解法供參考，如考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容參照評(píng)分標(biāo)準(zhǔn)酌情賦分．二、當(dāng)考生的解答在某一步出現(xiàn)錯(cuò)誤時(shí)，如果后繼部分的解答未改變?cè)擃}的內(nèi)容與難度，可視影響的程度決定后繼部分的給分，但不得超過該部分正確答案應(yīng)得分?jǐn)?shù)的一半；如果后繼部分的解答有較嚴(yán)重的錯(cuò)誤或又出現(xiàn)錯(cuò)誤，就不再給分．三、解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)．四、只給整數(shù)分?jǐn)?shù)，選擇題和填空題不給中間分．一、選擇題：本題共8小題，每小題5分，共40分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。題號(hào)12345678答案CBCABABD二、選擇題：本題共3小題，每小題6分，共18分。在每小題給出的四個(gè)選項(xiàng)中，有多項(xiàng)符合題目要求。全部選對(duì)的得6分，部分選對(duì)的得部分分，有選錯(cuò)的得0分。題號(hào)9答案ACACDACD三、填空題：本題共3小題，每小題5分，共15分。四、解答題：共77分.解答應(yīng)寫出文字說明、證明過程或演算步驟。因?yàn)閒(x)的圖象關(guān)于y軸對(duì)稱，所以f=2sin，······························································4分所以．····················································································6分（2）因?yàn)閒=2sin所以cos········································7分所以sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα,（1）因?yàn)镻A丄平面ABCD，CD平面ABCD，所以PA丄CD；······················2分因?yàn)锳E平面PAD，所以CD丄AE；··················································4分因?yàn)镻DCD=D，所以AE丄平面PCD；·············································6分（2）以點(diǎn)A為坐標(biāo)原點(diǎn)，AD為y軸，AP為z軸建立如圖所示的空間直角坐標(biāo)系，·································································································7分因?yàn)辄c(diǎn)G在平面AEF內(nèi)，所以n.=0，所以2λ-λ-2+2λ=0，所以λ=，因?yàn)锳B=2AG，所以上GAB························································2分延長(zhǎng)BG交AC于點(diǎn)D，所以DG丄AG；因?yàn)椋訥為△ABC的重心，····································4分所以BG=2DG；所以tan上GAC==.=；·························因?yàn)镃E為△ABC的中線，所以4CE2+AB2=2CA2+2CB2，在△ABC中，cos上ACB··························14分當(dāng)且僅當(dāng)a=b時(shí)取等號(hào)，（1）因?yàn)閒所以g(x)<g(0)=0，··········································································3分22設(shè)r(x)=ln(x+1)xax，注意到r(0)=0，設(shè)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，+∞)上單調(diào)遞增．則s(x)<s(0)=0，故r,(x)<0，r(x)單調(diào)遞減，所以s故存在x，使得s,(x2)=0，2)時(shí)，s,(x)<0，s(x)單調(diào)遞減，則s(x)<s(0)=0，所以當(dāng)x∈(一1，0)時(shí)，s,(x)<s,(0)=0，所以s(x)單調(diào)遞減，則s(x)>s(0)=0，故r,(x)>0，r(x)單調(diào)遞增，r(x)<r(0)=0，符合題意；當(dāng)x∈(0，+∞)時(shí)，s,(x)>s,(0)=0，所以s(x)單調(diào)遞增，則s(x)>s(0)=0，綜上ae．·················································································17分S5的含有5但不含1的“膨脹子集”···············································4分（2）由題意可知，集合Sn一1的“膨脹子集”均為集合Sn的“膨脹子集”；故集合Sn不含元素n的“膨脹子集”的個(gè)數(shù)為cn一1；·····································5分下面分析集合Sn含有元素n的“膨脹子集”A：將A中n去掉，剩余的所有元素減去1，得到新集合A*，所以A*為Sn一2的“膨脹子集”，共有cn一2個(gè)，由于任意一個(gè)滿足條件的A都對(duì)應(yīng)一個(gè)滿足條件的A*，且任意一個(gè)滿足條件的A*都對(duì)應(yīng)一個(gè)滿足條件的A，所以滿足條件的A共有cn一2個(gè)．·················9分所以cnn22025c20262025=(c3-c2)+(c5-c4)++(c2025-c2024)+(c2025-c2026)+(c3+23)++(c2023+22023)-c2024-22024=(c2+c3+c5