【数学大题解析】福建省泉州市2023届高中毕业班质量监测（二）

保密★使用前泉州市2023届高中毕业班质量监测（二）2023.01高三数学参考答案17．（10分）在梯形ABCD中，AD∥BC，ADCD，BC3，ACsinBCA3ABcosABC．（1）若△ABC的面积为33，求AC；（2）若CD3，求tanBAC．【命题意图】本小题主要考查解三角形、三角恒等变换等基础知识；考查运算求解能力等；考查函数与方程思想，化归与转化思想等；体现基础性，导向对数学运算、逻辑推理、直观想象等核心素养的关注．【试题解析】解法一：ACAB解：（1）在△ABC中，由正弦定理，得，················1分【正弦定理】sinABCsinACB化简得ACsinACBABsinABC，代入ACsinACB=3ABcosABC，得到sinABC3cosABC，即tanABC3．······························································2分【运算求解】因为ABC0,，所以ABC．·······················································3分3由1，得到．分【面积公式】S△ABBCsin33AB4························4ABC23在△ABC中，由余弦定理，得AC2AB2BC22ABBCcos42324313，3所以AC13．···································································5分【余弦定理】2（2）设BAC，因为AD∥BC，所以CAD．···································6分3ACBC33在△ABC中，由正弦定理，得，所以AC．·············7分sinABCsin2sin高三数学试题参考答案第1页CD3在Rt△ADC中，sinCAD，所以AC．···························8分AC2sin33332所以，即3sin2sin，2sin23sin331所以，·································9分【两角差正弦公式】3cossin2sin22得到tan33，即tanBAC33．······················································10分解法二：解：（1）同解法一.······························································································1分（2）过A作AEBC，垂足为E，AE在Rt△ABE中，AECD3，所以AB2，····························6分sinABC在△ABC中，由余弦定理，得1AC2AB2BC22ABBCcosABC22322237，2所以AC7，·····················································································7分AB2AC2BC21由余弦定理得cosBAC，·····································9分2ABAC2733所以sinBAC，tanBAC33.······················································10分27高三数学试题参考答案第2页解法三：解：（1）同解法一．··························································································5分CD3（2）在Rt△BCD中，tanCBD，所以CBD，BC36过A作AEBC，垂足为E，AE在Rt△ABE中，AECD3，所以AB2，··························6分sinABC在△ABD中，因为ABDADB，所以ADAB2，······7分【平几知识】6CD3在Rt△ADC中，tanCAD，··················································8分AD2332所以tanBACtan(CAD)233.33132················································10分【两角差正切公式和运算求解，各1分】解法四：解：（1）同解法一．··························································································5分（2）过A作AEBC，垂足为E，在Rt△ABE中，AECD3，AE所以BE1，CEBCBE2,BAE，································7分tanABC6CE223在Rt△AEC中，tanCAE，············································8分AE3332333在△ABC中,tanBACtan(BAECAE)tan(CAE)33.6323133················································10分【两角和正切公式和运算求解，各1分】高三数学试题参考答案第3页解法五：解：（1）同解法一．··························································································5分（2）过A作AEBC，垂足为E，在Rt△ABE中，AECD3，AE所以BE1，CEBCBE2，···············································6分tanABC以E为原点，BC所在直线为x轴建立平面直角坐标系，如图所示：则B(1,0),C(2,0),A(0,3)，所以AB(1,3),AC(2,3).························································7分ABAC1在△ABC中,cosBACcosAB,AC，····························9分ABAC2733所以sinBAC，tanBAC33.······················································10分27解法六：解：（1）同解法一．··························································································5分（2）过A作AEBC，垂足为E，AE在Rt△ABE中，AECD3，所以AB2，···························6分sinABC高三数学试题参考答案第4页1设BAa,BCb，则ab233，ACba，222AC(ba)2b2baa7，···········································7分ABACa(ba)1在△ABC中,cosBACcosAB,AC，········9分ABAC272733所以sinBAC，tanBAC33····················································10分27【说明：本题在无其他解答得分的情况下，若能正确作出满足题干条件的直角梯形，可以给1分.】高三数学试题参考答案第5页保密★使用前泉州市2023届高中毕业班质量监测（二）2023.01高三数学参考答案18．（12分）31已知数列a满足a，2aa．n12n1n2n1（）求，及的通项公式；1a2a3an（2）求数列|an|的前n项和Sn．【命题意图】本题主要考查数列的递推，数列前n项和等知识；考查运算求解能力，推理论证能力、抽象概括能力等；考查化归与转化思想，分类与整合思想等；体现考查逻辑推理，数学运算等核心素养的命题意图．【试题解析】解法一：3111解：（1）a，aa，可得a，··················································1分12n12n2n241a，·······························································································2分381n1n由2aa可得2a2a2，··················································3分n1n2n1n1nn所以数列2an是以2a13为首项，公差为2的等差数列，·························4分n于是2an3n122n5，····························································5分2n5*所以annN．·······································································6分2n2n5（2）a，n1,2时an0，n3时，an0，n2n3317于是S，S，···································································7分12224431132n72n5当时，，n…3Sn234n1n222222高三数学试题参考答案第1页(共23页)131132n72n5S，2n222324252n2n1132222n5两式相减得：S0，································9分2n222242n2n11111232n32n5512n552n1S1，·······················10分n1n1n1n1n121242242252n1所以S（n3），································································11分n22n7又S也符合上式，243,n1,2综上：S························12分（如果n2没有合并，不扣分）n52n1,n2.22n解法二：3111解：（）因为，，可得，1a1an1anna22224····································