泉州市2023届高三适应性练习卷 数学答案

保密★使用前泉州市2023届高三适应性练习卷一、二选择题答案2023.051-8DACBACBB9.ACD10.AC11.AD12.BCD三、填空题：本题共4小题，每小题5分，共20分。13．曲线ysin2x在点(0,0)处的切线方程为_________．【命题意图】本小题主要考查导数的几何意义等基础知识；考查运算求解等；体现基础性，导向对数学运算等核心素养的关注．【试题解析】因为，所以，所以，可得曲线在点ysin2xy2cos2xy|x02ysin2x(0,0)处的切线方程为：y2x．14．已知定义在R上的函数f(x)满足：f(x2)为偶函数；当x(,2]时，f(x)x2．写出f(x)的一个单调递增区间为_________．【命题意图】本小题主要考查函数的对称性和单调性等基础知识；考查推理论证能力等；考查数形结合思想、化归与转化思想等，体现基础性，导向对直观想象等核心素养的关注．【试题解析】因为f(x2)为偶函数，所以f(x2)f(x2)，因此f(x)的图象关于直线x2对称．当x(,2]时，f(x)x2，所以f(x)在(,0]单调递减，在[0,2]单调递增．由对称性，可得f(x)在[2,4]单调递减，在[4,)单调递增．故f(x)的一个单调递增区间为[0,2]，[4,)．注：只要写出一个答案就可以，同时区间端点除处只能用开区间外，其他的用开闭符号都可以。315．在平面直角坐标系xOy中，已知抛物线C:y26x的焦点为F，过D(,0)的直线l与C2|AF|交于A,B两点．若△ABF的面积等于△OAD的面积的2倍，则________．|BF|【命题意图】本小题主要考查抛物线的定义及标准方程、直线与抛物线的位置关系等基础知识；考查运算求解等；考查数形结合思想、化归与转化思想等，体现基础性，导向对数学运算、直观想象等核心素养的关注．高三数学试题第1页（共12页）33【试题解析】由题意，知抛物线C:y26x的焦点F(,0)，准线l:x．设O,F到直线AB22d|DF|1的距离分别为，则2，又，d1,d22S△OAD|DA|d1d1|DO|21S|AB|d，从而△FAB2，所以．分别过点S△FAB|AB|d22|DA||AB|2S△OAD|DA|d1作于，于，则，，∥，A,BAA1lA1BB1lB1|FA||AA1||FB||BB1|AA1BB1|FA||AA||DA|11所以1，故答案为．|FB||BB1||DB|2216．将0，1，2，3，10任意排成一行，可以组成个不同的6位数．（用数字作答）【命题意图】本小题主要考查分类加法计数原理、分步乘法计数原理、排列等基础知识；考查运算求解等能力；考查化归与转化等思想；体现综合性，导向对数学运算、数学建模等核心素养的关注．4【试题解析】解法一：将0，1，2，3，10任意排成一行，且数字0不在首位，有4A496种，4数字1和0相邻且1在0之前的排法有A424种，24故所求满足题意的6位数有9684个.2解法二：将0，1，2，3，10任意排成一行，可以组成满足题意的6位数的情况分为如下三类：2第一类，1和0相邻，且1在0之前，此时有A412个；3第二类，1和0相邻，且0在1之前，此时有3A318个；312第三类，1和0不相邻，此时有A3(A3A3)54个；故所求满足题意的6位数共有12185484个.高三数学试题第2页（共12页）四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。17．（10分）数列{an}中，a11，且an12ann1．（1）证明：数列{ann}为等比数列，并求出an；（2）记数列{bn}的前n项和为Sn．若anbn2Sn，求S11．【命题意图】本小题主要考查等比数列的定义与前n项和等基础知识，考查运算求解能力，考查函数与方程、化归与转化、分类与整合等思想，体现基础性和综合性，导向对发展数学运算等核心素养的关注．【试题解析】解法一：（1）由an12ann1可得an1n12(ann)，····································1分因为a11，所以a112，····························································2分所以数列{ann}是首项为2，公比为2的等比数列．···························3分nn所以ann2，即an2n．······················································4分（2）因为anbn2Sn，当n…2时，an1bn12Sn1，·························································5分所以anan1bnbn12(SnSn1)，即bn1bnanan1．·················6分nn1n1又anan12n[2(n1)]21，n1所以bn1bn21．··································································7分又因为a1b12S1，所以b1a11．················································8分故分S11b1b2b11b1(b2b3)(b10b11)······························91(22242628210)51360．············································································10分解法二：（1）因为an12ann1，an1n12ann1n1所以2，·············································1分annann高三数学试题参考答案第1页（共15页）因为a11，所以a112，····························································2分所以数列{ann}是首项为2，公比为2的等比数列．···························3分nn所以ann2，即an2n．······················································4分（2）因为anbn2Sn，当n…2时，anSnSn12Sn，即SnSn1an．·······························5分所以Sn1Snan1，因此Sn1Sn1an1an．··································6分n1nn又an1an2(n1)(2n)21，n所以Sn1Sn121．··································································7分又因为a1b12S1，所以S1b11．················································8分所以S11(S11S9)(S9S7)(S7S5)(S5S3)(S3S1)S1··············9分(2101)(281)(261)(241)(221)11360．·········································································10分解法三：（1）同解法一．··················································································4分（2）因为anbn2Sn，当n…2时，an1bn12Sn1，所以anan1bnbn12(SnSn1)，即bn1bnanan1．·················5分nn1n1n1又anan12n[2(n1)]21，所以bn1bn21，·······6分2n12n11所以b(b).····················································7分n32n132215又因为ab2S，所以ba1，b．···························8分1111113262n15所以数列{b}是首项为，公比为1的等比数列．n3262n1552n1因此b(1)n1，b(1)n1，··························9分n326n632nn5n1212n(1)n1这时ab25n1，Snn632(1)n1n22312242125111故S1360．·····················································10分1131224解法四：（1）同解法一．··················································································4分（2）因为anbn2Sn，高三数学试题参考答案第2页（共15页）当n…2时，anSnSn12Sn，n所以SnSn1an，即SnSn12n．········································5分2n1n12nn11所以S(S)．···································7分n324n132422115又因为ab2S，所以ba1，S．····················8分111111324122n1n15所以数列{S}是首项为，公比为1的等比数列．n324122n1n1552n1n1因此S(1)n1，S(1)n1，·········9分n32412n123245212111故S1360．······················································10分111232418．（12分）在平面四边形ABCD中，ABC60，ADC120，点B,D在直线AC的两侧，AB1，BC2.（1）求BAC；（2）求△ABD与△ACD的面积之和的最大值．【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，考查推理论证、运算求解等能力，考查数形结合和化归与转化等思想，体现综合性与应用性，导向对发展直观想象、逻辑推理及数学运算等核心素养的关注．【试题解析】（1）在△ABC中，由余弦定理，得AC2AB2BC22ABBCcosABC．·······2分因为AB1，BC2，ABC60，1所以AC212222123，即AC3．···········································3分2所以BC2AC2AB2，·························································