保密★使用前泉州市2025届高中毕业班适应性练习卷参考答案2025.04高三数学一、选择题：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。1~8：CDCABBDC二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的选项中，有多项符合题目要求。全部选对的得6分，有选错的得0分，部分选对的得部分分。9．AB10．BCD11．ACD三、填空题：本题共3小题，每小题5分，共15分。742212．0.7（也可）13．14．．109π四、解答题：本题共5小题，共77分。解答应写出文字说明，证明过程或演算步骤。15.（13分）某旅行社推出“文化古城游”旅游路线后，为了解游客的满意度，对该路线的游客进行随机抽样调查，得到如下满意度评分的频率分布直方图．（1）估计“文化古城游”路线游客满意度评分的众数和第80百分位数；（同一组中的数据用该区间的中点值为代表）（2）现从参与“文化古城游”的游客中随机抽取3人，设这3人中满意度评分不低于90分的人数为X，求X的分布列和数学期望．（以样本数据中游客的满意度评分位于各区间的频率作为游客的满意度评分位于该区间的概率）【命题意图】本题主要考查频率分步直方图、样本的数字特征、二项分布等基础知识；考查数据处理能力、运算求解能力等；考查数形结合思想、化归与转化思想、或然与必然思想等；体现基础性与应用性，导向对发展数学运算、数据分析、数学建模等核心素养的关注．【试题解析】9095（1）估计“文化古迹游”路线游客满意度评分的众数为92.5，···································1分2高三数学试题第1页（共13页）满意度评分在70,95内的频率为（0.0080.0120.0200.0400.070）50.750.8，故第80百分位数位于[95,100]，·····················································································2分0.80.75所以第80百分位数为95596．（列式1分，计算1分，只写答案扣1分）·········4分0.25（2）解法一：3由表可知，“文化古迹游”线路游客满意度评分不低于90分的概率为（0.0700.050）50.6，5分5依题意，X的可能取值为0,1,2,3，·················································································6分38则PX0(1)3，·························································································7分51253336PX1C1(1)2，······················································································8分3551253354PX2C2(）2（1），·················································································9分355125327PX3（）3，······························································································10分5125所以X的分布列为：X012327P83654125125125125·······························································································································11分83654279所以EX0123．（列式1分，计算1分）·························13分1251251251255解法二：设事件A“抽取一名‘文化古迹游’线路游客的满意度评分不低于90分”，3由表可知，P(A)（0.0700.050）50.6．·································································5分53依题意，X表示事件A发生的次数，则X~B(3,)，·························································7分533于是，X的分布列为P(Xk)Ck()k（1)3k，k0,1,2,3．·········································9分355即：X012327P83654125125125125（需要计算出每个X取值的概率，否则扣2分）······························································11分39所以E(X)3．（答案也可以写为1.8）································································13分55高三数学试题第2页（共13页）16.（15分）我们把公差不为零的等差数列称为一阶等差数列．若{an1an}是一阶等差数列，则称{an}为二阶等差数列．2（1）若annn1，判断{an}是否为二阶等差数列，并说明理由；（2）若{an}是二阶等差数列，且a11,a24,a39．（ⅰ）求{an}的通项公式；an（ⅱ）求数列的前n项和Sn．4an1【命题意图】本题主要考查数列的基本概念、等差数列的定义与通项公式、数列求和等基础知识；考查运算求解能力、推理论证能力等；考查函数与方程思想、化归与转化思想等；体现基础性和综合性，导向对发展数学抽象、数学运算、逻辑推理等核心素养的关注．【试题解析】（）因为2，22，分1annn1an1(n1)(n1)1n3n3·········································1所以an1an2n2，·································································································2分所以(an2an1)(an1an)20，nN，（不等于0没写不扣分）································3分所以{an1an}是一阶等差数列，····················································································4分所以{an}是二阶等差数列．（有判断是二阶等差数列得1分）··············································5分（2）（ⅰ）因为{an}是二阶等差数列，所以{an1an}是一阶等差数列．因为a11,a24,a39，所以a2a13,a3a25，所以{an1an}是首项为3，公差为2的等差数列，····························································6分所以an1an2n1，nN．·····················································································7分当n≥2时，an(anan1)(an1an2)(a2a1)a1(2n1)(2n3)31··························································································8分n(2n11)n2．···································································································9分2当n1时，a11也符合上式，（没检验扣1分）·····························································10分所以的通项公式是2．分{an}ann···················································································11an2（ⅱ）因为n24an14n111(1)·················································································12分44n211111()，····································································13分482n12n1n111111所以S[(1)()()]·····················································14分n483352n12n1n11nnn2n(1)．（后面这三个答案皆可）···················15分482n144(2n1)4n2高三数学试题第3页（共13页）17.（15分）已知函数f(x)ln(x1)aex1．（1）当a1时，求f(x)的单调区间；（2）若f(x)≤lna，求a的取值范围．【命题意图】本题主要考查导数的运算，利用导数研究函数的单调性、不等式恒成立等基础知识；考查运算求解能力、推理论证能力等；考查划归与转化思想、函数与方程思想、数形结合思想等；体现综合性，导向对数学运算、逻辑推理、数学抽象、直观想象等核心素养的关注．【试题解析】（1）当a1时，f(x)ln(x1)ex1，其定义域为(1,)，（写出定义域1分）············1分1则f(x)ex，································································································2分x111x令g(x)ex，则g(x)e0，x1(x1)21所以f(x)ex在区间(1,)内单调递减，（没证明直接写出单调递减扣1分）·········3分x1又f(0)0，··········································································································4分所以当x(1,0)，f(x)0；当x(0,)，f(x)0．所以f(x)的单调递增区间为(1,0)，单调递减区间为(0,)．·········································5分（备注：没写定义域且单调递增区间写成(,0)，扣2分）（2）解法一：11a(x1)ex由题意可知a0，f(x)aex,(x