2026年广东省深圳市高三数学月考提优高频考点诊断卷（聚焦函数压轴与几何综合含答案详解与评分标准）S4A99

2026年广东省深圳市高三数学月考提优高频考点诊断卷（聚焦函数压轴与几何综合，含答案详解与评分标准）S4A992026年广东省深圳市高三数学月考提优高频考点诊断卷（聚焦函数压轴与几何综合，含答案详解与评分标准）S4A99适用对象：广东省深圳市高三学生考试时间：120分钟满分：150分答题说明：本卷用于月考阶段诊断与提优训练，题目按由易到难设置。请在规定时间内独立完成，书写过程要完整、规范。学校：__________班级：__________姓名：__________考号：__________注意事项：选择题用规定符号作答；填空题只填写最终结果；解答题应写出必要的推理、计算与结论。

2026年广东省深圳市高三数学月考提优高频考点诊断卷（聚焦函数压轴与几何综合，含答案详解与评分标准）S4A99学校：__________班级：__________姓名：__________考号：__________考试时间：120分钟满分：150分答题说明：本卷共22题。题卷正文不得书写答案以外的提示；所有解答题须写明主要步骤，结果保留准确值。一、单项选择题（本题共8小题，每小题5分，共40分。每小题只有一个选项符合题意。）1.（5分）已知函数f(x)=√(x+1)+ln(2−x)，则f(x)的定义域为A.(−1,2)B.[−1,2)C.[−1,2]D.(−∞,2)2.（5分）已知f(x)=x²−2x+3，则f(x+1)−f(x)=A.2x+1B.−2x+1C.2x−1D.13.（5分）函数f(x)=x³−3x²+2的单调递减区间是A.(−∞,0)B.(0,2)C.(2,+∞)D.(−∞,+∞)4.（5分）圆C：(x−1)²+(y+2)²=4与直线x−y+m=0相切，则m的取值为A.−3B.−3+2√2C.−3−2√2D.−3±2√25.（5分）在△ABC中，AB=3，AC=4，∠A=60°，则BC=A.√13B.5C.7D.√376.（5分）等比数列{aₙ}满足a₁=2，公比q=1/2，则前n项和Sₙ=A.2(1−2⁻ⁿ)B.4(1−2⁻ⁿ)C.4(1+2⁻ⁿ)D.2ⁿ−17.（5分）若函数f(x)=eˣ−ax（a>0）的最小值为0，则a=A.e⁻¹B.1C.eD.e²8.（5分）函数h(x)=lnx+x−2（x>0）的零点个数为A.0B.1C.2D.3二、多项选择题（本题共4小题，每小题5分，共20分。全部选对得5分，部分选对得2分，有错选得0分。）9.（5分）设f(x)=lnx−x+1（x>0），下列结论正确的是A.f(1)=0B.f(x)≤0C.f′(x)=0的解为x=1D.f(x)=−1有两个正根10.（5分）已知向量a=(1,2)，b=(t,−1)，下列命题正确的是A.a⊥b时t=2B.|a+b|²=t²+2t+2C.a∥b时t=−1/2D.t=2时投影为3/√511.（5分）四棱锥P-ABCD的底面ABCD是边长为2的正方形，PA⊥平面ABCD，PA=2。下列说法正确的是A.体积为8/3B.PC=2√3C.PC与底面所成角为45°D.点P到BD的距离为√612.（5分）设fₐ(x)=x³−3ax。下列判断正确的是A.a>0时极值点为±√aB.a=1时fₐ(x)=0有三个不同实根C.a<0时fₐ(x)在R上单调递增D.a=0时x=1处切线为y=3x−2三、填空题（本题共4小题，每小题5分，共20分。请把答案填写在横线上。）13.（5分）曲线y=lnx在x=e处的切线方程为：____________________________14.（5分）椭圆x²/4+y²/b²=1（0<b<2）的焦距为2√3，则其离心率为：____________________________15.（5分）方程sin2x=√3sinx在区间[0,2π]上的解集为：____________________________16.（5分）函数f(x)=|x²−2x−3|，若方程f(x)=m有四个不同实根，则m的取值范围为：____________________________四、解答题（本题共6小题，共70分。解答应写出文字说明、演算步骤或证明过程。）17.（10分）已知函数f(x)=2sinxcosx+2cos²x−1。（1）求f(x)的最小正周期和最大值；（2）求不等式f(x)≥1在区间[0,π]上的解集。答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________18.（12分）某校高三年级组织函数与几何综合限时训练，随机抽取100名学生的达标情况如下表：类别几何达标几何未达标合计函数达标521870函数未达标201030合计7228100（1）从这100名学生中任选1人，求其至少一项达标的概率；（2）在几何达标的学生中任选1人，求其函数也达标的概率；（3）按函数达标与未达标分层抽取10人，再从这10人中任选2人，记X为被选2人中函数达标的人数，求X的分布列与数学期望。答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________19.（12分）如图形关系所示，四棱锥P-ABCD的底面ABCD为边长2的正方形，PA⊥平面ABCD，PA=2。（1）证明：BD⊥平面PAC；（2）求平面PBC与平面ABCD所成二面角的大小；（3）求点A到平面PBC的距离。答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________20.（12分）已知椭圆E：x²/4+y²/3=1，点P(1,0)。过P的直线l：y=k(x−1)与椭圆交于M、N两点。（1）求椭圆E的焦点坐标和离心率；（2）当k=1时，求弦MN的长度；（3）若直线OM与ON的斜率之积为−3，求k的值。答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________21.（12分）设函数f(x)=eˣ−ax−1，其中a为实数。（1）当a=1时，证明f(x)≥0；（2）讨论f(x)的单调性；（3）若方程f(x)=0恰有两个不同实根，求a的取值范围。答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________________________________________________________________________答：________________________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2026年广东省深圳市高三数学月考提优高频考点诊断卷（聚焦函数压轴与几何综合，含答案详解与评分标准）S4A99参考答案与解析一、答案速查题号12345678答案BCBDABCB题号910111213141516答案ABCDABCABDABCDy=x/e√3/2{0,π/6,π,11π/6,2π}0<m<4二、逐题解析、评分标准与易错提醒1.解析解析：定义域由x+1≥0与2−x>0同时确定，得x≥−1且x<2，故为[−1,2)。评分标准：5分：列出两个限制条件2分，求交集2分，选B1分。易错提醒：根式允许被开方数为0，ln的真数必须严格大于0。2.解析解析：f(x+1)=(x+1)²−2(x+1)+3=x²+2，故f(x+1)−f(x)=2x−1。评分标准：5分：展开2分，代入相减2分，选C1分。易错提醒：不要把f(x+1)误写为f(x)+1。3.解析解析：f′(x)=3x²−6x=3x(x−2)，当0<x<2时f′(x)<0，所以递减区间为(0,2)。评分标准：5分：求导2分，判断符号2分，选B1分。易错提醒：单调区间应写开区间，不能把两个递增区间合并。4.解析解析：圆心为(1,−2)，半径为2。圆心到直线x−y+m=0的距离为|1−(−2)+m|/√2=|m+3|/√2。相切得|m+3|=2√2，故m=−3±2√2。评分标准：5分：圆心半径1分，距离公式2分，解方程1分，选D1分。易错提醒：距离公式分母是√(1²+(−1)²)，不要漏掉绝对值。5.解析解析：由余弦定理，BC²=3²+4²−2·3·4·cos60°=13，故BC=√13。评分标准：5分：写出余弦定理2分，代入2分，选A1分。易错提醒：60°的余弦为1/2，不是√3/2。6.解析解析：Sₙ=a₁(1−qⁿ)/(1−q)=2[1−(1/2)ⁿ]/(1/2)=4(1−2⁻ⁿ)。评分标准：5分：公式2分，代入化简2分，选B1分。易错提醒：公比q=1/2时分母1−q=1/2。7.解析解析：f′(x)=eˣ−a。a>0时极小值点为x=lna，最小值为a−alna=a(1−lna)。令其为0，得lna=1，所以a=e。评分标准：5分：求导与极值点2分，最小值表达式2分，选C1分。易错提醒：不能把eˣ=ax的解误认为x=a。8.解析解析：h′(x)=1/x+1>0，h(x)在(0,+∞)上严格递增；且x→0⁺时h(x)→−∞，x→+∞时h(x)→+∞，故有且仅有1个零点。评分标准：5分：单调性2分，端点趋势2分，选B1分。易错提醒：判断零点个数时要同时说明存在性与唯一性。9.解析解析：f(1)=0；f′(x)=1/x−1，x=1为唯一驻点，且f在x=1处取最大值0，所以f(x)≤0；由图像或单调性知f(x)=−1在(0,1)、(1,+∞)各有一解。评分标准：5分：四项判断各1分，完整选择ABCD1分。易错提醒：f(x)≤0是全定义域结论，不能只代入特殊点。10.解析解析：a·b=t−2，故a⊥b时t=2；a+b=(t+1,1)，|a+b|²=(t+1)²+1=t²+2t+2；平行条件1·(−1)−2t=0，得t=−1/2；t=2时a·b=0，投影长度为0。评分标准：5分：A、B、C各1分，排除D1分，完整选择ABC1分。易错提醒：投影长度要用点积除以|b|，垂直时为0。11.解析解析：体积V=(1/3)·2²·2=8/3；AC=2√2，PC=√(PA²+AC²)=2√3；PC与底面夹角满足sinθ=PA/PC=1/√3，不是45°；用坐标或距离公式可得点P到BD的距离为√6。评分标准：5分：A、B、D各1分，排除C1分，完整选择ABD1分。易错提醒：线面角不能直接看空间线段示意，应转化为正弦比。12.解析解析：f′ₐ(x)=3x²−3a。a>0时驻点为±√a；a=1时f(x)=x(x²−3)，有三个不同实根；a<0时f′ₐ(x)=3(x²−a)>0，单调递增；a=0时f=x³，x=1处切线斜率3，方程y=3x−2。评分标准：5分：四项各1分，完整选择ABCD1分。易错提醒：参数a的符号会改变导数零点的存在性。13.解析解析：y′=1/x，在x=e处斜率为1/e，切点为(e,1)，切线y−1=(1/e)(x−e)，即y=x/e。评分标准：5分：斜率2分，切点1分，方程2分。易错提醒：切线方程要经过切点，不能只写斜率。14.解析解析：a²=4，c=√3，b²=a²−c²=1，离心率e=c/a=√3/2。评分标准：5分：由焦距得c1分，求a1分，求b1分，离心率2分。易错提醒：焦距是2c，不是c。15.解析解析：sin2x=2sinxcosx，原方程化为sinx(2cosx−√3)=0。故x=0,π,2π，或cosx=√3/2得x=π/6,11π/6。评分标准：5分：因式分解2分，求两类解2分，写全集1分。易错提醒：区间包含端点0与2π，两者都应保留。16.解析解析：x²−2x−3=(x−1)²−4。函数|(x−1)²−4|的图像关于x=1对称，最低值0，局部峰值对应横轴翻折高度4。水平线y=m与图像有四个交点当且仅当0<m<4。评分标准：5分：配方2分，图像特征2分，范围1分。易错提醒：m=0时只有两个不同根，m=4时有三个不同根，端点不能取。17.解析答案：最小正周期π，最大值√2；不等式解集为[0,π/4]∪{π}。解析：f(x)=2sinxcosx+2cos²x−1=sin2x+cos2x=√2sin(2x+π/4)。因此最小正周期T=π，最大值为√2。由√2sin(2x+π/4)≥1，得sin(2x+π/4)≥√2/2。令u=2x+π/4，x∈[0,π]时u∈[π/4,9π/4]，故u∈[π/4,3π/4]或u=9π/4，于是x∈[0,π/4]∪{π}。评分标准：评分标准：（1）化为sin2x+cos2x2分，化为√2sin(2x+π/4)2分，周期与最大值各1分；（2）转化不等式2分，结合区间求解2分。易错提醒：易错提醒：端点x=π对应u