2026年深圳高三数学高考三模冲刺卷：函数零点与参数分类讨论（校际联考版第5套）含参考答案、逐题解析与评分细则

2026年深圳高三数学高考三模冲刺卷：函数零点与参数分类讨论（校际联考版第5套）含参考答案、逐题解析与评分细则深圳市高三校际联考版第5套2026届高考三模冲刺数学试卷适用地区深圳/校际联考版考试节点2026届高考三模冲刺科目与重点数学：函数零点与参数分类讨论满分与时间150分/120分钟注意事项：1.本卷分为单项选择题、多项选择题、填空题和解答题四部分，共22题；满分150分，考试时间120分钟。2.单项选择题每题只有一个正确选项；多项选择题至少有两个正确选项，全部选对得5分，部分选对得2分，有错选得0分。3.填空题须写出最简、确定答案；解答题须写出必要推理、计算过程和结论，答案写在指定作答区域内。4.函数零点与参数分类讨论题要重视端点、定义域、单调性、极值和临界参数；书写保持规范。一、单项选择题（本大题共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.已知函数f(x)=lnx+x-2(x>0)，则它的零点所在区间是（）A.(0,1)B.(1,2)C.(2,3)D.(3,4)2.函数f(x)=eˣ-3x在实数集上的零点个数为（）A.0B.1C.2D.33.若二次函数f(x)=x²-2ax+1有两个不同的正零点，则实数a的取值范围是（）A.a>0B.a≥1C.a>1D.0<a<14.函数h(x)=ln(x+1)-ln(3-x)的定义域为(-1,3)，其零点为（）A.-1B.0C.1D.25.在区间[0,2π]内，方程2sin²x-3sinx+1=0的解的个数为（）A.1B.2C.3D.46.已知数列aₙ=n·2ⁿ⁻¹，则a₁/2+a₂/4+a₃/8+a₄/16+a₅/32的值为（）A.5B.7C.15/2D.87.椭圆x²/9+y²/5=1的焦距为（）A.2B.4C.2√5D.68.直线y=2x+a与抛物线y=x²相切，则a的值为（）A.-2B.-1C.0D.1二、多项选择题（本大题共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.关于函数fₐ(x)=x³-3x+a的零点，下列说法正确的是（）A.a=0时有三个不同零点B.a=2时有两个不同零点C.a>2时恰有一个零点D.a=-3时有三个不同零点10.关于方程lnx-ax=0(x>0)，下列结论正确的是（）A.a<0时恰有一个零点B.a=0时唯一零点为x=1C.0<a<1/e时恰有两个零点D.a≥1/e时无零点11.已知向量u=(1,λ,2)，v=(λ,1,0)，下列结论正确的是（）A.λ=1时，u·v=2B.u⊥v的充要条件是λ=-1C.u的模平方为λ²+5，v的模平方为λ²+1D.对任意λ，u与v都不可能垂直12.若随机变量X~B(4,p)且E(X)=2，则下列结论正确的是（）A.p=1/2B.D(X)=1C.P(X=2)=3/8D.P(X≥3)=1/2选择题答题栏：题号123456789101112答案三、填空题（本大题共4小题，每小题5分，共20分。请把答案填写在题后横线上）13.若函数f(x)=(x-1)(x-a)在区间(0,1)内有且只有一个零点，则a的取值范围是__________。14.若函数f(x)=x²-4x+a在区间[0,3]上有零点，则a的取值范围是__________。15.若函数f(x)=lnx+m在区间(1,e²)内有零点，则m的取值范围是__________。16.已知数列{aₙ}满足a₁=2，aₙ₊₁=aₙ+2n+1，则a₁₀=__________。四、解答题（本大题共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.（本小题10分）在锐角三角形ABC中，角A,B,C所对的边分别为a,b,c。已知2cosA=1，b=4，c=6。（1）求边a的长与三角形ABC的面积；（2）设数列{uₙ}满足u₁=S/√3，uₙ₊₁=uₙ+2n-1，其中S为（1）中的三角形面积。求uₙ的通项公式，并求满足uₙ≥100的最小正整数n。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（本小题12分）如图形文字描述：四棱锥P-ABCD的底面ABCD是矩形，AB=2，BC=√3，PA⊥平面ABCD，且PA=√3。点M是棱PC的中点。（1）证明：AD⊥平面PAB；（2）求直线BM与平面PCD所成角的正弦值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（本小题12分）某深圳高三校际联考阅卷组抽取60名学生，统计其“函数零点专题”和“导数专题”的达标情况。已知函数零点专题达标36人，导数专题达标30人，两项均达标22人。（1）完成2×2列联表，并判断“函数零点专题达标”与“导数专题达标”是否相互独立；（2）在“函数零点专题达标但导数专题未达标”的14人中，有6名男生、8名女生。现从这14人中随机抽取3人参加讲评，记抽到的女生人数为X，求X的分布列和数学期望。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（本小题12分）已知椭圆C:x²/a²+y²/b²=1(a>b>0)的焦距为2√3，且经过点P(2,1)。（1）求椭圆C的标准方程；（2）过点Q(0,1)的直线l:y=kx+1与椭圆C交于A,B两点，求弦长AB的表达式，并求其最大值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（本小题12分）已知函数fₐ(x)=lnx-a(x-1)，定义域为(0,+∞)。（1）证明：x=1恒为函数fₐ(x)的零点；（2）讨论函数fₐ(x)在(0,+∞)上零点个数随参数a的变化情况；（3）若函数fₐ(x)恰有两个零点，求a的取值范围。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（本小题12分）已知函数Fₐ(x)=eˣ-ax²，其中a∈R。（1）当a=1时，证明函数F₁(x)至少有一个负零点；（2）讨论Fₐ(x)在实数集上的零点个数；（3）若Fₐ(x)恰有三个零点，求a的取值范围，并说明两个正零点所在区间的判定方法。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与详解一、客观题参考答案题号123456答案BCCCCC题号789101112答案BBABCABCACABC二、填空题参考答案题号13141516答案(0,1)[0,4](-2,0)101三、逐题解析与评分细则评分总则与讲评口径题型题号主要考查给分口径单选题1-8函数零点判断、三角方程、数列基础、圆锥曲线与切线每题5分多选题9-12导数极值与零点个数、参数方程、向量、二项分布全对5分，漏选2分填空题13-16区间零点、参数范围、对数零点、递推数列每题5分解答题17-20三角与数列、立体几何、概率统计、椭圆弦长按步骤给分压轴题21-22对数型参数分类、指数函数零点分类关键临界值必须写明书写要求全卷定义域、单调区间、极值、端点和临界参数过程与结论一致阅卷补充一：函数零点题凡使用介值定理，必须说明函数在相应区间连续；仅写端点异号而不说明连续性，过程分酌情扣1分。阅卷补充二：参数分类讨论题必须列出临界参数的来源，例如极值、端点值、重根或辅助函数最值；只写最终范围但无来源，最多给结论分。阅卷补充三：导数法讨论单调性时，若导数符号表缺少区间端点或定义域限制，但结论未受影响，扣1分；若导致零点个数错误，则相应步骤不得分。阅卷补充四：概率统计题的分布列应包含随机变量全部可能取值，并能检查概率和为1；期望可以用公式直接求，但必须与抽样模型一致。阅卷补充五：立体几何题允许综合法或坐标法。坐标法中若法向量比例正确即可给分，最终角度关系必须区分线线角、线面角与面面角。阅卷补充六：圆锥曲线题若弦长表达式正确但最大值讨论不完整，可给表达式分；若遗漏斜率伸缩因子，弦长相关得分不超过一半。阅卷补充七：全卷书写中出现等价变形除以含变量表达式时，必须说明该表达式不为零；若遗漏并排除了合法解，按结果错误处理。第1题答案：B因为f(1)=ln1+1-2=-1，f(2)=ln2>0，且f'(x)=1/x+1>0，所以函数在(0,+∞)上单调递增，零点唯一且在(1,2)。评分细则：选B得5分；其他选项、不选或多选均得0分。第2题答案：Cf'(x)=eˣ-3，临界点为x=ln3。函数在(-∞,ln3)上递减，在(ln3,+∞)上递增；最小值f(ln3)=3-3ln3<0，而x趋向正负无穷时函数值均趋向+∞，故有两个零点。评分细则：选C得5分。误把极小值点当零点或只看一侧极限者不得分。第3题答案：C二次方程x²-2ax+1=0有两个不同正根，需要判别式4a²-4>0，且两根和2a>0、积1>0。由此a>1。评分细则：选C得5分；把“不同”遗漏而选a≥1得0分。第4题答案：C由ln(x+1)-ln(3-x)=0得(x+1)/(3-x)=1，解得x=1，且1属于定义域(-1,3)。评分细则：选C得5分；未检验定义域导致错误不得分。第5题答案：C方程可化为(2sinx-1)(sinx-1)=0。sinx=1在[0,2π]内有1个解；sinx=1/2在[0,2π]内有2个解，共3个。评分细则：选C得5分；把端点重复或漏算sinx=1的解不得分。第6题答案：C因为aₙ/2ⁿ=n/2，所以a₁/2+a₂/4+a₃/8+a₄/16+a₅/32=(1+2+3+4+5)/2=15/2。评分细则：选C得5分；指数下标处理错误不得分。第7题答案：B椭圆中a²=9，b²=5，所以c²=a²-b²=4，c=2，焦距2c=4。评分细则：选B得5分；把焦半距c当焦距得0分。第8题答案：B由x²=2x+a得x²-2x-a=0。相切要求判别式Δ=4+4a=0，因此a=-1。评分细则：选B得5分；仅写交点方程未用判别式者不得分。第9题答案：ABCfₐ'(x)=3x²-3，极值点为x=-1与x=1。fₐ(-1)=a+2，fₐ(1)=a-2。三不同零点等价于-2<a<2；a=2时x=1为重根且另有一个不同零点；a>2时函数只有一个零点；a=-3<-2，只有一个零点。故A、B、C正确。评分细则：全部选ABC得5分；只选其中正确选项且无错误得2分；含D或错选得0分。第10题答案：ABC设F(x)=lnx-ax。若a<0，则F'(x)=1/x-a>0，函数从-∞增至+∞，有一个零点。a=0时lnx=0，零点为x=1。若a>0，令φ(x)=lnx/x，φ在x=e处取最大值1/e；0<a<1/e有两个解，a=1/e有一个解，a>1/e无解。评分细则：全部选ABC得5分；漏选且无错选得2分；错选D得0分。第11题答案：ACu·v=1·λ+λ·1+2·0=2λ。λ=1时u·v=2，A正确；u⊥v等价于λ=0，B、D错误；u的模平方为1+λ²+4=λ²+5，v的模平方为λ²+1，C正确。评分细则：全部选AC得5分；只选A或只选C得2分；含错误选项得0分。第12题答案：ABCX~B(4,p)，E(X)=4p=2，所以p=1/2。D(X)=4·1/2·1/2=1，P(X=2)=C₄²(1/2)⁴=6/16=3/8，而P(X≥3)=C₄³/16+C₄⁴/16=5/16，不是1/2。评分细则：全部选ABC得5分；漏选且无错选得2分；含D得0分。第13题答案：(0,1)f(x)=(x-1)(x-a)的零点为1和a。区间(0,1)不含1，因此要在该区间内有且只有一个零点，只需a∈(0,1)。评分细则：写出(0,1)得5分；写成[0,1]或含端点得3分；只写a>0不得分。第14题答案：[0,4]方程x²-4x+a=0在[0,3]上有解，等价于a=4x-x²在[0,3]上有取值。函数4x-x²在[0,3]上的最小值为0，最大值为4，故a∈[0,4]。评分细则：答案[0,4]得5分；漏端点得3分；只用判别式得到a≤4但未考虑区间得2分。第15题答案：(-2,0)lnx+m=0得x=e⁻ᵐ。要求1<e⁻ᵐ<e²，取自然对数得0<-m<2，即-2<m<0。评分细则：写出(-2,0)得5分；方向反写不得分；端点写入扣2分。第16题答案：101由aₙ₊₁-aₙ=2n+1，累加得aₙ=2+∑ₖ₌₁ⁿ⁻¹(2k+1)=2+(n-1)n+(n-1)=n²+1，所以a₁₀=101。评分细则：结果101得5分；通项正确但计算失误得3分；只列递推未求值得1分。第17题10分解析：由2cosA=1得cosA=1/2。由于三角形为锐角三角形，A=60°，sinA=√3/2。（1）由余弦定理，a²=b²+c²-2bccosA=4²+6²-2·4·6·1/2=28，所以a=2√7。面积S=1/2·bc·sinA=1/2·4·6·√3/2=6√3。（2）u₁=S/√3=6。由uₙ₊₁-uₙ=2n-1，累加得uₙ=6+∑ₖ₌₁ⁿ⁻¹(2k-1)=6+(n-1)²。令6+(n-1)²≥100，得(n-1)²≥94，故最小正整数n=11。评分细则：求出A=60°得1分；余弦定理求a=2√7得3分；面积S=6√3得2分；写出递推累加式得2分；通项uₙ=6+(n-1)²及最小n=11得2分。易错点：面积中的sinA容易误写为cosA；数列累加上限是n-1，不是n。讲评提示：本题前半部分考查三角形基本计算，后半部分考查递推累加，教师讲评时可强调“几何量进入数列”的转化。第18题12分解析：建立空间直角坐标系，令A(0,0,0)，B(2,0,0)，D(0,√3,0)，C(2,√3,0)，P(0,0,√3)。（1）因为PA⊥平面ABCD，所以PA⊥AD；又底面为矩形，AD⊥AB。直线PA与AB相交于A，且均在平面PAB内，因此AD⊥平面PAB。（2）M为PC中点，M(1,√3/2,√3/2)，故向量BM=(-1,√3/2,√3/2)。平面PCD内有向量PC=(2,√3,-√3)，PD=(0,√3,-√3)，法向量可取n=(0,1,1)。设直线BM与平面PCD所成角为θ，则sinθ=|BM·n|/(|BM||n|)=(√3)/(√(5/2)