2026年金华高三数学高考三模冲刺卷：导数不等式与新定义函数（冲刺讲评版第2套）含参考答案、逐题解析与评分细则

第1页2026年金华高三数学高考三模冲刺卷：导数不等式与新定义函数（冲刺讲评版第2套）含参考答案、逐题解析与评分细则地区或学校簇金华/冲刺讲评版届别2026届考试节点高考三模冲刺科目数学试卷属性导数不等式与新定义函数重点突破版本冲刺讲评版第2套满分150分考试时间120分钟注意事项1.本卷共22题，满分150分，考试时间120分钟。请在规定位置填写姓名、班级与准考信息。2.单项选择题每题只有一个正确选项；多项选择题每题至少有两个正确选项，全对得5分，少选且正确得2分，有错选得0分。3.填空题只需写出最终结果，结果应化简；解答题必须写出必要的推理、计算和结论，按步骤给分。4.答题时保持书写规范，坐标、区间、参数范围和概率表达式不得漏写；作图题可用文字说明图形关系。5.本卷讲评重点为函数与导数、导数不等式、新定义函数，同时覆盖三角数列、概率统计、立体几何与圆锥曲线。题型题号题量每题分值小计单项选择题1-885分40分多项选择题9-1245分20分填空题13-1645分20分解答题17-226第17题10分，其余每题12分70分合计1-2222150分一、单项选择题：本大题共8小题，每小题5分，共40分1.已知复数z=((1+i)^2)/(1-i)，则|z|等于（）。A.1B.√2C.2D.2√22.函数f(x)=√(x+1)+ln(3-x)的定义域为（）。A.(-∞,3)B.[-1,3]C.[-1,3)D.(-1,3)3.曲线y=x^3-3x^2+2在点x=2处的切线方程为（）。A.y=-2B.y=2x-6C.y=3x-8D.y=04.若α∈(0,π/2)，且sin(α+π/6)=√3/2，则cosα等于（）。A.1/2B.√2/2C.√3/2D.15.展开式(1+2x)^5中x^2的系数为（）。A.20B.30C.40D.806.等差数列{a_n}中，a_3=4，a_7=12，则前10项和S_{10}等于（）。A.80B.90C.100D.1107.在正方体ABCD-A_1B_1C_1D_1中，直线AB_1与直线BD所成的角为（）。A.30°B.45°C.60°D.90°8.对可导函数f定义新运算Φ(f)(x)=f′(x)-f(x)。若f(x)=x^2+1，则方程Φ(f)(x)=0的实根个数为（）。A.0B.1C.2D.无穷多个单项选择题答题栏：题号12345678答案二、多项选择题：本大题共4小题，每小题5分，共20分每题至少有两个正确选项。全选对得5分，少选且正确得2分，有错选得0分。9.已知f(x)=lnx-x+1，下列结论正确的是（）。A.定义域为(0,+∞)B.在(0,1)上单调递增，在(1,+∞)上单调递减C.对任意x>0，有f(x)≤0D.方程f(x)=0有两个不同实根10.平面向量a,b满足|a|=2，|b|=1，a·b=-1，下列结论正确的是（）。A.a与b的夹角为120°B.|a+b|=√3C.a在b方向上的投影数量为-1D.|a-b|=√511.随机变量X服从二项分布B(3,1/3)，下列结论正确的是（）。A.P(X=0)=8/27B.E(X)=1C.D(X)=2/3D.P(X≥2)=8/2712.设f_a(x)=x^3-3ax，其中a为实数，下列结论正确的是（）。A.当a≤0时，f_a(x)在R上单调递增B.当a>0时，f_a(x)的驻点为x=±√aC.当a=1时，x=-1处取极大值，x=1处取极小值D.当a=4时，极大值为8多项选择题答题栏：题号9101112答案三、填空题：本大题共4小题，每小题5分，共20分13.已知f(x)=x^2e^{-x}，则f′(1)=__________。14.抛物线y^2=4x的焦点到准线的距离为__________。15.函数g(x)=|x-2|+|x+1|的最小值为__________。16.若m>0，函数h(x)=e^x-mx在R上的最小值为0，则m=__________。题号13141516答案四、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。函数与导数题请写明单调性、极值、参数范围的由来；概率统计题请写明事件含义。17.（10分）已知函数F(x)=2sinxcosx+2cos^2x-1。

（1）求F(x)的最小正周期和单调递增区间；

（2）在△ABC中，角A∈(0,π/2)，且F(A)=1，若b=2，c=2√2，求边a的长。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（12分）如图形关系用文字描述：在直三棱柱ABC-A_1B_1C_1中，底面ABC为等腰直角三角形，AB=AC=2，∠BAC=90°，侧棱AA_1⊥平面ABC，且AA_1=2。

（1）证明：AB⊥平面ACC_1A_1；

（2）求直线B_1C与平面ACC_1A_1所成角的正弦值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（12分）金华某校高三数学三模冲刺讲评后，抽取20名学生统计某道导数压轴题的得分，得分只可能为0,2,4,6，频数如下表。得分0246频数2585（1）求这20名学生该题得分的平均数与方差；

（2）从这20名学生中不放回地随机抽取2人，记X为抽到得6分学生的人数，求X的分布列和数学期望。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（12分）设椭圆E:x^2/a^2+y^2/b^2=1(a>b>0)的离心率为1/2，且过点P(1,√3/2)。

（1）求椭圆E的方程；

（2）设F为椭圆右焦点，直线l过F且斜率为-1，与椭圆交于M,N两点，求弦MN的中点坐标及|MN|。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（12分）已知函数f_a(x)=lnx-x+a(x>0)。

（1）求f_a(x)的单调区间及最大值；

（2）讨论方程f_a(x)=0的实根个数；

（3）当a=2时，设方程f_2(x)=0的两个实根为α,β，且0<α<1<β，证明：αβ<1。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（12分）定义新函数：对x>0，令Ψ(x)=(e^x-1-x)/x^2，并规定Ψ(0)=1/2，称Ψ为指数函数在0点的二阶余量函数。

（1）证明Ψ(x)在[0,+∞)上单调递增；

（2）求所有实数a，使不等式e^x≥1+x+ax^2对任意x≥0恒成立；

（3）若a>1/2，证明方程e^x=1+x+ax^2在(0,+∞)上有且仅有一个实根。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________解答题续答区（第17-22题）本区域供第17-22题继续书写推导、补充步骤与规范化结论。请标明题号后作答，保持推理顺序清晰。第17题续答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________第18题续答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________第19题续答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________第20题续答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________第21题续答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________第22题续答：_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与详解一、单项选择题答案、解析与评分细则1.答案：B解析：(1+i)^2=1+2i+i^2=2i，故z=2i/(1-i)=2i(1+i)/2=-1+i，所以|z|=√((-1)^2+1^2)=√2。A忽略了模长平方，C、D把分子模长直接当成结果。评分细则：选B得5分，错选、多选或不选得0分。2.答案：C解析：根式要求x+1≥0，对数要求3-x>0，联立得x≥-1且x<3，定义域为[-1,3)。B把对数端点x=3误纳入，D漏掉x=-1。评分细则：选C得5分，错选、多选或不选得0分。3.答案：A解析：令y=x^3-3x^2+2，则y′=3x^2-6x，在x=2处斜率为0，点坐标为(2,-2)，切线为y=-2。B、C为斜率计算错误，D误把切线当成x轴。评分细则：选A得5分，错选、多选或不选得0分。4.答案：C解析：由α∈(0,π/2)知α+π/6∈(π/6,2π/3)。方程sin(α+π/6)=√3/2在该范围内对应α+π/6=π/3，故α=π/6，cosα=√3/2。评分细则：选C得5分，错选、多选或不选得0分。5.答案：C解析：(1+2x)^5中x^2项系数为C_5^2·2^2=10·4=40。A漏乘一个2，B为组合数误算，D把2^2误作2^3。评分细则：选C得5分，错选、多选或不选得0分。6.答案：B解析：设公差为d，则a_7-a_3=4d=8，得d=2；a_1=a_3-2d=0，所以S_{10}=10(a_1+a_{10})/2=5×18=90。评分细则：选B得5分，错选、多选或不选得0分。7.答案：C解析：设正方体棱长为1，取向量AB_1=(1,0,1)，BD=(-1,1,0)，则|cosθ|=|AB_1·BD|/(|AB_1||BD|)=1/2，故所成角为60°。评分细则：选C得5分，错选、多选或不选得0分。8.答案：B解析：由定义Φ(f)(x)=f′(x)-f(x)=2x-(x^2+1)=-(x-1)^2。方程Φ(f)(x)=0只有实根x=1，所以实根个数为1。评分细则：选B得5分，错选、多选或不选得0分。二、多项选择题答案、解析与评分细则9.答案：ABC解析：函数f(x)=lnx-x+1的定义域为(0,+∞)，f′(x)=1/x-1。当0<x<1时f′(x)>0，当x>1时f′(x)<0，所以f(1)=0为最大值，故对任意x>0，f(x)≤0。方程f(x)=0仅有根x=1，D错误。评分细则：全选ABC得5分；只选A、B、C中的一项或两项且无错选得2分；含D或不选得0分。10.答案：ABC解析：由a·b=|a||b|cosθ得cosθ=-1/2，所以θ=120°，A对。|a+b|^2=4+1+2(a·b)=3，B对。a在b方向上的投影数量为a·b/|b|=-1，C对。|a-b|^2=4+1-2(a·b)=7，故D错。评分细则：全选ABC得5分；只选A、B、C中的一项或两项且无错选得2分；含D或不选得0分。11.答案：ABC解析：若X~B(3,1/3)，则P(X=0)=(2/3)^3=8/27，E(X)=np=1，D(X)=np(1-p)=2/3。而P(X≥2)=C_3^2(1/3)^2(2/3)+(1/3)^3=7/27，D错误。评分细则：全选ABC得5分；只选A、B、C中的一项或两项且无错选得2分；含D或不选得0分。12.答案：ABC解析：f_a′(x)=3x^2-3a=3(x^2-a)。当a≤0时导数非负且函数整体递增，A对；当a>0时驻点为x=±√a，B对；a=1时由导数符号可知x=-1为极大值点，x=1为极小值点，C对；a=4时极大值f_4(-2)=16，D错。评分细则：全选ABC得5分；只选A、B、C中的一项或两项且无错选得2分；含D或不选得0分。三、填空题答案、解析与评分细则13.答案：1/e解析：由f(x)=x^2e^{-x}，得f′(x)=2xe^{-x}-x^2e^{-x}=e^{-x}(2x-x^2)，故f′(1)=e^{-1}=1/e。评分细则：写1/e、e^{-1}均得5分；只写导数未代入得2分；计算符号错误不得分。14.答案：2解析：抛物线y^2=4x可写为y^2=2px，所以2p=4，得p=2。焦点为(1,0)，准线为x=-1，二者距离为2。评分细则：结果2得5分；写出焦点或准线但距离错误得2分；将p与焦距混淆不得分。15.答案：3解析：|x-2|+|x+1|表示数轴上点x到2与-1的距离之和。当x∈[-1,2]时，该和恒为3，故最小值为3。评分细则：结果3得5分；能说明区间[-1,2]但结果漏写得3分；只代一个点碰巧得到3得2分。16.答案：e解析：由h(x)=e^x-mx，得h′(x)=e^x-m。当m>0时，极小点满足e^x=m，即x=lnm，最小值为m-mlnm。令m-mlnm=0，得lnm=1，故m=e。评分细则：结果e得5分；建立最小值m-mlnm得3分；只写驻点方程得2分。四、解答题答案、逐题解析与评分细则17.答案：（1）最小正周期π，递增区间(-3π/8+kπ,π/8+kπ)(k∈Z)；（2）a=2。解析：（1）F(x)=2sinxcosx+2cos^2x-1=sin2x+cos2x=√2sin(2x+π/4)，故最小正周期为T=2π/2=π。当cos(2x+π/4)>0时函数递增，即-π/2+2kπ<2x+π/4<π/2+2kπ，解得-3π/8+kπ<x<π/8+kπ。（2）由F(A)=1，得√2sin(2A+π/4)=1。由于A∈(0,π/2)，所以2A+π/4∈(π/4,5π/4)，从而2A+π/4=3π/4，即A=π/4。由余弦定理，a^2=b^2+c^2-2bccosA=4+8-2×2×2√2×√2/2=4，故a=2。采分点给分化简为√2sin(2x+π/4)2分求出周期π1分求出递增区间3分由F(A)=1得A=π/42分用余弦定理求得a=22分讲评提示：本题易错在把2cos^2x-1写成sin2x，或解sinθ=√2/2时忽略A的取值范围。18.答案：（1）证明见解析；（2）sinθ=1/√3。解析：以A为坐标原点，取AB,AC,AA_1分别为x,y,z轴正方向，建立空间直角坐标系，则A(0,0,0)，B(2,0,0)，C(0,2,0)，A_1(0,0,2)，B_1(2,0,2)，平面ACC_1A_1即为平面x=0。（1）AB的方向向量为(2,0,0)，而平面ACC_1A_1内有不共线向量AC=(0,2,0)，AA_1=(0,0,2)。显然AB·AC=0，AB·AA_1=0，故AB垂直于该平面内两条相交直线，所以AB⊥平面ACC_1A_1。（2）直线B_1C的方向向量为C-B_1=(-2,2,-2)。平面ACC_1A_1的一个法向量为n=(1,0,0)。设直线与平面所成角为θ，则sinθ=|v·n|/(|v||n|)=2/(2√3)=1/√3。采分点给分正确建立坐标系或给出等价几何关系2分写出平面内两条相交直线方向并说明垂直4分写出B_1C方向向量和平面法向量3分套用线面角公式并得1/√33分讲评提示：线面角求正弦时使用“直线方向与平面法向量夹角的余弦绝对值”，不要误求成余弦。19.答案：（1）平均数18/5，方差86/25；（2）分布列见解析，E(X)=1/2。解析：（1）平均数ar{x}=(0×2+2×5+4×8+6×5)/20=72/20=18/5。方差s^2=[2(0-18/5)^2+5(2-18/5)^2+8(4-18/5)^2+5(6-18/5)^2]/20=86/25。（2）20人中得6分的有5人，非6分的有15人。不放回抽取2人，X可取0,1,2。P(X=0)=C_{15}^2/C_{20}^2=21/38；P(X=1)=C_5^1C_{15}^1/C_{20}^2=15/38；P(X=2)=C_5^2/C_{20}^2=1/19。因此E(X)=0×21/38+1×15/38+2×1/19=1/2。X012P21/3815/381/19采分点给分列出或正确使用平均数公式2分平均数结果18/51分列出方差计算式并得86/253分明确X的取值0,1,21分三个概率计算正确3分数学期望1/22分讲评提示：本题第（2）问是不放回抽样，不能直接套二项分布；若用超几何分布