【数学试卷+答案】2024-2025学年泉州高二上期末质检

保密★使用前保密★使用前2024-2025学年度上学期泉州市高中教学质量监测高二数学参考答案、12345678DDBACDBC9 四、1513分）列基本量法、等差数列的性质等基础知识，考查逻辑推理能力、运等，考查函数与方程思想、化归与转化思想等，体现基础性和综合由已知3d=14每个式子各1分）.................解得每个结果各1分）.....................................................................4分所以an=a1+(n-1)×d=3+(n-1)×2=2n+1（公式1分）.6分...............................3所以d=a3-a2=2，4分所以an=a2+(n-2)×d=5+(n-2)×2=2n+1（公式1分）.6分因为bn=代通项1分）.9分【命题意图】本小题主要考查直线与平面的位置关系、求异面直线所成角等基础知识；考查空间想象、推理论证、运算求解等能力；考查化归与转化等思想；体现基础性和综合性，导向对发展直观想象、逻辑推理、数学运算等核心素【试题解析】（1）解法一：因为M，N分别为PA，PB的中点，所以MN//△OCD为等边三角形，所以∠DCO=60。=∠AOC，所以CD//AB//MN，··································································································5分又MN丈平面PCD，CD平面PCD，所以MN//平面PCD，··············7分∠AOC=∠COD=60。，△OCD为等边三角形，所以∠DCO=60。=∠AOC，所以CD//AB，····································2分取CD中点Q，连结OQ，OP，则OQ⊥CD，OQ⊥AB，OP⊥平面ABDCOP所在的直线为x，y，z轴建立空间直角坐标系································································································3分又MN丈平面PCD，CD平面PCD，所以MN//平面PCD，··············7分所以∠AOC=∠COD=60o，△OCD为等边三角形，所以∠DCO=60o=∠AOC，所以CD//AB，····································2分取CD中点Q，连结OQ，OP，则OQ⊥CD，OQ⊥AB，OP⊥平面ABDCOP所在的直线为x，y，z轴建立空间直角坐标系MN=(0,2,0)，···············则CD=MN，所以CD//MN，由四点不共线，易得CD//MN，···········5分又MN丈平面PCD，CD平面PCD，所以MN//平面PCD，··············7分角坐标系，················································································设异面直线AC与BM所成角为θ,则cosθ= ································································· 即直线AC与BM所成角的余弦值为，·····································15分解法二：连结CB，∠AOC=60o，△OAC为等边三角形，所以∠CAO=60o，在△ABC中，由余弦定理可得BC2=AC2+AB2—2AC.AB.cos60o=22+422×2×4×=12知BC⊥AC说明：若未证明，扣1分）CB所在的直线为x，y轴，l为z轴建立空间直角坐标系，如图，································································································9分C(0,0,0)，A(2,0,0)，B(0,2,0)····························10分所以AC=(2,0,0)··················设异面直线AC与BM所成角为θ,则cosθ= ································································· 即直线AC与BM所成角的余弦值为，·····································15分解法三：连结OC，因为点C，D为AB的三等分点，所以∠AOC=60o，连结OP，则OP⊥平面ABDC，所以OP⊥AB，OP⊥OA，OP⊥OC，又OB=2，PB=2，易得OP=························································ 设异面直线AC与BM所成角为θ,则|cosθ|= 即直线AC与BM所成角的余弦值为，·······························∠AOC=∠COD=60o，△AOC为等边三角形，所以，∠ACO=60o=∠COD，所以AC//OD，取AM中点E，连结OE，所以OE为△ABM的中位线，OE//BM，所以直线AC与BM所成角即为∠EOD或其补角················9分易得OP=2=OA，∠PAO=45o，在△AOE中，由余弦定理可得·································································过点E作EF//OP，交AB于F，连结FD，ED，则EF⊥平面ABDC，所以EF⊥FD，FD=，所以················································在△DOE中，由余弦定理可得即直线AC与BM所成角的余弦值为········································15分【命题意图】本小题主要考查曲线的方程、直线的方程、点线距离、弦长公式等基础知识，体现基础性和综合性，导向对发展直观想象、逻辑推理、数学运算等【试题解析】（1）解法一：设圆Γ的标准方程为(x+a)2+2+(--b)2=r2 则由已知Γ过P(0,-3)，Q(0,3)，R(-3,0)，得{(0+a)2+(3-b)2=r2， ································································································3分所以圆Γ的方程为(x+1)2+y2=4；·················································5分解法二：设圆Γ的一般方程为x2+y2+Dx+Ey+F=0，······················1分································································································3分所以圆Γ的方程为x2+y2+2x-3=0；·············································5分（2）解法一：因为点P(0,-3)，设直线PE的方程为y= 22-0=7，············································7分解得，···················································································································xE22777 解法二：过P作直线x=2的垂线，垂足P，设直线EF的斜率为k， 3于是圆心(-1,0)到直线PE的距离为于是圆心(-1,0)到直线PE的距离为7\777\777 又点Q(0,)到直线PE的距离为d1=43，····································14分22777 22777 22-0=7，············································7分 解得k=±,·········································································· 又因为k<0，所以k=-，·········记轨迹Γ的圆心H(-1,0)，所以|FH|=9+12=21，过点F的切线长为|FH|=FH2-r2=································································· 解法四：过P作直线x=2的垂线，垂足P，设直线EF的斜率为k，则 k=-tan上PPF=-，·········于是圆心(-1,0)到直线PE的距离为d=，···································10分又点Q(0,)到直线PE的距离为，·······························解得，···················································································································解得点E的横坐标为xE=-，··········································直线与平面的位置关系、平面与平面的位置关系、求线面角及二面角等基础知识；考查空间想象、推理论证、运算求解等能力；考查化归与转化等思想；体现基础性和综合性，导向对发展直观想象、逻辑推理、数学运算等核心素养的AC丄BC.··························· 因为AC∥DF，所以DF丄平面BCFE VDEF=.S△EDF.BE=.2.因为三棱锥EBDF的体积为所以.·················设平面DEQ的法向量为，则所以取=1.所以，是平面DEQ的一个法向量.································································································8分所以平面DEQ丄平面BDF.···············VDEF=.S△EDF.BE=.2.2.BE=.························5分因为三棱锥EBDF的体积为所以.······························6分取BC的中点M，连接ME与BF交于点N，连接DN.因为Q为AC中点，所以QM∥AB.因为DE∥AB，所以QM∥DE，所以D,E,M,Q四点共面.··································································7分因为EF=2BM，EF∥BM，所以△EFN∽△MBN，且.所以BN2+MN2=BM2，所以BF丄ME.··········································又BFDF=F，BF，DF平面BDF，所以ME丄平面BDF，·········10分又因为ME平面DEQ，所以平面DEQ丄平面BDF.···························11分所以OE为PE在平面DEF内的射影，所以上OEP为直线PE与平面DEF所成的角，所以上OEP=α.··························································12分过O作OI⊥DF，垂足为I，连接PI，显然PI⊥DF，所以上OIP为二面角PDFE的平面角，所以上OIP=β.················13分所以tanα=，tanβ=.因为α=β,所以OE=OI,所以点O到点E的距离等于点O到直线DF的距离，所以点P到点B的距离等于点P到直线AC的距离，·························14分所以点P的轨迹是以B为焦点，直线AC为准线的抛物线在△ABC内的部分.·······························································································15分又点P到AC的距离为d，所以d=PB，··············· 所以PQ+d的最小值为·.·························································17分解法二：设P(x,y,0)(0<x,y<2，x+y<2)，不妨设CF=t（t>0则E(0,2,t)，显然i=(0,0,1)为平面DEF的一个法向量.则sinα=|cos<,i>|=.··························12分取y3=t,则z3=y.所以，j=(0,2,y)是平面PDF的一个法向量.所以cosβ=|cos<i,j>|=.·····································13分因为α=β,所以sin2α+cos2β=sin2α+cos2α=1，所以···································································所以点P的轨迹是以B为焦点，直线AC为准线的抛物线在△ABC内的部分.····································································································································································································又点P到AC的距离为d，所以d=PB，·············· 所以PQ+d的最小值为·5.····························································17分第三定义为背景考察椭圆、双曲线方程以及几何性质，直线与圆锥曲线的位置关系等基础知识；考查运算求解、推理论证等能力以及创新意识；考查化归与转化、数形结合、函数与方程、特殊与一般思想；体现基础性、综合性与创新当λ=2时，k1k2=2，所以，····································2分即y2=2|x21|,x≠±1，·······························································4分故没有分段不扣分··································5分即在|x|<1时，轨迹是椭圆2x2+y2=2去除左右顶点；当|x|>1时，轨迹是双曲线2x2—y2=2去除左右顶点注意到两直线关于x轴对称，·························································6分又曲线图形也关于x轴对称，故可得M1,P1关于x轴对称;N1,Q1关于x轴对称，所以T1必在x轴