【數(shù)學(xué)試卷+答案】2024-2025學(xué)年泉州高二上期末質(zhì)檢

保密★使用前保密★使用前2024-2025學(xué)年度上學(xué)期泉州市高中教學(xué)質(zhì)量監(jiān)測(cè)高二數(shù)學(xué)參考答案、12345678DDBACDBC9 四、1513分）列基本量法、等差數(shù)列的性質(zhì)等基礎(chǔ)知識(shí)，考查邏輯推理能力、運(yùn)等，考查函數(shù)與方程思想、化歸與轉(zhuǎn)化思想等，體現(xiàn)基礎(chǔ)性和綜合由已知3d=14每個(gè)式子各1分）.................解得每個(gè)結(jié)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?yàn)閎n=代通項(xiàng)1分）.9分【命題意圖】本小題主要考查直線與平面的位置關(guān)系、求異面直線所成角等基礎(chǔ)知識(shí)；考查空間想象、推理論證、運(yùn)算求解等能力；考查化歸與轉(zhuǎn)化等思想；體現(xiàn)基礎(chǔ)性和綜合性，導(dǎo)向?qū)Πl(fā)展直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等核心素【試題解析】（1）解法一：因?yàn)镸，N分別為PA，PB的中點(diǎn)，所以MN//△OCD為等邊三角形，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為等邊三角形，所以∠DCO=60。=∠AOC，所以CD//AB，····································2分取CD中點(diǎn)Q，連結(jié)OQ，OP，則OQ⊥CD，OQ⊥AB，OP⊥平面ABDCOP所在的直線為x，y，z軸建立空間直角坐標(biāo)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為等邊三角形，所以∠DCO=60o=∠AOC，所以CD//AB，····································2分取CD中點(diǎn)Q，連結(jié)OQ，OP，則OQ⊥CD，OQ⊥AB，OP⊥平面ABDCOP所在的直線為x，y，z軸建立空間直角坐標(biāo)系MN=(0,2,0)，···············則CD=MN，所以CD//MN，由四點(diǎn)不共線，易得CD//MN，···········5分又MN丈平面PCD，CD平面PCD，所以MN//平面PCD，··············7分角坐標(biāo)系，················································································設(shè)異面直線AC與BM所成角為θ,則cosθ= ································································· 即直線AC與BM所成角的余弦值為，·····································15分解法二：連結(jié)CB，∠AOC=60o，△OAC為等邊三角形，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說(shuō)明：若未證明，扣1分）CB所在的直線為x，y軸，l為z軸建立空間直角坐標(biāo)系，如圖，································································································9分C(0,0,0)，A(2,0,0)，B(0,2,0)····························10分所以AC=(2,0,0)··················設(shè)異面直線AC與BM所成角為θ,則cosθ= ································································· 即直線AC與BM所成角的余弦值為，·····································15分解法三：連結(jié)OC，因?yàn)辄c(diǎn)C，D為AB的三等分點(diǎn)，所以∠AOC=60o，連結(jié)OP，則OP⊥平面ABDC，所以O(shè)P⊥AB，OP⊥OA，OP⊥OC，又OB=2，PB=2，易得OP=························································ 設(shè)異面直線AC與BM所成角為θ,則|cosθ|= 即直線AC與BM所成角的余弦值為，·······························∠AOC=∠COD=60o，△AOC為等邊三角形，所以，∠ACO=60o=∠COD，所以AC//OD，取AM中點(diǎn)E，連結(jié)OE，所以O(shè)E為△ABM的中位線，OE//BM，所以直線AC與BM所成角即為∠EOD或其補(bǔ)角················9分易得OP=2=OA，∠PAO=45o，在△AOE中，由余弦定理可得·································································過(guò)點(diǎn)E作EF//OP，交AB于F，連結(jié)FD，ED，則EF⊥平面ABDC，所以EF⊥FD，F(xiàn)D=，所以················································在△DOE中，由余弦定理可得即直線AC與BM所成角的余弦值為········································15分【命題意圖】本小題主要考查曲線的方程、直線的方程、點(diǎn)線距離、弦長(zhǎng)公式等基礎(chǔ)知識(shí)，體現(xiàn)基礎(chǔ)性和綜合性，導(dǎo)向?qū)Πl(fā)展直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等【試題解析】（1）解法一：設(shè)圓Γ的標(biāo)準(zhǔn)方程為(x+a)2+2+(--b)2=r2 則由已知Γ過(guò)P(0,-3)，Q(0,3)，R(-3,0)，得{(0+a)2+(3-b)2=r2， ································································································3分所以圓Γ的方程為(x+1)2+y2=4；·················································5分解法二：設(shè)圓Γ的一般方程為x2+y2+Dx+Ey+F=0，······················1分································································································3分所以圓Γ的方程為x2+y2+2x-3=0；·············································5分（2）解法一：因?yàn)辄c(diǎn)P(0,-3)，設(shè)直線PE的方程為y= 22-0=7，············································7分解得，···················································································································xE22777 解法二：過(guò)P作直線x=2的垂線，垂足P，設(shè)直線EF的斜率為k， 3于是圓心(-1,0)到直線PE的距離為于是圓心(-1,0)到直線PE的距離為7\777\777 又點(diǎn)Q(0,)到直線PE的距離為d1=43，····································14分22777 22777 22-0=7，············································7分 解得k=±,·········································································· 又因?yàn)閗<0，所以k=-，·········記軌跡Γ的圓心H(-1,0)，所以|FH|=9+12=21，過(guò)點(diǎn)F的切線長(zhǎng)為|FH|=FH2-r2=································································· 解法四：過(guò)P作直線x=2的垂線，垂足P，設(shè)直線EF的斜率為k，則 k=-tan上PPF=-，·········于是圓心(-1,0)到直線PE的距離為d=，···································10分又點(diǎn)Q(0,)到直線PE的距離為，·······························解得，···················································································································解得點(diǎn)E的橫坐標(biāo)為xE=-，··········································直線與平面的位置關(guān)系、平面與平面的位置關(guān)系、求線面角及二面角等基礎(chǔ)知識(shí)；考查空間想象、推理論證、運(yùn)算求解等能力；考查化歸與轉(zhuǎn)化等思想；體現(xiàn)基礎(chǔ)性和綜合性，導(dǎo)向?qū)Πl(fā)展直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)的AC丄BC.··························· 因?yàn)锳C∥DF，所以DF丄平面BCFE VDEF=.S△EDF.BE=.2.因?yàn)槿忮FEBDF的體積為所以.·················設(shè)平面DEQ的法向量為，則所以取=1.所以，是平面DEQ的一個(gè)法向量.································································································8分所以平面DEQ丄平面BDF.···············VDEF=.S△EDF.BE=.2.2.BE=.························5分因?yàn)槿忮FEBDF的體積為所以.······························6分取BC的中點(diǎn)M，連接ME與BF交于點(diǎn)N，連接DN.因?yàn)镼為AC中點(diǎn)，所以QM∥AB.因?yàn)镈E∥AB，所以QM∥DE，所以D,E,M,Q四點(diǎn)共面.··································································7分因?yàn)镋F=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?yàn)镸E平面DEQ，所以平面DEQ丄平面BDF.···························11分所以O(shè)E為PE在平面DEF內(nèi)的射影，所以上OEP為直線PE與平面DEF所成的角，所以上OEP=α.··························································12分過(guò)O作OI⊥DF，垂足為I，連接PI，顯然PI⊥DF，所以上OIP為二面角PDFE的平面角，所以上OIP=β.················13分所以tanα=，tanβ=.因?yàn)棣?β,所以O(shè)E=OI,所以點(diǎn)O到點(diǎn)E的距離等于點(diǎn)O到直線DF的距離，所以點(diǎn)P到點(diǎn)B的距離等于點(diǎn)P到直線AC的距離，·························14分所以點(diǎn)P的軌跡是以B為焦點(diǎn)，直線AC為準(zhǔn)線的拋物線在△ABC內(nèi)的部分.·······························································································15分又點(diǎn)P到AC的距離為d，所以d=PB，··············· 所以PQ+d的最小值為·.·························································17分解法二：設(shè)P(x,y,0)(0<x,y<2，x+y<2)，不妨設(shè)CF=t（t>0則E(0,2,t)，顯然i=(0,0,1)為平面DEF的一個(gè)法向量.則sinα=|cos<,i>|=.··························12分取y3=t,則z3=y.所以，j=(0,2,y)是平面PDF的一個(gè)法向量.所以cosβ=|cos<i,j>|=.·····································13分因?yàn)棣?β,所以sin2α+cos2β=sin2α+cos2α=1，所以···································································所以點(diǎn)P的軌跡是以B為焦點(diǎn)，直線AC為準(zhǔn)線的拋物線在△ABC內(nèi)的部分.····································································································································································································又點(diǎn)P到AC的距離為d，所以d=PB，·············· 所以PQ+d的最小值為·5.····························································17分第三定義為背景考察橢圓、雙曲線方程以及幾何性質(zhì)，直線與圓錐曲線的位置關(guān)系等基礎(chǔ)知識(shí)；考查運(yùn)算求解、推理論證等能力以及創(chuàng)新意識(shí)；考查化歸與轉(zhuǎn)化、數(shù)形結(jié)合、函數(shù)與方程、特殊與一般思想；體現(xiàn)基礎(chǔ)性、綜合性與創(chuàng)新當(dāng)λ=2時(shí)，k1k2=2，所以，····································2分即y2=2|x21|,x≠±1，·······························································4分故沒有分段不扣分··································5分即在|x|<1時(shí)，軌跡是橢圓2x2+y2=2去除左右頂點(diǎn)；當(dāng)|x|>1時(shí)，軌跡是雙曲線2x2—y2=2去除左右頂點(diǎn)注意到兩直線關(guān)于x軸對(duì)稱，·························································6分又曲線圖形也關(guān)于x軸對(duì)稱，故可得M1,P1關(guān)于x軸對(duì)稱;N1,Q1關(guān)于x軸對(duì)稱，所以T1必在x軸