數(shù)學(xué)-福建省莆田市2025屆高中畢業(yè)班第二次教學(xué)質(zhì)量檢測(cè)試卷（莆田二檢）試題和答案

莆田市2025屆高中畢業(yè)班第二次教學(xué)質(zhì)量檢測(cè)數(shù)學(xué)試題參考解答及評(píng)分標(biāo)準(zhǔn)12345678CCADBBCB91011ACDACDABD 12.1213.2614.415．本小題主要考查等差數(shù)列、等比數(shù)列及數(shù)列求和等基礎(chǔ)知識(shí)；考查運(yùn)算求解能力，邏輯推理能力等；考查化歸與轉(zhuǎn)化思想，分類與整合思想，函數(shù)與方程思想等；考查邏輯推解1）解法一：依題意，設(shè)數(shù)列{an}的公差為d，因?yàn)镾3=15，所以a1+a2+a3=15，則3a1+3d=15，即a1+d=5①,···············1分因?yàn)?a4=a3+14，即2(a1+3d)=a1+2d+14，所以a1+4d=14②,·················2分由①②,得a1=2，·················································································3分d=3.·················································································4分?jǐn)?shù)學(xué)答案第1頁(yè)（共11頁(yè)）所以an=2+(n-1)×3.············································································5分所以an=3n-1.······················································································6分 ,50，···································8分其中5和17為數(shù)列{2n+1}的前5項(xiàng)與{an}的前17項(xiàng)的公共項(xiàng)， 10分=(3+5+9+17+33)+(a1+a2+…+a17)-5-17··············································11分=67+2×17+×3-5-17.·································································12分=487.··································································································13分解法二1）依題意，設(shè)數(shù)列{an}的公差因?yàn)镾3=15，所以=3a2=15，所以a2=5，·····································1分又因?yàn)?a4=a3+14，所以a3+a5=a3+14，所以a5=14，································2分所以3d=a5-a2=9，解得d=3.·································································3分所以a1=a2-d=2.·················································································4分所以an=2+(n-1)×3.············································································5分=3n-1.······················································································6分（2）設(shè)cn=2n+1，因?yàn)閍20=3×20-1=59，···········································································7分所以由cn=2n+1≤59可得n≤5，·································································8分因?yàn)閏1=3,c2=5,c3=9,c4=17,c5=33,···························································9分其中c2=5=a2，c4=17=a6，···································································10分?jǐn)?shù)學(xué)答案第2頁(yè)（共11頁(yè)）131217)································································11分=487.··································································································13分分.丄AB.····························1分所以AB1B.····················································3分B，所以AB1BC.················································································4分.···················································5分B1B，所以BC丄平面AA1B1B，···········································································7分軸的正方向，建立空間直角坐標(biāo)系.·······························8分C(2,0,0)，D(1,1,1)，·················································9分?jǐn)?shù)學(xué)答案第3頁(yè)（共11頁(yè)）取x=1，則z=-1，所以n=(1,0,-1)是平面ABD的一個(gè)法向量.···········----→ 所以直線A1C與平面ABD所成角的正弦值為.·········································15分丄BC.······························6分B1B，所以BC丄平面AA1B1B，即BC，BA，BB1兩兩垂直.······································7分 C=，·············································8分 所以△ABD的面積S=2.·······································································10分 所以CC,=2，·····················································································13分設(shè)直線A1C與平面ABD所成角為θ,則sinθ=.················································································14分?jǐn)?shù)學(xué)答案第4頁(yè)（共11頁(yè)） 所以直線A1C與平面ABD所成角的正弦值為.···············································15分性．滿分15分. 事件B=“測(cè)試結(jié)果語(yǔ)音識(shí)別成功”.····························································1分·······························································3分············································································4分=0.69.·····························································································5分條件下事件A發(fā)生的概率.··········································································6分（2）方案一的測(cè)試次數(shù)的數(shù)學(xué)期望為4.·····························································9分用X表示“方案二測(cè)試的次數(shù)”，由題意得X的可能取值為3，5.···················10分=0.512，·······································································11分P(X=5)=1P(X=3)=10.512=0.488.·····················································12分又因?yàn)镋(X)<4，···················································································14分所以，以測(cè)試次數(shù)的期望值大小為決策依據(jù)，應(yīng)選擇方案二.····························15分?jǐn)?shù)學(xué)答案第5頁(yè)（共11頁(yè)）解1）由已知，得b=1，············································································1分2，e==，···································································2分所以a=2.······························································································3分所以E1的方程為:+y2=1.······································································4分PyPAAOQBOQBx此時(shí)AP=BQ.·······················································································5分2所以弦AB中點(diǎn)的橫坐標(biāo)為22，········································································8分xPxP綜上所述，AP=BQ.·············································································10分?jǐn)?shù)學(xué)答案第6頁(yè)（共11頁(yè)）2，所以···········································12分又，·······································································14分 1所以km=②.····················································································15分 12由①②解得，經(jīng)檢驗(yàn)符合題意.·························································16分所以m=3.·····························································································17分AP=BQ.·····························································································5分?jǐn)?shù)學(xué)答案第7頁(yè)（共11頁(yè)）BBB化簡(jiǎn)得xA+xB=xp+xQ.············································································9分22綜上所述，AP=BQ.·············································································10分11則BPOR=PACS.··············11分11又因?yàn)锽P=2PA，所以BP=2PA，所以O(shè)R=CS，·····················12分PTPTyySSAARROBCx所以O(shè)為Rt△CST斜邊CT的中點(diǎn).····························································13分2=2=m②,·····················16分所以m=3.·····························································································17分?jǐn)?shù)學(xué)答案第8頁(yè)（共11頁(yè)）綜合性和創(chuàng)新性．滿分17分.···············································································2分············································3分故f(x)的單調(diào)遞增區(qū)間為(0，e)，單調(diào)遞減區(qū)間為(e，+∞).·····························4分因?yàn)閒(2)=f(4).······················································································6分所以，當(dāng)x∈[2，e]時(shí)，x+2∈[4，e+············································································································7分所以f(x+2)≤f(x)；················································································8分f(x+2)≤f(x).································9分f(x+2)≤f(x).即f(x)在[2，+∞)上具有性質(zhì)P(2).···················