保密★使用前保密★使用前泉州市2025屆高中畢業班適應性練習卷參考答案2025.04一、選擇題：選擇題：本題共8小題，每小題5分，共40分。在每小題給出的四個選項中，只有一項是符合題1~8：CDCABBDC二、選擇題：本題共3小題，每小題6分，共18分。在每小題給出的選項中，有多項符合題目要求。全部選對的得6分，有選錯的得0分，部分選對的得部分分。三、填空題：本題共3小題，每小題5分，共15分。．．四、解答題：本題共5小題，共77分。解答應寫出文字說明，證明過程或演算步驟。15.（13分）某旅行社推出“文化古城游”旅游路線后，為了解游客的滿意度，對該路線的游客進行隨機抽樣調查，得到如下滿意度評分的頻率分布直方圖．（1）估計“文化古城游”路線游客滿意度評分的眾數和第80百分位數同一組中的數據用該區間的中點值為代表）（2）現從參與“文化古城游”的游客中隨機抽取3人，設這3人中滿意度評分不低于90分的人數為X，求X的分布列和數學期望以樣本數據中游客的滿意度評分位于各區間的頻率作為游客的滿意度評分位于該區間的概率）【命題意圖】本題主要考查頻率分步直方圖、樣本的數字特征、二項分布等基礎知識；考查數據處理能力、運算求解能力等；考查數形結合思想、化歸與轉化思想、或然與必然思想等；體現基礎性與應用性，導向對發展數學運算、數據分析、數學建模等核心素養的關注．【試題解析】（1）估計“文化古跡游”路線游客滿意度評分的眾數為分故第80百分位數位于[95,100]，·····················································································2分所以第80百分位數為列式1分，計算1分，只寫答案扣1分）·········4分（2）解法一：由表可知，“文化古跡游”線路游客滿意度評分不低于90分的概率為（0.070+0.050）×5=0.6=，5分依題意，X的可能取值為0,1,2,3，·················································································6分······························································································10分所以X的分布列為：XX01238 36 54 27所以·························13分解法二：設事件A=“抽取一名‘文化古跡游’線路游客的滿意度評分不低于90分”，依題意，X表示事件A發生的次數，則·························································7分于是，X的分布列為分XX01238 36 54 2716.（15分）我們把公差不為零的等差數列稱為一階等差數列．若{an+1-an}是一階等差數列，則稱{an}為二階等差數列．+1，判斷{an}是否為二階等差數列，并說明理由；(ⅰ)求{an}的通項公式；(ⅱ)求數列{l4ann-1,}的前n項和Sn．【命題意圖】本題主要考查數列的基本概念、等差數列的定義與通項公式、數列求和等基礎知識；考查運算求解能力、推理論證能力等；考查函數與方程思想、化歸與轉化思想等；體現基礎性和綜合性，導向對發展數學抽象、數學運算、邏輯推理等核心素養的關注．【試題解析】22所以an+1-an=2n+2，·································································································2分所以{an+1-an}是一階等差數列，····················································································4分（2）(ⅰ)因為{an}是二階等差數列，所以{an+1-an}是一階等差數列．=(2n-1)+(2n-3)+…+3+1··························································································8分···································································································9分當n=1時，a1=1也符合上式沒檢驗扣1分）·····························································10分 17.（15分）（1）當a=1時，求f(x)的單調區間；（2）若f(x)≤lna，求a的取值范圍．【命題意圖】本題主要考查導數的運算，利用導數研究函數的單調性、不等式恒成立等基礎知識；考查運算求解能力、推理論證能力等；考查劃歸與轉化思想、函數與方程思想、數形結合思想等；體現綜合性，導向對數學運算、邏輯推理、數學抽象、直觀想象等核心素養的關注．【試題解析】xxf（備注：沒寫定義域且單調遞增區間寫成(—∞,0)，扣2分）（2）解法一：由題意可知分x所以f(x)在區間(—1,x0)內單調遞增；在區間(x0,+∞)內單調遞減，注意到注意到等號兩邊皆為正數，故取自然對數得x0+lna=—ln(x0+1)+1)，····························································································10分所以題意等價于分故G(t)在區間(0,+∞)單調遞增，·················································································12分則所以h為關于t的單調遞減函數，值域為[0,+∞)解法二：2故可得，題意等價于t1+et≤t2+et．·········································································9分從而題意等價于ln(x+1)≤x+lna，即ln(x+1)—x≤lna，······························而在區間x∈(0,+∞)，G/(x)<0,G(x)單調遞減，··························································13分所以當x=0時，G(x)取到最大值G(0)，···································································14分從而可得題意的充要條件為lna≥G(0)=0，解得a≥1．解法三：由f(x)≤lna的必要性可知，當x=0時，f(x)≤lna，整理得—a+1≤lna，即lna+a≥1，下面證明充分性：當a≥1時，f(x)≤lna，x導函數G，可得當x∈(1,0)，H(x)<0，H(x)單調遞減；當x∈(0,+∞)，H(x)>0，H(x)單調遞增；18.（17分）點F在線段PD上（F與P,D不重合）．（1）若平面AFB∩平面PCD=l，證明：l∥平面ABCD；（3）在（2）的條件下，若點F1,F2,…,Fn是線段PF的n+1等分點，分別過點F1,F2,…,Fn在四棱錐上作平行于平面AFB的截面，記相應的截面面積為Si(i=1,2,3,...,n)，證明：．（參考公式【命題意圖】本題主要考查線面平行判定定理和性質定理、二面角的求解，以及數列求和與不等式證明等基礎知識；考查推理論證能力、運算求解能力、空間想象能力等；考查數形結合思想、化歸與轉化思想等；體現基礎性、綜合性與創新性；導向對直觀想象、數學運算、邏輯推理等核心素養的關注．【試題解析】則CD∥平面ABF．······································································································2分又因為CD平面PCD，且平面ABF∩平面PCD=l，所以CD∥l．········································3分因為CD平面ABCD，l丈平面ABCD，所以l∥平面ABCD漏掉部分條件酌情扣1-2分）···4分（2）(ⅰ)解法一：因為四邊形ABCD為正方形，則BC丄AB，因為平面ABCD丄平面ABP，平面ABCD∩平面ABP=AB,BC平面ABCD，所以BP=，則AB2+BP2=AP2，即AB丄BP．································································6分由（1）得l∥平面ABCD，不妨設l交PC于點G，連結BG,FG即平面ABF∩平面PCD=FG,平面ABF∩平面PBC=BG，如圖1,則CD∥FG∥AB，所以點F到AB的距離等于點G到AB的距離．又因為BG平面PBC，所以AB丄BG．因為BC丄平面ABP，所以BC丄BP以B為原點，分別以BP，BA，BC所在的直線為x,y,z軸，建立如圖所示的空間直角坐標系B—xyz．由AB丄平面PBC,PC平面PBC，所以AB丄PC．又因為BG丄PC，AB∩BG=B，AB,BG平面AFGB，所以PC丄平面AFGB，························9分又因為BG丄PC，AB∩BG=B，AB,BG平面AFGB，所以PC丄平面AFGB，························9分則cos<n,>=···················································································12分設二面角P—AFB的平面角為解法二：因為四邊形ABCD為正方形，則BC丄AB，因為平面ABCD丄平面ABP，平面ABCD∩平面ABP=AB，BC平面ABCD，所以BP=，則AB2+BP2=AP2，即AB丄BP．································································6分以B為原點，分別以BP,BA,BC所在的直線為x,y,z軸，建立如圖所示的空間直角坐標系B—xyz．點F到AB的距離為分當時，點F到AB的距離最小，即△AFB的面積最小．·················································8分此時······························································································9分,y1,z1)為平面ABF的一個法向量，=(x,y,z)為平面PAD的一個法向量，,設二面角P—AFB的平面角為(ⅱ)因為AB丄BG，所以四邊形ABGF為直角梯形．又所以S梯形分由題意，不妨設F1,F2,...,Fn是距離點P由近及遠的n個等分點這句不寫不扣分）設過點Fi截四棱錐P—ABCD的截面交PC于點Gi，交BP于點Bi，交AP于點Ai，則所以．所以分19.（17分）已知橢圓的離心率為且過點(,1)．（1）求E的方程；（2）若直線與橢圓交于兩點，當以這兩點和橢圓的中心為頂點的三角形面積達到最大值時，稱該直線為橢圓的“好直線”．(ⅰ)設O為坐標原點，若斜率存在的直線l是E的“好直線”，l與E交于A,B兩點，求△AOB的面積；(ⅱ)已知四邊形MNPQ為平行四邊形，若直線MN,NP,PQ,QM均為E的“好直線”，且不與y軸平行，求四邊形MNPQ的面積的最小值．【命題意圖】本題主要考查橢圓基本性質、直線與橢圓的位置關系等基礎知識；考查運算求解能力、推理論證能力等；考查化歸與轉化思想、數形結合思想、函數與方程思想等；體現綜合性與創新性，導向對直觀想象、邏輯推理、數學運算、數學抽象等核心素養的關注．【試題解析】（2）(ⅰ)解法一：設直線l:y=kx+m，l與E的交點為A(x1,y1),B(x2,y2)．Δ=(4km)24(2m24)(1+2k2)=8(4k2m2+2)>0時，由韋達定理分原點O到直線l的距離，·············································································7分解法二：設A(x1,y1),B(x2,y2)，O為原點，直線OB的方程為y2x—x2y=0，····································4分 （用面積公式S△AOB=×|OB|×|OA|證明也可，直接用公式扣3分）·············6分點A(x1,y1),B(x2,y2)在橢圓上，滿足分EQ\*jc3\*hps20\o\al(\s\up5(y),2)·····························································································································9分因此當且僅當x1x2+2y1y2=0時等號成立．所以△AOB的面積為．························································································10分解法三：設A(2cosα,sinα),B(2cosβ,sinβ)，0<α,β≤2π,O為原點．··································4分由解法二可知S△AOB=|2cosαsinβ—2sinαcosβ|（利用該結論必須要證明，否則扣3分）······································································································