福建泉州2025屆高中畢業(yè)班適應(yīng)性練習(xí)卷數(shù)學(xué)評分標(biāo)準(zhǔn)

評分標(biāo)準(zhǔn)制定相應(yīng)的評分細(xì)則.嚴(yán)重的錯誤，就不再給分.3．解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得4．只給整數(shù)分?jǐn)?shù)．填空題和解答題不給中間分.在ABC中，角A，B，C所對的邊分別為a，b，c，已知a<b<C且tanA,tanB,tanc均為整數(shù).【命題意圖】本小題主要考查解三角形的正、余弦定運算求解能力，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想，考查數(shù)學(xué)運算、邏輯推理等核心素養(yǎng)，體現(xiàn)基礎(chǔ)性、綜合性與應(yīng)用性.【試題簡析】（1）在/\ABC中，tanA,tanB,tanc均為整數(shù)，a<b<C42:A·B·C(r·")，且A-、B-、C，····················42,則，·······························································2分且tanA為整數(shù)，即tanBtanc1tanBtanc.由tanB,tanc均為整數(shù)，且B<c,tanA1，得mns"=2,mnczs，可得tanB2,tanc3，···················故tan2B13tanAtanc.則.··················································由正弦定理，·························可得，···················································2在BCD中，由余弦定理，得BD2a25 在BCD中，由余弦定理，得BD2a25.····································個得4分．某數(shù)學(xué)小組研究發(fā)現(xiàn)，多選題為1p(0<p<1)．現(xiàn)有一道多選題，學(xué)生李華完全不會，此時他有三種答題方案：若且學(xué)生李華選擇方案Ⅰ,求本題得分的數(shù)學(xué)期望;【命題意圖】本小題主要考查概率等知識；考查運算求解能力、數(shù)據(jù)處理能力等；考查化歸與轉(zhuǎn)化思想等；體現(xiàn)綜合性和應(yīng)用性，導(dǎo)向?qū)Πl(fā)展數(shù)學(xué)運算、數(shù)據(jù)分析、數(shù)學(xué)建模等核心素養(yǎng)的關(guān)注.··········································································································1分所以x的分布列為,·······················所以x的分布列為X023P884················································································································4（2）記為“從四個選項中隨機(jī)選擇一個選項的得分”，則P(:0)P(p，,1P3131P313所以E()o(3pi2)i42(ip)icp2P；·······················所以E(n)o(pi)tip)(ip).··································11分已知拋物線E:y24X，點C在E的準(zhǔn)線上，過E焦點F的直線與E相交于A，B為正三角形.的余弦值.【命題意圖】本小題主要考查拋物線的定義及方程、直線與拋物線的位置關(guān)系、直線與平面垂直等基礎(chǔ)知,弦AB的中點，如圖所示，····················································································································所以.··································································3分為等邊三角形，則，················································································································4又，····························································所以·······································6分（2）由題PABC為正三棱錐，即ABACBC,PAPBPC，由（1）知ABACBC12，則PAPBPC6/2又PQnBQQ，BQc平面BSQ，SQC平面BPQ，（注：直線屬于平面可不寫，關(guān)注線面垂直..ACIPB，又AMnACA，AMc平面PAC，ACc平面PAC，··························（注：直線屬于平面可不寫，關(guān)注線面垂直證明中的交點即可）11分又正三棱錐各側(cè)面三角形都全等，所以PA即PA,PB,PC兩兩垂直，故可將PABC補(bǔ)成如圖所示的正方體，··············································································V2xPA·n0V2xPN·n03\2Y3\2Z取y1，則n····························································································【命題意圖】本題主要考查數(shù)列、常用邏輯用語等知識；考查推理論證能力、運算求解能力等；考查特殊與一般思想、有限與無限思想、化歸與轉(zhuǎn)化思想等；體現(xiàn)綜合性、創(chuàng)新性，導(dǎo)向?qū)Πl(fā)展數(shù)學(xué)運算、邏輯推理、數(shù)學(xué)建模等核心素養(yǎng)的關(guān)注.所以數(shù)列an}的通項公式為··········································n················································································································3················································································································4··············································6分綜上所述，···············································································7分即（注：求數(shù)列也可有其他解法，酌情給分．該步寫出通項公式的化簡················································································································8故···········································································9分···································所以（注：錯位相減過程可簡略，但要有表示．無過程且答案錯誤的此2步不給分）·····················································（3）由題，數(shù)列滿足，即··················所以·······························································································（1）證明：曲線是軸對稱圖形；（2）若直線與函數(shù)和的圖象共有三個交點，設(shè)這三個交點的橫坐,證明：；恒成立，求的最大值.【命題意圖】本小題主要考查函數(shù)與導(dǎo)數(shù)等知識；考查運算求解能力等；考查數(shù)形結(jié)合思想、分類與整合思想、函數(shù)與方程思想、化歸與轉(zhuǎn)化思想等；體現(xiàn)觀想象、邏輯推理、數(shù)學(xué)運算等核心素養(yǎng)的關(guān)注.【試題簡析，··············································································································································2所以g(x)為偶函數(shù)，····················································································3分圖形或?qū)ΨQ軸的該步給分，反之只寫到偶函數(shù)該步不給分）·································4分22當(dāng)x>0，D'(x)>0；當(dāng)x<0，D'(x)<0，x→-0,D(x)→-o注：進(jìn)行D(x)的單調(diào)性分析或極致分析即可給分）················································································································6致分析即可給分）························································································7分yt與圖象有三個交點，顯然t>1，令整理得22ei-o，····················