湖南省2025屆新高考教學(xué)教研聯(lián)盟(長(zhǎng)郡二十校)高三第二次預(yù)熱演練數(shù)學(xué)答案

2025屆新高考教學(xué)教研聯(lián)盟高三第二次預(yù)熱演練數(shù)學(xué)參考答案123456789ACDCBDCAACDABBCD2-2DE=2CD=4a．連接AE，由橢圓的定義：AE=2a-AB=2a-2a=2a,BD=2a-ED=4a；設(shè)B(-c,0)，則E(c,0)，故BE=2c， cc26 cc26 a\a23,,,t1-t，g,t1-t a2f(x)的最大值大于2”的充分不必要條件.2bc2bc2b(2,2bc2bc2b(2,數(shù)學(xué)參考答案-1|2|2|ll πsinAsinBsinAsinBsinAsinBsinB，sinAsinBsinAsinBsinAsinBsinB， , 對(duì)D，c=sin(π-3A)=sin3A=sin2AcosA+cos2AsinA2sinAcos2A+(1-2sin2A)sinA3sinA-4sin3A3-4sin2Ab-a對(duì)D，c=sin(π-3A)=sin3A=sin2AcosA+cos2AsinA2sinAcos2A+(1-2sin2A)sinA3sinA-4sin3A3-4sin2Ac(2,，c(2,，z12z12Ja2+25Tr+1Tf(x)=0與f(f(x))=0有相同的解，不妨設(shè)為x0，, 設(shè)三棱錐S-ABC的外接球半徑為R，則有R=AO1+OO1=x-x則f,(x)=(x-1)ex-1-2e(x-1)=(x-1)(ex-1-2e)，..................................................................................................................2分?jǐn)?shù)學(xué)參考答案-2數(shù)學(xué)參考答案-3令f′令f′(x-2)ex-1-ax2+2ax求導(dǎo)得f,(x)=(x-1)ex-1-2a(x-1)=(x-1)(ex-1-2a)，x-1-2a>0恒成立，.令f′令f′X)>0，解得x>1，即f(x)在1,+∞上單調(diào)遞增，′′X>0，函數(shù)f(x)在(1,+∞)上單調(diào)遞增，所以函數(shù)f(x)在x=1處取得極小值，符合題意．.................................2′X≥0恒成立且f′f(x)單調(diào)遞增，故函數(shù)f(x)無(wú)極值，不符合題意．................................................................................................2，′X)>0，函數(shù)f(x)在(-∞,1)上單調(diào)遞增；′所以函數(shù)f(x)在x=1處取得極大值，不符合題意．..................綜上，實(shí)數(shù)a的取值范圍是(-∞,)．..................................................... 2...................................................... （2）(i)設(shè)直線MN的方程為y=k(x-1),M(x1,y1),N(x2,y2)ly=k(x-1)，消去y得(3+4k2)x2-8kly=k(x-1)|x12|x122|4k-122則y1y22-12所以O(shè)M.ON=x1x2+y1y2=3+4k2=-2，................................................................................................................................ -x2-y22點(diǎn)O到直線的距離d==，......................................），..............................................................................................................................），即=72222F與平面ABCD的夾角的余弦值為．...........=1或2．.............................................................），,f2,f3,...,f2k2,f2k1,f2k，....................................................................5分2s故f1≥4，...............................................................................s所以f2≥2，同理可得f2k一1≥2，n*，所以klJ,一lJlJ一lJ數(shù)學(xué)參考答案-4lJ一—lJ(x,y,z,l)有94組；........................................lJ一—lJ(x,y,z,l,p)有29組；...................................15分{4,2,1,0,1,2}................n2n2nnn1nn．.....................................iik1k1lk．.....................................................................................................lliki，jikilijikjlkjikilijikjlkkj.2kj.2lk1llk．同理可證2i+2k≠2i1ki，....................................................................................ikijikjikijikjikj因?yàn)?kjkjikjk1不成立.j1k1．........