2025屆高三年級8月新起點(diǎn)調(diào)研模擬試卷（一）·數(shù)學(xué)參考答案、提示及評分標(biāo)準(zhǔn)23-1=7.故選C.g,(x)≥0恒成立，g,(x)=x2+2x-2m=(x+1)2-1-2m要使g,(x)≥0恒成立，則(x+1)2-1-2m≥0恒成23.C賦值法知道，解得4.B設(shè)A(x1,y1)，B(x2,y2)，可得兩式相減可得25.A設(shè)正四棱錐P-ABCD的內(nèi)切球OH=R，取CD的中點(diǎn)F，連接HF,PF，設(shè)O點(diǎn)在側(cè)面PCD上的投影為Q點(diǎn)，則Q點(diǎn)在PF上，且OQ丄PF，POQ∽PFH，球心到四棱錐頂點(diǎn)的距離為故選A.7.C因?yàn)閒(x)=x2-2-lnx，所以f,(x)=2x-lnx-1，令g(x)=2x-lnx-1對其求導(dǎo)當(dāng)時，gI(x)>0，故函數(shù)g(x)單調(diào)遞增所以 解得故A正確；對于B，則將點(diǎn)A(0,-1)代入得， 對于C，因?yàn)榱罱獾?，k∈Z，取k=k=1，則f(x)在[8,11]上單調(diào)遞減，故D錯誤.故選AC.f(x)單調(diào)遞增，故A正確；對于B，f(x)是定義在R上的函數(shù)，取x=y=0，則f(0)=0，取y=-x，則f(x)+f(-x)=0，即f(-x)=-f(x)，故f(x)是奇函數(shù)，故B正確；對于C、D，設(shè)f(x)=kx，代入 各至少存在一個零點(diǎn)，故D正確；故選ABD.n,則n的最大值為21，,則n的最大值為21，0?2sinBsinC，........................................................................2分即sin2A=sin2B+sin2C?sinBsinC..................................................................................................3分22?bc，................................................................................................4分........................................................................6分 16.解1）取弧BC中點(diǎn)H，則OH丄BC，以O(shè)為坐標(biāo)原點(diǎn)，直線OB,OH..............................................................2分設(shè)F(x,y,0)，則G(x,y,1)，其中x2+y2=4,y>0，CG=(x+2,y,1),BF=(x?2,y,0)，.......................3分...........................................................4分（2）由BE丄平面ABC,ACC平面ABC，得BE丄AC，...............................................................5分又AB2+AC2=BC2，則AB丄AC，而AB∩BE=B,AB,BEC平面ABE，........................................6分.................................................7分DF=(x+2,y,-1)，由DF//平面ABE，得DF.AC=-2x-4+2y=0，.............................................8分.........................................................9分設(shè)n=(a,b,c)是平面FOD的法向量，則取a=1，得，..................10分則平面FOD與平面GOD夾角的余弦值為分...............................................................................................13分則點(diǎn)G到直線OD的距離分所以整理得，...........................................................................2分所以E的標(biāo)準(zhǔn)方程為................................................................................................3分（2）由題意可知直線l1和直線l2斜率若存在則且曲線E的漸近線方程為x±y=0，................................4分,y1)、B(x2,y2)，..............................................5分2+8>0，.......................................................................................6分 ,................................................................................................7分............................................9分同理可得.................................................................................................10分所以P到兩漸近線的距離分別為...............................................................................................................................................11分...............................................................................................................................................12分則四邊形PMQN面積為SPMQN=S矩形=QMQN-QNd1-QM分 .........................................................................15分 ....................................................1分 ...........................................................................2分因?yàn)橘|(zhì)量指標(biāo)值X近似服從正態(tài)分布N(69,112)，所以P(X.........4分.........................................5分..........................................6分..........................7分相應(yīng)的概率為：隨機(jī)變量η的分布列為：η0123P38所以η的數(shù)學(xué)期望分設(shè)每箱產(chǎn)品的利潤為Z元，................................................................................................12分由題意知：Z=mY+(100-Y)ln(25-m)=(m-ln(25-m))Y+100ln(25-m)，.......................................13分所以Y~B(100,0.16)，所以E(Y)=100×0.16=16，.......................................................................14分所以E(Z)=E[(m-ln(25-m))Y+100ln(25-m)]=(m-ln(25-m))EY+100ln(25-m)=16(m-ln(25-m))+100ln(25-m)=16m+84ln(25-m)....................................................................................................................15分令=16x+84ln，由得.....................................16分又，f(x)>0，f(x)單調(diào)遞增f(x)<0，f(x)單調(diào)遞減，所以當(dāng)取得最大值.所以當(dāng)時，每箱產(chǎn)品利潤最大......................17分.............................................................1分4n}中的項(xiàng)，故數(shù)列{an}不是“乘法封閉數(shù)列”；...................2分2<8，則且，..................................3分聯(lián)立①②解得，a2=2,a3=4；................................................................................................5分.......................................................................................................6分24，........................................................................7分 由數(shù)列{an}為“除法封閉數(shù)列”，則,,,,這5個數(shù)都是數(shù)列{an}中的項(xiàng)，..................9分所以有則有......10分由③④可得故是等比數(shù)列.............................................................12分k-1