2025年2月診斷性測(cè)試數(shù)學(xué)答案

符合題目要求的.12345678BCDDBBDD9AC令f,(x)=0，得x=1（舍負(fù)）·················································所以f(x)的極小值為····································································5分（2）原不等式等價(jià)于x2+6x?3lnx?令x2+6x?3lnx?··································································6分得h(x)≥0的解集為(0,1]，即原不等式的解集為(0,1]···································13分解1）設(shè)兩個(gè)骰子的點(diǎn)數(shù)之和是4或5或6的事件為A，所以···························································（2）設(shè)被測(cè)試中的一個(gè)學(xué)生吸煙的概率為P，則被計(jì)數(shù)器計(jì)數(shù)的學(xué)生人數(shù)為：所以估計(jì)某高校吸煙的人數(shù)為12000P=220人·····································解1）設(shè)直線PQ的方程為x=my?3，因?yàn)橹本€PQ與拋物線Γ相切，所以=16m2?48=0→（2）直線AB的方程為x=my?3，設(shè)A(x1,y1),B(x2,y2)， 因?yàn)镻B=2PA，所以y2=2y1，解得y1=±·6············································9分 (3)當(dāng)y1=6時(shí)，A|(2,6,,F(1,0),AF:y=26(x?1)································10 (3)因?yàn)辄c(diǎn)G為ΔBCP的重心，所以解1）連接B1C交BC1于E點(diǎn)，連接D1E，得E為B1C的中點(diǎn)，因?yàn)锳1C丈平面BC1D1，D1E平面BC1D1，故A1C平面BC1D1···················································································4分（2）因?yàn)锳1DBD1，所以上C1BD1為A1D與BC1所成角········································5分設(shè)AB1BD1=F，所以BF=,AF=AB1，BB·······································································8分B1B·································································AF2+FB2=AB2，即為異面直線A1D與BC1所成角正切值··················（3）作BH丄AA1于H點(diǎn)，作HM丄CC1于M點(diǎn)，連接BM，則AA1丄平面BHM，故三棱柱ABC?A1B1C1內(nèi)切球半徑等于ΔBHM的內(nèi)切圓半徑，×····················································因?yàn)槠矫鍭A1B1B丄平面ABC， 所以BM=5sin上B1BC=23，故ΔBHM為正三角形··································15分設(shè)內(nèi)接球半徑為r，則SΔBHM=2(BM+BH+HM)r=4BM2，解得r=1，所以VABC?A1B1C1=SΔABC.2r=24··································································17分解1）(1,2,3,4)；(1,2,4,3)；(1,4,2,3)；(1,4,3,2)；(4,1,2,3)；(4,1,3,2)；(4,3,1,2)；(4,3,2,1)····················································（2）假設(shè)a1=k,k∈{2,3,4,,n?1},an=t，由題意得存在m10由題意得存在m10mn?k?1所以有a1=1或n成立·················································································10分（3）設(shè)n元凝聚排列的個(gè)數(shù)為xn，則x2=2，下面考慮n+1元凝聚排列的個(gè)數(shù)為xn+1，=n+1···········································①當(dāng)a1=n+1時(shí)，(a2,a3,,an+1)為1,2,,n的一個(gè)凝聚排列，所以此時(shí)有xn種不同排列···········································································13分2所以此時(shí)a2,a3,,an+1有xn種不同排列·······