山東省煙臺(tái)市期末學(xué)業(yè)水平診斷2025屆高三數(shù)學(xué)答案

DCADBCBB 又因?yàn)椋訠=.····································所以ΔABC的周長(zhǎng)l=2+································8分解得a=1,b=2或a=?··················································6分（2）f(x)的定義域?yàn)?，···········?分令g(x)=ax2+2(a?1)x+a，x∈(0,+∞).①當(dāng)a≤0時(shí)，f′(x)<0，函數(shù)f(x)在(0,+∞)綜上，當(dāng)a≤0時(shí)，函數(shù)f(x)在(0,+∞)上單調(diào)遞減； 當(dāng)0<a<時(shí)，f(x)在(0,)和(,+∞)上單調(diào)遞增，在單減；當(dāng)a≥在單增.··15分17.解1）連接AC，取BC中點(diǎn)O，連接AO,OP， 在直角梯形ABCD中，由已知可得AD=3，AC=2，·································2分又AB=BC=2，O為等邊ΔABC的邊BC中點(diǎn)，所以A又因?yàn)椤鱌BC為等邊三角形，O為BC中點(diǎn)，所以O(shè)P丄BC.···············因?yàn)锳O∩OP=O，AO?平面AOP，PO?平面AOP，所以BC丄平面AOP，·········································5分以O(shè)為坐標(biāo)原點(diǎn)，以O(shè)A,OB以及垂直于平面AOB的方向分別為x,y,z軸的正方向，建立空間直角坐標(biāo)系O?xyz如圖所示，則A(,0,0)，設(shè)m=(x,y,z)為平面APD的一個(gè)法向量，故直線CP與平面APD所成角的正弦值為、.······2218.解1）設(shè)雙曲線C:x?y=λ,·······43將點(diǎn)(4,3)代入方程可得=λ,解得λ=1，22所以，雙曲線方程為················（2i）設(shè)直線l與C的右支交于點(diǎn)A(x1,y1),B(x2,y2).當(dāng)l的斜率存在時(shí)，不妨設(shè)l:y=kx+m，2因?yàn)橐訟B為直徑的圓過雙曲線的右頂點(diǎn)(2,0)，x2當(dāng)直線為y=k(x?2)時(shí)，過定點(diǎn)(2,0)，不滿足條件；又當(dāng)l的斜率不存在時(shí)，由x1?2=得直線x=14，亦過x12x21x2+y1y2x222+12k2+1222+12k2+1222*，所以數(shù)列{an}是以2為首項(xiàng)、2為公比的等比數(shù)列，所以an=2n.···················4分,b3,…,bn各項(xiàng)分別除以2后，恰是a1,a2,…,an?1滿足條件①②的排列，滿足條件①②的排列，其個(gè)數(shù)為cn?3；···························≥16，設(shè)bk+1是該排列中第一個(gè)出現(xiàn)的2的偶數(shù)次冪，則前k個(gè)數(shù)應(yīng)是 因?yàn)閏4n+3)