南京一中等五校聯(lián)考四模高三數(shù)學(xué)試卷1

高三數(shù)學(xué)試題第高三數(shù)學(xué)試題第#頁共16頁由題知二面角16由題知二面角1625.A—BC1—B1為銳角，所以二面角A1—BC1—B1的余弦值為(2)設(shè)D(x,y,z)是直線BC1上一點，且BD二九BC.所以(x,y—3,z)二九(4,—3,4).解得x二4九,y=3—3九1所以AD=(4九,3—3九,4%).9由AD.々B=0,即9—25九二0.解得九二25.因為[同0,1],所以在線段BC1上存在點D,使得AD±A1B.BD9 910分此時， =x=10分BC25123.(本小題滿分分(1)證明：①Cr+Cr+1=Cr+1:②Cn+1=2Cn (其中n/￡N*,0<r<n—1,)；nn n+1 2n+2 2n+1 -Y ( 1) (2)某個比賽的決賽在甲、乙兩名運動員之間進(jìn)行，比賽共設(shè)2n+1局，每局比賽甲獲勝的概率均為pp>-，首先贏1 2J滿n+1局者獲勝(n￡N*)①若n=2，求甲獲勝的概率；②證明：總局?jǐn)?shù)越多，甲獲勝的可能性越大(即甲獲勝的概率越大)解：(1)①Cr+Cr+1=nn77+r!(n—r)! (r+1j!(n—r—1)!n!(n—r)!―(n+1)!(r+1)!((n+1)—(r+1))!=Cr+1n+12分33分②由①Cn+1=Cn+Cn+1=2Cn2n+2 2n+1 2n+1 2n+1⑵①若n二2，甲獲勝的概率p=p3+pC32p2(1-p)+pC2p2(1—p)2二p2—15p+105分②證明：設(shè)乙每一局獲勝的概率為q，則p+q=1,0<q<1。乙記在甲最終獲勝的概率為P，則nP=pn+1+pCnpnq+pCnpnq2+...+pCnpnqnn n+1 n+2=pn+11+Cnq+Cnq2+...+Cnqn)n+1 n+2

TOC\o"1-5"\h\zP—P=pn+1i+Cnq+Cnq2+...+Cnqn)—pn+2i+Cn+1q+Cn+1q2+…+Cn.qn+1n n+1 n+1 n+2 2n n+2 n+3 2計2=pn+11+Cnq+Cnq2+…+Cnqn)—\1—q^1+Cn+1q+Cn+1q2+…+Cn+1qn+1)-n+1 n+2 2n n+2 n+3 2n用n+1W+Cnq+Cnq2+...+Cnqn)—'1+Cn+1q+Cn+1q2+...+C?+1_qn+1)

n+1 n+2 2n n+2 n+3 ? -+q+Cn+1q2+Cn+1q3+…+Cn+1qn+2)n+2 n+3 2n+2所以，=pn+1(1-1)+q(Cn—Cn+1+1)+q2(Cn—Cn+1+1)n+1n+2n+2 n+32n+12n+2+…+qn(Cn—Cn+1+Cn+1)—qn+1(Cn+1—Cn+1)+qn+2Cn+12n 2n+2n+12n+2=pn+1—qn+1(Cn+1—Cn+1)+qn+2Cn+1」2n+2 2n+1 212=pn+1qn+1、Cn+1—Cn =pn+1qn+1YqCn —Cn2n+2 2n+1