所以sinAcosC+cosAsinC=sinA(sinC+cosC(所以tanA=1?????????6分所以(b+c(2=b2+c2+2bc=9+42=(1+22(2故b+c=1+22所以三角形的周長為a+b+c=2+22.?????????BD2=AD2+AB2-2AD?AB?cosθ=CD2+CB2-2CD?CB?cosθ?????????4分因為所以從而所以BD?AD.又AD?BD=D，所以AD⊥平面SBD，所以AD⊥SB.?????????9分2=且?????????·1·EQ\*jc3\*hps21\o\al(\s\up6(3),2),EQ\*jc3\*hps21\o\al(\s\up9(→),A)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)EQ\*jc3\*hps21\o\al(\s\up5(3),2)r3y-z=0-EQ\*jc3\*hps21\o\al(\s\up5(3),2)x-23y=0EQ\*jc3\*hps21\o\al(\s\up5(x),2)-2x+lnx，fl(x(=x-2+EQ\*jc3\*hps21\o\al(\s\up5(1),x)所以fl(1(=0,f(1(=-EQ\*jc3\*hps21\o\al(\s\up5(3),2)，函數f(x(在x=1處的切線方程為y=-EQ\*jc3\*hps21\o\al(\s\up5(3),2)；(Ⅱ)f(x(的定義域為(0,+∞(，fl(x(=x-2+EQ\*jc3\*hps21\o\al(\s\up4(a),x)=x2-EQ\*jc3\*hps21\o\al(\s\up4(2),x)x+a當a=0時f(x(=EQ\*jc3\*hps21\o\al(\s\up5(x),2)2-2x，f(x(有且僅有一個零點4；5分6分當a≥1時，fl(x(≥0，函數f(x(遞增，由f(1(<0,f(4(=aln4>0，知f(x(存在唯一零點x0∈當0<a<1時，令fl(x(=0得x1=1-、1-a,x2=1+、1-a，0<x1<1<x2時fl(x(>0，函數f(x(遞增；當x∈(x1,x2(時fl(x(<0，函數f(x(遞減；當x∈(x2,+∞(時fl(x(>0，函數f(x(遞增；且f(x2(<f(1(<0，f(4(=aln4>0，所以f(綜上所述,當a≥0時，f(x(有且僅有一個零點.PAPB=得解得a2=4，00)，M(m,0)，顯然-2<m<2，設直線PM：x=ty+m，與x2-4y2-4=2-4)y2+2tmy+m2-4=0，則又P(-4,3)在直線PM：x=ty+m上，得3t=-4-m，代入上式得整理得5m2+4m-4=0·2·解得進而0整理得-3x0-6y0+23yEQ\*jc3\*hps12\o\al(\s\up3(2),0)+x0y0=00+x0)=0由y0<0知聯立2=1+2+22=7，……2分(n(2=1+2+21+…+2n-1=2n-1EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(1),1)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(1),1)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(1),1)(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(1),1)(q-(EQ\*jc3\*hps12\o\al(\s\up3(n),k)+1(q=qn-k(EQ\*jc3\*hps12\o\al(\s\up3(n),k)(q(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(2),1)(q-(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(1),1)(q=qn-k+1(EQ\*jc3\*hps12\o\al(\s\up3(n),k)+1(q(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(3),1)(q-(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(2),1)(q=qn-k+2(EQ\*jc3\*hps12\o\al(\s\up3(n),k)+2(q…(EQ\*jc3\*hps12\o\al(\s\up3(n),k)EQ\*jc3\*hps12\o\al(\s\up3(m),1)+1(q-(EQ\*jc3\*hps12\o\al