2023吉安一模理综

17.解:(1)设{an}的公比为q,∵{an}是等比数列,且a1=e,lna2+na3=8..Ina2+Ina,=In(a2a3)=8,a3=eq=e,解得q=e,{an}的通项公式为a=eX(e2)1=e2n-1(2):an=2-1,lnan=2n-1,又S.是数列{lnan}的前n项和,∴S=2(1+2n-1)=n2,又Sn+Sn+2=5m+4m2+(m+2)2=(m+4)2(m∈N),解得;m=6m的值为6.

I=t22.解:(1)由题意得:,消去t,∴+2=1-y2(y≥0)=1(y≥0)pcos0+3p8-3=0#pp+2p'sin8-3=0化简为:P2(2-cos20)-3=0,0∈[0,元]∴C1的极坐标方程为:P(2-c0s20)-3=0,B∈[0,x],由ma(+)一得(2sIn2y+2x-3=0即:3y+3x-6=0C2的直角坐标方程为:3y+3x-6=0(2)由得:=2,A(,)2(2-cs20)-3=0im(+)=得p=3,B(,吾)02Am(一5)=12.③sm票一

22.[选修4-4:坐标系与参数方程](10分)x=cos 2+3【解析】(1)因为曲线C的参数方程为(q为参数)y=sin+I所以x=cos2q+3=4-2sin2g,y=sin2g+1,所以x=4-2(y-1),即x+2y-6=0,(3分)由-1scos20≤1,可得2≤x≤4,所以曲线C1的普通方程为x+2y-6=02≤x≤4).(5分)(2)由 acos=x, psin 8=y,可得曲线C1的极坐标方程为 pcos+2 psin 0=6,且≤tan≤1,因为射线6=a(P>0)与曲线C1交于点A,所以OA|=≤tana≤1,(7分)cos a+2sin a 4因为射线6=a(P>0)与曲线C2交于点O,B,所以OB|=2cosa,12 cosa所以|OA||OB(9分)coS a +2sin a 1+2 tan a因为元 s tana s1,所以s1+2 tan as3,所以4s,121+2taa△≤8,所以OAOB|的取值范围为[4,8].(10分)

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