t)2+(y﹣t+2)2=1(t∈R)的切线,切点分别为A,B,则•的最小值为( )РA.?B.?C.?D.2﹣3Р【解答】解:圆C:(x﹣t)2+(y﹣t+2)2=1的圆心坐标为(t,t﹣2),半径为1,Р∴|PC|2=(t+1)2+(t﹣3)2=2t2﹣4t+10,Р∴|PA|2=|PB|2=|PC|2﹣1=(t+1)2+(t﹣3)2﹣1=2t2﹣4t+9,cos∠APC==,Р∴cos∠PAB=2cos2∠APC﹣1=2×()﹣1==Р∴•=||•||cos∠PAB=(2t2﹣4t+9)•=[(t2﹣2t+5)+(t2﹣2t+4)]•,Р设t2﹣2t+4=x,则x≥3,Р则•=f(x)=(x+x+1)•=,Р∴f′(x)=>0恒成立,Р∴f(x)在[3,+∞)单调递增,Р∴f(x)min=f(3)=,Р∴•的最小值为Р故选:CР Р12.(5分)已知数列{an}与{bn}的前n项和分别为Sn,Tn,且an>0,6Sn=an2+3an,n∈N*,bn=,若∀n∈N*,k>Tn恒成立,则k的最小值是( )РA.?B.49?C.?D.Р【解答】解:∵6Sn=an2+3an,∴6Sn+1=an+12+3an+1,Р∴6an+1=(an+1+an)(an+1﹣an)+3(an+1﹣an)Р∴(an+1+an)(an+1﹣an)=3(an+1+an),Р∵an>0,Р∴an+1+an>0,Р∴an+1﹣an=3,Р又6a1=a12+3a1,a1>0,∴a1=3.Р∴{an}是以3为首项,以3为公差的等差数列,Р∴an=3n,Р∴bn==(﹣)=(﹣),Р∴Tn=(﹣+﹣+…+﹣)Р=(﹣)<=.Р∴k≥.Р故选C.Р Р二.填空题(本题共4小题,共20分.把答案填写在答题卡相应的横线上)Р13.(5分)公差不为0的等差数列{an}的前n项和为Sn,若a2,a5,a14成等比数列,
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