(1)设=x得:(1-b)x2+cx+a=0,由根与系数的关系,得:, 解得,代入解析式f(x)=,由f(-2)=<-, 得c<3,又c∈N,b∈N,若c=0,b=1,则f(x)=x不止有两个不动点,∴c=2,b=2,于是f(x)=,(x≠1). (2)由题设,知4Sn•=1,所以,2Sn=an-an2 ①; 且an≠1,以n-1代n得:2Sn-1=an-1-an-12,②; 由①-②得:2an=(an-an-1)-(an2-an-12),即(an+an-1)(an-an-1+1)=0, ∴an=-an-1或an-an-1=-1,以n=1代入①得:2a1=a1-a12, 解得a1=0(舍去)或a1=-1;由a1=-1,若an=-an-1得a2=1,这与an≠1矛盾, ∴an-an-1=-1,即{an}是以-1为首项,-1为公差的等差数列,∴an=-n; (3)证法(一):运用反证法,假设an>3(n≥2),则由(1)知an+1=f(an)=, ∴==•(1+)<(1+)=<1,即an+1<an(n≥2,n∈N) ∴an<an-1<…<a2,而当n=2时,a2===<3; 这与假设矛盾,故假设不成立,∴an<3. 证法(二):由an+1=f(an)得an+1=,=-2(-)2+≤ 得an+1<0或an+1≥2,若an+1<0,则an+1<0<3,结论成立; 若an+1≥2,此时n≥2,从而an+1-an=≤0, 即数列{an}在n≥2时单调递减,由a2=2,可知an≤a2=2<3,在n≥2上成立. |