题目内容

(本题满分12分)设函数f (x) =(bc∈N*),若方程f(x) = x的解为0,2,且f (–2)<–.(Ⅰ)试求函数f(x)的单调区间;(Ⅱ)已知各项不为零的数列{an}满足4Sn·f () = 1,其中Sn为{an}的前n项和.求证:
(Ⅰ)f(x)的单调递增区间为(–∞,0),(2,+∞)单调递减区间为(0,1),(1,2)(Ⅱ) 略
(Ⅰ)解
.------(2分)
f (–2) =又∵bc∈N*   ∴c = 2,b = 2
f (x) =.-------(4分)
f′(x)>0得:x<0或x>2,令f′(x)<0得:0<x<2∴f(x)的单调递增区间为(–∞,0),
(2,+∞)f(x)的单调递减区间为(0,1),(1,2)--------(6分)
(Ⅱ)证明:由已知可得:2Sn = an 
两式相减得:(an + an – 1) (anan – 1+1) =" 0" (n≥2)∴an = –an–1anan–1 =" –1 " -(7分)
n ="1" 时,2a1 = a1
an = –an–1,则a2 = –a1 = 1与an≠1矛盾.(定义域要求an≠1)∴anan–1 = 1,∴an= –n.(8分)
要证的不等式转化为

先证不等式g (x) = x –ln(1 + x),h(x) = ln(x +1) –-----(10分)
g′(x) =h′(x) =x>0  ∴g′(x)>0,h′(x)>0∴g (x), h(x)在(0,+∞)上单调递增,
g (x)>g (0) = 0,h(x)>h(0) =" 0  " ∴
,即-----(12分)
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