题目内容
(本题满分12分)设函数f (x) =(b,c∈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)
由f (–2) =又∵b,c∈N* ∴c = 2,b = 2
∴f (x) =.-------(4分)
令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) (an – an – 1+1) =" 0" (n≥2)∴an = –an–1或an –an–1 =" –1 " -(7分)
当n ="1" 时,2a1 = a1 –
若an = –an–1,则a2 = –a1 = 1与an≠1矛盾.(定义域要求an≠1)∴an – an–1 = 1,∴an= –n.(8分)
要证的不等式转化为
先证不等式令g (x) = x –ln(1 + x),h(x) = ln(x +1) –-----(10分)
则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 " ∴。
故,即-----(12分)
.------(2分)
由f (–2) =又∵b,c∈N* ∴c = 2,b = 2
∴f (x) =.-------(4分)
令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) (an – an – 1+1) =" 0" (n≥2)∴an = –an–1或an –an–1 =" –1 " -(7分)
当n ="1" 时,2a1 = a1 –
若an = –an–1,则a2 = –a1 = 1与an≠1矛盾.(定义域要求an≠1)∴an – an–1 = 1,∴an= –n.(8分)
要证的不等式转化为
先证不等式令g (x) = x –ln(1 + x),h(x) = ln(x +1) –-----(10分)
则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 " ∴。
故,即-----(12分)
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