题目内容
(本题满分14分)
已知函数
的图象经过点
和
,记
(
)
(1)求数列
的通项公式;
(2)设
,若
,求
的最小值;
(3)求使不等式
对一切
均成立的最大实数
.
已知函数
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(1)求数列
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(2)设
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(3)求使不等式
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(1)
;
(2)
; (3)
.
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(2)
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本试题主要是借助于函数为背景求解数列的通项公式,并利用错位相减法得到数列的和,同时利用放缩法得到不等式的证明。
(1)因为函数
的图象经过点
和
,记
,联立方程组得到a,b的值。
(2)由(1)得
,然后利用错位相减法得到数列的和。
(3)要使不等式
对一切
均成立,则可以分离参数p,得到关于n的表达式,进而求解数列的最值,得到参数p的范围。
解:(1)由题意得
,解得
, …………2分
…………4分
(2)由(1)得
,
①
② ①-②得
.
, …………7分
设
,则由
得
随
的增大而减小,
随
的增大而增大。
时,
又
恒成立,
………10分
(3)由题意得
恒成立
记
,则
…………12分
是随
的增大而增大
的最小值为
,
,即
. …………14分
(1)因为函数
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(2)由(1)得
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(3)要使不等式
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解:(1)由题意得
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(2)由(1)得
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设
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得
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又
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(3)由题意得
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记
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