2024四川绵阳高中一诊：结合数列、放缩、迭代

已知函数\(f(x)=\ln x+x^2-3x+a,f(x)\)在\((0,1]\)上的最大值为\(\dfrac{3}{4}-\ln 2\)
(1) 求实数\(a\)的值;

(2) 若数列\(\{a_n\}\)满足\(2a_na_{n+1}=f(a_n)+3a_n-1\)，且\(a_1=\dfrac{4}{3}\)

(I) 当\(n\geq ,n\in\mathbb{Z}\)，比较\(a_n\)与\(1\)的大小，并说明理由;

(II) 求证：\(3\displaystyle\sum_{i=1}^n|1-a_i|<2\)

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解.(1) \(f^\prime(x)=\dfrac{1}{x}+2x-3\)，有\(f^\prime\left(\dfrac{1}{2}\right)=0,f^\prime(1)=0\)

由题\(f\left(\dfrac{1}{2}\right)=-\ln 2+\dfrac{1}{4}-\dfrac{3}{2}+a\)得\(a=2\)

(2) \(f(x)=\ln x+x^2-3x+2\)，则

\(2a_na_{n+1}=\ln a_n+a^2_n-3a_n+2+3a_{n+1}=\ln a_n+a_n^2+1\)

(I) 由题\(2a_{n+1}=\dfrac{\ln a_n}{a_n}+a_n+\dfrac{1}{a_n}\)

做数学归纳法：\(a_1>1,a_2>1\)，设\(a_n>1\)，则

\(2a_{n+1}=\dfrac{\ln a_n}{a_n}+a_n+\dfrac{1}{a_n}>2+\dfrac{\ln a_n}{a_n}>2\)

即\(a_n>1\)，从而归纳得证明.

(II)由(I)得
\(3\displaystyle\sum_{i=1}^n|1-a_1|=3\displaystyle\sum_{i=1}^n(a_i-1)\)

而\(2a_{n+1}=\dfrac{\ln a_n}{a_n}+a_n+\dfrac{1}{a_n}<\dfrac{a_n-1}{a_n}+a_n+\dfrac{1}{a_n}=a_n+1\)

从而\(a_{n+1}<\dfrac{a_n+1}{2}\)

待定系数得到

\(a_{n+1}-1<\dfrac{a_n-1}{2}\)

从而\(a_{n+1}-1<\dfrac{a_n-1}{2}<\dfrac{a_{n-1}}{2^2}<\cdots<\dfrac{a_1-1}{2^n}\)

\(\displaystyle\sum_{i=1}^n=(a_i-1)=(a_1-1)+(a_2-1)+\cdots+(a_n-1)<(a_1-1)+\dfrac{(a_1-1)}{2}+\dfrac{(a_1—1)}{2^2}+\cdots+\dfrac{(a_1-1)}{2^{n-1}}\)
\(=(a_1-1)\left[1+\dfrac{1}{2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{2^{n-1}}\right]\)

\(=\dfrac{1}{3}\cdot\dfrac{1-\dfrac{1}{2^n}}{1-\dfrac{1}{2}}=\dfrac{2}{3}\left(1-\dfrac{1}{2^n}\right)\)

则\(3\displaystyle\sum_{i=1}^n(a_i-1)<2\left(1-\dfrac{1}{2^n}\right)<2\)