MT【259】2016天津压轴题之最佳逼近

(2016天津压轴题)设函数$f(x)=(x-1)^3-ax-b,xin R$, 其中$a,bin R$
(1)求$f(x)$的单调区间.
(2)若$f(x)$存在极值点$x_0$,且$f(x_1)=f(x_0),$其中$x_1 e x_0$,求证:$x_1+2x_0=3$;
(3)设$a>0$,函数$g(x)=|f(x)|,$求证:$g(x)$在区间$[0,2]$上的最大值不小于$dfrac{1}{4}$

分析:
(1)
当$ale0,f(x)$在$(-infty,+infty)$单调递增.
当$a>0,f(x)$在$left(-infty,-dfrac{sqrt{3a}}{3}+1 ight) earrow,left(-dfrac{sqrt{3a}}{3}+1,dfrac{sqrt{3a}{3}}+1 ight)searrow,left(dfrac{sqrt{3a}}{3}+1,+infty ight) earrow$
(2)由于$x_0,$是$f(x)$的极值点,故由(1)知$a>0$,且$a=3(x_0-1)^2$,由题意$f(x)=f(x_0)$有且仅有两根$x_0,x_1$,容易验证$f(3-2x_0)-f(x_0)=0$

故$3-2x_0=x_0( extbf{舍去，此时}a=0) $或$3-2x_0=x_1$即$2x_0+x_1=3$

(3)记$M(a,b)=maxlimits_{a>0,bin R}|f(x)|$则
$$egin{cases}
M(a,b)&ge|f(0)|=|-1-b|\
M(a,b)&ge|f(dfrac{1}{2})|=|-dfrac{1}{8}-dfrac{1}{2}a-b|\
M(a,b)&ge|f(dfrac{3}{2})|=|dfrac{1}{8}-dfrac{3}{2}a-b|\
M(a,b)&ge|f(2)|=|1-2a-b|\
end{cases}$$
则egin{align*}
6M(a,b)&ge|f(0)|+2|f(dfrac{1}{2})|+2|f(dfrac{3}{2})|+|f(2)|\
&=|-1-b|+2|-dfrac{1}{8}-dfrac{1}{2}a-b|+2|dfrac{1}{8}-dfrac{3}{2}a-b|+|1-2a-b|\
&=|-1-b-2(-dfrac{1}{8}-dfrac{1}{2}a-b)+2(dfrac{1}{8}-dfrac{3}{2}a-b)-(1-2a-b)|\
&=dfrac{3}{2}
end{align*}
故$M(a,b)ge dfrac{1}{4}$,当$f(x)=(x-1)^3-dfrac{3}{4}(x-1)$时取到等号.

注：通过画图，两条直线“夹紧”曲线，得到0，1/2，3/2或者1/2，3/2，2都可以。

$3Mge |f(0)|+dfrac{3}{2}|f(dfrac{1}{2})|+dfrac{1}{2}|f(dfrac{3}{2})|$

或者$3Mge dfrac{1}{2}|f(dfrac{1}{2})|+dfrac{3}{2}|f(dfrac{3}{2})|+|f(2)|$  

两者并起来写就是$6M(a,b)ge | f(0)|+2|f(dfrac{1}{2})|+2|f(dfrac{3}{2})|+|f(2)|$