题160904(14分)若对任意实数$x$都有$left| 2x-a ight|+left| 3x-2a ight|ge {{a}^{2}}$,求$a$的取值范围.
试题来源:2016年中科大自招
参考答案:$left[-dfrac 13,dfrac 13 ight]$
解法1:直接分类讨论去绝对值,方法简单,但计算繁琐
当$a=0$时,不等式化为$left| 2x ight|+left| 3x ight|ge 0$,显然成立;
当$a>0$时,有$dfrac{a}{2}<dfrac{2a}{3}$,
(i)若$xle dfrac{a}{2}$,则不等式化为$left( a-2x ight)+left( 2a-3x ight)ge {{a}^{2}}$恒成立,
即对任意$xle dfrac{a}{2}$,${{a}^{2}}-3ale -5x$恒成立,
$ herefore $${{a}^{2}}-3ale -dfrac{5}{2}a$,即${{a}^{2}}-dfrac{1}{2}ale 0$,解得$0<ale dfrac{1}{2}$;
(ii)若$dfrac{a}{2}<xle dfrac{2a}{3}$,则不等式化为$left( 2x-a ight)+left( 2a-3x ight)ge {{a}^{2}}$恒成立,
即对任意$dfrac{a}{2}<xle dfrac{2a}{3}$,${{a}^{2}}-ale -x$恒成立,
$ herefore $${{a}^{2}}-ale -dfrac{2a}{3}$,即${{a}^{2}}-dfrac{1}{3}ale 0$,解得$0<ale dfrac{1}{3}$;
(iii)若$x>dfrac{2a}{3}$,则不等式化为$left( 2x-a ight)+left( 3x-2a ight)ge {{a}^{2}}$恒成立,
即对任意$x>dfrac{2a}{3}$,${{a}^{2}}+3ale 5x$恒成立,
$ herefore $${{a}^{2}}+3ale dfrac{10a}{3}$,即${{a}^{2}}-dfrac{1}{3}ale 0$,解得$0<ale dfrac{1}{3}$;
故$ain (0,dfrac{1}{3}]$;
同理可得,当$a<0$时,$-ain (0,dfrac{1}{3}]$,即$ain left[ -dfrac{1}{3},0 ight]$;
综上,$a$的取值范围是$ain left[ -dfrac{1}{3},dfrac{1}{3} ight]$.
解法2:利用不等式性质
由$left| 2x-a ight|+left| 3x-2a ight|$$=2left( left| x-dfrac{a}{2} ight|+left| x-dfrac{2}{3}a ight| ight)+left| x-dfrac{2}{3}a ight|$
$ge 2left| x-dfrac{a}{2}+dfrac{2a}{3}-x ight|+left| x-dfrac{2}{3}a ight|=2left| dfrac{a}{6} ight|+left| x-dfrac{2}{3}a ight|ge left| dfrac{a}{3} ight|$,
当且仅当$x=dfrac{2}{3}a$时取等,
即${{left( left| 2x-a ight|+left| 3x-2a ight| ight)}_{min }}=left| dfrac{a}{3} ight|$,
$ herefore $$left| dfrac{a}{3} ight|ge {{a}^{2}}$,解得$ain left[ -dfrac{1}{3},dfrac{1}{3} ight]$
综上,$a$的取值范围是$ain left[ -dfrac{1}{3},dfrac{1}{3} ight]$.
解法3:将题目转化为恒成立问题
取$kin R$,令$x=ka$,则不等式化为$left| 2ka-a ight|+left| 3ka-2a ight|ge {{left| a ight|}^{2}}$,
即$left( left| 2k-1 ight|+left| 3k-2 ight| ight)left| a ight|ge {{left| a ight|}^{2}}$,
当$a=0$时,显然成立;
当$a e 0$时,原不等式转化为$left( left| 2k-1 ight|+left| 3k-2 ight| ight)ge left| a ight|$,对$forall kin R$恒成立.
设$gleft( k ight)=$$left| 2k-1 ight|+left| 3k-2 ight|=left{ egin{array}{*{35}{l}} 5k-3, & kge dfrac{2}{3}, \ 1-k, & dfrac{1}{2}le k<dfrac{2}{3}, \ -5k+3, & k<dfrac{1}{2} \end{array} ight.$
则当$k=dfrac{2}{3}$时,$gleft( k ight)$有最小值$dfrac{1}{3}$,所以$0<left| a ight|le dfrac{1}{3}$,
即$ain left[ -dfrac{1}{3},0 ight)igcup left( 0,dfrac{1}{3} ight]$.
综上,$a$的取值范围是$ain left[ -dfrac{1}{3},dfrac{1}{3} ight]$.