己知函数f(x)=|2x-a|+a,a€R,g(x)=|2x-1|.(1)若当g(x)

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  • 己知函数f(x)=|2x-a|+a,a€R,g(x)=|2x-1|.(1)若当g(x)<=5时,恒有f(x)<=6,求a的最大值;(2)若当x€R时,恒有f(x)+g(x)>=3,求a的取值范围.

    (1)

    |2x-1|<=5

    -5<=2x-1<=5

    -4<=2x<=6

    2x=t

    f(x)=|t-a|+a t>a f(x)=t t<=a f(x)=2a-t

    f(t)有最大值,f(a)=a, f(x)<<=6 故,a<=6

    另t=-4 t=6时,

    f(x)=|-4-a|+a<=6 |4+a|<=6-a 按以上画图得:a<=1

    f(x)=|6-a|+a<=6 |6-a|<=6-a 得:a<=6

    综上:得a<=1

    (2) 己知函数f(x)=|2x-a|+a,a€R,g(x)=|2x-1|.(1)若当g(x)<=5时,恒有f(x)<=6,求a的最大值;(2)若当x€R时,恒有f(x)+g(x)>=3

    f(x)+g(x)=y=|2x-1|+|2x-a|+a>=3

    |2x-1|+|2x-a|>=3-a 首先3-a<=0时,a>=3时,是恒成立的.

    2x-1=t 2x=t+1

    |t|+a-3>=-|t+1-a|

    画图:

    看上图,只有a-3>=0时,才成立.

    故:a-3>=0

    a>=3