已知f(x)=-2asin(2x+(π/6))+2a+b,x属于[(π/4),(3π/4)],是否存在常数有理数a,b,

3个回答

  • (1)π/4<x<3π/4

    2π/3<2x+ π/6<5π/3

    当2x+ π/6=2π/3,即x=π/4时,sin(2x+ π/6)取得最大值√3 /2

    当2x+ π/6=3π/2,即x=2π/3时,sin(2x+ π/6)取得最小值-1

    当a>0时,f(π/4)=-2a(√3 /2)+2a+b=-3

    f(2π/3)=-2a(-1)+2a+b=√3-1

    两式联立解得:a=1满足a>0,但b=√3-5不是有理数,故舍去;

    当a<0时,f(π/4)=-2a(√3 /2)+2a+b=√3-1

    f(2π/3)=-2a(-1)+2a+b=-3 ;

    两式联立解得:a=-1满足a<0,b=1.

    综上,a=-1,b=1

    (2)此函数周期为4π,

    28π/5=4π+8π/5≤x≤a,

    2π+17π/15≤1/2x+π/3≤a/2+π/3,

    单调区间右端点的最大值是4π=a/2+π/3,得a=22π/3.

    (3)y=sinx+cosx

    =√2*(√2/2*sinx+√2/2*cosx)

    =√2*sin(x+π/4)

    T=2π/1=2π.

    (4) y=|tanx|+tanx

    =0,(kπ-π/2<x<kπ)或

    =2tanx(kπ<x<kπ+π/2),k为整数.

    由图象可得T=π.