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.
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