举报
karlsui
不好意思,上面两道题有错误现在更正一下! f(x)=√3(sin²x-cos²x)+2sinxcosx+1=-√3cos2x+sin2x+1=2sin(2x-π/3)+1 当2x-π/3∈[π/2+2kπ,3π/2+2kπ]时f(x)单调递减 解得 x∈[5π/12+kπ,11π/12+kπ] ∴函数f(x)单调递减区间为x∈[5π/12+kπ,11π/12+kπ](k∈Z) 若x∈[0,π/2]则2x-π/3∈[-π/3,2π/3] sin(2x+π/3)∈[-√3/2,1] ∴f(x)的值域为[1-√3,3] f(0.5x+π/6)=2sinx+1 g(x)=2asinx+cos2x+a=2asinx+1-2sin²x+a=2asinx-2sin²x-0.5a²+0.5a²+a+1=-2(sinx-0.5a)²+0.5a²+a+1 当-2≤a≤2时h(a)=0.5a²+a+1 当a≤-2时h(a)=-a-1 当a≥2时h(a)=3a-1 累死我了,请求加二十分!