Question

已知a＞0,函数f（x）=-a（2cos²x+√3 sin2x）+3a+b,当x∈[0,π/2]时,-5≤f（

Answer 1

f（x）=-a（2cos²x+√3 sin2x）+3a+b

=-a（cso2x+√3sin2x）+4a+b

=-a/2·sin(2x+π/6)+4a+b

x∈[0,π/2] ,

2x+π/6∈[π/6,7π/6]

sin(2x+π/6）∈[-1/2,1]

f(x)∈[15a/4+b,7a/2+b] -5≤f（x）≤1

所以15a/4+b=-5,7a/2+b=1

a=-16,b=57