f(x)=-a[1-2(sinx^2)]-√3*a*sin2x+2a+b
=-a*cos2x-√3*a*sin2x+2a+b
=-2a[1/2*cos2x+(√3)/2*sin2x]+2a+b
=-2a*sin(2x+π/6)+2a+b
因为定义域为[0,π/2],值域为[-5,1]
所以2x∈[0,π],2x+π/6∈[π/6,7π/6]
所以sin(2x+π/6)∈[-1/2,1/2]
则(1){f(-1/2)=3a+b=-5
{f(1/2)=a+b=1
{a=-3
{b=4
或(2){f(-1/2)=3a+b=1
{f(1/2)=a+b=-5
{a=3
{b=-8
PS:题中的∏/2是二分之派的意思吧.
π即派
{ 即联立的大括号
{