已知函数f(x)=-acos2x-(2√3)asinxcosx+2a+b的定义域为[0,π/2],值域为[-5,1],求常数a、b的值
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![已知函数f(x)=-acos2x-(2√3)asinxcosx+2a+b的定义域为[0,π/2],值域为[-5,1],求常数a、b的值](/uploads/image/z/2691656-8-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D-acos2x-%282%E2%88%9A3%29asinxcosx%2B2a%2Bb%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%5B0%2C%CF%80%2F2%5D%2C%E5%80%BC%E5%9F%9F%E4%B8%BA%5B-5%2C1%5D%2C%E6%B1%82%E5%B8%B8%E6%95%B0a%E3%80%81b%E7%9A%84%E5%80%BC)
已知函数f(x)=-acos2x-(2√3)asinxcosx+2a+b的定义域为[0,π/2],值域为[-5,1],求常数a、b的值
已知函数f(x)=-acos2x-(2√3)asinxcosx+2a+b的定义域为[0,π/2],值域为[-5,1],求常数a、b的值
已知函数f(x)=-acos2x-(2√3)asinxcosx+2a+b的定义域为[0,π/2],值域为[-5,1],求常数a、b的值
f(x)=-acos2x-(2√3)asinxcosx+2a+b
= -acos2x-√3asin2x+2a+b
= -2a(sin2x*cosπ/6+cos2x*sinπ/6) +2a+b
= -2a sin(2x+π/6) +2a+b
∵0≤x≤π/2
∴π/6 ≤ 2x+π/6 ≤ 7π/6
2x+π/6 = 7π/6时,sin(2x+π/6)有最小值-1/2;2x+π/6 =π/2时,sin(2x+π/6)有最大值1
∴-1/2≤sin(2x+π/6) ≤1
设a>0:
2x+π/6 =π/2时,最小值f(x)min=-a+2a+b=-5
2x+π/6 = 7π/6时,最大值f(x)max=1/2a+2a+b=1
解得:a=4,b=-9
设a<0:
2x+π/6 = 7π/6时,最小值f(x)min=1/2a+2a+b=-5
2x+π/6 =π/2时,最大值f(x)max=-a+2a+b=1
解得:a=-4,b=5
综上:a=4,b=-9;或a=-4,b=5
原式=-2asin(2x+π/6)+2a+b
2x+π/6 在π/6到7π/6之间
将sin(2x+π/6)=1/2代入
得到a+b=1
同理代入sin(2x+π/6)=1
得到b=-5
a=6
关键用到的是sin2x=2sinxcosx
和cos2x+√3sinx=2sin(2x+π/6)
可能有计算错误哦,我不太仔细的,再检查一下吧