函数f(x)=-1/2-a/4+acosx+(sinx)^2(0≤x≤π/2)的最大值为2,求实数a

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f(x)=-(cosx)^2+acosx-a/4+1/2
=-[(cosx)^2-acosx]-a/4+1/2
=-[(cosx)^2-acosx+a^2/4]+a^2/4-a/4+1/2
=-(cosx-a/2)^2+a^2/4-a/4+1/2
=-(cosx-a/2)^2+1/4[a^2-a]+1/2
=-(cosx-a/2)^2+1/4[a^2-a+1/4]-1/16+1/2
=-(cosx-a/2)^2+1/4(a-1/2)^2-1/16+1/2
=-(cosx-a/2)^2+1/4(a-1/2)^2+7/16
因为f(x)=-A(平方,可变)+B(平方,不可变)+C(常数)
所以只有A足够小(即cosx-a/2足够小),Max[f(x)]=2
所以当且仅当x=π/2时 cosx-a/2足够小 为-a/2.
所以 -(-a/2)^2 +1/4(a-1/2)^2+7/16=2
-a^2/4+a^2/4-a/4+1/2=2
a/4=-3/2
a=-6======以下答案可供参考======
供参考答案1:
f(x)=-(cosx-a/2)^2+a^2/4-a/4+1/2
0==1a=-6或10/3
供参考答案2:
1供参考答案3:
f(x)=(sinx)^2+acosx+0.5-0.25a
=[1-(cosx)^2]+acosx+0.5-0.25a
=-(cosx)^2+acosx+1.5-0.25a
Beacause x belongs to [0,π/2]
cosx belongs to [0,1]
t=cosx
g(t)=-t^2+at+1.5-0.25a
1' If 0.5a belongs to [0,1](a belongs to [0,2])
max[g(t)]=a-6-a^2/(-4)=0.25a^2-0.25a+1.5=2
a^2-a-2=0 a1=-1 (abandoned) a2=2
2'If 0.5a belongs to (1,+max) (a>2)max[g(t)]=g(1)=0.75a+0.5=2
a=2(abandoned)
3'If 0.5a belongs to (-max,0)(amax[g(t)]=g(0)=1.5-0.25a=2
a=-2