Question

已知函数f(x)=-sinx^2+sinx+a,(1)当f(x)=0有实数解时，求a的取值范围（2）若x属于R，有1≤f（x）≤17/4，

求a的取值范围

Answer 1

1.
f(x)=-(sin^2x-sinx+1/4)+1/4+a
=-(sinx-1/2)^2+(4a+1)/4
0<= (sinx-1/2)^2<=9/4
f(x)=0有实数解，则：(4a+1)/4=(sinx-1/2)^2，
0<=(4a+1)/4<=9/4
0<=4a+1<=9
-1<=4a<=8
-1/4<=a<=2

2.
f(x)= -sin²x+sinx+a
令唯庆sinx=t，-1≤肢山塌t≤1
g(t)=-t²+t+a= -(t - 1/2)²+a+1/4
∴g(t)的值域为[g(-1)，g(1/2)]，即：[a-2，a+1/4]
∴f(x)的历圆值域为[a-2，a+1/4]
∴[a-2，a+1/4]包含于[1，17/4]
∴a-2≥1，a+1/4≤17/4
∴3≤a≤4.

Answer 2

1】a<=1/4