Question

已知f（x）=-sinx*sinx+sinx+a，若1≤f（x）≤17/4对任意的实数R恒成立，求实数a的取值范围

Answer 1

令t=sinx，则t属于[-1,1]，根据题意F(t)=-t^2+t+a=-(t-1/2)^2+a+1/4的
delta=1+4a>0，最大值a+1/4<=17/4
函数值F(-1)>=1，F(1)>=1，实际上只要满足F(-1)>=1就够了
全写上不过出于完整性而已
即a>-1/4
a<=4
F(-1)=a-2>=1，即a>=3
F(1)=a>=1，即a>=1
所以a的取值范围[3,4]

Answer 2

F(X)=-SIN^2X+SINX+A
=-(sinx-1/2)^2+A+1/4
因为:-1<=sinx<=1,所以-3/2<=sinx-1/2<=1/2,
-9/4<=-(sinx-1/2)^2<=0,
a-2<=f(x)<=a+1/4
又有：1=<F(X)=<17/4
则:
1<=a-2,
a+1/4<=17/4
解得
a>=3,
a<=4
实数A取值范围3<=A<=4.

Answer 3

f(x)=a+1/4-(sinx-1/2)^2 => 0<=a+1/4-f(x)<=9/4 => a-2<=f(x)<=a+1/4 => 3<=a<=4