已知fx=cos2x+sinx+a-1

题目

已知fx=cos2x+sinx+a-1
1,当fx=0有实数解时,求实数a的取值范围
2,若x属于R,恒有1小于等于fx小于等于17|4成立,求实数a的取值范围

答案

f(x)=cos(2x)+sinx+a-1
=1-2sin²x+sinx+a-1
=-2sin²x+sinx+a
=-2(sinx -1/4)²+a+ 1/8
1.
f(x)=0
-2(sinx- 1/4)²+a+1/8=0
(sinx -1/4)²=(8a+1)/16
-1≤sinx≤1 -5/4≤sinx -1/4≤3/4 0≤(sinx -1/4)²≤25/16
要方程有解,
0≤(8a+1)/16≤25/16
0≤8a+1≤25
-1≤8a≤24
-1/8≤a≤3
a的取值范围为[-1/8,3]
2.
1≤f(x)≤17/4
1≤-2(sinx -1/4)²+a+1/8≤17/4
0≤(sinx -1/4)²≤25/16 -25/8≤-2(sinx -1/4)²≤0
a-3≤-2(sinx -1/4)²+a +1/8≤a+1/8
a+1/8≤17/4 a-3≥1
a+1/8≤17/4 解得a≤33/8
a-3≥1 解得a≥4
综上,得4≤a≤33/8,a的取值范围为[4,33/8]