题目:
已知函数f(x)=asinx•cosx-
(1)求函数的单调递减区间; (2)设x∈[0,
|
答案:
(1)f(x)=asinx•cosx-
a cos2x+3
a+b(a>0)=3 2
sin2x-a 2
a(1+cos2x)+3 2
+b
a3 2
=
sin2x-a 2
a•cos2x+b=asin(2x-3 2
)+b.π 3
由 2kπ+
≤2x-π 2
≤2kπ+π 3
,k∈z,解得 kπ+3π 2
≤x≤kπ+5π 12
,k∈z,11π 12
故函数的单调递减区间为[kπ+
,kπ+5π 12
],k∈z.11π 12
(2)∵x∈[0,
],∴-π 2
≤2x-π 3
≤π 3
,∴-2π 3
≤sin(2x-3 2
)≤1.π 3
∴f(x)min =-
+ b=-2,f(x)max =a+b=
a3 2
,3
解得 a=2,b=-2+
.3