Question

急急急急 函数f（x）=sin x^2+√3 sinx cosx在区间【π/4,π/2】上有最小值是___

Answer 1

f(x) = sin²x + √3 sinx cosx
= (1 - cos2x)/2 + √3/2 * sin2x
= √3/2 * sin2x - (1/2)cos2x + 1/2
= √[(√3/2)² + (1/2)²]sin(2x - π/6) + 1/2
= sin(2x - π/6) + 1/2
当2x - π/6 = π/2，即x = π/3时，极大值f(π/3) = 1 + 1/2 = 3/2 = 1.5
当2x - π/6 = -π/2，即x = -π/6时，极小值f(-π/6) = -1 + 1/2 = -1/2，不在[π/4，π/2]上，舍去
在端点[π/4，π/2]，
f(π/4) = 1/2 + √3/2 = 1.366...
f(π/2) = 1
∴最小值是1。

Answer 2

f（x）=sin x^2+√3 sinx cosx=1/2+sin(2x-π/6)，区间【π/4,π/2】则2x-π/6的区间[π/3，5π/6]，当sin(5π/6)时最小1/2，函数f（x）最小值是1.

追问

能不能详细些

追答

正弦函数在第四至第一象限是增函数，在第二至象三限为减函数，区间[π/3，5π/6]，在第一至第二象限间，比较π/3与5π/6的正弦值即可。

Answer 3

f（x）=sin x^2+√3 sinx cosx=(1/2)(1-cos2x)+(√3 /2)sin2x =1/2+sin(2x-π/6)
x∈[π/4,π/2]则(2x-π/6)∈[π/3，5π/6]
sin(5π/6)时最小1/2，函数f（x）最小值是1.

Answer 4

f（x）=sin x^2+√3 sinx cosx
=（1-cos2x)/2+√3/2sin2x
=√3/2sin2x-1/2cos2x-1/2
=cosπ/6sin2x-sinπ/6cos2x-1/2
=sin(2x-π/6）-1/2
∴π/4≤x≤π/2
则：π/2≤2x≤π
∴π/3≤2x-π/6≤5π/6
∴当2x-π/6=5π/6时
f(x)min=sin5π/6-1/2=0

追问

答案是 1 耶。麻烦再算一下 分常感谢

追答

如果你上面的提问没错，那么我的解答就没问题