Question

已知函数f(x)=sinx+√3cosx,定义域为A=(0,2π)

⑴求值域 ⑵求单调递减区间 ⑶若x1、x2∈A,且x1≠x2,f(x1)=f(x2),求x1+x2

Answer 1

f(x) = 2( 1/2 *sinx + √3/2 * cosx)
= 2 (cos(π/3) sinx + sin(π/3)cosx)
= 2 Sin(x + π/3) 逆用和差化积

∴ 值域是 [-2,2]

单调减区间
2kπ + π/2 <= x + π/3 <= 2kπ +3/2 * π
化简一下就是了

(3)
相当于
sin(x1 + π/3) = sin (x2 + π/3)
=> sin(x1 + π/3) - sin (x2 + π/3) = 0
=> 2 cos(x1 + π/3 + x2 + π/3)*sin(x1 + π/3 - x2 - π/3) = 0
=> cos(x1 + x2 + 2/3 * π) *sin(x1 - x2) = 0
所以有
cos(x1 + x2 + 2/3 * π) = 0
或者
sin(x1 - x2) = 0
又因为x1不等于x2
且x1，x2 ∈ A
所以sin(x1 - x2) ≠ 0
那么只有cos(x1 + x2 + 2/3 * π) =0
所以x1 + x2 + 2/3 * π = 3/2 * π
x1 + x2 = 5/6 * π

Answer 2

原式=2sinx（x+π/3)
1) 值域为（-2，2)
2)因为sinx减区间为[2kπ+π/2，2kπ+3π/2]，所以f(x)减区间为[π/6，7π/6]
3)满足条件的有π/6和7π/6