Question

高一数学 必修四 求解

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Answer 1

1. f(x)=sin30cosx+cos30sinx+2
   

        =sin(x+30)+2

所以最小正周期为2π。

2. f(x)的值域为[2.5,3]

3. 单调增区间为 -π/2+2kπ≤x+π/6≤π/2+2kπ

                         -2π/3+2kπ≤x≤π/3+2kπ

3. f(x)=2sinxcos(π/6)+cosx+a

         =2(sinxcos(π/6)+sin(π/6)cosx)+a

         =2sin(x+π/6)+a

  最大值=2+a=1，所以a=-1

f(x)≥0成立，则2sin(x+π/6)-1≥0

sin(x+π/6)≥0.5

2kπ≤x≤2π/3+2kπ

追问

值域是怎么求的？

追答

由第一问可知
f(x)=sin(x+30)+2
x∈[0,π/2],则π/6≤x+π/6≤2π/3
所以0.5≤sin(x+30)≤1，所以值域为[2.5,3]

Answer 2

1、f(x)=sin(π/6)cosx + cos(π/6)sinx + 2
=sin(x + π/6) + 2
(1)T=2π/1=2π
(2)∵0≤x≤π/2
∴π/6≤x + π/6≤2π/3
则1/2≤sin(x + π/6)≤1
∴f(x)∈[2.5，3]
(3)2kπ - π/2≤x + π/6≤2kπ + π/2
2kπ - 2π/3≤x≤2kπ + π/3，k∈Z

追答