Question

快！急用！

己知f(x)=2cos^2+根号3sin2x+a(a属于R)(1)f(x)的最大值为4,求a的值？ (2)若x属于R,求f(x)的单调减区间？

Answer 1

f(x)=2(cosx)²＋根号3sin2x＋a＝蔽棚[cos(2x)＋1]＋√(3)sin(2x)＋a＝2[½ cos(2x)＋（√(3)/2）sin(2x)]＋1＋燃亮a＝宏段则2sin(2x＋π/6)＋1＋a∵)f(x)的最大值为4∴a＝1

Answer 2

2cos^2,这是神马？

Answer 3

f(x)=cos2x+1+√3sin2x+a =2sin(2x+30)+a+1
x=30 f(x)最大值芹袜链雹=2+a+1=4 a=1
正弦 单调减区间在第二，三象限，∴2x+π/6∈〔嫌唤激2kπ+π/2，2kπ+3π/2]
∴ x∈〔kπ+π/6，kπ+2π/3〕为单调减区间

Answer 4

f(x)=2(cosx)²＋根号3sin2x＋a＝[cos(2x)＋1]＋√(3)sin(2x)＋a＝2[½ cos(2x)＋（√(3)/2）sin(2x)]＋1＋a＝2sin(2x＋π/凯简6)＋1＋a∵)f(x)的最大值为4∴a＝1 （2）令π/逗孙没2＋2kπ≤2x＋π/6≤3π/2＋2kπ得π/6＋kπ≤x≤2π/3＋kπ∴f(x)的山纳单调减区间为[π/6＋kπ，2π/3＋kπ](k∈Z)

Answer 5

a=2,x=sp

Answer 6

f(x)=cos2x+√3sin2x+a+1=2[½ cos(2x)＋(√(3)/2)sin(2x)]＋1＋a＝2sin(2x＋毁隐π/6)＋1+a; max(f(x))=4,2+1+a=4∴a＝1;令π/2＋2kπ≤2x＋π/6≤3π/2＋2kπ∴f(x)的单调减区间为纤敬厅[π/6＋kπ，2π/3＋kπ](k=±整稿困数)