Question

数学题解析 已知f(x)=2根号3sinxcosx+2cos2x-1求t及在[0,2分之派]

最大值,最小值

Answer 1

答：
f(x)=2√3sinxcosx+2(cosx)^2 -1
=√3sin(2x)+cos(2x)
=2*[(√3/2)*sin(2x)+(1/2)cos(2x) ]
=2sin(2x+π/6)

最小正周期T=2π/2=π
周期为kT=kπ，k为非零整数

0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
所以：
2x+π/6=π/2时取得最大值f(x)=2
2x+π/6=7π/6时取得最小值f(x)=-1

Answer 2

f(x)=√3sin2x+cos2x=2sin(2x+π/6),所以T=2π/2=π
又x∈[0.π/2]所以2x+6/π∈[π/6.7π/6].所以当2x+π/6=π/2 时取得最大为2..当2x+π/6=7π/6时取得最小为-1

Answer 3