Question

已知函数f(x)=sin²x+2sinxcosx+3cos²x,x∈R,问

求f(x)的最大值与最小值，并分别写出最大值与最小值是自变量x的集合

Answer 1

f(x)=(sin²x+cos²x)+2sinxcosx+2cos²x
=1+sin2x+2cos²x
=sin2x+(2cos²x-1)+2
=sin2x+cos2x+2
=2sin(2x+π/4)+2
当2x+π/4=2kπ+π/2即x=kπ+π/8
f(x)max=2+√2
当2x+π/4=2kπ-π/2即x=kπ-3π/8
f(x)min=2-√2

Answer 2

f(x)=1+sin2x+2cos²x
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
f(x) 最大值为 2+√2
最小值 2-√2

Answer 3

f(x)=1+sin2x+2cos²x
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
当2x+π/4=2kπ+1/2π时取到最大值2+√2，此时x=kπ+π/8

当2x+π/4=2kπ-1/2π时取到最小值2-√2，此时x=kπ-3π/8