已知a=(sinx,1),b=(1,cosx)且函数f(x)=ab,f'(x)是f(x)的导函数

Question

求函数F(x)=f(x)f'(x)+f^2(x)的最大值和最小正周期

Answer 1

a = (sinx，1)，b = (1，cosx)
ƒ(x) = a • b
= (sinx)(1) + (1)(cosx)
= √2sin(x + π/4)
ƒ'(x) = √2cos(x + π/4)
F(x) = ƒ(x)ƒ'(x) + ƒ²(x)
= [√2sin(x + π/4)][√2cos(x + π/4)] + [√2sin(x + π/4)]²
= sin[2(x + π/4)] + 1 - cos[2(x + π/4)]
= sin(2x + π/4) - cos(2x + π/2) + 1
= cos(2x) - [- sin(2x)] + 1
= sin(2x) + cos(2x) + 1
= √2sin(2x + π/4) + 1
最小正周期T = (2π)/2 = π
最大值 = 1 + √2
最小值 = 1 - √2

Answer 2

f(x)=sinx+cosx
f'(x)=cosx-sinx
F(x)=f(x)f'(x)+f²(x)
=(sinx+cosx)(sinx-cosx)+(sinx+cosx)²
=sin²x-cos²x+1+2sinxcosx
=-cos2x+sin2x+1
=-√2(√2/2sin2x-√2/2cos2x)+1
=-√2sin(2x-π/4)+1
F(x)的最大值=√2+1 (当sin(2x-π/4)=-1
最小正周期：2π/2=π