Question

已知向量a=(根号3sin3/2,cosx/2)，b=( cosx/2,cosx/2),令fx=a*b,试求fx+f'x的值域

Answer 1

f(x) =a.b
= (√3sinx/2,cosx/2).(cosx/2,cosx/2)
=√3sinx/2cosx/2+cosx/2cosx/2
= (√3/2)sinx + (1/2)cosx + 1/2
= sin(x+π/6) +1/2
f'(x) = cos(x+π/6)
f(x) + f'(x) = sin(x+π/6) +1/2 + cos(x+π/6)

consider
g(x)= sin(x+π/6) + cos(x+π/6)
g'(x) = cos(x+π/6) - sin(x+π/6) =0
tan(x+π/6) = 1
x = π/源纳袜12(max) or 13π/12(min)
max f(x) +f'(x) = sin(π/4) +1/茄旁2 + cos(π/4)
= √2 +1/雹激2
min f(x) +f'(x) = sin(5π/4) +1/2 + cos(5π/4)
= -√2 +1/2
域值=[-√2 +1/2,√2 +1/2]

Answer 2

f(x)=根3sinx/2cosx/2+cosx/2cosx/2=sinπ/3sinx+1/2+cosπ/3cosx=cos(π/液伏3-x)+1/2
f`(x)=sin(π/3-x)
f(x)+f`(x)=根2sin(π/闷漏3-x+π/4)+1/2属于[1/2-根2，1/蚂埋烂2+根2]