Question

已知向量a=(√3sinx,cosx),b=(cosx,cosx),c=(2√3,1)

1.若a//b，求sinx•cosx
2.若0＜x≤Π/3，求函数f（x）=a•b的值域

Answer 1

(1)
a//b
=> cosx/(√3sinx) = cosx/cosx
cosx = √3sinx
tanx = 1/√3
sinx. cosx = (1/2) .(√早氏3/2)
= √3/4
(2)
f(x) = a.b
= (√3sinx,cosx).(cosx,cosx)
= √3sinx(cosx)+(cosx)^2
f'(x) = √3[ -(sinx)^2 + (cosx)^2] - 2cosxsinx
= √3cos2x -sin2x >0 ( for 0<x ≤π/3)
f(x) is increasing on (for 0<x ≤山睁派π/3)

= f(π/逗贺6)
= 3/4+ 3/4
= 3/2
a.b 的值域 = (0, 3/2]