Question

已知a=(cosx,sinx),b=1,且a与b满足|ka+b|=根号3|a-kb|(k>0).

Answer 1

1)易知|a|=1,|b|=1

由|ka+b|=√3|a-kb|,平方,得

(ka+b)²=3(a-kb)²

k²a²+2kab+b²=3a²-6kab+3k^2b²

2k²-8kab+2=0

ab=(2+2k²)/(8k)=(1+k^2)/4k=1/(4k)+k/4≥2√(1/16)=1/2

即ab的最小值是1／2.

(2) ab=1／2cosx＋√3／2sinx=sin(x+π/6),

0≤x≤π,π/6≤x+π/6≤7π/6

-1/2≤sin(x+π/6)≤1

从而 ab的最大值为1,此时有x+π/6=π/2,

即有x=π/3