∵|b|²=sin²2x+(1-cos2x)²
=sin²2x+1+cos²2x-2co2x
=2-2cos2x
=2-2(1-2sin²x)
=4sin²x
又 x∈(0,π)
∴|b|=2sinx
∵(a+b)*c=(cosx+sin2x)×0+(sinx+1-co2x)×1
=sinx+1-cos2x
=sinx+1-(1-2sin²x)
=2sin²x+sinx
∴函數(shù)f(x)=|b|-(a+b)*c
=2sinx-(2sin²x+sinx)
=-2sin²x+sinx
=-2(sinx-1/4)²+1/8
函數(shù)f(x)可看成是自變量為sinx的二次函數(shù),定義域?yàn)?0,1]
且在區(qū)間(0,1/4]為增函數(shù),在[1/4,1]內(nèi)為減函數(shù)
所以,當(dāng)sinx=1,即x=π/2時(shí),函數(shù)f(x)取得最小值,最小值f(π/2)=-1