已知函数f(x)=asinwx+bcoswx(a,b,w为正常数)最小正周期为π/2,当x=π/3时,f(x)取最小值-4

显示全部楼层 · 2012-10-3 20:08:56

f(x)=asinwx+bcoswx=√a2+b2 sin （wx+φ）最小正周期为π/2， 2π/ w = π/2,即w=4当x=π/3时,f(x)取最小值-4，即加减1/4个周期（π/8）与x轴相交，即在5π/24处或 11π/24即 asin(4π/3)+bcos(4π/3)=4asin5π/6+bcos5π/6=0解得 a=2√3b=2因为当x= 5π/24处或 11π/24，y=0若函数f(x)在区间[π/4,m]上存在零点故m最小值是11π/24...

千问 · 2012-10-3 20:08:56

1.令c=根号下(a^2+b^2),sin(y)=b/c,cos(y)=a/c,(0<y<π/2),则f(x)=c*sin(wx+y).因为最小正周期为π/2，所以w*π/2=2π,故w=4.当x=π/3时,f(x)取最小值-4,则wx+y=3π/2.故y=π/6.同事可得c=4,所以a=1/2,b=(根号3)/2. 2.依题意知，存...