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

题目

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
1.求a,b的值.
2.若函数f(x)在区间[π/4,m]上存在零点,求m的最小值

答案

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)=4...