Question

已知函数f(x)=asinx+(2-b)cosx(a＞0,b＞0)关于直线x=π/4对称,则1/

Answer 1

解f(x)=asinx+(2-b)cosx

=√[a^2+(2-b)^2](a/[a^2+(2-b)^2]sinx+((2-b)/[a^2+(2-b)^2]cosx))

=√[a^2+(2-b)^2]sin(x+θ)

则tanθ=(2-b)/a

又由函数f(x)=asinx+(2-b)cosx(a＞0,b＞0)关于直线x=π/4对称

故θ=kπ+π/4,k属于Z

故tanθ=1

即(2-b)/a=1

即a=2-b

即a+b=2

故1/a+1/b

=(1/a+1/b)×1

=(1/a+1/b)×(a+b)/2

=1/2[1+1+b/a+a/b]

=1/2(2+b/a+a/b)

≥1+1/2×2√b/a×a/b

=1+1

=2

故1/a+1/b的最小值为2.