设函数f(x)在[0,π/4]上连续,在(0,π/4)可导,且f(π/4)=0,证明:2f(c)+sin2cf'(c)=

Question

设函数f(x)在[0,π/4]上连续,在(0,π/4)可导,且f(π/4)=0,证明:2f(c)+sin2cf'(c)=

Answer 1

证明：令 g(x) = f(x) tanx,g(0) = g(π/4) = 0 ,
g(x)在[0,π/4]上连续,在(0,π/4)可导 .
g '(x) = f '(x) tanx + f(x) sec²x = (1/2) sec²x [ f '(x) sin(2x) + 2 f(x) ]
由罗尔定理得：存在 c∈(0,π/4),使得 g '(c) = 0
于是 2f(c) + sin2c f '(c) = 0