题目

用函数单调性证明Sinx>2x/π

提问时间：2020-10-08

答案

证明：构造函数f(x)=πsinx-2x.(x∈(0,π/2)).求导得：f'(x)=πcosx-2.f''(x)=-πsinx.∴在(0,π/2)上,恒有f''(x)＜0,∴在(0,π/2)上,导函数f'(x)在(0,π/2)上递减,又f'(0)=π-2＞0,f'(π/2)=-2＜0,∴存在a,0＜a＜π/2.f'(a)=0.且在(0,a)上,f(x)递增,在[a,π/2)上,f(x)递减.∴在(0,π/2)上,恒有f(x)＞f(0)=0,且f(x)＞f(π/2)=0.即在(0,π/2)上,恒有f(x)＞0.===>恒有πsinx-2x＞0.===>sinx＞2x/π.