已知函数f(x)=tan x,x∈(0,π/2),若x1,x2∈ (0,π/2),且x1≠x2 求证:1/2[f(x1)+f(x2)]>f[(x1+x2)/2]

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 02:08:02
已知函数f(x)=tanx,x∈(0,π/2),若x1,x2∈(0,π/2),且x1≠x2求证:1/2[f(x1)+f(x2)]>f[(x1+x2)/2]已知函数f(x)=tanx,x∈(0,π/2)

已知函数f(x)=tan x,x∈(0,π/2),若x1,x2∈ (0,π/2),且x1≠x2 求证:1/2[f(x1)+f(x2)]>f[(x1+x2)/2]
已知函数f(x)=tan x,x∈(0,π/2),若x1,x2∈ (0,π/2),且x1≠x2 求证:1/2[f(x1)+f(x2)]>f[(x1+x2)/2]

已知函数f(x)=tan x,x∈(0,π/2),若x1,x2∈ (0,π/2),且x1≠x2 求证:1/2[f(x1)+f(x2)]>f[(x1+x2)/2]
左边=(1/2)*(tanx1+tanx2)=(1/2)*(sinx1/cosx1+sinx2/cosx2)
=(1/2)*[(sinx1cosx2+cosx1sinx2)/(cosx1cosx2)] ……分母利用积化和差公式
=sin(x1+x2)/[cos(x1+x2)+cos(x1-x2)]
右边=tan((x1+x2)/2)= sin((x1+x2)/2) /cos((x1+x2)/2)
=[2 sin((x1+x2)/2 ) cos((x1+x2)/2)] /[ 2cos²((x1+x2)/2)]
=sin(x1+x2)/[1+cos(x1+x2)]
∵cos(x1-x2)<1,∴分母 cos(x1+x2)+cos(x1-x2)<1+cos(x1+x2)]
左边>右边.
即1/2[f(x1)+f(x2)]>f[(x1+x2)/2].