已知函数f(x)=(1/2x-1+1/2)sinx (-π2<x<π2且x≠0) (1)判断f(x)的奇偶性; (2)求证f(x)>0.
问题描述:
已知函数f(x)=(
+1
2x-1
)sinx (-1 2
<x<π 2
且x≠0)π 2
(1)判断f(x)的奇偶性;
(2)求证f(x)>0.
答
(1)∵f(-x)=(12-x-1+12)sin(-x)=-(112x-1+12)sinx=-(2x1-2x+12)sinx=(2x2x-1-12)sinx=[(1+12x-1)-12]sinx=(12x-1+12)sinx=f(x),∴f(x)是偶函数.(2)当0<x<π2时,2x>1, 2x-1>0 ,...