Question

急解高中不等式奇函数f(x)在定义域内为R,满足f'(x)＜0,试解：f(1-m)+f(1-m^2)＞0

Answer 1

因为f(x)在定义域内为奇函数,所以f(1-m^2)=-f(m^2-1)

又f'(x)＜0,所以函数在R上单调递减

f(1-m)+f(1-m^2)＞0 即可转化为f(1-m)-f(m^2-1)>0

所以1-m1或m