m∈(0,1),1-m^2∈(0,1),且f(x)与f(1-x^2)单调性相反
∴在(0,1)上必然存在m,使得:[f(m)f(1-m^2)]'=0
即:f'(m)f(1-m^2)+f(m)f'(1-m^2)(1-m^2)'=0
f'(m)f(1-m^2)-2mf(m)f'(1-m^2)=0
f'(m)f(1-m^2)=2mf(m)f'(1-m^2)
m∈(0,1),1-m^2∈(0,1),且f(x)与f(1-x^2)单调性相反
∴在(0,1)上必然存在m,使得:[f(m)f(1-m^2)]'=0
即:f'(m)f(1-m^2)+f(m)f'(1-m^2)(1-m^2)'=0
f'(m)f(1-m^2)-2mf(m)f'(1-m^2)=0
f'(m)f(1-m^2)=2mf(m)f'(1-m^2)