f(x) = x^2 +(1/2 - 2m) x + m^2-1 = 0
判别式 Δ = 17/4 - 2m > 0 => m < 17/8 @
两个实根在区间【0,2】内 =>
f(0) = m^2-1 > 0, f(2) = m^2 - 4m + 4 > 0 @@
综上, 1< m < 2 或 2 < m < 17/8
f(x) = x^2 +(1/2 - 2m) x + m^2-1 = 0
判别式 Δ = 17/4 - 2m > 0 => m < 17/8 @
两个实根在区间【0,2】内 =>
f(0) = m^2-1 > 0, f(2) = m^2 - 4m + 4 > 0 @@
综上, 1< m < 2 或 2 < m < 17/8