TH1: \(-\frac{b}{2a}=\frac{-2m-1}{2}\in\left[0;1\right]\)
\(\Leftrightarrow0\le\frac{-2m-1}{2}\le1\) \(\Leftrightarrow\) \(-\frac{3}{2}\le m\le-\frac{1}{2}\)
Khi đó \(f\left(x\right)_{min}=f\left(-\frac{b}{2a}\right)=f\left(\frac{-2m-1}{2}\right)=\frac{-4m-5}{4}\)
\(\Rightarrow-\frac{4m+5}{4}=1\Rightarrow m=-\frac{9}{4}\notin\left[-\frac{3}{2};-\frac{1}{2}\right]\) (loại)
TH2: \(-\frac{b}{2a}=\frac{-2m-1}{2}< 0\Leftrightarrow m>-\frac{1}{2}\)
Khi đó \(f\left(x\right)\) đồng biến trên \(\left[0;1\right]\Rightarrow f\left(x\right)_{min}=f\left(0\right)=m^2-1=1\)
\(\Rightarrow\left[{}\begin{matrix}m=-\sqrt{2}\left(l\right)\\m=\sqrt{2}\end{matrix}\right.\)
TH3: \(-\frac{b}{2a}=\frac{-2m-1}{2}>1\Leftrightarrow m< -\frac{3}{2}\)
Khi đó \(f\left(x\right)\) nghịch biến trên \(\left[0;1\right]\Rightarrow f\left(x\right)_{min}=f\left(1\right)=m^2+2m+1=1\)
\(\Rightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=-2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=-2\\m=\sqrt{2}\\\end{matrix}\right.\)