Question 2: The function (y = frac{1}{3}{x^3} – left( {m + 1} right){x^2} + left( {2{m^2} + 1} right)x + m) is minimized at x = 1 when

y’ = x² – 2(m + 1)x + (2m .)² + 1)

y” = 2x – 2(m + 1)

y = – (m + 1)x² + (2m² + 1)x + m is minimized at x = 1

\(\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
y’\left( 1 \right) = 0\\
y”\left( 1 \right) > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2{m^2} – 2m = 0\\
– 2m > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[\begin{array}{l}[\begin{array}{l}
m = 0\\
m = 1
\end{array} \right.\\
m < 0
\end{array} \right.
\end{array}\)

So there is no value of m for the function to be minimized at x = 1.