Answer is B)
$$\begin{array}{|cc|cc|c|c|}\hline
\bf{Q_1}& \bf{Q_0}& \bf{{Q_1}^+}&\bf{ {Q_0}^+}& \bf{T_1}& \bf{T_2} \\\hline
0&0&0&1&0 &1 \\ 0&1&1&0&1&1\\ 1& 0&1&1&0&1 \\ 1&1&1&1&0&0 \\ \hline
\end{array}$$
By using above excitation table,
$T_1={Q_1}'{Q_0},$
$T_2={(Q_1Q_0)}'={Q_1}'+{Q_0}'$