Answer: Option B
$$\eqalign{
& \therefore {\text{When}}\left( {x - 1} \right) = 0 \cr
& x = 1 \cr
& {x^2} + {k_1}x + {k_2}{\text{ }} = 0 \cr
& \Rightarrow 1 + {k_1} + {k_2}{\text{ }} = 0 \cr
& \Rightarrow {k_1} + {k_2} = - 1\,.......(i) \cr
& {\text{When}}\left( {x + 3} \right) = 0 \cr
& x = - 3 \cr
& 9 - 3{k_1} + {k_2}{\text{ }} = 0 \cr
& \Rightarrow - 3{k_1} + {k_2}{\text{ }} = - 9\,.......(ii) \cr
& {\text{From equation (i) and (ii)}} \cr
& {k_1} = 2,\,\,\,\,\,\,{\text{ }}{k_2} = - 3 \cr} $$