$$\eqalign{
& y = {{m\cos x + 1} \over {\sin x + 3}}\,\,\left( {D = R} \right) \cr
& \Leftrightarrow y\sin x + 3y = m\cos x + 1 \cr
& \Leftrightarrow y\sin x - m\cos x = 1 - 3y \cr
& PT\,\,co\,\,nghiem \cr
& \Leftrightarrow {y^2} + {m^2} \ge {\left( {1 - 3y} \right)^2} \cr
& \Leftrightarrow {y^2} + {m^2} \ge 9{y^2} - 6y + 1 \cr
& \Leftrightarrow 8{y^2} - 6y + 1 - {m^2} \le 0\,\,\left( * \right) \cr
& \Delta ' = 9 - 8\left( {1 - {m^2}} \right) = 8{m^2} + 1 > 0\,\,\forall m \cr
& 8{y^2} - 6y + 1 - {m^2} = 0\,\,co\,\,2\,\,nghiem\,\,pb \cr
& \Rightarrow \left[ \matrix{
{y_1} = {{3 + \sqrt {8{m^2} + 1} } \over 8} \hfill \cr
{y_2} = {{3 - \sqrt {8{m^2} + 1} } \over 8} \hfill \cr} \right. \cr
& \Rightarrow {{3 - \sqrt {8{m^2} + 1} } \over 8} \le y \le {{3 + \sqrt {8{m^2} + 1} } \over 8} \cr
& \Rightarrow \max y = {{3 + \sqrt {8{m^2} + 1} } \over 8} = 1 \cr
& \Leftrightarrow 3 + \sqrt {8{m^2} + 1} = 8 \cr
& \Leftrightarrow \sqrt {8{m^2} + 1} = 5 \cr
& \Leftrightarrow 8{m^2} + 1 = 25 \cr
& \Leftrightarrow 8{m^2} = 24 \cr
& \Leftrightarrow {m^2} = 3 \cr
& \Leftrightarrow m = \pm \sqrt 3 \cr
& Chon\,\,B. \cr} $$
