Lời giải:Ta có \(f\left( x \right) > \ln \left( {\cos x} \right) - {e^{\pi x}} + m \Leftrightarrow f\left( x \right) - \ln \left( {\cos x} \right) + {e^{\pi x}} > m\,\,\forall x \in \left( {0;\dfrac{\pi }{2}} \right)\)
Đặt \(g\left( x \right) = f\left( x \right) - \ln \left( {\cos x} \right) + {e^{\pi x}} \Rightarrow g\left( x \right) > m\,\,\forall x \in \left( {0;\dfrac{\pi }{2}} \right) \Leftrightarrow m \le \mathop {\min }\limits_{\left[ {0;\frac{\pi }{2}} \right]} g\left( x \right)\)
Ta có \(g'\left( x \right) = f'\left( x \right) + \dfrac{{\sin x}}{{\cos x}} + \pi {e^{\pi x}}\)
Với \(x \in \left( {0;\dfrac{\pi }{2}} \right) \Rightarrow \left\{ \begin{array}{l}\sin x > 0\\\cos x > 0\end{array} \right.\), theo giả thiết ta có \(f'\left( x \right) > 0\,\,\forall x \in \left( {0;\dfrac{\pi }{2}} \right) \Rightarrow g'\left( x \right) > 0\,\,\forall x \in \left( {0;\dfrac{\pi }{2}} \right)\)
\( \Rightarrow \) Hàm số \(y = g\left( x \right)\) đồng biến trên \(\left( {0;\dfrac{\pi }{2}} \right)\).
\( \Rightarrow \mathop {\min }\limits_{\left[ {0;\frac{\pi }{2}} \right]} g\left( x \right) = g\left( 0 \right) = f\left( 0 \right) - \ln \left( {\cos 0} \right) + {e^0} = f\left( 0 \right) + 1 \Leftrightarrow m \le f\left( 0 \right) + 1\).
Chọn A.
