Phương pháp giải:

\(m < g\left( x \right)\,\,\forall x \in \left( {0;1} \right) \Leftrightarrow m \le \mathop {\min }\limits_{\left[ {0;1} \right]} g\left( x \right)\).

Lời giải chi tiết:

Ta có : \(\dfrac{{f\left( x \right)}}{{36}} + \dfrac{{\sqrt {x + 3}  - 2}}{{x - 1}} > m \Leftrightarrow \dfrac{{f\left( x \right)}}{{36}} + \dfrac{{x + 3 - 4}}{{\left( {\sqrt {x + 3}  + 2} \right)\left( {x - 1} \right)}} > m \Leftrightarrow \dfrac{{f\left( x \right)}}{{36}} + \dfrac{1}{{\sqrt {x + 3}  + 2}} > m\)

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right)}}{{36}} + \dfrac{1}{{\sqrt {x + 3}  + 2}} > m\,\,\forall x \in \left( {0;1} \right) \Rightarrow m \le \mathop {\min }\limits_{\left[ {0;1} \right]} f\left( x \right)\)

Xét hàm số \(g\left( x \right) = \dfrac{{f\left( x \right)}}{{36}} + \dfrac{1}{{\sqrt {x + 3}  + 2}}\) ta có:

 \(g'\left( x \right) = \dfrac{{f'\left( x \right)}}{{36}} - \dfrac{{\dfrac{1}{{2\sqrt {x + 3} }}}}{{{{\left( {\sqrt {x + 3}  + 2} \right)}^2}}} = \dfrac{{f'\left( x \right)}}{{36}} - \dfrac{1}{{2\sqrt {x + 3} {{\left( {\sqrt {x + 3}  + 2} \right)}^2}}}\)

Dựa vào đồ thị hàm số \(y = f'\left( x \right)\) ta thấy

\(\begin{array}{l}f'\left( x \right) \le 1\,\forall x \in \mathbb{R} \Rightarrow \sqrt {x + 3}  < 2 \Rightarrow \sqrt {x + 3}  + 2 < 4\\ \Rightarrow 2\sqrt {x + 3} \left( {\sqrt {x + 3}  + 2} \right) < 2.2.4 = 16 \Rightarrow \dfrac{1}{{2\sqrt {x + 3} \left( {\sqrt {x + 3}  + 2} \right)}} > \dfrac{1}{{16}}\end{array}\)

\( \Rightarrow g'\left( x \right) \le \dfrac{1}{{36}} - \dfrac{1}{{16}} < 0 \Rightarrow \) Hàm số nghịch biến trên \(\left( {0;1} \right)\)

\( \Rightarrow \mathop {\min }\limits_{\left[ {0;1} \right]} g\left( x \right) = g\left( 1 \right) = \dfrac{{f\left( 1 \right)}}{{36}} + \dfrac{1}{4} \Rightarrow m \le \dfrac{{f\left( 1 \right)}}{{36}} + \dfrac{1}{4} = \dfrac{{f\left( 1 \right) + 9}}{{36}}\).

Chọn D.