Câu hỏi 2

TXĐ : \(D = \left[ { - 2; + \infty } \right)\).

Ta có : \(y' = \dfrac{1}{2} - \dfrac{1}{{2\sqrt {x + 2} }} = \dfrac{{\sqrt {x + 2}  - 1}}{{2\sqrt {x + 2} }}\).

Cho \(y' = 0 \Leftrightarrow \sqrt {x + 2}  - 1 = 0\)\( \Leftrightarrow \sqrt {x + 2}  = 1 \Leftrightarrow x =  - 1 \in \left[ { - 1;34} \right]\).

Lại có : \(y\left( { - 1} \right) =  - \dfrac{3}{2},y\left( {34} \right) = 11\) nên \(m = \mathop {\min }\limits_{\left[ { - 1;34} \right]} y = y\left( { - 1} \right) =  - \dfrac{3}{2};\,\,M = \mathop {\max }\limits_{\left[ { - 1;34} \right]} y = y\left( {34} \right) = 11\).

Vậy \(3m + M = 3.\left( { - \dfrac{3}{2}} \right) + 11 = \dfrac{{13}}{2}\).

Chọn A.