a) Đ, b) Đ, c) S, d) S

a) Ta có \(f\left( x \right)\) liên tục trên \(\mathbb{R}\)nên \(f\left( x \right)\) liên tục tại \(x = 1\).

Do đó \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = f\left( 1 \right)\)\( \Leftrightarrow m + 5 = 3 \Leftrightarrow m = - 2\).

b) Ta có \(F\left( x \right) = \left\{ \begin{array}{l}{x^3} + {x^2} + mx + {C_1}\;{\rm{khi}}\;x \ge 1\\5x - {x^2} + {C_2}\;\;\;\;\;\;\;\;{\rm{khi}}\;x < 1\end{array} \right.\).

Ta có \(F\left( { - 2} \right) = 5.\left( { - 2} \right) - {\left( { - 2} \right)^2} + {C_2} \Rightarrow {C_2} = - 10 + 14 = 4\).

Có \(\mathop {\lim }\limits_{x \to {1^ + }} F\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^3} + {x^2} + mx + {C_1}} \right) = m + 2 + {C_1}\).

\(\mathop {\lim }\limits_{x \to {1^ - }} F\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {5x - {x^2} + {C_2}} \right) = 4 + {C_2}\).

Ta lại có \(F\left( x \right)\) liên tục tại \(x = 1\) nên

\(\mathop {\lim }\limits_{x \to {1^ - }} F\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} F\left( x \right) = F\left( 1 \right)\) \( \Leftrightarrow m + 2 + {C_1} = 4 + {C_2} \Leftrightarrow {C_1} = 6 - m\).

Mà \(m = - 2\) nên \({C_1} = 8\).

Vậy \(F\left( x \right) = \left\{ \begin{array}{l}{x^3} + {x^2} - 2x + 8\;{\rm{khi}}\;x \ge 1\\5x - {x^2} + 4\;\;\;\;\;\;\;\;{\rm{khi}}\;x < 1\end{array} \right.\).

c) \(F\left( 3 \right) = {3^3} + {3^2} - 2.3 + 8 = 38\).

d) \(\int\limits_1^{{e^2}} {f\left( {\ln x} \right)\frac{1}{x}dx} \)\( = \int\limits_1^{{e^2}} {f\left( {\ln x} \right)d\left( {\ln x} \right)} \)\( = \int\limits_0^2 {f\left( t \right)dt} \)\( = \int\limits_0^1 {f\left( x \right)dx + \int\limits_1^2 {f\left( x \right)dx} } \)

\( = \int\limits_0^1 {\left( {5 - 2x} \right)dx + \int\limits_1^2 {\left( {3{x^2} + 2x - 2} \right)dx} } = 12\).

Tính tích phân I = 2 ∫ 0 ( 2x + 1 ) dx