a) ĐÚNG

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

Do đó \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = f\left( 1 \right)\)\( \Leftrightarrow m + 1 = - 3 \Leftrightarrow m = - 4\).

b) ĐÚNG

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

\(F\left( { - 2} \right) = \left( { - 2} \right) - 2.{\left( { - 2} \right)^2} + {C_2} = {C_2} - 10 \Rightarrow {C_2} = 10 - 6 = 4\).

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

\(\mathop {\lim }\limits_{x \to {1^ - }} F\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {x - 2{x^2} + {C_2}} \right) = - 1 + {C_2} = 3\).

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

Do đó \(\mathop {\lim }\limits_{x \to {1^ - }} F\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} F\left( x \right) = F\left( 1 \right)\)\[ \Leftrightarrow m + {C_1} = 3 \Leftrightarrow {C_1} = 3 - m = 7\].

Vậy \(F\left( x \right) = \left\{ \begin{array}{l}{x^3} - {x^2} - 4x + 7\,\,\,\,\,{\rm{khi}}\,\,x \ge 1\\x - 2x{}^2\,\, + 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x < 1\end{array} \right.\).

c) SAI

Ta có \[\int\limits_{ - 1}^5 {f\left( x \right)dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} + \int\limits_1^5 {f\left( x \right)dx} = \int\limits_{ - 1}^1 {\left( {1 - 4x} \right)dx} + \int\limits_1^5 {\left( {3{x^2} - 2x - 4} \right)dx} = 86\]

d) SAI

Đặt \(t = \ln x \Rightarrow dt = \frac{1}{x}dx\).

Khi \(x = 1 \Rightarrow t = 0\);

Khi \(x = {e^2} \Rightarrow t = 2\).

Do đó

\[\int\limits_1^{{e^2}} {f\left( {\ln x} \right)\frac{1}{x}dx} = \int\limits_0^2 {f\left( t \right)dt} = \int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} = \int\limits_0^1 {\left( {1 - 4x} \right)dx} + \int\limits_1^2 {\left( {3{x^2} - 2x - 4} \right)dx} = - 1\].

Diện tích hình phẳng được giới hạn bởi đồ thị hàm số \(y = {x^3}