60 bài tập tính đạo hàm bằng các quy tắc đạo hàm mức độ vận dụng, vận dụng cao

\(\eqalign{  & f\left( x \right) = {\left( {\sqrt x  - {1 \over {\sqrt x }}} \right)^3} = {\left( {\sqrt x } \right)^3} - 3{\left( {\sqrt x } \right)^2}.{1 \over {\sqrt x }} + 3\sqrt x {\left( {{1 \over {\sqrt x }}} \right)^2} - {\left( {{1 \over {\sqrt x }}} \right)^3}  \cr   & f\left( x \right) = {x^{{3 \over 2}}} - 3\sqrt x  + {3 \over {\sqrt x }} - {1 \over {{x^{{3 \over 2}}}}}  \cr   & f\left( x \right) = {x^{{3 \over 2}}} - 3\sqrt x  + 3{x^{ - {1 \over 2}}} - {x^{ - {3 \over 2}}}  \cr   & f'\left( x \right) = {3 \over 2}{x^{{3 \over 2} - 1}} - {3 \over {2\sqrt x }} + 3.\left( { - {1 \over 2}} \right){x^{ - {1 \over 2} - 1}} + {3 \over 2}{x^{ - {3 \over 2} - 1}}  \cr   & f'\left( x \right) = {3 \over 2}\sqrt x  - {3 \over {2\sqrt x }} - {3 \over 2}{x^{ - {3 \over 2}}} + {3 \over 2}{x^{ - {5 \over 2}}}  \cr   & f'\left( x \right) = {3 \over 2}\left( {\sqrt x  - {1 \over {\sqrt x }} - {1 \over {x\sqrt x }} + {1 \over {{x^2}\sqrt x }}} \right) \cr} \)

Chọn D.

\(f(x)=1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n}}=\frac{{{x}^{n+1}}-1}{x-1},\,\,\,(x\ne 1)\)

\(\begin{align}  f'(x)=1+2x+3{{x}^{2}}+...+n{{x}^{n-1}}=\frac{(n+1){{x}^{n}}(x-1)-({{x}^{n+1}}-1)}{{{\left( x-1 \right)}^{2}}}\\=\frac{(n+1){{x}^{n+1}}-(n+1){{x}^{n}}-{{x}^{n+1}}+1}{{{\left( x-1 \right)}^{2}}}=\frac{n{{x}^{n+1}}-(n+1){{x}^{n}}+1}{{{\left( x-1 \right)}^{2}}} \\ \end{align}\)

Cho \(x=2,\,\,n=2018\), ta có: \(S=1+2.2+{{3.2}^{2}}+{{4.2}^{3}}+...+{{2018.2}^{2017}}=\frac{{{2018.2}^{2019}}-{{2019.2}^{2018}}+1}{{{\left( 2-1 \right)}^{2}}}\\={{2}^{2018}}\left( 2018.2-2019 \right)+1={{2017.2}^{2018}}+1\)

Chọn: A

Ta có:

\(f'\left( x \right) = 1.\sqrt x  + x.\dfrac{1}{{2\sqrt x }} + 2x \Rightarrow f'\left( 1 \right) = 1 + \dfrac{1}{2} + 2 = \dfrac{7}{2}\).

Chọn C.

Ta có \(f'\left( x \right) = 1 + \dfrac{{2x}}{{2\sqrt {{x^2} + 1} }} = 1 + \dfrac{x}{{\sqrt {{x^2} + 1} }}\).

Theo bài ra ta có \(2x.f'\left( x \right) - f\left( x \right) \ge 0\)

\(\begin{array}{l} \Leftrightarrow 2x\left( {1 + \dfrac{x}{{\sqrt {{x^2} + 1} }}} \right) - x - \sqrt {{x^2} + 1}  \ge 0\\ \Leftrightarrow 2x + \dfrac{{2{x^2}}}{{\sqrt {{x^2} + 1} }} - x - \sqrt {{x^2} + 1}  \ge 0\\ \Leftrightarrow \dfrac{{2x\sqrt {{x^2} + 1}  + 2{x^2} - x\sqrt {{x^2} + 1}  - {x^2} - 1}}{{\sqrt {{x^2} + 1} }} \ge 0\\ \Leftrightarrow g\left( x \right) = {x^2} - 1 + x\sqrt {{x^2} + 1}  \ge 0\end{array}\)

Thử các đáp án:

Với \(x = \dfrac{1}{{\sqrt 3 }} \Rightarrow g\left( x \right) = 0 \Rightarrow x = \dfrac{1}{{\sqrt 3 }}\) thuộc tập nghiệm của BPT \( \Rightarrow \) Loại đáp án A, B và C.

Chọn D.

Ta có \(g'\left( x \right) = f\left( x \right) + xf'\left( x \right) + 1\).

\( \Rightarrow g'\left( 3 \right) = f\left( 3 \right) + 3f'\left( 3 \right) + 1 \Leftrightarrow 2 = f\left( 3 \right) + 3.\left( { - 1} \right) + 1 \Leftrightarrow f\left( 3 \right) = 4\)

\( \Rightarrow g\left( 3 \right) = 3f\left( 3 \right) + 3 = 3.4 + 3 = 15\) .

Chọn D.

Ta có: \(y' = \dfrac{{ - {m^2} - {m^2} - 10m + 12}}{{{{\left( {x - m} \right)}^2}}} = \dfrac{{ - 2{m^2} - 10m + 12}}{{{{\left( {x - m} \right)}^2}}}\).

\(y' \ge 0\,\,\forall x \in \left( { - \infty ; - 4} \right) \Leftrightarrow \left\{ \begin{array}{l} - 2{m^2} - 10m + 12 \ge 0\\m \notin \left( { - \infty ; - 4} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 6 \le m \le 1\\m \ge  - 4\end{array} \right. \Leftrightarrow  - 4 \le m \le 1\).

Mà \(m \in \mathbb{Z} \Rightarrow m \in \left\{ { - 4; - 3; - 2; - 1;0;1} \right\}\). Vậy có 6 giá trị của \(m\) thỏa mãn. Chọn C.

Ta có \(f'\left( x \right) = m{x^2} - mx + 3 - m\).

\(\begin{array}{l}f'\left( x \right) > 0\,\,\forall x \in \mathbb{R} \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}m = 0\\f'\left( x \right) = 3 > 0\,\,\forall x \in \mathbb{R}\end{array} \right.\\\left\{ \begin{array}{l}m \ne 0\\1 > 0\,\,\left( {luon\,\,dung} \right)\\\Delta  = {m^2} - 4m\left( {3 - m} \right) < 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}m = 0\\5{m^2} - 12m < 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}m = 0\\0 < m < \dfrac{{12}}{5}\end{array} \right. \Leftrightarrow 0 \le m < \dfrac{{12}}{5}\end{array}\).

Chọn C.

Xét khoảng \(\left( {0;\,3} \right)\) ta có: \(\left\{ \begin{array}{l}F\left( x \right) = {x^2} - 4x + 7\\G\left( x \right) = \dfrac{1}{2}x + 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}F'\left( x \right) = 2x - 4\\G'\left( x \right) = \dfrac{1}{2}\end{array} \right.\)

Ta có: \(P\left( x \right) = F\left( x \right).G\left( x \right)\)

\(\begin{array}{l} \Rightarrow P'\left( x \right) = F'\left( x \right).G\left( x \right) + F\left( x \right).G'\left( x \right)\\ \Rightarrow P'\left( 2 \right) = F'\left( 2 \right).G\left( 2 \right) + F\left( 2 \right).G'\left( 2 \right)\\ = \left( {2.2 - 4} \right).2 + 3.\dfrac{1}{2} = \dfrac{3}{2}.\end{array}\)

Chọn C.

Ta có \(a\left( t \right) = v'\left( t \right) = \left( {s'\left( t \right)} \right)' = s''\left( t \right)\)

\(v\left( t \right) = S'\left( t \right) = 3{t^2} + 6t - 9 \Rightarrow a\left( t \right) = S''\left( t \right) = 6t + 6\).

Giả sử \({t_0}\) là thởi điểm vận tốc của vật triệt tiêu \( \Rightarrow v\left( {{t_0}} \right) = 0\)

\( \Leftrightarrow 3t_0^2 + 6{t_0} - 9 = 0 \Leftrightarrow {t_0} = 1\).

Vậy giá tốc của vật tại thời điểm \({t_0} = 1\) là \(a\left( 1 \right) = 6.1 + 6 = 12\,\,\left( {m/{s^2}} \right)\).

Chọn D

\(y' = 3{\left( {2{x^2} + 1} \right)^2}\left( {2{x^2} + 1} \right)' = 12x{\left( {2{x^2} + 1} \right)^2}\).

\(y' \ge 0 \Leftrightarrow 2x{\left( {2{x^2} + 1} \right)^2} \ge 0\). Do \({\left( {2{x^2} + 1} \right)^2} \ge 0\,\,\forall x \Rightarrow y' \ge 0 \Leftrightarrow 2x \ge 0 \Leftrightarrow x \ge 0\).

Vậy \(x \in \left[ {0; + \infty } \right)\).

Chọn B

ĐK : \( - \sqrt 3  \le x \le \sqrt 3 \)

Ta có \(f'\left( x \right) = 2 + \dfrac{{ - 2x}}{{2\sqrt {3 - {x^2}} }} = 2 - \dfrac{x}{{\sqrt {3 - {x^2}} }}\)

Xét  

\( \Rightarrow 2\sqrt {3 - {x^2}}  = x \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\4\left( {3 - {x^2}} \right) = {x^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\5{x^2} = 12\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\\left[ \begin{array}{l}x = \sqrt {\dfrac{{12}}{5}} \\x =  - \sqrt {\dfrac{{12}}{5}} \end{array} \right.\end{array} \right. \Rightarrow x = \dfrac{{2\sqrt {15} }}{5}\,\left( {TM} \right)\)

Vậy \(x = \dfrac{{2\sqrt {15} }}{5}.\)

Ta có:

\(\begin{array}{l}y' = \dfrac{{\dfrac{{2x + 2}}{{2\sqrt {{x^2} + 2x + 3} }}.x - \sqrt {{x^2} + 2x + 3} }}{{{x^2}}}\\y' = \dfrac{{{x^2} + x - {x^2} - 2x - 3}}{{{x^2}\sqrt {{x^2} + 2x + 3} }} = \dfrac{{ - x - 3}}{{{x^2}\sqrt {{x^2} + 2x + 3} }}\\ \Rightarrow \left\{ \begin{array}{l}a =  - 1\\b =  - 3\end{array} \right. \Rightarrow \max \left\{ {a;b} \right\} = \max \left\{ { - 1; - 3} \right\} =  - 1\end{array}\)

Chọn B.

\(\begin{array}{l}DKXD:\,\, - 1 \le x \le 1\\f'\left( x \right) = 2 + \dfrac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} = 2 - \dfrac{x}{{\sqrt {1 - {x^2}} }}\\f'\left( x \right) > 0 \Leftrightarrow 2 - \dfrac{x}{{\sqrt {1 - {x^2}} }} > 0 \Leftrightarrow \dfrac{{2\sqrt {1 - {x^2}}  - x}}{{\sqrt {1 - {x^2}} }} > 0\,\,\left( {x \in \left( { - 1;1} \right)} \right)\\ \Leftrightarrow 2\sqrt {1 - {x^2}}  - x > 0 \Leftrightarrow 2\sqrt {1 - {x^2}}  > x\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}1 - {x^2} > 0\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\4\left( {1 - {x^2}} \right) > {x^2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} - 1 < x < 1\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\5{x^2} < 4\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\\left\{ \begin{array}{l}x \ge 0\\\dfrac{{ - 2}}{{\sqrt 5 }} < x < \dfrac{2}{{\sqrt 5 }}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\0 \le x < \dfrac{2}{{\sqrt 5 }}\end{array} \right. \Leftrightarrow  - 1 < x < \dfrac{2}{{\sqrt 5 }}\,\,\left( {tm\,\,DKXD} \right)\end{array}\)

Vậy nghiệm của BPT là:  \(x \in \left( { - 1;\dfrac{2}{{\sqrt 5 }}} \right).\)

Chọn C.

\(f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to \left( { - 1} \right)} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}\)

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{\frac{{{x^2} + x + 1}}{x} + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{{x^2} + 2x + 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{x + 1}}{x} = 0\\\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{\frac{{{x^2} - x - 1}}{x} + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{{x^2} - 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{x - 1}}{x} = 2\\ \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} \ne \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}\end{array}\)

Do đó không tồn tại  \(\mathop {\lim }\limits_{x \to \left( { - 1} \right)} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}\), vậy không tồn tại đạo hàm của hàm số tại \({x_0} =  - 1\).

Chọn D

\(\begin{array}{l}f\left( x \right) = {\left( {\sqrt x  - \frac{1}{{\sqrt x }}} \right)^3} = {\left( {\sqrt x } \right)^3} - 3{\left( {\sqrt x } \right)^2}.\frac{1}{{\sqrt x }} + 3\sqrt x {\left( {\frac{1}{{\sqrt x }}} \right)^2} - {\left( {\frac{1}{{\sqrt x }}} \right)^3}\\f\left( x \right) = {x^{\frac{3}{2}}} - 3\sqrt x  + \frac{3}{{\sqrt x }} - \frac{1}{{{x^{\frac{3}{2}}}}}\\f\left( x \right) = {x^{\frac{3}{2}}} - 3\sqrt x  + 3{x^{ - \frac{1}{2}}} - {x^{ - \frac{3}{2}}}\\f'\left( x \right) = \frac{3}{2}{x^{\frac{3}{2} - 1}} - \frac{3}{{2\sqrt x }} + 3.\left( { - \frac{1}{2}} \right){x^{ - \frac{1}{2} - 1}} + \frac{3}{2}{x^{ - \frac{3}{2} - 1}}\\f'\left( x \right) = \frac{3}{2}\sqrt x  - \frac{3}{{2\sqrt x }} - \frac{3}{2}{x^{ - \frac{3}{2}}} + \frac{3}{2}{x^{ - \frac{5}{2}}}\\f'\left( x \right) = \frac{3}{2}\left( {\sqrt x  - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\end{array}\)

Chọn D.

TXĐ: \(D = \mathbb{R}\).

Ta có: \(y = \dfrac{1}{3}{x^3} - 2{x^2} + 3x + \dfrac{1}{3}\)\( \Rightarrow y' = {x^2} - 4x + 3\) 

Khi đó \(y' \le 0 \Leftrightarrow {x^2} - 4x + 3 \le 0 \Leftrightarrow 1 \le x \le 3\).

Vậy tập nghiệm của bất phương trình \(y' \le 0\) là \(\left[ {1;3} \right]\).

Chọn D.

\(y = \frac{{{{\left( {2x - 1} \right)}^9}}}{{{{\left( {x + 3} \right)}^8}}}\)

\(\begin{array}{l}y' = \frac{{9{{\left( {2x - 1} \right)}^8}.2.{{\left( {x + 3} \right)}^8} - {{\left( {2x - 1} \right)}^9}.8{{\left( {x + 3} \right)}^7}}}{{{{\left( {x + 3} \right)}^{16}}}}\\\,\,\,\,\,\, = \frac{{18{{\left( {2x - 1} \right)}^8}.\left( {x + 3} \right) - 8{{\left( {2x - 1} \right)}^9}}}{{{{\left( {x + 3} \right)}^9}}}\\\,\,\,\,\,\, = \frac{{2{{\left( {2x - 1} \right)}^8}.\left[ {9\left( {x + 3} \right) - 4\left( {2x - 1} \right)} \right]}}{{{{\left( {x + 3} \right)}^9}}}\\\,\,\,\,\,\, = \frac{{2{{\left( {2x - 1} \right)}^8}.\left( {x + 31} \right)}}{{{{\left( {x + 3} \right)}^9}}}\end{array}\)

Chọn C.

\(y = \frac{{4x + 1}}{{\sqrt {{x^2} - x + 2} }}\)

\(\begin{array}{l}y' = \frac{{4\sqrt {{x^2} - x + 2}  - \left( {4x + 1} \right).\frac{{2x - 1}}{{2\sqrt {{x^2} - x + 2} }}}}{{{x^2} - x + 2}}\\\,\,\,\,\, = \frac{{8\left( {{x^2} - x + 2} \right) - \left( {4x + 1} \right).\left( {2x - 1} \right)}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\\\,\,\,\,\, = \frac{{8{x^2} - 8x + 16 - 8{x^2} + 4x - 2x + 1}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\\\,\,\,\,\, = \frac{{ - 6x + 17}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\end{array}\)

Chọn A.

\(g\left( x \right) = x - \sqrt {{x^2} + 2x - 3} \)

ĐKXĐ: \({x^2} + 2x - 3 \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \ge 1\\x \le  - 3\end{array} \right.\)

Ta có: \(g'\left( x \right) = 1 - \frac{{x + 1}}{{\sqrt {{x^2} + 2x - 3} }}\)

\(\begin{array}{l}g'\left( x \right) \ge 0\\ \Leftrightarrow 1 - \frac{{x + 1}}{{\sqrt {{x^2} + 2x - 3} }} \ge 0\\ \Leftrightarrow \frac{{x + 1}}{{\sqrt {{x^2} + 2x - 3} }} \le 1\\ \Leftrightarrow x + 1 \le \sqrt {{x^2} + 2x - 3} \\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 1 < 0\\{x^2} + 2x - 3 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}x + 1 \ge 0\\{\left( {x + 1} \right)^2} \le {x^2} + 2x - 3\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x <  - 1\\\left[ \begin{array}{l}x \ge 1\\x \le  - 3\end{array} \right.\end{array} \right.\\\left\{ \begin{array}{l}x \ge  - 1\\1 \le  - 3\,\,\,\left( {Loai} \right)\end{array} \right.\end{array} \right. \Leftrightarrow x \le  - 3\end{array}\).

Vậy tập nghiệm của bất phương trình là \(\left( { - \infty ; - 3} \right]\).

Ta có: \(f'\left( x \right) = 3m{x^2} - 6mx + 3\).

TH1: \(3m = 0 \Leftrightarrow m = 0\).

\( \Rightarrow f'\left( x \right) = 3 > 0\,\,\,\forall x \in \mathbb{R}\), do đó \(m = 0\) không thỏa mãn.

TH2: \(3m \ne 0 \Leftrightarrow m \ne 0\)

\(f'\left( x \right) \le 0\,\,\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}3m < 0\\\Delta ' \le 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}m < 0\\9{m^2} - 9m \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 0\\0 \le m \le 1\end{array} \right. \Leftrightarrow m \in \emptyset \). 

Vậy \(m \in \emptyset \).

Chọn D.

Ta có: \(f'\left( x \right) = 3\left( {m + 1} \right){x^2} - 2\left( {m + 1} \right)x + 3\).

TH1: \(m + 1 = 0 \Leftrightarrow m =  - 1\).

\( \Rightarrow f'\left( x \right) = 3 > 0\,\,\,\forall x \in \mathbb{R}\), do đó \(m =  - 1\) không thỏa mãn.

TH2: \(m + 1 \ne 0 \Leftrightarrow m \ne  - 1\)

\(f'\left( x \right) \le 0\,\,\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}m + 1 < 0\\\Delta ' \le 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}m <  - 1\\{m^2} - 7m - 8 \le 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}m <  - 1\\ - 1 \le m \le 8\end{array} \right. \Leftrightarrow m \in \emptyset \). 

Vậy \(m \in \emptyset \).

Chọn D.

TXĐ: \(D = \mathbb{R}\).

Ta có:

\(\begin{array}{l}y = \dfrac{1}{3}{x^3} + \left( {2m + 1} \right){x^2} - mx - 4\\ \Rightarrow y' = {x^2} + 2\left( {2m + 1} \right)x - m\\ \Rightarrow y' \ge 0\,\,\forall x \in \mathbb{R}\\ \Leftrightarrow {x^2} + 2\left( {2m + 1} \right)x - m \ge 0\,\,\forall x \in \mathbb{R}\\ \Leftrightarrow \left\{ \begin{array}{l}1 > 0\\\Delta ' = {\left( {2m + 1} \right)^2} + m \le 0\end{array} \right.\\ \Leftrightarrow 4{m^2} + 4m + 1 + m \le 0\\ \Leftrightarrow 4{m^2} + 5m + 1 \le 0\\ \Leftrightarrow  - 1 \le m \le  - \dfrac{1}{4}\end{array}\)

Vậy \(m \in \left[ { - 1; - \dfrac{1}{4}} \right]\).

Chọn C.

Ta có:

\(\begin{array}{l}y = {\left( {{x^3} - 2{x^2}} \right)^{2019}}\\ \Rightarrow y' = 2019.{\left( {{x^3} - 2{x^2}} \right)^{2018}}.\left( {{x^3} - 2{x^2}} \right)'\\\,\,\,\,\,\,\,y' = 2019.{\left( {{x^3} - 2{x^2}} \right)^{2018}}.\left( {3{x^2} - 4x} \right)\end{array}\)

Lại có \(f\left( 2 \right) = m\).

Do đó để hàm số liên tục tại \(x = 2\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\)\( \Leftrightarrow m = 3\).

Chọn C.

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

\(\begin{array}{l} \Rightarrow \dfrac{{y'}}{y} = \dfrac{{\sqrt {{x^2} + 1}  + x}}{{\sqrt {{x^2} + 1} }}:\left( {x + \sqrt {{x^2} + 1} } \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt {{x^2} + 1}  + x}}{{\sqrt {{x^2} + 1} }}.\dfrac{1}{{x + \sqrt {{x^2} + 1} }} = \dfrac{1}{{\sqrt {{x^2} + 1} }}\end{array}\)

Chọn A.

Ta có: \(y' = f'\left( x \right) = 3m{x^2} + 2x + 1\).

Để \(f'\left( x \right) = 0\) có hai nghiệm trái dấu thì \(\left\{ \begin{array}{l}3m \ne 0\\3m.1 < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne 0\\m < 0\end{array} \right. \Leftrightarrow m < 0\).

Chọn C.

\(\eqalign{  & y = {{m{x^3}} \over 3} - m{x^2} + \left( {3m - 1} \right)x + 1  \cr   &  \Rightarrow y' = m{x^2} - 2mx + 3m - 1  \cr   & y' \le 0\,\,\forall x \in R \Rightarrow m{x^2} - 2mx + 3m - 1 \le 0\,\,\forall x \in R \cr} \)

TH1: m = 0, khi đó \(BPT \Leftrightarrow  - 1 \le 0\), đúng \(\forall x \in R\)

TH2: \(\eqalign{  & m \ne 0 \Leftrightarrow y' \le 0\,\,\forall x \in R \Leftrightarrow \left\{ \matrix{  a = m < 0 \hfill \cr   \Delta ' = {m^2} - m\left( {3m - 1} \right) \le 0 \hfill \cr}  \right.  \cr   &  \Leftrightarrow \left\{ \matrix{  m < 0 \hfill \cr    - 2{m^2} + m \le 0 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  m < 0 \hfill \cr   \left[ \matrix{  m \le 0 \hfill \cr   m \ge {1 \over 2} \hfill \cr}  \right. \hfill \cr}  \right. \Leftrightarrow m < 0 \cr} \)

Kết hợp cả 2 trường hợp ta có \(m \le 0\) là những giá trị cần tìm.

Chọn C.

Do \(P\left( x \right)\) bậc 3 và có 3 nghiệm phân biệt \({x_1},{x_2},{x_3}\) nên \(P\left( x \right)\) được biểu diễn dưới dạng \(P\left( x \right) = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\,\,\left( {a \ne 0} \right)\).

Ta có: \(P'\left( x \right) = a\left( {x - {x_2}} \right)\left( {x - {x_3}} \right) + a\left( {x - {x_1}} \right)\left( {x - {x_3}} \right) + a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}P\left( {{x_1}} \right) = a\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right)\\P\left( {{x_2}} \right) = a\left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right)\\P\left( {{x_3}} \right) = a\left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)\end{array} \right.\\ \Rightarrow \dfrac{1}{{P'\left( {{x_1}} \right)}} + \dfrac{1}{{P'\left( {{x_2}} \right)}} + \dfrac{1}{{P'\left( {{x_3}} \right)}}\\ = \dfrac{1}{{a\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right)}} + \dfrac{1}{{a\left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right)}} + \dfrac{1}{{a\left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)}}\\ = \dfrac{{ - {x_2} + {x_3} - {x_3} + {x_1} - {x_1} + {x_2}}}{{a\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}} = 0 \end{array}\)

\(\begin{array}{l}y = \frac{{m{x^3}}}{3} - m{x^2} + \left( {3m - 1} \right)x + 1\\ \Rightarrow y' = m{x^2} - 2mx + 3m - 1\\y' \le 0\,\,\forall x \in R \Rightarrow m{x^2} - 2mx + 3m - 1 \le 0\,\,\forall x \in R\end{array}\)

TH1: m = 0, khi đó \(BPT \Leftrightarrow  - 1 \le 0\) , đúng \(\forall x \in R\)

TH2: \(\begin{array}{l}m \ne 0 \Leftrightarrow y' \le 0\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}a = m < 0\\\Delta ' = {m^2} - m\left( {3m - 1} \right) \le 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m < 0\\ - 2{m^2} + m \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 0\\\left[ \begin{array}{l}m \le 0\\m \ge \frac{1}{2}\end{array} \right.\end{array} \right. \Leftrightarrow m < 0\end{array}\)

Kết hợp cả 2 trường hợp ta có \(m \le 0\) là những giá trị cần tìm.

Chọn C.