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Câu hỏi: Cho hàm số $f\left( x \right)=m+\int\limits_{0}^{x}{\dfrac{3{{t}^{4}}+4{{t}^{3}}}{{{\left( t+1 \right)}^{2}}}dt}$ với $x\in \left[ 1 ;2 \right]$ và $m$ là tham số. Có bao nhiêu giá trị nguyên của tham số $m$ để $\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|$.
A. $9$
B. $7$
C. $10$
D. $8$
A. $9$
B. $7$
C. $10$
D. $8$
$\begin{aligned}
& f\left( x \right)=m+\int\limits_{0}^{x}{\dfrac{3{{t}^{4}}+4{{t}^{3}}}{{{\left( t+1 \right)}^{2}}}dt}=m+\int\limits_{0}^{x}{\left( 3{{t}^{2}}-2t+1-\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right)dt} \\
& =m+\left. \left( {{t}^{3}}-{{t}^{2}}+t+\dfrac{1}{t+1} \right) \right|_{t=0}^{t=x}=m+{{x}^{3}}-{{x}^{2}}+x+\dfrac{1}{x+1}-1 \\
\end{aligned}$
$\Rightarrow {f}'\left( x \right)=\dfrac{3{{x}^{4}}+4{{x}^{3}}}{{{\left( x+1 \right)}^{2}}}>0,\forall x\in \left[ 1 ;2 \right]$ $\Rightarrow f\left( 1 \right)\le f\left( x \right)\le f\left( 2 \right)$ $\Leftrightarrow m+\dfrac{1}{2}\le f\left( x \right)\le m+\dfrac{16}{3}$
+ Trường hợp 1: $m+\dfrac{1}{2}\ge 0\Leftrightarrow m\ge -\dfrac{1}{2}$ $\Rightarrow m+\dfrac{1}{2}\le \left| f\left( x \right) \right|\le m+\dfrac{16}{3}$
$\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|\Leftrightarrow m+\dfrac{16}{3}\ge 3\left( m+\dfrac{1}{2} \right)\Leftrightarrow m\le \dfrac{23}{12}$ $\Rightarrow -\dfrac{1}{2}\le m\le \dfrac{23}{12}$.
Mà $m$ nguyên nên $m\in \left\{ 0 ;1 \right\}$.
+ Trường hợp 2: $m+\dfrac{16}{3}\le 0\Leftrightarrow m\le -\dfrac{16}{3}$ $\Rightarrow -m-\dfrac{16}{3}\le \left| f\left( x \right) \right|\le -m-\dfrac{1}{2}$
$\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|\Leftrightarrow -m-\dfrac{1}{2}\ge 3\left( -m-\dfrac{16}{3} \right)\Leftrightarrow m\ge \dfrac{-31}{4}$ $\Rightarrow -\dfrac{31}{4}\le m\le \dfrac{-16}{3}$.
Mà $m$ nguyên nên $m\in \left\{ -7 ;-6 \right\}$.
+ Trường hợp 3: $m+\dfrac{1}{2}<0<m+\dfrac{16}{3}\Leftrightarrow -\dfrac{16}{3}<m<-\dfrac{1}{2}$ $\Rightarrow 0\le \left| f\left( x \right) \right|\le Max\left\{ \left| m+\dfrac{1}{2} \right| ;\left| m+\dfrac{16}{3} \right| \right\}$
$\Rightarrow \underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=0$ nên luôn có $\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|$.
$\Rightarrow -\dfrac{16}{3}<m<-\dfrac{1}{2}$ thỏa mãn. Mà $m$ nguyên nên $m\in \left\{ -5 ;-4;-3 ;-2 ;-1 \right\}$.
Vậy có tất cả 9 giá trị nguyên của $m$ thỏa mãn yêu cầu đề bài
& f\left( x \right)=m+\int\limits_{0}^{x}{\dfrac{3{{t}^{4}}+4{{t}^{3}}}{{{\left( t+1 \right)}^{2}}}dt}=m+\int\limits_{0}^{x}{\left( 3{{t}^{2}}-2t+1-\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right)dt} \\
& =m+\left. \left( {{t}^{3}}-{{t}^{2}}+t+\dfrac{1}{t+1} \right) \right|_{t=0}^{t=x}=m+{{x}^{3}}-{{x}^{2}}+x+\dfrac{1}{x+1}-1 \\
\end{aligned}$
$\Rightarrow {f}'\left( x \right)=\dfrac{3{{x}^{4}}+4{{x}^{3}}}{{{\left( x+1 \right)}^{2}}}>0,\forall x\in \left[ 1 ;2 \right]$ $\Rightarrow f\left( 1 \right)\le f\left( x \right)\le f\left( 2 \right)$ $\Leftrightarrow m+\dfrac{1}{2}\le f\left( x \right)\le m+\dfrac{16}{3}$
+ Trường hợp 1: $m+\dfrac{1}{2}\ge 0\Leftrightarrow m\ge -\dfrac{1}{2}$ $\Rightarrow m+\dfrac{1}{2}\le \left| f\left( x \right) \right|\le m+\dfrac{16}{3}$
$\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|\Leftrightarrow m+\dfrac{16}{3}\ge 3\left( m+\dfrac{1}{2} \right)\Leftrightarrow m\le \dfrac{23}{12}$ $\Rightarrow -\dfrac{1}{2}\le m\le \dfrac{23}{12}$.
Mà $m$ nguyên nên $m\in \left\{ 0 ;1 \right\}$.
+ Trường hợp 2: $m+\dfrac{16}{3}\le 0\Leftrightarrow m\le -\dfrac{16}{3}$ $\Rightarrow -m-\dfrac{16}{3}\le \left| f\left( x \right) \right|\le -m-\dfrac{1}{2}$
$\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|\Leftrightarrow -m-\dfrac{1}{2}\ge 3\left( -m-\dfrac{16}{3} \right)\Leftrightarrow m\ge \dfrac{-31}{4}$ $\Rightarrow -\dfrac{31}{4}\le m\le \dfrac{-16}{3}$.
Mà $m$ nguyên nên $m\in \left\{ -7 ;-6 \right\}$.
+ Trường hợp 3: $m+\dfrac{1}{2}<0<m+\dfrac{16}{3}\Leftrightarrow -\dfrac{16}{3}<m<-\dfrac{1}{2}$ $\Rightarrow 0\le \left| f\left( x \right) \right|\le Max\left\{ \left| m+\dfrac{1}{2} \right| ;\left| m+\dfrac{16}{3} \right| \right\}$
$\Rightarrow \underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=0$ nên luôn có $\underset{\left[ 1;2 \right]}{\mathop{\text{max}}} \left| f\left( x \right) \right|\ge 3\underset{\left[ 1;2 \right]}{\mathop{\min }} \left| f\left( x \right) \right|$.
$\Rightarrow -\dfrac{16}{3}<m<-\dfrac{1}{2}$ thỏa mãn. Mà $m$ nguyên nên $m\in \left\{ -5 ;-4;-3 ;-2 ;-1 \right\}$.
Vậy có tất cả 9 giá trị nguyên của $m$ thỏa mãn yêu cầu đề bài
Đáp án A.