Google
Câu hỏi: Cho hàm số $f\left( x \right)=ax-\left( a-3 \right)\text{ln}\left( {{x}^{2}}+3x \right)$ với $a$ là tham số thực Biết rằng nếu $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)$ thì $\underset{\left[ 1;3 \right]}{\mathop{\text{min}}} f\left( x \right)=m$. Khẳng định nào sau đây đúng?
A. $m\in \left( 6;7 \right)$.
B. $m\in \left( 7;8 \right)$.
C. $m\in \left( 8;9 \right)$.
D. $m\in \left( 9;10 \right)$.
A. $m\in \left( 6;7 \right)$.
B. $m\in \left( 7;8 \right)$.
C. $m\in \left( 8;9 \right)$.
D. $m\in \left( 9;10 \right)$.
$f\left( x \right)=ax-\left( a-3 \right)\text{ln}\left( {{x}^{2}}+3x \right)\Rightarrow {f}'\left( x \right)=a-\left( a-3 \right)\dfrac{2x+3}{{{x}^{2}}+3x}$
Vì $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)$ nên ${f}'\left( 2 \right)=0$.
$\Rightarrow a-\left( a-3 \right)\dfrac{7}{10}=0\Leftrightarrow a=-7$
$\Rightarrow {f}'\left( x \right)=-7+10\dfrac{2x+3}{{{x}^{2}}+3x}$
${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=2 \\
& x=\dfrac{-15}{7} \\
\end{aligned} \right.$.
$f\left( 1 \right)=-7+10\ln 4; f\left( 2 \right)=-14+10\ln 10; f\left( 3 \right)=-21+10\ln 18$
Vậy $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)$ và $m=\underset{\left[ 1;3 \right]}{\mathop{\text{min}}} f\left( x \right)=f\left( 1 \right)\approx 6,86$.
Vì $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)$ nên ${f}'\left( 2 \right)=0$.
$\Rightarrow a-\left( a-3 \right)\dfrac{7}{10}=0\Leftrightarrow a=-7$
$\Rightarrow {f}'\left( x \right)=-7+10\dfrac{2x+3}{{{x}^{2}}+3x}$
${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=2 \\
& x=\dfrac{-15}{7} \\
\end{aligned} \right.$.
$f\left( 1 \right)=-7+10\ln 4; f\left( 2 \right)=-14+10\ln 10; f\left( 3 \right)=-21+10\ln 18$
Vậy $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=f\left( 2 \right)$ và $m=\underset{\left[ 1;3 \right]}{\mathop{\text{min}}} f\left( x \right)=f\left( 1 \right)\approx 6,86$.
Đáp án A.