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Câu hỏi: Cho hàm số ${f\left( x \right) = \dfrac{{{x^2} + \left( {x + 2} \right)\sqrt {x - 2} + m}}{{\sqrt {6 - x} + 2}}}$. Biết hàm số có giá trị nhỏ nhất bằng ${10}$, tìm giá trị lớn nhất của hàm số ${f\left( x \right)}$.
A. ${14}$.
B. ${24}$.
C. ${34}$.
D. ${44}$.
A. ${14}$.
B. ${24}$.
C. ${34}$.
D. ${44}$.
TXÐ: $D=\left[ 2;6 \right]$
Ta có $f\left( x \right)=\dfrac{{{x}^{2}}+\left( x+2 \right)\sqrt{x-2}+m}{\sqrt{6-x}+2}$ $\Leftrightarrow f\left( x \right)\left( \sqrt{6-x}+2 \right)={{x}^{2}}+\left( x+2 \right)\sqrt{x-2}+m$
$\Leftrightarrow \dfrac{-1}{\sqrt{6-x}}f\left( x \right)+\left( \sqrt{6-x}+2 \right)f'\left( x \right)=2x+\sqrt{x-2}+\dfrac{x+2}{2\sqrt{x-2}}$
$\Leftrightarrow f'\left( x \right)=\dfrac{2x+\sqrt{x-2}+\dfrac{x+2}{2\sqrt{x-2}}+\dfrac{1}{\sqrt{6-x}}f\left( x \right)}{\sqrt{6-x}+2}$
Vì $\underset{\left[ 2;6 \right]}{\mathop{\min }} f\left( x \right)=10$ nên suy ra $f'\left( x \right)>0 \forall x\in \left[ 2;6 \right]$
Vậy $\underset{\left[ 2;6 \right]}{\mathop{\min }} f\left( x \right)=f\left( 2 \right)\Rightarrow m=36$
Do đó $\underset{\left[ 2;6 \right]}{\mathop{\max }} f\left( x \right)=f\left( 6 \right)=44$
Ta có $f\left( x \right)=\dfrac{{{x}^{2}}+\left( x+2 \right)\sqrt{x-2}+m}{\sqrt{6-x}+2}$ $\Leftrightarrow f\left( x \right)\left( \sqrt{6-x}+2 \right)={{x}^{2}}+\left( x+2 \right)\sqrt{x-2}+m$
$\Leftrightarrow \dfrac{-1}{\sqrt{6-x}}f\left( x \right)+\left( \sqrt{6-x}+2 \right)f'\left( x \right)=2x+\sqrt{x-2}+\dfrac{x+2}{2\sqrt{x-2}}$
$\Leftrightarrow f'\left( x \right)=\dfrac{2x+\sqrt{x-2}+\dfrac{x+2}{2\sqrt{x-2}}+\dfrac{1}{\sqrt{6-x}}f\left( x \right)}{\sqrt{6-x}+2}$
Vì $\underset{\left[ 2;6 \right]}{\mathop{\min }} f\left( x \right)=10$ nên suy ra $f'\left( x \right)>0 \forall x\in \left[ 2;6 \right]$
Vậy $\underset{\left[ 2;6 \right]}{\mathop{\min }} f\left( x \right)=f\left( 2 \right)\Rightarrow m=36$
Do đó $\underset{\left[ 2;6 \right]}{\mathop{\max }} f\left( x \right)=f\left( 6 \right)=44$
Đáp án D.