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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 2;4 \right]$ và ${f}'\left( x \right)>0,\forall x\in \left[ 2;4 \right]$. Biết $4{{x}^{3}}f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{3}}-{{x}^{3}},\forall x\in \left[ 2;4 \right]$, $f\left( 2 \right)=\dfrac{7}{4}$. Giá trị $f\left( 4 \right)$ bằng
A. $\dfrac{40\sqrt{5}-1}{2}$
B. $\dfrac{20\sqrt{5}-1}{4}$
C. $\dfrac{20\sqrt{5}-1}{2}$
D. $\dfrac{40\sqrt{5}-1}{4}$
A. $\dfrac{40\sqrt{5}-1}{2}$
B. $\dfrac{20\sqrt{5}-1}{4}$
C. $\dfrac{20\sqrt{5}-1}{2}$
D. $\dfrac{40\sqrt{5}-1}{4}$
Ta có $4{{x}^{3}}.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{3}}-{{x}^{3}}\Leftrightarrow {{x}^{3}}.\left[ 4f\left( x \right)+1 \right]={{\left[ {f}'\left( x \right) \right]}^{3}}$
$\Leftrightarrow {{x}^{3}}=\dfrac{{{\left[ {f}'\left( x \right) \right]}^{3}}}{4f\left( x \right)+1}\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt[3]{4f\left( x \right)+1}}=x\Leftrightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt[3]{4f\left( x \right)+1}}dx}=\int{xdx}=\dfrac{{{x}^{2}}}{2}+C$
$\Leftrightarrow \dfrac{3}{8}\sqrt[3]{{{\left[ 4f\left( x \right)+1 \right]}^{2}}}=\dfrac{{{x}^{2}}}{2}+C$ mà $f\left( 2 \right)=\dfrac{7}{4}\xrightarrow{{}}\dfrac{3}{2}=2+C\Rightarrow C=-\dfrac{1}{2}$.
Do đó $f\left( x \right)=\dfrac{\sqrt{{{\left[ \dfrac{4}{3}\left( {{x}^{2}}-1 \right) \right]}^{3}}}-1}{4}\Rightarrow f\left( 4 \right)=\dfrac{40\sqrt{5}-1}{4}$.
$\Leftrightarrow {{x}^{3}}=\dfrac{{{\left[ {f}'\left( x \right) \right]}^{3}}}{4f\left( x \right)+1}\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt[3]{4f\left( x \right)+1}}=x\Leftrightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt[3]{4f\left( x \right)+1}}dx}=\int{xdx}=\dfrac{{{x}^{2}}}{2}+C$
$\Leftrightarrow \dfrac{3}{8}\sqrt[3]{{{\left[ 4f\left( x \right)+1 \right]}^{2}}}=\dfrac{{{x}^{2}}}{2}+C$ mà $f\left( 2 \right)=\dfrac{7}{4}\xrightarrow{{}}\dfrac{3}{2}=2+C\Rightarrow C=-\dfrac{1}{2}$.
Do đó $f\left( x \right)=\dfrac{\sqrt{{{\left[ \dfrac{4}{3}\left( {{x}^{2}}-1 \right) \right]}^{3}}}-1}{4}\Rightarrow f\left( 4 \right)=\dfrac{40\sqrt{5}-1}{4}$.
Đáp án D.