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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 4;8 \right]$ và $f\left( x \right)\ne 0,\forall x\in \left[ 4;8 \right].$ Biết rằng $\int\limits_{4}^{8}{\dfrac{{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{\left[ f\left( x \right) \right]}^{4}}}dx=1,} f\left( 4 \right)=\dfrac{1}{4},f\left( 8 \right)=\dfrac{1}{2}.$ Tính $f\left( 6 \right).$
A. $f\left( 6 \right)=\dfrac{5}{8}.$
B. $f\left( 6 \right)=\dfrac{2}{3}.$
C. $f\left( 6 \right)=\dfrac{3}{8}.$
D. $f\left( 6 \right)=\dfrac{1}{3}.$
A. $f\left( 6 \right)=\dfrac{5}{8}.$
B. $f\left( 6 \right)=\dfrac{2}{3}.$
C. $f\left( 6 \right)=\dfrac{3}{8}.$
D. $f\left( 6 \right)=\dfrac{1}{3}.$
Ta có $\int\limits_{4}^{8}{\dfrac{{{\left[ {f}'\left( x \right) \right]}^{2}}}{{{\left[ f\left( x \right) \right]}^{4}}}dx}=1\Leftrightarrow \int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)} \right)}^{2}}dx}=1.$
Tìm $k\in \mathbb{R}$ để $\int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-k \right)}^{2}}dx=0.}$
$\begin{aligned}
& \int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-k \right)}^{2}}dx=0}\Leftrightarrow \int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)} \right)}^{2}}dx-2k\int\limits_{4}^{8}{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}}+{{k}^{2}}\int\limits_{4}^{8}{dx}}=0 \\
& \Leftrightarrow \left( 1+2k\dfrac{1}{f\left( x \right)} \right)\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.+{{k}^{2}}x\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.=0\Leftrightarrow 4{{k}^{2}}-4k+1=0\Leftrightarrow k=\dfrac{1}{2}. \\
\end{aligned}$
Vậy $\int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-\dfrac{1}{2} \right)}^{2}}dx=0}\Rightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-\dfrac{1}{2}=0\Leftrightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=\dfrac{1}{2}.$
$\int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=\int{\dfrac{1}{2}dx}\Leftrightarrow -\dfrac{1}{f\left( x \right)}=\dfrac{1}{2}x+C\Leftrightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}x+C}.$
Theo giả thiết $f\left( 4 \right)=\dfrac{1}{4}\Leftrightarrow -\dfrac{1}{\dfrac{1}{2}.4+C}=\dfrac{1}{4}\Leftrightarrow C=-6\Rightarrow f\left( x \right)=-\dfrac{2}{x-12}\Rightarrow f\left( 6 \right)=\dfrac{1}{3}.$
Tìm $k\in \mathbb{R}$ để $\int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-k \right)}^{2}}dx=0.}$
$\begin{aligned}
& \int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-k \right)}^{2}}dx=0}\Leftrightarrow \int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)} \right)}^{2}}dx-2k\int\limits_{4}^{8}{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}}+{{k}^{2}}\int\limits_{4}^{8}{dx}}=0 \\
& \Leftrightarrow \left( 1+2k\dfrac{1}{f\left( x \right)} \right)\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.+{{k}^{2}}x\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.=0\Leftrightarrow 4{{k}^{2}}-4k+1=0\Leftrightarrow k=\dfrac{1}{2}. \\
\end{aligned}$
Vậy $\int\limits_{4}^{8}{{{\left( \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-\dfrac{1}{2} \right)}^{2}}dx=0}\Rightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-\dfrac{1}{2}=0\Leftrightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=\dfrac{1}{2}.$
$\int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=\int{\dfrac{1}{2}dx}\Leftrightarrow -\dfrac{1}{f\left( x \right)}=\dfrac{1}{2}x+C\Leftrightarrow f\left( x \right)=-\dfrac{1}{\dfrac{1}{2}x+C}.$
Theo giả thiết $f\left( 4 \right)=\dfrac{1}{4}\Leftrightarrow -\dfrac{1}{\dfrac{1}{2}.4+C}=\dfrac{1}{4}\Leftrightarrow C=-6\Rightarrow f\left( x \right)=-\dfrac{2}{x-12}\Rightarrow f\left( 6 \right)=\dfrac{1}{3}.$
Đáp án D.