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Câu hỏi: Cho hàm số $y=f\left( x \right)$ luôn nhận giá trị dương và có đạo hàm đến cấp hai trên khoảng $\left( 1;+\infty \right)$ đồng thời thỏa mãn các điều kiện $f\left( 1 \right)={f}'\left( 1 \right)=2$ và ${{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right)\left[ {f}''\left( x \right)-\dfrac{{f}'\left( x \right)}{x} \right]=x\left( 2x+1 \right)$. Tính giá trị $f\left( 2 \right)$.
A. $f\left( 2 \right)=\dfrac{\sqrt{82}}{2}$.
B. $f\left( 2 \right)=\dfrac{133}{6}$.
C. $f\left( 2 \right)=\dfrac{\sqrt{123}}{4}$.
D. $f\left( 2 \right)=\dfrac{\sqrt{798}}{6}$.
A. $f\left( 2 \right)=\dfrac{\sqrt{82}}{2}$.
B. $f\left( 2 \right)=\dfrac{133}{6}$.
C. $f\left( 2 \right)=\dfrac{\sqrt{123}}{4}$.
D. $f\left( 2 \right)=\dfrac{\sqrt{798}}{6}$.
Theo bài ra ta có:
${{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right)\left[ {f}''\left( x \right)-\dfrac{{f}'\left( x \right)}{x} \right]=x\left( 2x+1 \right)$
$\Rightarrow \dfrac{\left( {{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right).{f}''\left( x \right) \right).x-f\left( x \right).{f}'\left( x \right)}{x}=x\left( 2x+1 \right)$
$\Rightarrow \dfrac{{{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}x-f\left( x \right).{f}'\left( x \right)}{{{x}^{2}}}=2x+1$
$\Rightarrow \dfrac{f\left( x \right){f}'\left( x \right)}{x}={{x}^{2}}+x+C$.
Do $f\left( 1 \right)={f}'\left( 1 \right)=2\Rightarrow C=2$ $\Rightarrow f\left( x \right).{f}'\left( x \right)={{x}^{3}}+{{x}^{2}}+2x$.
Suy ra $\int\limits_{1}^{2}{f\left( x \right).{f}'\left( x \right)dx=\int\limits_{1}^{2}{\left( {{x}^{3}}+{{x}^{2}}+2x \right)dx}}$ $\Rightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.=\dfrac{109}{12}$
${{f}^{2}}\left( 2 \right)-{{f}^{2}}\left( 1 \right)=\dfrac{109}{6}\Rightarrow {{f}^{2}}\left( 2 \right)=\dfrac{133}{6}\Rightarrow f\left( 2 \right)=\dfrac{\sqrt{798}}{6}$ ( Do $f\left( x \right)$ luôn nhận giá trị dương trên khoảng $(1 ;+\infty)$.
${{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right)\left[ {f}''\left( x \right)-\dfrac{{f}'\left( x \right)}{x} \right]=x\left( 2x+1 \right)$
$\Rightarrow \dfrac{\left( {{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right).{f}''\left( x \right) \right).x-f\left( x \right).{f}'\left( x \right)}{x}=x\left( 2x+1 \right)$
$\Rightarrow \dfrac{{{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}x-f\left( x \right).{f}'\left( x \right)}{{{x}^{2}}}=2x+1$
$\Rightarrow \dfrac{f\left( x \right){f}'\left( x \right)}{x}={{x}^{2}}+x+C$.
Do $f\left( 1 \right)={f}'\left( 1 \right)=2\Rightarrow C=2$ $\Rightarrow f\left( x \right).{f}'\left( x \right)={{x}^{3}}+{{x}^{2}}+2x$.
Suy ra $\int\limits_{1}^{2}{f\left( x \right).{f}'\left( x \right)dx=\int\limits_{1}^{2}{\left( {{x}^{3}}+{{x}^{2}}+2x \right)dx}}$ $\Rightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.=\dfrac{109}{12}$
${{f}^{2}}\left( 2 \right)-{{f}^{2}}\left( 1 \right)=\dfrac{109}{6}\Rightarrow {{f}^{2}}\left( 2 \right)=\dfrac{133}{6}\Rightarrow f\left( 2 \right)=\dfrac{\sqrt{798}}{6}$ ( Do $f\left( x \right)$ luôn nhận giá trị dương trên khoảng $(1 ;+\infty)$.
Đáp án D.