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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left[ 0;2 \right],$ thỏa các điều kiện $f\left( 2 \right)=1$ và $\int\limits_{0}^{2}{f\left( x \right)dx}=\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}}dx=\dfrac{2}{3}.$ Giá trị của $\int\limits_{1}^{2}{\dfrac{f\left( x \right)}{x}dx:}$
A. 1.
B. 2.
C. $\dfrac{1}{4}.$
D. $\dfrac{1}{3}.$
A. 1.
B. 2.
C. $\dfrac{1}{4}.$
D. $\dfrac{1}{3}.$
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=x \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=x.f\left( x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{2}{x.{f}'\left( x \right)dx=2-\int\limits_{0}^{2}{x.{f}'\left( x \right)dx}\Rightarrow -\int\limits_{0}^{2}{x.{f}'\left( x \right)dx}=\dfrac{2}{3}-2=-\dfrac{4}{3}}$
Ta lại có: $\int\limits_{0}^{2}{\dfrac{1}{4}{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{12}\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.=\dfrac{2}{3}.$
Do đó: $\int\limits_{0}^{2}{\left[ {f}'\left( x \right) \right]dx}-\int\limits_{0}^{2}{x.{f}'\left( x \right)}dx+\int\limits_{0}^{2}{\dfrac{1}{4}{{x}^{2}}dx}=\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{2}{3}\Leftrightarrow \int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)-\dfrac{1}{2}x \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)-\dfrac{1}{2}x=0$ (vì $\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)-\dfrac{1}{2}x \right]}^{2}}dx\ge 0,\forall x\in \left[ 0;2 \right]}$ )
$\Rightarrow f\left( x \right)=\dfrac{1}{4}{{x}^{2}}+C\Rightarrow f\left( 2 \right)=1+C\Leftrightarrow C=0$
Vậy $f\left( x \right)=\dfrac{1}{4}{{x}^{2}}\Rightarrow \int\limits_{1}^{2}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int\limits_{1}^{2}{\dfrac{1}{4}dx}=\dfrac{1}{4}x\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=\dfrac{1}{4}.$
& u=f\left( x \right) \\
& dv=dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=x \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=x.f\left( x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{2}{x.{f}'\left( x \right)dx=2-\int\limits_{0}^{2}{x.{f}'\left( x \right)dx}\Rightarrow -\int\limits_{0}^{2}{x.{f}'\left( x \right)dx}=\dfrac{2}{3}-2=-\dfrac{4}{3}}$
Ta lại có: $\int\limits_{0}^{2}{\dfrac{1}{4}{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{12}\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.=\dfrac{2}{3}.$
Do đó: $\int\limits_{0}^{2}{\left[ {f}'\left( x \right) \right]dx}-\int\limits_{0}^{2}{x.{f}'\left( x \right)}dx+\int\limits_{0}^{2}{\dfrac{1}{4}{{x}^{2}}dx}=\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{2}{3}\Leftrightarrow \int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)-\dfrac{1}{2}x \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)-\dfrac{1}{2}x=0$ (vì $\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)-\dfrac{1}{2}x \right]}^{2}}dx\ge 0,\forall x\in \left[ 0;2 \right]}$ )
$\Rightarrow f\left( x \right)=\dfrac{1}{4}{{x}^{2}}+C\Rightarrow f\left( 2 \right)=1+C\Leftrightarrow C=0$
Vậy $f\left( x \right)=\dfrac{1}{4}{{x}^{2}}\Rightarrow \int\limits_{1}^{2}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int\limits_{1}^{2}{\dfrac{1}{4}dx}=\dfrac{1}{4}x\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=\dfrac{1}{4}.$
Đáp án C.