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Câu hỏi: Cho hàm số $f\left( x \right)$ đồng biến và có đạo hàm lên tục trên đoạn $\left[ 1;4 \right]$ thỏa mãn $f\left( 1 \right)=1$ và ${{\left[ f\left( x \right)+x{f}'\left( x \right) \right]}^{2}}=4f\left( x \right), \forall x\in \left[ 1;4 \right]$. Tính diện tích $S$ của hình phẳng giới hạn bởi đồ thị hàm số $y=f\left( x \right)$, trục hoành và hai đường thẳng $x=1,x=4.$
A. $4-2\ln 2.$
B. $4+2\ln 2.$
C. $4+\ln 2.$
D. $4-\ln 2.$
$\Rightarrow \dfrac{f\left( x \right)+x{f}'\left( x \right)}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow \dfrac{{{\left( x \right)}^{\prime }}f\left( x \right)+x{f}'\left( x \right)}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow \dfrac{{{\left( xf\left( x \right) \right)}^{\prime }}}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow {{\left( \sqrt{xf\left( x \right)} \right)}^{\prime }}=\dfrac{1}{\sqrt{x}}$ $\Rightarrow \sqrt{xf\left( x \right)}=\int{\dfrac{1}{\sqrt{x}}dx}\Rightarrow \sqrt{xf\left( x \right)}=2\sqrt{x}+C.$
+) Lại có $f\left( 1 \right)=1\Rightarrow C=-1\Rightarrow \sqrt{xf\left( x \right)}=2\sqrt{x}-1\Rightarrow f\left( x \right)=\dfrac{{{\left( 2\sqrt{x}-1 \right)}^{2}}}{x}.$
+) Do đó $S=\int\limits_{1}^{4}{\dfrac{{{\left( 2\sqrt{x}-1 \right)}^{2}}}{x}dx}=\int\limits_{1}^{4}{\left( 4-\dfrac{4}{\sqrt{x}}+\dfrac{1}{x} \right)dx=}4x\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.-8\sqrt{x}\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.+\ln x\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.=4+2\ln 2.$
A. $4-2\ln 2.$
B. $4+2\ln 2.$
C. $4+\ln 2.$
D. $4-\ln 2.$
+) Ta có ${{\left( f\left( x \right)+x{f}'\left( x \right) \right)}^{2}}=4f\left( x \right)\Rightarrow \dfrac{{{\left( f\left( x \right)+x{f}'\left( x \right) \right)}^{2}}}{4f\left( x \right)}=1\Rightarrow \dfrac{{{\left( f\left( x \right)+x{f}'\left( x \right) \right)}^{2}}}{4xf\left( x \right)}=\dfrac{1}{x}$ $\Rightarrow \dfrac{f\left( x \right)+x{f}'\left( x \right)}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow \dfrac{{{\left( x \right)}^{\prime }}f\left( x \right)+x{f}'\left( x \right)}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow \dfrac{{{\left( xf\left( x \right) \right)}^{\prime }}}{2\sqrt{xf\left( x \right)}}=\dfrac{1}{\sqrt{x}}\Rightarrow {{\left( \sqrt{xf\left( x \right)} \right)}^{\prime }}=\dfrac{1}{\sqrt{x}}$ $\Rightarrow \sqrt{xf\left( x \right)}=\int{\dfrac{1}{\sqrt{x}}dx}\Rightarrow \sqrt{xf\left( x \right)}=2\sqrt{x}+C.$
+) Lại có $f\left( 1 \right)=1\Rightarrow C=-1\Rightarrow \sqrt{xf\left( x \right)}=2\sqrt{x}-1\Rightarrow f\left( x \right)=\dfrac{{{\left( 2\sqrt{x}-1 \right)}^{2}}}{x}.$
+) Do đó $S=\int\limits_{1}^{4}{\dfrac{{{\left( 2\sqrt{x}-1 \right)}^{2}}}{x}dx}=\int\limits_{1}^{4}{\left( 4-\dfrac{4}{\sqrt{x}}+\dfrac{1}{x} \right)dx=}4x\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.-8\sqrt{x}\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.+\ln x\left| \begin{aligned}
& 4 \\
& 1 \\
\end{aligned} \right.=4+2\ln 2.$
Đáp án B.