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Câu hỏi: Cho hàm số $y=f\left( x \right)$ đồng biến trên $\left( 0;+\infty \right)$, $y=f\left( x \right)$ liên tục, nhận giá trị dương trên $\left( 0;+\infty \right)$ và thỏa mãn $f\left( 3 \right)=\dfrac{4}{9}$ và ${{\left[ {f}'\left( x \right) \right]}^{2}}=\left( x+1 \right)f\left( x \right)$. Tính $f\left( 8 \right)$
A. $f\left( 8 \right)=49$.
B. $f\left( 8 \right)=\dfrac{49}{64}$.
C. $f\left( 8 \right)=256$.
D. $f\left( 8 \right)=\dfrac{1}{16}$.
A. $f\left( 8 \right)=49$.
B. $f\left( 8 \right)=\dfrac{49}{64}$.
C. $f\left( 8 \right)=256$.
D. $f\left( 8 \right)=\dfrac{1}{16}$.
Ta có:
${{\left[ {f}'\left( x \right) \right]}^{2}}=\left( x+1 \right)f\left( x \right)\quad \left( 1 \right)$
Vì hàm số $y=f\left( x \right)$ đồng biến trên $\left( 0;+\infty \right)$ $\Rightarrow {f}'\left( x \right)\ge 0,\ \forall x\in \left( 0;+\infty \right)$
Và $f\left( x \right)>0,\ \forall x\in \left( 0;+\infty \right)$
$\begin{aligned}
& \left( 1 \right)\Leftrightarrow {f}'\left( x \right)=\sqrt{\left( x+1 \right)f\left( x \right)} \\
& \Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=\sqrt{x+1} \\
& \Rightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}\text{d}x=\int{\sqrt{x+1}\text{d}x}} \\
& \Leftrightarrow \int{\dfrac{\text{d}\left( f\left( x \right) \right)}{\sqrt{f\left( x \right)}}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+2C} \\
& \Leftrightarrow 2\sqrt{f\left( x \right)}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+2C \\
& \Leftrightarrow \sqrt{f\left( x \right)}=\dfrac{1}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+C\quad \left( 2 \right) \\
\end{aligned}$
Mà: $f\left( 3 \right)=\dfrac{4}{9}$. Thay vào (2) ta được: ${{\left( \dfrac{8}{3}+C \right)}^{2}}=\dfrac{4}{9}\Leftrightarrow \dfrac{8}{3}+C=\dfrac{2}{3}\Leftrightarrow C=-2$
Với $C=-2\Rightarrow f\left( x \right)={{\left( \dfrac{1}{3}\sqrt{{{\left( x+1 \right)}^{3}}}-2 \right)}^{2}}\Rightarrow f\left( 8 \right)=49$
${{\left[ {f}'\left( x \right) \right]}^{2}}=\left( x+1 \right)f\left( x \right)\quad \left( 1 \right)$
Vì hàm số $y=f\left( x \right)$ đồng biến trên $\left( 0;+\infty \right)$ $\Rightarrow {f}'\left( x \right)\ge 0,\ \forall x\in \left( 0;+\infty \right)$
Và $f\left( x \right)>0,\ \forall x\in \left( 0;+\infty \right)$
$\begin{aligned}
& \left( 1 \right)\Leftrightarrow {f}'\left( x \right)=\sqrt{\left( x+1 \right)f\left( x \right)} \\
& \Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=\sqrt{x+1} \\
& \Rightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}\text{d}x=\int{\sqrt{x+1}\text{d}x}} \\
& \Leftrightarrow \int{\dfrac{\text{d}\left( f\left( x \right) \right)}{\sqrt{f\left( x \right)}}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+2C} \\
& \Leftrightarrow 2\sqrt{f\left( x \right)}=\dfrac{2}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+2C \\
& \Leftrightarrow \sqrt{f\left( x \right)}=\dfrac{1}{3}\sqrt{{{\left( x+1 \right)}^{3}}}+C\quad \left( 2 \right) \\
\end{aligned}$
Mà: $f\left( 3 \right)=\dfrac{4}{9}$. Thay vào (2) ta được: ${{\left( \dfrac{8}{3}+C \right)}^{2}}=\dfrac{4}{9}\Leftrightarrow \dfrac{8}{3}+C=\dfrac{2}{3}\Leftrightarrow C=-2$
Với $C=-2\Rightarrow f\left( x \right)={{\left( \dfrac{1}{3}\sqrt{{{\left( x+1 \right)}^{3}}}-2 \right)}^{2}}\Rightarrow f\left( 8 \right)=49$
Đáp án A.