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Câu hỏi: Giả sử hàm số $y=f\left( x \right)$ liên tục, nhận giá trị dương trên $\left( 0;+\infty \right)$ và thỏa mãn $f\left( 1 \right)=e,$ $f\left( x \right)={f}'\left( x \right).\sqrt{3x+1},$ với mọi $x>0$. Mệnh đề nào sau đây là đúng?
A. $10<f\left( 5 \right)<11$.
B. $4<f\left( 5 \right)<5$.
C. $11<f\left( 5 \right)<12$.
D. $3<f\left( 5 \right)<4$.
A. $10<f\left( 5 \right)<11$.
B. $4<f\left( 5 \right)<5$.
C. $11<f\left( 5 \right)<12$.
D. $3<f\left( 5 \right)<4$.
Xét $x\in \left( 0;+\infty \right)$ và $f\left( x \right)>0$ ta có: $f\left( x \right)={f}'\left( x \right).\sqrt{3x+1}\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{3x+1}}.$
$\Rightarrow \int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{1}{\sqrt{3x+1}}dx}\Leftrightarrow \int{\dfrac{1}{f\left( x \right)}d\left( f\left( x \right) \right)}=\dfrac{2}{3}\int{\dfrac{1}{2\sqrt{3x+1}}d\left( 3x+1 \right)}$
$\Rightarrow \ln \left( f\left( x \right) \right)=\dfrac{2}{3}\sqrt{3x+1}+C\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}+C}}$
Theo bài $f\left( 1 \right)=e$ nên ${{e}^{\dfrac{4}{3}+C}}=e\Rightarrow C=-\dfrac{1}{3}\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}-\dfrac{1}{3}}}$
Do đó $f\left( 5 \right)\approx 10,3123\Rightarrow 10<f\left( 5 \right)<11.$
$\Rightarrow \int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{1}{\sqrt{3x+1}}dx}\Leftrightarrow \int{\dfrac{1}{f\left( x \right)}d\left( f\left( x \right) \right)}=\dfrac{2}{3}\int{\dfrac{1}{2\sqrt{3x+1}}d\left( 3x+1 \right)}$
$\Rightarrow \ln \left( f\left( x \right) \right)=\dfrac{2}{3}\sqrt{3x+1}+C\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}+C}}$
Theo bài $f\left( 1 \right)=e$ nên ${{e}^{\dfrac{4}{3}+C}}=e\Rightarrow C=-\dfrac{1}{3}\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}-\dfrac{1}{3}}}$
Do đó $f\left( 5 \right)\approx 10,3123\Rightarrow 10<f\left( 5 \right)<11.$
Đáp án A.