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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 $f\left( 1 \right)=e,f\left( x \right)=f'\left( x \right).\sqrt{3\text{x}+1},$ 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: $f\left( x \right)=f'\left( x \right).\sqrt{3\text{x}+1}\Leftrightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{3\text{x}+1}}$
$\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}}dx=\int{\dfrac{1}{\sqrt{3\text{x}+1}}}dx=\int{\dfrac{1}{f\left( x \right)}}d\left( f\left( x \right) \right)=\dfrac{2}{3}\int{\dfrac{1}{2\sqrt{3\text{x}+1}}}d\left( 3\text{x}+1 \right)$
$\Rightarrow \ln \left( f\left( x \right) \right)=\dfrac{2}{3}\sqrt{3\text{x}+1}+C\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3\text{x}+1}+C}}$
Theo bài ra ta có: $f\left( 1 \right)=e$ nên ${{e}^{\dfrac{4}{3}+C}}=e\Rightarrow C=-\dfrac{1}{3}\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3\text{x}+1}-\dfrac{1}{3}}}$
Do đó $f\left( 5 \right)\approx 10,3123\Rightarrow 10<f\left( 5 \right)<11$
$\Rightarrow \int{\dfrac{f'\left( x \right)}{f\left( x \right)}}dx=\int{\dfrac{1}{\sqrt{3\text{x}+1}}}dx=\int{\dfrac{1}{f\left( x \right)}}d\left( f\left( x \right) \right)=\dfrac{2}{3}\int{\dfrac{1}{2\sqrt{3\text{x}+1}}}d\left( 3\text{x}+1 \right)$
$\Rightarrow \ln \left( f\left( x \right) \right)=\dfrac{2}{3}\sqrt{3\text{x}+1}+C\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3\text{x}+1}+C}}$
Theo bài ra ta có: $f\left( 1 \right)=e$ nên ${{e}^{\dfrac{4}{3}+C}}=e\Rightarrow C=-\dfrac{1}{3}\Rightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3\text{x}+1}-\dfrac{1}{3}}}$
Do đó $f\left( 5 \right)\approx 10,3123\Rightarrow 10<f\left( 5 \right)<11$
Đáp án A.