Giả sử hàm số y=f(x) liên tục, nhận giá trị

* Hướng dẫn giải

Đáp án A

Xét $x\in \left(0;+\infty \right)$ và f(x)>0 ta có: $f\left(x\right)=f\text{'}\left(x\right).\sqrt{3\text{x}+1}\iff \frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{\sqrt{3\text{x}+1}}$

$\Rightarrow \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\int \frac{1}{\sqrt{3\text{x}+1}}dx=\int \frac{1}{f\left(x\right)}d\left(f\left(x\right)\right)=\frac{2}{3}\int \frac{1}{2\sqrt{3\text{x}+1}}d\left(3\text{x}+1\right)$

$\Rightarrow \ln \left(f\left(x\right)\right)=\frac{2}{3}\sqrt{3\text{x}+1}+C\Rightarrow f\left(x\right)=e^{\frac{2}{3}\sqrt{3\text{x}+1}+C}$

Theo bài ra ta có: $f\left(1\right)=e$ nên $e^{\frac{4}{3}+C}=e\Rightarrow C=-\frac{1}{3}\Rightarrow f\left(x\right)=e^{\frac{2}{3}\sqrt{3\text{x}+1}-\frac{1}{3}}$

Do đó $f\left(5\right)\approx 10,3123\Rightarrow 10<f\left(5\right)<11$