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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 $f\left( 1 \right)=1$, $f\left( x \right)={f}'\left( x \right).\sqrt{3x+1}$ với mọi $x>0$. Mệnh đề nào sau đây đúng?
A. $3<f\left( 5 \right)<4$.
B. $1<f\left( 5 \right)<2$
C. $4<f\left( 5 \right)<5$.
D. $2<f\left( 5 \right)<3$.
A. $3<f\left( 5 \right)<4$.
B. $1<f\left( 5 \right)<2$
C. $4<f\left( 5 \right)<5$.
D. $2<f\left( 5 \right)<3$.
Từ $f\left( x \right)={f}'\left( x \right).\sqrt{3x+1}$ ta có $\dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{3x+1}}$
Suy ra: $\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{1}{\sqrt{3x+1}}dx}\Rightarrow \ln f\left( x \right)=\dfrac{2}{3}\sqrt{3x+1}+C$.
Ta có $\ln f\left( 1 \right)=\dfrac{2}{3}\sqrt{3.1+1}+C\Leftrightarrow \ln 1=\dfrac{4}{3}+C\Leftrightarrow C=-\dfrac{4}{3}$.
Nên $\ln f\left( x \right)=\dfrac{2}{3}\sqrt{3x+1}-\dfrac{4}{3}\Leftrightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}-\dfrac{4}{3}}}$.
Vậy $f\left( 5 \right)={{e}^{\dfrac{2}{3}\sqrt{3.5+1}-\dfrac{4}{3}}}={{e}^{\dfrac{4}{3}}}\in \left( 3;4 \right)$.
Suy ra: $\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx}=\int{\dfrac{1}{\sqrt{3x+1}}dx}\Rightarrow \ln f\left( x \right)=\dfrac{2}{3}\sqrt{3x+1}+C$.
Ta có $\ln f\left( 1 \right)=\dfrac{2}{3}\sqrt{3.1+1}+C\Leftrightarrow \ln 1=\dfrac{4}{3}+C\Leftrightarrow C=-\dfrac{4}{3}$.
Nên $\ln f\left( x \right)=\dfrac{2}{3}\sqrt{3x+1}-\dfrac{4}{3}\Leftrightarrow f\left( x \right)={{e}^{\dfrac{2}{3}\sqrt{3x+1}-\dfrac{4}{3}}}$.
Vậy $f\left( 5 \right)={{e}^{\dfrac{2}{3}\sqrt{3.5+1}-\dfrac{4}{3}}}={{e}^{\dfrac{4}{3}}}\in \left( 3;4 \right)$.
Đáp án A.