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Câu hỏi: Cho hàm số $f\left( x \right)>0,x\in \mathbb{R},f\left( 0 \right)=1$ và $f\left( x \right)={f}'\left( x \right).\sqrt{x+1},\forall x\in \mathbb{R}$. Mệnh đề nào dưới đây đúng?
A. $f\left( 3 \right)<2$
B. $2<f\left( 3 \right)<4$
C. $4<f\left( 3 \right)<6$
D. $f\left( 3 \right)>f\left( 6 \right)$
A. $f\left( 3 \right)<2$
B. $2<f\left( 3 \right)<4$
C. $4<f\left( 3 \right)<6$
D. $f\left( 3 \right)>f\left( 6 \right)$
Ta có $f\left( x \right)={f}'\left( x \right)\sqrt{x+1}\Leftrightarrow \dfrac{{f}'\left( x \right)}{f\left( x \right)}=\dfrac{1}{\sqrt{x+1}}\Leftrightarrow \int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx=\int{\dfrac{dx}{\sqrt{x+1}}}}$
$\Leftrightarrow \ln \left| f\left( x \right) \right|=2\sqrt{x+1}+C\Leftrightarrow f\left( x \right)={{e}^{2\sqrt{x+1}+C}}$ mà $f\left( 0 \right)=1\Rightarrow C=-2$
Vậy $f\left( x \right)={{e}^{2\sqrt{x+1}-2}}\xrightarrow{{}}f\left( 3 \right)={{e}^{2}}\approx 7,4.$
$\Leftrightarrow \ln \left| f\left( x \right) \right|=2\sqrt{x+1}+C\Leftrightarrow f\left( x \right)={{e}^{2\sqrt{x+1}+C}}$ mà $f\left( 0 \right)=1\Rightarrow C=-2$
Vậy $f\left( x \right)={{e}^{2\sqrt{x+1}-2}}\xrightarrow{{}}f\left( 3 \right)={{e}^{2}}\approx 7,4.$
Đáp án D.