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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm xác định trên $\mathbb{R}$ thỏa mãn $f\left( 0 \right)=2\sqrt{2}$, $f\left( x \right)>0$ và $f\left( x \right).{f}'\left( x \right)=\left( 2x+1 \right)\sqrt{1+{{f}^{2}}\left( x \right)} ,\ \forall x\in \mathbb{R}$. Giá trị $f\left( 2 \right)$ là
A. $5\sqrt{4}$.
B. $4\sqrt{5}$.
C. $3\sqrt{5}$.
D. 9.
A. $5\sqrt{4}$.
B. $4\sqrt{5}$.
C. $3\sqrt{5}$.
D. 9.
Ta có $f\left( x \right).{f}'\left( x \right)=\left( 2x+1 \right)\sqrt{1+{{f}^{2}}\left( x \right)} ,\ \forall x\in \mathbb{R}$.
$\Leftrightarrow \dfrac{f\left( x \right).{f}'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}=2x+1 ,\ \forall x\in \mathbb{R}$ $\Leftrightarrow \dfrac{2f\left( x \right).{f}'\left( x \right)}{2\sqrt{1+{{f}^{2}}\left( x \right)}}=2x+1 ,\ \forall x\in \mathbb{R}$
$\Rightarrow \int{\dfrac{2f\left( x \right).{f}'\left( x \right)}{2\sqrt{1+{{f}^{2}}\left( x \right)}}\text{d}x}=\int{\left( 2x+1 \right)\text{d}x}$ $\Rightarrow \sqrt{1+{{f}^{2}}\left( x \right)}={{x}^{2}}+x+C$.
Cho $x=0$ ta được: $C=\sqrt{1+{{f}^{2}}\left( 0 \right)}=\sqrt{1+{{\left( 2\sqrt{2} \right)}^{2}}}=3$.
Do đó $\sqrt{1+{{f}^{2}}\left( x \right)}={{x}^{2}}+x+3$.
Lại cho $x=2$ ta được: $\sqrt{1+{{f}^{2}}\left( 2 \right)}=4+2+3=9$ $\Rightarrow 1+{{f}^{2}}\left( 2 \right)=81$ $\Rightarrow {{f}^{2}}\left( 2 \right)=80$
$\Rightarrow f\left( 2 \right)=4\sqrt{5}$ (do $f\left( x \right)>0$ ).
Vậy $f\left( 2 \right)=4\sqrt{5}$.
$\Leftrightarrow \dfrac{f\left( x \right).{f}'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}=2x+1 ,\ \forall x\in \mathbb{R}$ $\Leftrightarrow \dfrac{2f\left( x \right).{f}'\left( x \right)}{2\sqrt{1+{{f}^{2}}\left( x \right)}}=2x+1 ,\ \forall x\in \mathbb{R}$
$\Rightarrow \int{\dfrac{2f\left( x \right).{f}'\left( x \right)}{2\sqrt{1+{{f}^{2}}\left( x \right)}}\text{d}x}=\int{\left( 2x+1 \right)\text{d}x}$ $\Rightarrow \sqrt{1+{{f}^{2}}\left( x \right)}={{x}^{2}}+x+C$.
Cho $x=0$ ta được: $C=\sqrt{1+{{f}^{2}}\left( 0 \right)}=\sqrt{1+{{\left( 2\sqrt{2} \right)}^{2}}}=3$.
Do đó $\sqrt{1+{{f}^{2}}\left( x \right)}={{x}^{2}}+x+3$.
Lại cho $x=2$ ta được: $\sqrt{1+{{f}^{2}}\left( 2 \right)}=4+2+3=9$ $\Rightarrow 1+{{f}^{2}}\left( 2 \right)=81$ $\Rightarrow {{f}^{2}}\left( 2 \right)=80$
$\Rightarrow f\left( 2 \right)=4\sqrt{5}$ (do $f\left( x \right)>0$ ).
Vậy $f\left( 2 \right)=4\sqrt{5}$.
Đáp án B.