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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$. Biết $f\left( x \right)>-1,\forall x\in \mathbb{R},f\left( 0 \right)=0$ và thoả mãn ${f}'\left( x \right)\sqrt{{{x}^{2}}+1}=2x\sqrt{f\left( x \right)+1}$. Khi đó, $\int\limits_{0}^{\sqrt{3}}{{f}'\left( x \right) }\text{d}x$ bằng
A. 0.
B. 3.
C. 9.
D. 5.
A. 0.
B. 3.
C. 9.
D. 5.
Ta có: ${f}'\left( x \right)\sqrt{{{x}^{2}}+1}=2x\sqrt{f\left( x \right)+1}$
$\Leftrightarrow \dfrac{{f}'\left( x \right)}{2\sqrt{f\left( x \right)+1}}=\dfrac{x}{\sqrt{{{x}^{2}}+1}}\Leftrightarrow {{\left[ \sqrt{f\left( x \right)+1} \right]}^{\prime }}={{\left( \sqrt{{{x}^{2}}+1} \right)}^{\prime }}$
Suy ra $\sqrt{f\left( x \right)+1}=\sqrt{{{x}^{2}}+1}+C$. Do $f\left( 0 \right)=0$ nên $C=0$.
Do đó, $f\left( x \right)={{x}^{2}}$. Khi đó, $\int\limits_{0}^{\sqrt{3}}{{f}'\left( x \right) }\text{d}x=\int\limits_{0}^{\sqrt{3}}{2x }\text{d}x=3$.
$\Leftrightarrow \dfrac{{f}'\left( x \right)}{2\sqrt{f\left( x \right)+1}}=\dfrac{x}{\sqrt{{{x}^{2}}+1}}\Leftrightarrow {{\left[ \sqrt{f\left( x \right)+1} \right]}^{\prime }}={{\left( \sqrt{{{x}^{2}}+1} \right)}^{\prime }}$
Suy ra $\sqrt{f\left( x \right)+1}=\sqrt{{{x}^{2}}+1}+C$. Do $f\left( 0 \right)=0$ nên $C=0$.
Do đó, $f\left( x \right)={{x}^{2}}$. Khi đó, $\int\limits_{0}^{\sqrt{3}}{{f}'\left( x \right) }\text{d}x=\int\limits_{0}^{\sqrt{3}}{2x }\text{d}x=3$.
Đáp án B.