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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$, thỏa mãn điều kiện $\left\{ \begin{aligned}
& 2x\left[ 1+f\left( x \right) \right]={{\left[ {f}'\left( x \right) \right]}^{2}},\forall x\in \mathbb{R} \\
& f\left( 0 \right)=-1 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{1}{4}$
B. $-\dfrac{5}{6}$
C. $\dfrac{-17}{18}$
D. $-\dfrac{2}{3}$
& 2x\left[ 1+f\left( x \right) \right]={{\left[ {f}'\left( x \right) \right]}^{2}},\forall x\in \mathbb{R} \\
& f\left( 0 \right)=-1 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{1}{4}$
B. $-\dfrac{5}{6}$
C. $\dfrac{-17}{18}$
D. $-\dfrac{2}{3}$
Ta có: $2x\left[ 1+f\left( x \right) \right]={{\left[ {f}'\left( x \right) \right]}^{2}}\Rightarrow \dfrac{{{\left[ {f}'\left( x \right) \right]}^{2}}}{1+f\left( x \right)}=2x$
Với $x\in \left[ 0;1 \right]\Rightarrow \left\{ \begin{aligned}
& 2x>0 \\
& {{\left[ {f}'\left( x \right) \right]}^{2}}\ge 0 \\
\end{aligned} \right.\Rightarrow 1+f\left( x \right)>0$
Do đó $\dfrac{\pm {f}'\left( x \right)}{\sqrt{1+f\left( x \right)}}=\sqrt{2x}$ với $x\in \left[ 0;1 \right]$
Lấy nguyên hàm 2 vế ta được: $\int{\dfrac{\pm {f}'\left( x \right)}{\sqrt{1+f\left( x \right)}}dx}=\int{\sqrt{2x}dx}$
$\Leftrightarrow \pm 2\int{\dfrac{d\left[ 1+f\left( x \right) \right]}{2\sqrt{1+f\left( x \right)}}=\sqrt{2}.\dfrac{2}{3}\sqrt{{{x}^{3}}}+C\Leftrightarrow \pm \sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C}$
* TH1: Với $\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C\Rightarrow f\left( 0 \right)=-1\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{2{{x}^{3}}}{9}-1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{-17}{18}$
* TH2: Với $-\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C\Rightarrow f\left( 0 \right)=-1\Rightarrow C=0\Rightarrow -\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}$ (loại)
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{-17}{18}$.
Với $x\in \left[ 0;1 \right]\Rightarrow \left\{ \begin{aligned}
& 2x>0 \\
& {{\left[ {f}'\left( x \right) \right]}^{2}}\ge 0 \\
\end{aligned} \right.\Rightarrow 1+f\left( x \right)>0$
Do đó $\dfrac{\pm {f}'\left( x \right)}{\sqrt{1+f\left( x \right)}}=\sqrt{2x}$ với $x\in \left[ 0;1 \right]$
Lấy nguyên hàm 2 vế ta được: $\int{\dfrac{\pm {f}'\left( x \right)}{\sqrt{1+f\left( x \right)}}dx}=\int{\sqrt{2x}dx}$
$\Leftrightarrow \pm 2\int{\dfrac{d\left[ 1+f\left( x \right) \right]}{2\sqrt{1+f\left( x \right)}}=\sqrt{2}.\dfrac{2}{3}\sqrt{{{x}^{3}}}+C\Leftrightarrow \pm \sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C}$
* TH1: Với $\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C\Rightarrow f\left( 0 \right)=-1\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{2{{x}^{3}}}{9}-1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{-17}{18}$
* TH2: Với $-\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}+C\Rightarrow f\left( 0 \right)=-1\Rightarrow C=0\Rightarrow -\sqrt{1+f\left( x \right)}=\dfrac{\sqrt{2{{x}^{3}}}}{3}$ (loại)
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{-17}{18}$.
Đáp án C.