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Câu hỏi: Tính $T=a-3b$ biết hàm số $y=f\left( x \right)$ liên tục và có đạo hàm trên $\mathbb{R}$ thỏa mãn $3{{f}^{2}}\left( x \right).{f}'\left( x \right)-4x{{e}^{-{{f}^{3}}\left( x \right)+2{{x}^{2}}+x+1}}=1=f\left( 0 \right)$. Biết rằng $I=\int\limits_{0}^{\dfrac{-1+\sqrt{4089}}{4}}{\left( 4x+1 \right)f\left( x \right)dx}=\dfrac{a}{b}$ là phân số tối giản.
A. $T=6123$.
B. $T=12279$.
C. $T=6125$.
D. $T=12273$.
A. $T=6123$.
B. $T=12279$.
C. $T=6125$.
D. $T=12273$.
Ta có: $3{{f}^{2}}\left( x \right).{f}'\left( x \right)-4x{{e}^{-{{f}^{3}}\left( x \right)+2{{x}^{2}}+x+1}}=1=f\left( 0 \right)$
$\Leftrightarrow {{\left( {{f}^{3}}\left( x \right) \right)}^{\prime }}{{e}^{{{f}^{3}}\left( x \right)}}-{{e}^{{{f}^{3}}\left( x \right)}}=\left( 4x-1 \right).{{e}^{2{{x}^{2}}+x+1}}-{{e}^{2{{x}^{2}}+x+1}}$
$\Rightarrow {{\left[ {{f}^{3}}\left( x \right)-x \right]}^{\prime }}.{{e}^{{{f}^{2}}\left( x \right)-x}}={{\left( 2{{x}^{2}}+1 \right)}^{\prime }}.{{e}^{2{{x}^{2}}+1}}\Rightarrow {{e}^{{{f}^{3}}\left( x \right)-x}}={{e}^{2{{x}^{2}}+1}}+C$
Mà $f\left( 0 \right)=1\Rightarrow C=0\Rightarrow {{f}^{3}}\left( x \right)-x=2{{x}^{2}}+1$
$\Rightarrow {{f}^{3}}\left( x \right)=2{{x}^{2}}+x+1\Rightarrow f\left( x \right)=\sqrt[3]{2{{x}^{2}}+x+1}$
$\Rightarrow \int\limits_{0}^{\dfrac{-1+\sqrt{4089}}{4}}{\left( 4x+1 \right)f\left( x \right)dx}=\dfrac{12285}{4}\Rightarrow \left\{ \begin{aligned}
& a=12285 \\
& b=4 \\
\end{aligned} \right.\Rightarrow T=a-3b$.
$\Leftrightarrow {{\left( {{f}^{3}}\left( x \right) \right)}^{\prime }}{{e}^{{{f}^{3}}\left( x \right)}}-{{e}^{{{f}^{3}}\left( x \right)}}=\left( 4x-1 \right).{{e}^{2{{x}^{2}}+x+1}}-{{e}^{2{{x}^{2}}+x+1}}$
$\Rightarrow {{\left[ {{f}^{3}}\left( x \right)-x \right]}^{\prime }}.{{e}^{{{f}^{2}}\left( x \right)-x}}={{\left( 2{{x}^{2}}+1 \right)}^{\prime }}.{{e}^{2{{x}^{2}}+1}}\Rightarrow {{e}^{{{f}^{3}}\left( x \right)-x}}={{e}^{2{{x}^{2}}+1}}+C$
Mà $f\left( 0 \right)=1\Rightarrow C=0\Rightarrow {{f}^{3}}\left( x \right)-x=2{{x}^{2}}+1$
$\Rightarrow {{f}^{3}}\left( x \right)=2{{x}^{2}}+x+1\Rightarrow f\left( x \right)=\sqrt[3]{2{{x}^{2}}+x+1}$
$\Rightarrow \int\limits_{0}^{\dfrac{-1+\sqrt{4089}}{4}}{\left( 4x+1 \right)f\left( x \right)dx}=\dfrac{12285}{4}\Rightarrow \left\{ \begin{aligned}
& a=12285 \\
& b=4 \\
\end{aligned} \right.\Rightarrow T=a-3b$.
Đáp án D.