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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 1-x \right)+{{x}^{2}}{{f}'}'\left( x \right)=5{{x}^{3}}+3{{x}^{2}}-3x$ với mọi $x\in \mathbb{R}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ bằng
A. $-\dfrac{3}{2}$.
B. $\dfrac{3}{2}$.
C. $-\dfrac{3}{4}$.
D. $\dfrac{3}{4}$.
A. $-\dfrac{3}{2}$.
B. $\dfrac{3}{2}$.
C. $-\dfrac{3}{4}$.
D. $\dfrac{3}{4}$.
Nhận xét: $f\left( x \right)$ là hàm số bậc $3$.
Giả sử $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ $\left( a\ne 0 \right)$, ta có ${f}'\left( x \right)=3a{{x}^{2}}+2bx+c$ ; ${{f}'}'\left( x \right)=6ax+2b$.
Mặt khác $f\left( 1-x \right)=-a{{x}^{3}}+\left( 3a+b \right){{x}^{2}}+\left( -3a-2b-c \right)x+\left( a+b+c+d \right)$ suy ra
$VT=5a{{x}^{3}}+3\left( a+b \right){{x}^{2}}-\left( 3a+2b+c \right)x+\left( a+b+c+d \right)=VP$ với mọi $x\in \mathbb{R}$
$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& a+b=1 \\
& 3a+2b+c=3 \\
& a+b+c+d=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=c=0 \\
& d=-1 \\
\end{aligned} \right. $. Do đó $ f\left( x \right)={{x}^{3}}-1$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{3}}-1 \right)\text{d}x}=-\dfrac{3}{4}$.
Giả sử $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ $\left( a\ne 0 \right)$, ta có ${f}'\left( x \right)=3a{{x}^{2}}+2bx+c$ ; ${{f}'}'\left( x \right)=6ax+2b$.
Mặt khác $f\left( 1-x \right)=-a{{x}^{3}}+\left( 3a+b \right){{x}^{2}}+\left( -3a-2b-c \right)x+\left( a+b+c+d \right)$ suy ra
$VT=5a{{x}^{3}}+3\left( a+b \right){{x}^{2}}-\left( 3a+2b+c \right)x+\left( a+b+c+d \right)=VP$ với mọi $x\in \mathbb{R}$
$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& a+b=1 \\
& 3a+2b+c=3 \\
& a+b+c+d=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=c=0 \\
& d=-1 \\
\end{aligned} \right. $. Do đó $ f\left( x \right)={{x}^{3}}-1$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{3}}-1 \right)\text{d}x}=-\dfrac{3}{4}$.
Đáp án C.