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Câu hỏi: Hàm số $f\left( x \right)$ có đạo hàm đến cấp hai trên $\mathbb{R}$ thoả mãn: ${{f}^{2}}\left( 1-x \right)=\left( {{x}^{2}}+3 \right)f\left( x+1 \right)$. Biết rằng $f\left( x \right)\ne 0,\forall x\in \mathbb{R}$, tính $I=\int\limits_{0}^{2}{\left( 2x-1 \right){f}''\left( x \right)dx}$.
A. 8.
B. 0.
C. -4.
D. 4
A. 8.
B. 0.
C. -4.
D. 4
Ta có: $\left\{ \begin{aligned}
& {{f}^{2}}\left( 1-x \right)=\left( {{x}^{2}}+3 \right).f\left( x+1 \right)\Rightarrow {{f}^{4}}\left( 1-x \right)={{\left( {{x}^{2}}+3 \right)}^{2}}.{{f}^{2}}\left( x+1 \right)\text{ }\left( 1 \right) \\
& {{f}^{2}}\left( 1+x \right)=\left( {{x}^{2}}+3 \right).f\left( 1-x \right)\text{ }\left( 2 \right) \\
\end{aligned} \right.$
Từ (1) và (2) $\Rightarrow f\left( 1-x \right)={{x}^{2}}+3={{\left( 1-x-1 \right)}^{2}}+3\Rightarrow f\left( x \right)={{\left( x-1 \right)}^{2}}+3\Rightarrow {f}''\left( x \right)=2$
$\Rightarrow I=\int\limits_{0}^{2}{\left( 4x-2 \right)dx}=\left. \left( 2{{x}^{2}}-2x \right) \right|_{0}^{2}=4$.
& {{f}^{2}}\left( 1-x \right)=\left( {{x}^{2}}+3 \right).f\left( x+1 \right)\Rightarrow {{f}^{4}}\left( 1-x \right)={{\left( {{x}^{2}}+3 \right)}^{2}}.{{f}^{2}}\left( x+1 \right)\text{ }\left( 1 \right) \\
& {{f}^{2}}\left( 1+x \right)=\left( {{x}^{2}}+3 \right).f\left( 1-x \right)\text{ }\left( 2 \right) \\
\end{aligned} \right.$
Từ (1) và (2) $\Rightarrow f\left( 1-x \right)={{x}^{2}}+3={{\left( 1-x-1 \right)}^{2}}+3\Rightarrow f\left( x \right)={{\left( x-1 \right)}^{2}}+3\Rightarrow {f}''\left( x \right)=2$
$\Rightarrow I=\int\limits_{0}^{2}{\left( 4x-2 \right)dx}=\left. \left( 2{{x}^{2}}-2x \right) \right|_{0}^{2}=4$.
Đáp án D.