Google
Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn hai điều kiện ${{\left[ f\left( x \right) \right]}^{2}}+3{{x}^{2}}+2x-1\le 4x.f\left( x \right)$ ; $\forall x\in \mathbb{R}$ và $\int\limits_{-1}^{3}{f\left( x \right)\text{d}x=12}$. Giá trị $\int\limits_{0}^{2}{f\left( x \right)\text{d}x}$ bằng
A. $6$.
B. $8$.
C. $7$.
D. $5$.
A. $6$.
B. $8$.
C. $7$.
D. $5$.
Ta có ${{f}^{2}}\left( x \right)+3{{x}^{2}}+2x-1\le 4x.f\left( x \right)$ $\Leftrightarrow \left[ f\left( x \right)-\left( x+1 \right) \right]\left[ f\left( x \right)-\left( 3x-1 \right) \right]\le 0$.
Nếu $x\ge 1$ thì $x+1\le f\left( x \right)\le 3x-1$
$\Rightarrow \int\limits_{1}^{3}{\left( x+1 \right)\text{d}x}\le \int\limits_{1}^{3}{f\left( x \right)\text{d}x}\le \int\limits_{1}^{3}{\left( 3x-1 \right)\text{d}x}$ $\Leftrightarrow 6\le \int\limits_{1}^{3}{f\left( x \right)\text{d}x}\le 10$ $\left( * \right)$.
Nếu $x\le 1$ thì $3x-1\le f\left( x \right)\le x+1$
$\Rightarrow \int\limits_{-1}^{1}{\left( 3x-1 \right)\text{d}x}\le \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}\le \int\limits_{-1}^{1}{\left( x+1 \right)\text{d}x}$ $\Leftrightarrow -2\le \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}\le 2$ $\left( ** \right)$.
Từ $\left( * \right)$ và $\left( ** \right)$ ta có $\Leftrightarrow 8\le \int\limits_{-1}^{3}{f\left( x \right)\text{d}x}\le 12$ mà $\int\limits_{-1}^{3}{f\left( x \right)\text{d}x=12}$.
$\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& 3x-1 khi x\ge 1 \\
& x+1 khi x\le 1 \\
\end{aligned} \right.$.
Vậy $\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=5$.
Nếu $x\ge 1$ thì $x+1\le f\left( x \right)\le 3x-1$
$\Rightarrow \int\limits_{1}^{3}{\left( x+1 \right)\text{d}x}\le \int\limits_{1}^{3}{f\left( x \right)\text{d}x}\le \int\limits_{1}^{3}{\left( 3x-1 \right)\text{d}x}$ $\Leftrightarrow 6\le \int\limits_{1}^{3}{f\left( x \right)\text{d}x}\le 10$ $\left( * \right)$.
Nếu $x\le 1$ thì $3x-1\le f\left( x \right)\le x+1$
$\Rightarrow \int\limits_{-1}^{1}{\left( 3x-1 \right)\text{d}x}\le \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}\le \int\limits_{-1}^{1}{\left( x+1 \right)\text{d}x}$ $\Leftrightarrow -2\le \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}\le 2$ $\left( ** \right)$.
Từ $\left( * \right)$ và $\left( ** \right)$ ta có $\Leftrightarrow 8\le \int\limits_{-1}^{3}{f\left( x \right)\text{d}x}\le 12$ mà $\int\limits_{-1}^{3}{f\left( x \right)\text{d}x=12}$.
$\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& 3x-1 khi x\ge 1 \\
& x+1 khi x\le 1 \\
\end{aligned} \right.$.
Vậy $\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=5$.
Đáp án D.