Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ và có đồ thị như hình bên. Đặt $K=\int\limits_{0}^{1}{x.f\left( x \right).{f}'\left( x \right)dx}$, khi đó K thuộc khoảng nào sau đây?
A. $\left( -3;-2 \right)$
B. $\left( -2;-\dfrac{3}{2} \right)$
C. $\left( -\dfrac{3}{2};-\dfrac{2}{3} \right)$
D. $\left( -\dfrac{2}{3};0 \right)$
A. $\left( -3;-2 \right)$
B. $\left( -2;-\dfrac{3}{2} \right)$
C. $\left( -\dfrac{3}{2};-\dfrac{2}{3} \right)$
D. $\left( -\dfrac{2}{3};0 \right)$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=f\left( x \right).{f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{{{f}^{2}}\left( x \right)}{2} \\
\end{aligned} \right.$
Khi đó $K=\int\limits_{0}^{1}{x.f\left( x \right).{f}'\left( x \right)dx}=\dfrac{x{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\dfrac{1}{2}-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$
Từ đồ thị, ta thấy:
* $f\left( x \right)>2-x,\forall x\in \left[ 0;1 \right]\Rightarrow \int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>\int\limits_{0}^{1}{\dfrac{{{\left( 2-x \right)}^{2}}}{2}dx}=\dfrac{7}{6}\Rightarrow K=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}<-\dfrac{2}{3}$
* $f\left( x \right)<2,\forall x\in \left[ 0;1 \right]\Rightarrow \int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>\int\limits_{0}^{1}{2dx}=2\Rightarrow K=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>-\dfrac{3}{2}$.
& u=x \\
& dv=f\left( x \right).{f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{{{f}^{2}}\left( x \right)}{2} \\
\end{aligned} \right.$
Khi đó $K=\int\limits_{0}^{1}{x.f\left( x \right).{f}'\left( x \right)dx}=\dfrac{x{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\dfrac{1}{2}-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$
Từ đồ thị, ta thấy:
* $f\left( x \right)>2-x,\forall x\in \left[ 0;1 \right]\Rightarrow \int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>\int\limits_{0}^{1}{\dfrac{{{\left( 2-x \right)}^{2}}}{2}dx}=\dfrac{7}{6}\Rightarrow K=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}<-\dfrac{2}{3}$
* $f\left( x \right)<2,\forall x\in \left[ 0;1 \right]\Rightarrow \int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>\int\limits_{0}^{1}{2dx}=2\Rightarrow K=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{f}^{2}}\left( x \right)}{2}dx}>-\dfrac{3}{2}$.
Đáp án C.