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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right), G\left( x \right)$ là hai nguyên hàm của hàm số $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 1 \right)+G\left( 1 \right)=-2$ và $F\left( -1 \right)+G\left( -1 \right)=0$. Tính $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin x-2\sin 2x f\left( \cos 2x \right) \right]} \text{d}x$.
A. $2$.
B. $-2$.
C. $3$.
D. $-1$.
A. $2$.
B. $-2$.
C. $3$.
D. $-1$.
Ta có: $G\left( x \right)=F\left( x \right)+C$
$\left\{ \begin{aligned}
& F\left( 1 \right)+G\left( 1 \right)=-2 \\
& F\left( -1 \right)+G\left( -1 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(1)+C=-2 \\
& 2F(-1)+C=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ F(1)-F\left( -1 \right)=-1 \right.$.
Do đó $\int\limits_{-1}^{1}{f}\left( x \right)\text{d}x=F\left( 1 \right)-F\left( -1 \right)=-1$.
Lại có $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin x-2\sin 2x f\left( \cos 2x \right) \right]}\text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x} \text{d}x-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x} f\left( \cos 2x \right) \text{d}x$ $=1-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x} f\left( \cos 2x \right) \text{d}x$.
Mà $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x f\left( \cos 2x \right)} \text{d}x=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \cos 2x \right)} \text{d}\left( \cos 2x \right)=-\dfrac{1}{2}\int\limits_{1}^{-1}{f\left( u \right)} \text{d}u=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( u \right)} \text{d}u=-\dfrac{1}{2}$.
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin x-2\sin 2x f\left( \cos 2x \right) \right]} \text{d}x=1-2.\left( -\dfrac{1}{2} \right)=2$.
$\left\{ \begin{aligned}
& F\left( 1 \right)+G\left( 1 \right)=-2 \\
& F\left( -1 \right)+G\left( -1 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(1)+C=-2 \\
& 2F(-1)+C=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ F(1)-F\left( -1 \right)=-1 \right.$.
Do đó $\int\limits_{-1}^{1}{f}\left( x \right)\text{d}x=F\left( 1 \right)-F\left( -1 \right)=-1$.
Lại có $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin x-2\sin 2x f\left( \cos 2x \right) \right]}\text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x} \text{d}x-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x} f\left( \cos 2x \right) \text{d}x$ $=1-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x} f\left( \cos 2x \right) \text{d}x$.
Mà $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x f\left( \cos 2x \right)} \text{d}x=-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \cos 2x \right)} \text{d}\left( \cos 2x \right)=-\dfrac{1}{2}\int\limits_{1}^{-1}{f\left( u \right)} \text{d}u=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( u \right)} \text{d}u=-\dfrac{1}{2}$.
Vậy $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \sin x-2\sin 2x f\left( \cos 2x \right) \right]} \text{d}x=1-2.\left( -\dfrac{1}{2} \right)=2$.
Đáp án A.