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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ 0 ;\dfrac{\pi }{2} \right]$ thỏa mãn:
A. 2.
B. 0.
C. $8$.
D. $16$.
Lấy tích phân từ $0$ đến $\dfrac{\pi }{2}$ hai vế của $(*)$ ta được:
$\int\limits_{0}^{\dfrac{\pi }{2}}{2\cos x.f\left( 1+4\sin x \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x.f\left( 3-2\cos 2x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 4x+4\sin 2x -4\cos x \right)dx}$
$\begin{aligned}
& \Leftrightarrow \dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 1+4\sin x \right)d(1+4\sin x)}-\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 3-2\cos 2x \right)d(3-2\cos 2x)}=0 \\
& \Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)dt}-\dfrac{1}{4}\int\limits_{1}^{5}{f\left( t \right)dt}=0\Leftrightarrow \int\limits_{1}^{5}{f\left( t \right)dt}=0\Leftrightarrow \int\limits_{1}^{5}{f\left( x \right)dx}=0 \\
\end{aligned}$
Vậy $I=\int\limits_{1}^{5}{f\left( x \right)dx}$ = 0.
$2\cos x.f\left( 1+4\sin x \right)-\sin 2x.f\left( 3-2\cos 2x \right)=\sin 4x+4\sin 2x -4\cos x$, $\forall x\in \left[ 0 ;\dfrac{\pi }{2} \right]$.
Khi đó $I=\int\limits_{1}^{5}{f\left( x \right)dx}$ bằngA. 2.
B. 0.
C. $8$.
D. $16$.
Ta có: $2\cos x.f\left( 1+4\sin x \right)-\sin 2x.f\left( 3-2\cos 2x \right)=\sin 4x+4\sin 2x -4\cos x (*)$ Lấy tích phân từ $0$ đến $\dfrac{\pi }{2}$ hai vế của $(*)$ ta được:
$\int\limits_{0}^{\dfrac{\pi }{2}}{2\cos x.f\left( 1+4\sin x \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x.f\left( 3-2\cos 2x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 4x+4\sin 2x -4\cos x \right)dx}$
$\begin{aligned}
& \Leftrightarrow \dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 1+4\sin x \right)d(1+4\sin x)}-\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 3-2\cos 2x \right)d(3-2\cos 2x)}=0 \\
& \Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)dt}-\dfrac{1}{4}\int\limits_{1}^{5}{f\left( t \right)dt}=0\Leftrightarrow \int\limits_{1}^{5}{f\left( t \right)dt}=0\Leftrightarrow \int\limits_{1}^{5}{f\left( x \right)dx}=0 \\
\end{aligned}$
Vậy $I=\int\limits_{1}^{5}{f\left( x \right)dx}$ = 0.
Đáp án B.