Google
Câu hỏi: Cho $f\left( x \right)$ thỏa mãn $f'\left( x \right)=x.\cos 2x,\forall x\in \mathbb{R}$ và $f\left( 0 \right)=\dfrac{1}{4}$. Hàm số $f\left( x \right)$ là
A. $\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
B. $\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+\dfrac{1}{4}$.
C. $-\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
D. $-\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+\dfrac{1}{4}$.
A. $\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
B. $\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+\dfrac{1}{4}$.
C. $-\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
D. $-\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+\dfrac{1}{4}$.
$f\left( x \right)=\int{f'\left( x \right)\text{d}x=\int{x.\cos 2x.\text{d}x}}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v=\cos 2x.\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=\dfrac{1}{2}\sin 2x \\
\end{aligned} \right.$
$f\left( x \right)=x.\dfrac{1}{2}\sin 2x-\int{\dfrac{1}{2}\sin 2x.\text{d}x}$ $\Leftrightarrow f\left( x \right)=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+C$.
$f\left( 0 \right)=\dfrac{1}{4}$ $\Rightarrow C=0$.
Vậy $f\left( x \right)=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v=\cos 2x.\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=\dfrac{1}{2}\sin 2x \\
\end{aligned} \right.$
$f\left( x \right)=x.\dfrac{1}{2}\sin 2x-\int{\dfrac{1}{2}\sin 2x.\text{d}x}$ $\Leftrightarrow f\left( x \right)=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+C$.
$f\left( 0 \right)=\dfrac{1}{4}$ $\Rightarrow C=0$.
Vậy $f\left( x \right)=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x$.
Đáp án A.