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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 0 \right)=0$ và ${f}'\left( x \right)=\cos x.{{\cos }^{2}}2x,\forall x\in \mathbb{R}$. Biết ${F\left( x \right)}$ là nguyên hàm của ${f\left( x \right)}$ thỏa mãn $F\left( 0 \right)=-\dfrac{121}{225}$, khi đó $F\left( \pi \right)$ bằng
A. $\dfrac{242}{225}$.
B. $\dfrac{208}{225}$.
C. $\dfrac{121}{225}$.
D. $\dfrac{149}{225}$.
A. $\dfrac{242}{225}$.
B. $\dfrac{208}{225}$.
C. $\dfrac{121}{225}$.
D. $\dfrac{149}{225}$.
Ta có ${f}'\left( x \right)=\cos x.{{\cos }^{2}}2x,\forall x\in \mathbb{R}$ nên $f\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right)$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\cos x.{{\cos }^{2}}2x\text{d}x}=\int{\cos x.\dfrac{1+\cos 4x}{2}\text{d}x}=\int{\dfrac{\cos x}{2}\text{d}x+\int{\dfrac{\cos x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\cos x}\text{d}x+\dfrac{1}{4}\int{\left( \cos 5x+\cos 3x \right)\text{d}x=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x+C}$.
Suy ra $f\left( x \right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x+C,\forall x\in \mathbb{R}$. Mà $f\left( 0 \right)=0\Rightarrow C=0$.
Do đó $f\left( x \right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( \dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\cos x-\dfrac{1}{100}\cos 5x-\dfrac{1}{36}\cos 3x \right) \right|_{0}^{\pi }=\dfrac{242}{225} \\
& \Rightarrow F\left( \pi \right)=F\left( 0 \right)+\dfrac{242}{225}=-\dfrac{121}{225}+\dfrac{242}{225}=\dfrac{121}{225} \\
\end{aligned}$.
Có $\int{{f}'\left( x \right)\text{d}x}=\int{\cos x.{{\cos }^{2}}2x\text{d}x}=\int{\cos x.\dfrac{1+\cos 4x}{2}\text{d}x}=\int{\dfrac{\cos x}{2}\text{d}x+\int{\dfrac{\cos x.\cos 4x}{2}\text{d}x}}$
$=\dfrac{1}{2}\int{\cos x}\text{d}x+\dfrac{1}{4}\int{\left( \cos 5x+\cos 3x \right)\text{d}x=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x+C}$.
Suy ra $f\left( x \right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x+C,\forall x\in \mathbb{R}$. Mà $f\left( 0 \right)=0\Rightarrow C=0$.
Do đó $f\left( x \right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x,\forall x\in \mathbb{R}$. Khi đó:
$\begin{aligned}
& F\left( \pi \right)-F\left( 0 \right)=\int\limits_{0}^{\pi }{f\left( x \right)\text{d}x}=\int\limits_{0}^{\pi }{\left( \dfrac{1}{2}\sin x+\dfrac{1}{20}\sin 5x+\dfrac{1}{12}\sin 3x \right)\text{d}x} \\
& =\left. \left( -\dfrac{1}{2}\cos x-\dfrac{1}{100}\cos 5x-\dfrac{1}{36}\cos 3x \right) \right|_{0}^{\pi }=\dfrac{242}{225} \\
& \Rightarrow F\left( \pi \right)=F\left( 0 \right)+\dfrac{242}{225}=-\dfrac{121}{225}+\dfrac{242}{225}=\dfrac{121}{225} \\
\end{aligned}$.
Đáp án C.