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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là ${f}'(x)=4\cos 2x+\sin x,\forall x\in \mathbb{R}$ và $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ thỏa mãn $F\left( \dfrac{\pi }{4} \right)=-\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}-2$, khi đó $F\left( 0 \right)$ bằng
A. $-1$.
B. $1$.
C. $-3$.
D. $3$.
A. $-1$.
B. $1$.
C. $-3$.
D. $3$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}$ $=\int{\left( 4\cos 2x+\sin x \right)\text{d}x=2\sin 2x-\cos x+C}$.
Do $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}$ $\Leftrightarrow C=\dfrac{\pi }{2}$. Suy ra $f\left( x \right)=2\sin 2x-\cos x+\dfrac{\pi }{2}$.
Ta lại có: $\left. F\left( x \right) \right|_{0}^{\dfrac{\pi }{4}}=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( \dfrac{\pi }{4} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2\sin 2x-\cos x+\dfrac{\pi }{2} \right)\text{d}x}$
$\Leftrightarrow -\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}-2-F\left( 0 \right)=\left. \left( -\cos 2x-\sin x+\dfrac{\pi }{2}x \right) \right|_{0}^{\dfrac{\pi }{4}}$
$\Leftrightarrow -\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}-2-F\left( 0 \right)=-\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}+1$ $\Leftrightarrow F\left( 0 \right)=-3$. Vậy $F\left( 0 \right)=-3$.
Do $f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}$ $\Leftrightarrow C=\dfrac{\pi }{2}$. Suy ra $f\left( x \right)=2\sin 2x-\cos x+\dfrac{\pi }{2}$.
Ta lại có: $\left. F\left( x \right) \right|_{0}^{\dfrac{\pi }{4}}=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}$ $\Leftrightarrow F\left( \dfrac{\pi }{4} \right)-F\left( 0 \right)=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2\sin 2x-\cos x+\dfrac{\pi }{2} \right)\text{d}x}$
$\Leftrightarrow -\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}-2-F\left( 0 \right)=\left. \left( -\cos 2x-\sin x+\dfrac{\pi }{2}x \right) \right|_{0}^{\dfrac{\pi }{4}}$
$\Leftrightarrow -\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}-2-F\left( 0 \right)=-\dfrac{\sqrt{2}}{2}+\dfrac{{{\pi }^{2}}}{8}+1$ $\Leftrightarrow F\left( 0 \right)=-3$. Vậy $F\left( 0 \right)=-3$.
Đáp án C.