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Câu hỏi: Cho hàm số $f\left( x \right)=3{{x}^{2}}+\sin x-\cos 2x$. Nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)$ thỏa mãn $F\left( 0 \right)=2$ là
A. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+2$.
B. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+3$.
C. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x-3$.
D. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x-2$.
A. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+2$.
B. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+3$.
C. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x-3$.
D. $F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x-2$.
Ta có: $F\left( x \right)=\int{\left( 3{{x}^{2}}+\sin x-\cos 2x \right)\text{d}x}={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+C.$
$F\left( 0 \right)=2\Leftrightarrow {{0}^{3}}-1-\dfrac{1}{2}.0+C=2\Leftrightarrow C=3\Rightarrow F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+3.$.
$F\left( 0 \right)=2\Leftrightarrow {{0}^{3}}-1-\dfrac{1}{2}.0+C=2\Leftrightarrow C=3\Rightarrow F\left( x \right)={{x}^{3}}-\cos x-\dfrac{1}{2}\sin 2x+3.$.
Đáp án B.