Google
Câu hỏi: Cho $F\left( x \right)=\int{\cos 2x\text{d}x}$,biết rằng $F\left( \dfrac{\pi }{4} \right)=3$. Mệnh đề nào dưới đây đúng?
A. $F\left( \dfrac{\pi }{12} \right)\in \left( 0;2 \right)$
B. $F\left( \dfrac{\pi }{12} \right)\in \left( 2;3 \right)$
C. $F\left( \dfrac{\pi }{12} \right)\in \left( 3;4 \right)$ $$
D. $F\left( \dfrac{\pi }{12} \right)\in \left( -2;0 \right)$
A. $F\left( \dfrac{\pi }{12} \right)\in \left( 0;2 \right)$
B. $F\left( \dfrac{\pi }{12} \right)\in \left( 2;3 \right)$
C. $F\left( \dfrac{\pi }{12} \right)\in \left( 3;4 \right)$ $$
D. $F\left( \dfrac{\pi }{12} \right)\in \left( -2;0 \right)$
$F\left( x \right)=\int{\cos 2x\text{d}x}=\dfrac{1}{2}\sin 2x+C$
Mà $F\left( \dfrac{\pi }{4} \right)=3\Leftrightarrow \dfrac{1}{2}\sin \dfrac{\pi }{2}+C=3\Leftrightarrow C=\dfrac{5}{2}$
Vậy $F\left( \dfrac{\pi }{12} \right)=\dfrac{1}{2}\sin \dfrac{\pi }{6}+\dfrac{5}{2}=\dfrac{11}{4}\in \left( 2;3 \right)$.
Mà $F\left( \dfrac{\pi }{4} \right)=3\Leftrightarrow \dfrac{1}{2}\sin \dfrac{\pi }{2}+C=3\Leftrightarrow C=\dfrac{5}{2}$
Vậy $F\left( \dfrac{\pi }{12} \right)=\dfrac{1}{2}\sin \dfrac{\pi }{6}+\dfrac{5}{2}=\dfrac{11}{4}\in \left( 2;3 \right)$.
Đáp án B.