Đáp án:
321
Giải thích các bước giải:
$$\eqalign{
& y = f\left( x \right) = \sin 2x \cr
& y' = 2\cos 2x = 0 \Leftrightarrow 2x = {\pi \over 2} + k\pi \cr
& \Leftrightarrow x = {\pi \over 4} + {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \Leftrightarrow \left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = - {\pi \over 4} + k\pi \hfill \cr} \right. \cr
& y'' = - 4\sin 2x \cr
& y''\left( {{\pi \over 4} + k\pi } \right) = - 4\sin \left( {{\pi \over 2} + k2\pi } \right) = - 4 < 0 \cr
& y''\left( { - {\pi \over 4} + k\pi } \right) = - 4\sin \left( { - {\pi \over 2} + k2\pi } \right) = 4 > 0 \cr
& \Rightarrow Ham\,\,so\,\,co\,\,{x_{CT}} = - {\pi \over 2} + k2\pi \,\,\left( {k \in Z} \right) \cr
& 0 < - {\pi \over 2} + k2\pi < 2018 \cr
& \Leftrightarrow {1 \over 4} < k < 321,42 \cr
& k \in Z \Rightarrow k \in \left\{ {1;2;3;...;321} \right\} \cr
& Vay\,\,ham\,\,so\,\,co\,\,321\,\,cuc\,\,tieu\,\,tm\,\,ycbt. \cr} $$
