Hàm số f(x) = 2sin2x - 3 đạt cực tiểu tại: A. x = pi/4 + kpi B. x = pi/4 + kpi/2

TXĐ:  $D=R.$

$f^{\text{'}}\left(x\right)=4\cos 2x,f^{\text{'}}\left(x\right)=0\iff \cos 2x=0\iff 2x=\frac{\pi }{2}+k\pi \iff x=\frac{\pi }{4}+\frac{k\pi }{2},k\in Z$

$f^{\text{'}\text{'}}\left(x\right)=-8\sin 2x$

Ta có: $f^{\text{'}\text{'}}\left(\frac{\pi }{4}+\frac{k\pi }{2}\right)=-8\sin \left(\frac{\pi }{2}+k\pi \right),k\in Z$

Khi $k=2n$ thì $\sin \left(\frac{\pi }{2}+2n\pi \right)=\sin \frac{\pi }{2}=1$ nên $f^{\text{'}\text{'}}\left(\frac{\pi }{4}+\frac{2n\pi }{2}\right)=-8<0$

Khi $k=2n+1$ thì $\sin \left(\frac{\pi }{2}+\left(2n+1\right)\pi \right)=\sin \frac{3\pi }{2}=-1$ nên $f^{\text{'}\text{'}}\left(\frac{\pi }{4}+\frac{\left(2n+1\right)\pi }{2}\right)=8>0$