Đáp án:
$C. 1$
Giải thích các bước giải:
$ y=3\sin x-4\sin^3x\\ y'=3\cos x-12\sin^2 x.(\sin x)'\\ =3\cos x-12\sin^2 x.\cos x\\ =3\cos x(1-4\sin^2 x)\\ y'=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=0\\ 1-4\sin^2 x=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x=\dfrac{\pi}{2}+k\pi &(k\in \mathbb{Z}\\ \sin x=\pm\dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x=\dfrac{\pi}{2}+k\pi &(k\in \mathbb{Z})\\ x=-\dfrac{\pi}{6}+l2\pi&(l\in \mathbb{Z})\\ x=\dfrac{\pi}{6}+m2\pi&(m\in \mathbb{Z})\end{array} \right.\\ x \in \left(-\dfrac{\pi}{2};\dfrac{\pi}{2}\right)\\ \Rightarrow x \in \left\{-\dfrac{\pi}{6};\dfrac{\pi}{6}\right\}\\ \\ f\left(-\dfrac{\pi}{6}\right)=-1\\ f\left(\dfrac{\pi}{6}\right)=1\\\displaystyle\lim_{x \to -\tfrac{\pi}{2}} f(x)=f\left(-\dfrac{\pi}{2}\right)=1\\\displaystyle\lim_{x \to \tfrac{\pi}{2}} f(x)=f\left(\dfrac{\pi}{2}\right)=-1\\ BBT:\\ \begin{array}{|c|ccccccccc|} \hline x&-\dfrac{\pi}{2}&&-\dfrac{\pi}{6}&&\dfrac{\pi}{6}&&\dfrac{\pi}{2}\\\hline y'&&-&0&+&0&-&\\\hline &1&&&&1\\y&&\searrow&&\nearrow&&\searrow\\&&&-1&&&&-1\\\hline\end{array} \\ \Rightarrow \underset{\left(-\dfrac{\pi}{2};\dfrac{\pi}{2}\right)}{max f(x) }=1$
$\Rightarrow C$
