Đáp án:
$2 \, A.\, 4$
$3 \, B.\, -\dfrac{13}{4}$
Giải thích các bước giải:
Câu 2:
$\sin4x - \sqrt3\cos4x = 2\sin x$
$\Leftrightarrow \dfrac{1}{2}\sin4x - \dfrac{\sqrt3}{2}\cos4x = \sin x$
$\Leftrightarrow \sin\left(4x -\dfrac{\pi}{3}\right) = \sin x$
$\Leftrightarrow \left[\begin{array}{l}4x - \dfrac{\pi}{3} = x + k2\pi\\4x - \dfrac{\pi}{3} = \pi - x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{3}+ k2\pi\\5x = \dfrac{4\pi}{3}+ k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{9}+ k\dfrac{2\pi}{3}\\x = \dfrac{4\pi}{15}+ k\dfrac{2\pi}{5}\end{array}\right. \quad (k\in\Bbb Z)$
Ta có:
$0 \leq x \leq \pi$
$\Leftrightarrow \left[\begin{array}{l}0 \leq \dfrac{\pi}{9}+ k\dfrac{2\pi}{3}\leq \pi \\0 \leq \dfrac{4\pi}{15}+ k\dfrac{2\pi}{5}\leq \pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}-\dfrac{1}{6} k\leq \dfrac{4}{3}\\-\dfrac{2}{3} \leq k\leq \dfrac{11}{6}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}k = 0;1\\k = 0;1\end{array}\right.$
$4$ giá trị $k$ tương ứng $4$ nghiệm
Câu 3:
$y = (\sin x -\cos x)^2 + 2\cos2x + 3\sin x\cos x$
$\to y = 1 - \sin2x + 2\cos2x + \dfrac{3}{2}\sin2x$
$\to y - 1 = \dfrac{1}{2}\sin2x + 2\cos2x$
Phương trình có nghiệm
$\Leftrightarrow \dfrac{1}{4} + 4 \geq (y - 1)^2$
$\Leftrightarrow (y -1)^2 \leq \dfrac{17}{4}$
$\Leftrightarrow -\dfrac{\sqrt{17}}{2} \leq y - 1 \leq \dfrac{\sqrt{17}}{2}$
$\Leftrightarrow \dfrac{2 - \sqrt{17}}{2} \leq y \leq \dfrac{2 + \sqrt{17}}{2}$
$\Rightarrow \begin{cases}M = \dfrac{2 + \sqrt{17}}{2}\\m = \dfrac{2 -\sqrt{17}}{2}\end{cases}$
$\Rightarrow M.n = \dfrac{4 - 17}{4} = -\dfrac{13}{4}$
