1 câu trả lời
Đáp án:
$\begin{cases}\min y = 0\Leftrightarrow x = k2\pi\\\max y = \dfrac92\Leftrightarrow x =\pm \dfrac{2\pi}{3} + k2\pi\end{cases}\quad (k\in\Bbb Z)$
Giải thích các bước giải:
Cách 1: Lớp 11
$\quad y = 3 - 2\cos x - \cos2x$
$\Leftrightarrow y = 3 - 2\cos x -(2\cos^2x - 1)$
$\Leftrightarrow y = -2\cos^2x - 2\cos x + 4$
$\Leftrightarrow y = -2\left(\cos x+ \dfrac12\right)^2 + \dfrac92$
Ta có:
$\quad -1 \leqslant \cos x \leqslant 1$
$\Leftrightarrow -\dfrac12 \leqslant \cos x + \dfrac12 \leqslant \dfrac32$
$\Leftrightarrow 0 \leqslant \left(\cos x+ \dfrac12\right)^2 \leqslant \dfrac94$
$\Leftrightarrow - \dfrac92 \leqslant -2\left(\cos x+ \dfrac12\right)^2 \leqslant 0$
$\Leftrightarrow 0 \leqslant -2\left(\cos x+ \dfrac12\right)^2 + \dfrac92 \leqslant \dfrac92$
$\Leftrightarrow 0 \leqslant y \leqslant \dfrac92$
Do đó:
$\bullet\quad \min y = 0$
$\Leftrightarrow \cos x = 1$
$\Leftrightarrow x = k2\pi\quad (k\in\Bbb Z)$
$\bullet\quad \max y = \dfrac92$
$\Leftrightarrow \cos x +\dfrac12= 0$
$\Leftrightarrow \cos x = -\dfrac12$
$\Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$
Vậy $\min y = 0\Leftrightarrow x = k2\pi;\ \max y = \dfrac92\Leftrightarrow x =\pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$
Cách 2: Lớp 12
$\quad y = -2\cos^2x - 2\cos x + 4$
Đặt $t = \cos x\quad (t\in [-1;1])$
Hàm số trở thành:
$\quad y = f(t) = -2t^2 - 2t + 4$
$\Rightarrow y' = -4t - 2$
$y' = 0 \Leftrightarrow t = -\dfrac12$
Bảng biến thiên:
$\begin{array}{|c|cr|}
\hline
t & -1 & &- \dfrac{1}{2} & & & 1\\
\hline
y' & & + & 0& & - & \\
\hline
&&&\dfrac92\\
y & &\nearrow& & &\searrow\\
&4&&&&&0\\
\hline
\end{array}$
Dựa vào bảng biến thiên, ta được:
$\bullet\quad \min y = 0 \Leftrightarrow t = 1 \Leftrightarrow \cos x =1 \Leftrightarrow x = k2\pi\quad (k\in\Bbb Z)$
$\bullet\quad \max y = \dfrac92\Leftrightarrow t =-\dfrac12 \Leftrightarrow \cos x = - \dfrac12 \Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$
Vậy $\min y = 0\Leftrightarrow x = k2\pi;\ \max y = \dfrac92\Leftrightarrow x =\pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$