a) $\begin{cases}\min y = – 1\Leftrightarrow x = k2\pi\\\max y = 5 \Leftrightarrow x = \pi + k2\pi\end{cases}\quad (k \in \Bbb Z)$
b) $\begin{cases}\min y = – 1\Leftrightarrow x = k\dfrac{\pi}{3}\\\max y = 1 \Leftrightarrow x = \dfrac{\pi}{6} +k\dfrac{\pi}{3}\end{cases}\quad (k \in \Bbb Z)$
c) $\begin{cases}\min y = 1 \Leftrightarrow x = k\dfrac{\pi}{2}\\\max y = 3 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{cases} \quad (k \in \Bbb Z)$
d) $\begin{cases}\min y = 1\Leftrightarrow x = k2\pi\\\max y = 2\Leftrightarrow x = \dfrac{\pi}{2} + k\pi\end{cases}\quad (k \in Bbb Z)$
Giải thích các bước giải:
a) $y = 2 – 3\cos x$
Ta có:
$-1 \leq \cos x \leq 1$
$\Leftrightarrow -3 \leq -3\cos x \leq 3$
$\Leftrightarrow -1 \leq 2 – 3\cos x \leq 5$
Hay $ -1 \leq y \leq 5$
Vậy $\min y = – 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi$
$\max y = 5 \Leftrightarrow \cos x = -1 \Leftrightarrow x = \pi + k2\pi \quad (k \in \Bbb Z)$
b) $y = 2\sin^23x – 1$
Ta có: $0 \leq \sin^23x \leq 1$
$\Leftrightarrow 0 \leq 2\sin^23x \leq 2$
$\Leftrightarrow -1 \leq 2\sin^23x \leq 1$
Hay $-1 \leq y \leq 1$
Vậy $\min y = – 1\Leftrightarrow \sin3x = 0 \Leftrightarrow x = k\dfrac{\pi}{3}$
$\max y = 1 \Leftrightarrow \sin3x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{6} +k\dfrac{\pi}{3}\quad (k \in \Bbb Z)$
c) $y = 2|\sin2x| + 1$
Ta có:
$0\leq |\sin2x| \leq 1$
$\Leftrightarrow 0 \leq 2|\sin2x| \leq 2$
$\Leftrightarrow 1 \leq 2|\sin2x| + 1 \leq 3$
Hay $1 \leq y \leq 3$
Vậy $\min y = 1 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2}$
$\max y = 3 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
d) $y = 2 – \sqrt{\cos x}$
Ta có:
$0 \leq \sqrt{\cos x} \leq 1$
$\Leftrightarrow – 1 \leq – \sqrt{\cos x} \leq 0$
$\Leftrightarrow 1 \leq 2 – \sqrt{\cos x} \leq 2$
Hay $1 \leq y \leq 2$
Vậy $\min y = 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi$
$\max y = 2 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in Bbb Z)$
Answers ( )
Đáp án:
a) $\begin{cases}\min y = – 1\Leftrightarrow x = k2\pi\\\max y = 5 \Leftrightarrow x = \pi + k2\pi\end{cases}\quad (k \in \Bbb Z)$
b) $\begin{cases}\min y = – 1\Leftrightarrow x = k\dfrac{\pi}{3}\\\max y = 1 \Leftrightarrow x = \dfrac{\pi}{6} +k\dfrac{\pi}{3}\end{cases}\quad (k \in \Bbb Z)$
c) $\begin{cases}\min y = 1 \Leftrightarrow x = k\dfrac{\pi}{2}\\\max y = 3 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{cases} \quad (k \in \Bbb Z)$
d) $\begin{cases}\min y = 1\Leftrightarrow x = k2\pi\\\max y = 2\Leftrightarrow x = \dfrac{\pi}{2} + k\pi\end{cases}\quad (k \in Bbb Z)$
Giải thích các bước giải:
a) $y = 2 – 3\cos x$
Ta có:
$-1 \leq \cos x \leq 1$
$\Leftrightarrow -3 \leq -3\cos x \leq 3$
$\Leftrightarrow -1 \leq 2 – 3\cos x \leq 5$
Hay $ -1 \leq y \leq 5$
Vậy $\min y = – 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi$
$\max y = 5 \Leftrightarrow \cos x = -1 \Leftrightarrow x = \pi + k2\pi \quad (k \in \Bbb Z)$
b) $y = 2\sin^23x – 1$
Ta có: $0 \leq \sin^23x \leq 1$
$\Leftrightarrow 0 \leq 2\sin^23x \leq 2$
$\Leftrightarrow -1 \leq 2\sin^23x \leq 1$
Hay $-1 \leq y \leq 1$
Vậy $\min y = – 1\Leftrightarrow \sin3x = 0 \Leftrightarrow x = k\dfrac{\pi}{3}$
$\max y = 1 \Leftrightarrow \sin3x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{6} +k\dfrac{\pi}{3}\quad (k \in \Bbb Z)$
c) $y = 2|\sin2x| + 1$
Ta có:
$0\leq |\sin2x| \leq 1$
$\Leftrightarrow 0 \leq 2|\sin2x| \leq 2$
$\Leftrightarrow 1 \leq 2|\sin2x| + 1 \leq 3$
Hay $1 \leq y \leq 3$
Vậy $\min y = 1 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2}$
$\max y = 3 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
d) $y = 2 – \sqrt{\cos x}$
Ta có:
$0 \leq \sqrt{\cos x} \leq 1$
$\Leftrightarrow – 1 \leq – \sqrt{\cos x} \leq 0$
$\Leftrightarrow 1 \leq 2 – \sqrt{\cos x} \leq 2$
Hay $1 \leq y \leq 2$
Vậy $\min y = 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi$
$\max y = 2 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in Bbb Z)$