Hộ em với em cảm ơn nhiều ạ !!!

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Hộ em với em cảm ơn nhiều ạ !!!
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Neala 8 tháng 2020-11-01T14:39:29+00:00 1 Answers 70 views 0

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  1. 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 $\min y = 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi$

    $\max y = 2 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in Bbb Z)$

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