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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: tìm max và min của hàm số y=2sin^2x-2cosx+5

Toán Lớp 11: tìm max và min của hàm số y=2sin^2x-2cosx+5

Comments ( 2 )

  1. Giải đáp:
    $\begin{cases}\min y = 3 \Leftrightarrow x = k2\pi\\\max y = \dfrac{15}{2}\Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\end{cases}\quad (k\in\Bbb Z)$
    Lời giải và giải thích chi tiết:
    $\quad y = 2\sin^2x – 2\cos x +5$
    $\Leftrightarrow y = 2(1-\cos^2x) – 2\cos x +5$
    $\Leftrightarrow y = – 2\cos^2x – 2\cos x + 7$
    $\Leftrightarrow y = -2\left(\cos x +\dfrac12\right)^2+ \dfrac{15}{2}$
    Ta có:
    $\quad -1 \leqslant \cos x \leqslant 1$
    $\Leftrightarrow -\dfrac12 \leqslant \cos x +\dfrac12 \leqslant \dfrac32$
    $\Leftrightarrow 0\leqslant \left(\cos x +\dfrac12\right)^2 \leqslant \dfrac94$
    $\Leftrightarrow – \dfrac92 \leqslant -2\left(\cos x +\dfrac12\right)^2 \leqslant 0$
    $\Leftrightarrow 3 \leqslant -2\left(\cos x +\dfrac12\right)^2 +\dfrac{15}{2}\leqslant \dfrac{15}{2}$
    $\Leftrightarrow 3 \leqslant y\leqslant \dfrac{15}{2}$
    Do đó:
    $+)\quad \min y = 3$
    $\Leftrightarrow \cos x = 1$
    $\Leftrightarrow x = k2\pi\quad (k\in\Bbb Z)$
    $+)\quad \max y = \dfrac{15}{2}$
    $\Leftrightarrow \cos x = -\dfrac12$
    $\Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$
    Vậy $\min y = 3 \Leftrightarrow x = k2\pi;\ \max y = \dfrac{15}{2}\Leftrightarrow x = \pm \dfrac{2\pi}{3} + k2\pi\quad (k\in\Bbb Z)$

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222-9+11+12:2*14+14 = ? ( )

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