Toán Lớp 11: tìm GTLN GTNN CỦA HÀM SỐ y=cosx +cos(x-pi/2)
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Toán Lớp 11: tìm GTLN GTNN CỦA HÀM SỐ y=cosx +cos(x-pi/2)
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\min y = -\sqrt2 \Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\\
\max y = \sqrt2\ \ \Leftrightarrow x = \dfrac{\pi}{4} + k2\pi
\end{cases}\quad (k\in\Bbb Z)\)
\quad y = \cos x + \cos\left(x – \dfrac{\pi}{2}\right)\\
\Leftrightarrow y = \cos x + \sin x\\
\Leftrightarrow y = \sqrt2\cos\left(x – \dfrac{\pi}{4}\right)\\
\text{Ta có:}\\
\quad -1 \leqslant \cos\left(x – \dfrac{\pi}{4}\right) \leqslant 1\\
\Leftrightarrow -\sqrt2\leqslant \sqrt2\cos\left(x – \dfrac{\pi}{4}\right) \leqslant \sqrt2\\
\Leftrightarrow -\sqrt2 \leqslant y \leqslant \sqrt2\\
\text{Do đó:}\\
+)\quad \min y = -\sqrt2\\
\Leftrightarrow \cos\left(x – \dfrac{\pi}{4}\right) = -1\\
\Leftrightarrow x – \dfrac{\pi}{4} = \pi + k2\pi\\
\Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\quad (k\in\Bbb Z)\\
+)\quad \max y = \sqrt2 \\
\Leftrightarrow \cos\left(x – \dfrac{\pi}{4}\right) = 1\\
\Leftrightarrow x – \dfrac{\pi}{4} = k2\pi\\
\Leftrightarrow x = \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ \min y = -\sqrt2 \Leftrightarrow x = \dfrac{5\pi}{4} + k2\pi\\
\max y = \sqrt2 \Leftrightarrow x = \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)
\end{array}\)