Toán Lớp 11: Giúp mình với ạ
Tìm Min, Max của hàm số
1) y=-sin^2x-cosx+2
2) y= 2+sin^2.cos^2x
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Toán Lớp 11: Giúp mình với ạ
Tìm Min, Max của hàm số
1) y=-sin^2x-cosx+2
2) y= 2+sin^2.cos^2x
Comments ( 2 )
1)\quad y = -\sin^2x – \cos x+ 2\\
\Leftrightarrow y = \cos^2x – 1 – \cos x + 2\\
\Leftrightarrow y = \left(\cos x – \dfrac12\right)^2 + \dfrac{3}{4}\\
\text{Ta có:}\\
\quad -1 \leqslant \cos x \leqslant 1\\
\Leftrightarrow -\dfrac{3}{2} \leqslant \cos x – \dfrac12\leqslant \dfrac12\\
\Leftrightarrow 0 \leqslant \left(\cos x – \dfrac12\right)^2 \leqslant \dfrac94\\
\Leftrightarrow \dfrac34 \Leftrightarrow \left(\cos x – \dfrac12\right)^2 + \dfrac{3}{4}\leqslant 3\\
\Leftrightarrow \dfrac34 \leqslant y \leqslant 3\\
\text{Do đó:}\\
+)\quad \min y = \dfrac34\\
\Leftrightarrow \cos x = \dfrac12\\
\Leftrightarrow x = \pm \dfrac{\pi}{3} + k2\pi\quad (k\in\Bbb Z)\\
+)\quad \max y = 3\\
\Leftrightarrow \cos x = -1\\
\Leftrightarrow x = \pi + k2\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ \begin{cases}\min y = \dfrac34 \Leftrightarrow x = \pm \dfrac{\pi}{3} + k2\pi\\\max y = 3 \Leftrightarrow x = \pi + k2\pi\end{cases}\quad (k\in\Bbb Z)\\
2)\quad y = 2 + \sin^2x\cos^2x\\
\Leftrightarrow y = 2 +\dfrac14\sin^22x\\
\Leftrightarrow y = 2 + \dfrac14\cdot \dfrac{1 – \cos4x}{2}\\
\Leftrightarrow y = \dfrac{17}{8} – \dfrac18\cos4x\\
\text{Ta có:}\\
\quad – 1 \leqslant \cos4x \leqslant 1\\
\Leftrightarrow -\dfrac18 \leqslant – \dfrac18\cos4x \leqslant \dfrac18\\
\Leftrightarrow 2 \leqslant \dfrac{17}{8} – \dfrac18\cos4x \leqslant \dfrac{9}{4}\\
\Leftrightarrow 2 \leqslant y \leqslant \dfrac{9}{4}\\
\text{Do đó:}\\
+)\quad \min y = 2\\
\Leftrightarrow \cos4x = 1\\
\Leftrightarrow 4x = k2\pi\\
\Leftrightarrow x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z0\\
+)\quad \max y = \dfrac94\\
\Leftrightarrow \cos4x = -1\\
\Leftrightarrow 4x = \pi + k2\pi\\
\Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\quad (k\in\Bbb Z)\\
\text{Vậy}\ \begin{cases}\min y = 2\Leftrightarrow x = k\dfrac{\pi}{2}\\\max y = \dfrac94\Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{cases}\quad (k\in\Bbb Z)
\end{array}\)