y= sin^4 x + cos^4 x- sin^2 x – cos^2 x .Tìm GTLN GTNN

Question

y= sin^4 x + cos^4 x- sin^2 x – cos^2 x .Tìm GTLN GTNN

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Claire 5 tháng 2021-07-30T21:34:12+00:00 1 Answers 4 views 0

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    2021-07-30T21:36:08+00:00

    $y = \sin^4x + \cos^4x – \sin^2x – \cos^2x$

    $= (\sin^2 + \cos^2x)^2 -2\sin^2x\cos^2x – (\sin^2x + \cos^2x)$

    $= 1 – \dfrac{1}{2}\sin^22x – 1$

    $= -\dfrac{1}{2}\sin^22x$

    Ta có:

    $0 \leq \sin^22x \leq 1$

    $-\dfrac{1}{2} \leq -\dfrac{1}{2}\sin^22x \leq 0$

    Hay $-\dfrac{1}{2} \leq y \leq 0$

    Vậy $\min y = -\dfrac{1}{2} \Leftrightarrow \sin^22x = 1 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$

    $\max y = 0 \Leftrightarrow \sin2x = 0\Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$

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