y= sin^4 x + cos^4 x- sin^2 x – cos^2 x .Tìm GTLN GTNN 30/07/2021 Bởi Claire y= sin^4 x + cos^4 x- sin^2 x – cos^2 x .Tìm GTLN GTNN
$y = \sin^4x + \cos^4x – \sin^2x – \cos^2x$ $= (\sin^2 + \cos^2x)^2 -2\sin^2x\cos^2x – (\sin^2x + \cos^2x)$ $= 1 – \dfrac{1}{2}\sin^22x – 1$ $= -\dfrac{1}{2}\sin^22x$ 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 $\min y = -\dfrac{1}{2} \Leftrightarrow \sin^22x = 1 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$ $\max y = 0 \Leftrightarrow \sin2x = 0\Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$ Bình luận
$y = \sin^4x + \cos^4x – \sin^2x – \cos^2x$
$= (\sin^2 + \cos^2x)^2 -2\sin^2x\cos^2x – (\sin^2x + \cos^2x)$
$= 1 – \dfrac{1}{2}\sin^22x – 1$
$= -\dfrac{1}{2}\sin^22x$
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 $\min y = -\dfrac{1}{2} \Leftrightarrow \sin^22x = 1 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$
$\max y = 0 \Leftrightarrow \sin2x = 0\Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$