Tìm max, min của hàm số a. y=2sin^2x-cos2x b. y=3-2|sinx| c. y=cosx+cos(x-pi/3) 01/08/2021 Bởi Rylee Tìm max, min của hàm số a. y=2sin^2x-cos2x b. y=3-2|sinx| c. y=cosx+cos(x-pi/3)
a) $y = 2\sin^2x – \cos2x$ $= 1 – 2\cos2x$ Ta có: $- 1 \leq \cos2x \leq 1$ $\Leftrightarrow -2\leq -2\cos2x \leq 2$ $\Leftrightarrow – 1 \leq 1 – 2\cos2x \leq 3$ Hay $-1 \leq y \leq 3$ Vậy $\min y = – 1 \Leftrightarrow \cos2x = 1\Leftrightarrow x = k\pi$ $\max y = 3 \Leftrightarrow \cos2x = – 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in \Bbb Z)$ b) $y = 3 – 2|\sin x|$ Ta có: $0 \leq |\sin x| \leq 1$ $\Leftrightarrow – 2 \leq -2|\sin x| \leq 0$ $\Leftrightarrow 1 \leq 3 -2|\sin x| \leq 3$ Hay $1 \leq y \leq 3$ Vậy $\min y = 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi$ $\max y = 3 \Leftrightarrow \sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi \quad (k \in \Bbb Z)$ c) $y = \cos x + \cos\left(x – \dfrac{\pi}{3}\right)$ $= \cos x + \cos x.\cos\dfrac{\pi}{3} + \sin x.\sin\dfrac{\pi}{3}$ $= \dfrac{3}{2}\cos x + \dfrac{\sqrt3}{2}\sin x$ $= \sqrt3.\left(\dfrac{\sqrt3}{2}\cos x + \dfrac{1}{2}\sin x\right)$ $= \sqrt3.\cos\left(x – \dfrac{\pi}{6}\right)$ Ta có: $-1 \leq \cos\left(x – \dfrac{\pi}{6}\right) \leq 1$ $\Leftrightarrow -\sqrt3 \leq \sqrt3.\cos\left(x – \dfrac{\pi}{6}\right) \leq \sqrt3$ Hay $-\sqrt3 \leq y \leq \sqrt3$ Vậy $\min y = – \sqrt3 \Leftrightarrow \cos\left(x – \dfrac{\pi}{6}\right) = -1 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi$ $\max y = \sqrt3 \Leftrightarrow \cos\left(x – \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi \quad (k \in \Bbb Z)$ Bình luận
a) $y = 2\sin^2x – \cos2x$
$= 1 – 2\cos2x$
Ta có: $- 1 \leq \cos2x \leq 1$
$\Leftrightarrow -2\leq -2\cos2x \leq 2$
$\Leftrightarrow – 1 \leq 1 – 2\cos2x \leq 3$
Hay $-1 \leq y \leq 3$
Vậy $\min y = – 1 \Leftrightarrow \cos2x = 1\Leftrightarrow x = k\pi$
$\max y = 3 \Leftrightarrow \cos2x = – 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in \Bbb Z)$
b) $y = 3 – 2|\sin x|$
Ta có: $0 \leq |\sin x| \leq 1$
$\Leftrightarrow – 2 \leq -2|\sin x| \leq 0$
$\Leftrightarrow 1 \leq 3 -2|\sin x| \leq 3$
Hay $1 \leq y \leq 3$
Vậy $\min y = 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi$
$\max y = 3 \Leftrightarrow \sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi \quad (k \in \Bbb Z)$
c) $y = \cos x + \cos\left(x – \dfrac{\pi}{3}\right)$
$= \cos x + \cos x.\cos\dfrac{\pi}{3} + \sin x.\sin\dfrac{\pi}{3}$
$= \dfrac{3}{2}\cos x + \dfrac{\sqrt3}{2}\sin x$
$= \sqrt3.\left(\dfrac{\sqrt3}{2}\cos x + \dfrac{1}{2}\sin x\right)$
$= \sqrt3.\cos\left(x – \dfrac{\pi}{6}\right)$
Ta có:
$-1 \leq \cos\left(x – \dfrac{\pi}{6}\right) \leq 1$
$\Leftrightarrow -\sqrt3 \leq \sqrt3.\cos\left(x – \dfrac{\pi}{6}\right) \leq \sqrt3$
Hay $-\sqrt3 \leq y \leq \sqrt3$
Vậy $\min y = – \sqrt3 \Leftrightarrow \cos\left(x – \dfrac{\pi}{6}\right) = -1 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi$
$\max y = \sqrt3 \Leftrightarrow \cos\left(x – \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi \quad (k \in \Bbb Z)$
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