Tìm max,min của hàm số: a. $y=\sqrt{5-2sin^2xcos^2x}$ b. $y=\sqrt{2(1+cosx)}+1$ c. $y=-\sqrt{2sinx}$

$a) \quad y = \sqrt{5 – 2\sin^2x\cos^2x}$

$=\sqrt{5 – \dfrac{1}{2}\sin^22x}$

Ta có:

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

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

$\Leftrightarrow \dfrac{9}{2} \leq 5 -\dfrac{1}{2}\sin^22x \leq 5$

$\Leftrightarrow \dfrac{3\sqrt2}{2} \leq \sqrt{5 – \dfrac{1}{2}\sin^22x}\leq \sqrt5$

Vậy $\min y = \dfrac{3\sqrt2}{2}\Leftrightarrow \sin^22x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$

$\max y = \sqrt5 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$

$b) \quad y = \sqrt{2(1 + \cos x)} + 1$

Ta có:

$-1 \leq \cos x \leq 1$

$\Leftrightarrow 0 \leq 1 + \cos x \leq 2$

$\Leftrightarrow 0 \leq 2(1 + \cos x) \leq 4$

$\Leftrightarrow 0 \leq \sqrt{2(1 + \cos x)} \leq 2$

$\Leftrightarrow 1 \leq \sqrt{2(1 + \cos x)} + 1\leq 3$

Vậy $\min y = 1\Leftrightarrow\cos x = -1\Leftrightarrow x = \pi + k2\pi$

$\max y = 3 \Leftrightarrow \cos x =1 \Leftrightarrow x = k\pi\quad (k \in \Bbb Z)$

$c) \quad y = -\sqrt{2\sin x}$

Ta có:

$0 \leq \sqrt{2\sin x} \leq \sqrt2$

$\Leftrightarrow -\sqrt2 \leq -\sqrt{2\sin x}\leq 0$

Vậy $\min y = -\sqrt2 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi$

$\max y = 0 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$