Đáp án:
Giải thích các bước giải:
a) `A=2sin(x+\frac{\pi}{3})-5`
Ta có:
`-1 \le sin\ (x+\frac{\pi}{3}) \le 1`
`⇔ -2 \le 2sin\ (x+\frac{\pi}{3}) \le 2`
`⇔ -7 \le 2sin\ (x+\frac{\pi}{3})-5 \le -3`
`⇒ -7 \le A \le -3`
Vậy `A_{min}=-7` khi `sin (x+\frac{\pi}{3})=-1⇔x+\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})⇔x=-\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
`A_{max}=-3` khi `sin (x+\frac{\pi}{3})=1⇔x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})⇔x=\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
b) `B=5-\sqrt{2cos\ x-1}`
Ta có:
`-1 \le cos\ x \le 1`
`⇔ -2 \le 2cos\ x \le 2`
`⇔ -3 \le 2cos\ x -1 \le 1`
`⇔ 0 \le \sqrt{2cos\ x-1} \le 1`
`⇔ 0 \ge -\sqrt{2cos\ x-1} \ge -1`
`⇔ 5 \ge 5-\sqrt{2cos\ x -1} \ge 4`
`⇒ 5 \ge B \ge 4`
Vậy `B_{min}=4` khi `cos\ x=1⇔x=k2\pi\ (k \in \mathbb{Z})`
`B_{max} = 5` khi `cos x = 1 /2 \Leftrightarrow x = ± \frac{\pi}{ 3} + k2π\ (k \in \mathbb{Z})`
