Question

Tìm GTNN, GTLN

Answer 1

b, Do x \(x\in\left[0;\dfrac{\pi}{2}\right]\) nên \(\dfrac{\pi}{4}\le x+\dfrac{\pi}{4}\le\dfrac{3\pi}{4}\)

⇔ \(cos\left(x+\dfrac{\pi}{4}\right)\in\left[\dfrac{-\sqrt{2}}{2};\dfrac{\sqrt{2}}{2}\right]\)

⇔ \(cos\left(x+\dfrac{\pi}{4}\right)+1\in\left[\dfrac{2-\sqrt{2}}{2};\dfrac{2+\sqrt{2}}{2}\right]\)

⇔ \(y\in\left[\dfrac{2-\sqrt{2}}{2};\dfrac{2+\sqrt{2}}{2}\right]\)

Vậy y_min = \(\dfrac{2-\sqrt{2}}{2}\). DBXR ⇔ \(x=\pm\dfrac{3\pi}{4}+k2\pi\) , k ∈ Z

y_max = \(\dfrac{2+\sqrt{2}}{2}\). DBXR ⇔ \(x=\pm\dfrac{\pi}{4}+k2\pi\) , k ∈ Z

c, y = sinx + cos2x - 3 = - 2sin²x + sinx - 2

d, y = -cos²x + cosx - 1

c,d dùng bảng biến thiên của hs bậc 2 là được