Tìm giá trị min max của y=2sin(x+pi/4)+1m ạ mai e ktra rùi tks mn nhiều <3 03/10/2021 Bởi Adeline Tìm giá trị min max của y=2sin(x+pi/4)+1m ạ mai e ktra rùi tks mn nhiều <3
Đáp án: Max y=3 khi $x=\dfrac{\pi}2+k2\pi$ Min y=-1 khi $x=\dfrac{{ – 3\pi }}{4} + k2\pi$ Giải thích các bước giải: \(\begin{array}{l} – 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\ \Leftrightarrow – 2 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) \le 2\\ \Leftrightarrow – 2 + 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 2 + 1\\ \Leftrightarrow – 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 3\\ \Rightarrow Min\,y = – 1 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = – 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{{ – \pi }}{2} + k2\pi \Leftrightarrow \dfrac{{ – 3\pi }}{4} + k2\pi \\ Max\,\,y = 3 \Leftrightarrow \,\,\sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \Leftrightarrow \dfrac{\pi }{4} + k2\pi \end{array}\) Bình luận
$$\eqalign{ & – 1 \le \sin \left( {x + {\pi \over 4}} \right) \le 1 \cr & \Leftrightarrow – 2 \le 2\sin \left( {x + {\pi \over 4}} \right) \le 2 \cr & \Leftrightarrow – 1 \le 2\sin \left( {x + {\pi \over 4}} \right) + 1 \le 3 \cr & \Rightarrow \min y = – 1 \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = – 1 \cr & \Leftrightarrow x + {\pi \over 4} = – {\pi \over 2} + k2\pi \Leftrightarrow x = – {{3\pi } \over 4} + k2\pi \cr & \max y = 3 \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = 1 \cr & \Leftrightarrow x + {\pi \over 4} = {\pi \over 2} + k2\pi \Leftrightarrow x = {\pi \over 4} + k2\pi \cr} $$ Bình luận
Đáp án:
Max y=3 khi $x=\dfrac{\pi}2+k2\pi$
Min y=-1 khi $x=\dfrac{{ – 3\pi }}{4} + k2\pi$
Giải thích các bước giải:
\(\begin{array}{l} – 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\ \Leftrightarrow – 2 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) \le 2\\ \Leftrightarrow – 2 + 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 2 + 1\\ \Leftrightarrow – 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 3\\ \Rightarrow Min\,y = – 1 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = – 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{{ – \pi }}{2} + k2\pi \Leftrightarrow \dfrac{{ – 3\pi }}{4} + k2\pi \\ Max\,\,y = 3 \Leftrightarrow \,\,\sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \Leftrightarrow \dfrac{\pi }{4} + k2\pi \end{array}\)
$$\eqalign{
& – 1 \le \sin \left( {x + {\pi \over 4}} \right) \le 1 \cr
& \Leftrightarrow – 2 \le 2\sin \left( {x + {\pi \over 4}} \right) \le 2 \cr
& \Leftrightarrow – 1 \le 2\sin \left( {x + {\pi \over 4}} \right) + 1 \le 3 \cr
& \Rightarrow \min y = – 1 \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = – 1 \cr
& \Leftrightarrow x + {\pi \over 4} = – {\pi \over 2} + k2\pi \Leftrightarrow x = – {{3\pi } \over 4} + k2\pi \cr
& \max y = 3 \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = 1 \cr
& \Leftrightarrow x + {\pi \over 4} = {\pi \over 2} + k2\pi \Leftrightarrow x = {\pi \over 4} + k2\pi \cr} $$