Toán Lớp 11: y = sin 5x + cos 5x -2 . Tìm GTLN, NN

Question

Toán Lớp 11:  y = sin 5x + cos 5x -2 . Tìm GTLN, NN

bichhhathu · Answer

$y= sin5x+ cos5x-2$   $= sqrt2 sin left(5x+ dfrac{ pi}{4} right)-2$   Ta c&oacute;:   $-1 le  sin left(5x+ dfrac{ pi}{4} right) le 1$   $ to - sqrt2-2 le y le  sqrt2-2$   Vậy $ min y=- sqrt2-2; max y= sqrt2-2$

kimnguenoanh · Answer

$ begin{array}{l} y =  sin 5x +  cos 5x - 2   y =  sqrt 2  sin  left( {5x +  dfrac{ pi }{4}}  right) - 2    - 1  le  sin  left( {5x +  dfrac{ pi }{4}}  right)  le 1  Rightarrow  -  sqrt 2   le  sqrt 2  sin  left( {5x +  dfrac{ pi }{4}}  right)  le  sqrt 2      Rightarrow  - 2 -  sqrt 2   le y  le  sqrt 2  - 2     Rightarrow  left {  begin{array}{l}  max y =  sqrt 2  - 2  Rightarrow  sin  left( {5x +  dfrac{ pi }{4}}  right) = 1    min y =  - 2 -  sqrt 2   Rightarrow  sin  left( {5x +  dfrac{ pi }{4}}  right) =  - 1  end{array}  right.     Leftrightarrow  left {  begin{array}{l}  max y =  sqrt 2  - 2  Rightarrow 5x +  dfrac{ pi }{4} =  dfrac{ pi }{2} + k2 pi     min y =  - 2 -  sqrt 2   Rightarrow 5x +  dfrac{ pi }{4} =  -  dfrac{ pi }{2} + k2 pi   end{array}  right.     Leftrightarrow  left {  begin{array}{l}  max y =  sqrt 2  - 2  Leftrightarrow x =  dfrac{ pi }{{20}} +  dfrac{{k2 pi }}{5}    min y =  - 2 -  sqrt 2   Leftrightarrow x =  -  dfrac{{3 pi }}{{20}} +  dfrac{{k2 pi }}{5}  end{array}  right. left( {k  in  mathbb{Z}}  right)  end{array}$