Toán Lớp 11: Tìm GTLN-GTNN a) y=cos^2x+2sinx+2 b) y=sin^4x-2cos^2x+1 c) y=sinx+cosx+2

Question

Toán Lớp 11:  Tìm GTLN-GTNN a) y=cos^2x+2sinx+2 b) y=sin^4x-2cos^2x+1 c) y=sinx+cosx+2

diemchau · Answer

$ begin{array}{l} y = { cos ^2}x + 2 sin x + 2   y = 1 - { sin ^2}x + 2 sin x + 2   y =  - { sin ^2}x + 2 sin x + 3   y =  - { sin ^2}x + 2 sin x + 3   t =  sin x left( { - 1  le t  le 1}  right)     Rightarrow y = f left( t  right) =  - {t^2} + 2t + 3 left( {t  in  left[ { - 1;1}  right]}  right)   f left( 1  right) = 4,f left( { - 1}  right) = 0     to  max y = 4, min y = 0    left {  begin{array}{l}  max y = 4  to  sin x = 1  to x =  dfrac{ pi }{2} + k2 pi     min y = 0  to  sin x =  - 1  to x =  -  dfrac{ pi }{2} + k2 pi   end{array}  right. left( {k  in  mathbb{Z}}  right)   b)y = { sin ^4}x - 2{ cos ^2}x + 1   y = { sin ^4}x - 2 left( {1 - {{ sin }^2}x}  right) + 1   y = { sin ^4}x - 2 + 2{ sin ^2}x + 1 = { sin ^4}x + 2{ sin ^2}x - 1   t = { sin ^2}x left( {t  in  left[ {0;1}  right]}  right)     Rightarrow y = f left( t  right) = {t^2} + 2t - 1   f left( 0  right) =  - 1,f left( 1  right) = 2     Rightarrow  left {  begin{array}{l}  max y = 2  to { sin ^2}x = 1  to  sin x =   pm 1    min y =  - 1  to { sin ^2}x = 0  to  sin x = 0  end{array}  right.     Rightarrow  left {  begin{array}{l}  max y = 2  to x =  dfrac{ pi }{2} + k pi     min y =  - 1  to x = k pi   end{array}  right. left( {k  in  mathbb{Z}}  right)   c)y =  sin x +  cos x + 2   y =  sqrt 2  sin  left( {x +  dfrac{ pi }{4}}  right) + 2    - 1  le  sin  left( {x +  dfrac{ pi }{4}}  right)  le 1     to  -  sqrt 2   le  sqrt 2  sin  left( {x +  dfrac{ pi }{4}}  right)  le  sqrt 2      to 2 -  sqrt 2   le y  le 2 +  sqrt 2      to  left {  begin{array}{l}  max y = 2 +  sqrt 2   Rightarrow  sin  left( {x +  dfrac{ pi }{4}}  right) = 1  Rightarrow x +  dfrac{ pi }{4} =  dfrac{ pi }{2} + k2 pi     min y = 2 -  sqrt 2   Rightarrow  sin  left( {x +  dfrac{ pi }{4}}  right) =  - 1  Rightarrow x +  dfrac{ pi }{4} =  -  dfrac{ pi }{2} + k2 pi   end{array}  right.     to  left {  begin{array}{l}  max y = 2 +  sqrt 2   Rightarrow x =  dfrac{ pi }{4} + k2 pi     min y = 2 -  sqrt 2   Rightarrow x =  -  dfrac{{3 pi }}{4} + k2 pi   end{array}  right. left( {k  in  mathbb{Z}}  right)  end{array}$

hothaibinh · Answer

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