Question Number 114765 by dw last updated on 21/Sep/20

Find the maximum. and minimum value of ⌊1+sinx⌋+⌊1+sin3x⌋+⌊1+sin2x⌋

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

let   f(x)=⌊1+sin x⌋+⌊1+sin 2x⌋+⌊1+sin 3x⌋  Now,     −1≤sin nx ≤ 1    0≤ ⌊1+sin nx ⌋ ≤ 2  ∴ the min. value of f(x) = 0  now for the max. value  the period of f(x)= L.C.M (2π,((2π)/2) ,((2π)/3))= 2π  we know that the fundamental period of    sin x ∈ [−(π/2), (π/2)]  ∴ the max. value of                                f(x) = max. {f((𝛑/2)),f((𝛑/4)),f((𝛑/6))}   {∵  sinx is an increasing function in [−(𝛑/2),(𝛑/2)] }    f((𝛑/2))= ⌊1+sin (π/2)⌋+⌊1+sin ((2π)/2)⌋+⌊1+sin ((3π)/2)⌋               = 2+1+0=3    f((π/4)) = ⌊1+sin (π/4)⌋+⌊1+sin ((2π)/4)⌋+⌊1+sin ((3π)/4)⌋                = 1+2+1 = 4    f((π/6)) = ⌊1+sin (π/6)⌋+⌊1+sin ((2π)/6)⌋+⌊1+sin ((3π)/6)⌋                = 1+1+2 = 4  ∴ the max. value of f(x) = 4

Commented bydw last updated on 21/Sep/20

Thank you Sir

Commented byPRITHWISH SEN 2 last updated on 21/Sep/20

welcome