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