Question Number 114765 by dw last updated on 21/Sep/20
$${Find}\:{the}\:\boldsymbol{{maximum}}.\:{and}\:\boldsymbol{{minimum}}\:{value}\:{of}\:\lfloor\mathrm{1}+{sinx}\rfloor+\lfloor\mathrm{1}+{sin}\mathrm{3}{x}\rfloor+\lfloor\mathrm{1}+{sin}\mathrm{2}{x}\rfloor \\ $$$$ \\ $$$$ \\ $$
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](https://www.tinkutara.com/question/Q114776.png)
$$\mathrm{let}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\lfloor\mathrm{1}+\mathrm{sin}\:\mathrm{x}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\mathrm{2x}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\mathrm{3x}\rfloor \\ $$$$\mathrm{Now},\: \\ $$$$\:\:−\mathrm{1}\leqslant\mathrm{sin}\:\mathrm{nx}\:\leqslant\:\mathrm{1} \\ $$$$\:\:\mathrm{0}\leqslant\:\lfloor\mathrm{1}+\mathrm{sin}\:\mathrm{nx}\:\rfloor\:\leqslant\:\mathrm{2} \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{min}}.\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{now}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{max}}.\:\boldsymbol{\mathrm{value}} \\ $$$$\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{L}.\mathrm{C}.\mathrm{M}\:\left(\mathrm{2}\pi,\frac{\mathrm{2}\pi}{\mathrm{2}}\:,\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\:\mathrm{2}\pi \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{the}\:\mathrm{fundamental}\:\mathrm{period}\:\mathrm{of} \\ $$$$\:\:\mathrm{sin}\:\mathrm{x}\:\in\:\left[−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{max}}.\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\boldsymbol{\mathrm{max}}.\:\left\{\boldsymbol{\mathrm{f}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right),\boldsymbol{\mathrm{f}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right),\boldsymbol{\mathrm{f}}\left(\frac{\boldsymbol{\pi}}{\mathrm{6}}\right)\right\} \\ $$$$\:\left\{\because\:\:\boldsymbol{\mathrm{sinx}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{increasing}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{in}}\:\left[−\frac{\boldsymbol{\pi}}{\mathrm{2}},\frac{\boldsymbol{\pi}}{\mathrm{2}}\right]\:\right\} \\ $$$$\:\:\boldsymbol{\mathrm{f}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)=\:\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}+\mathrm{1}+\mathrm{0}=\mathrm{3} \\ $$$$\:\:\mathrm{f}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{4}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{4}}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}+\mathrm{2}+\mathrm{1}\:=\:\mathrm{4} \\ $$$$\:\:\mathrm{f}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{6}}\rfloor+\lfloor\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{6}}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}+\mathrm{1}+\mathrm{2}\:=\:\mathrm{4} \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{max}}.\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\mathrm{4} \\ $$
Commented by dw last updated on 21/Sep/20
$${Thank}\:{you}\:{Sir} \\ $$
Commented by PRITHWISH SEN 2 last updated on 21/Sep/20
$$\mathrm{welcome} \\ $$