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Question Number 182332 by mnjuly1970 last updated on 07/Dec/22

$$$$$$\mathrm{If},f\left(x\right)=2{cos}^2\left(\frac{x}{2}\right)-\lfloor \frac{1}{3}+cos\left(x\right)\rfloor$$$$thenfindtherangeof:R_f$$

Answered by floor(10²Eta[1]) last updated on 07/Dec/22

$$-1⩽\mathrm{cosx}⩽1\Rightarrow \frac{-2}{3}⩽\frac{1}{3}+\mathrm{cosx}⩽\frac{4}{3}$$$$\Rightarrow -1⩽\lfloor \frac{1}{3}+\mathrm{cosx}\rfloor ⩽1\Rightarrow -1⩽-\lfloor \frac{1}{3}+\mathrm{cosx}\rfloor ⩽1$$$$0⩽{2\cos }^2\left(\frac{\mathrm{x}}{2}\right)⩽2$$$$\Rightarrow -1⩽\mathrm{f}\left(\mathrm{x}\right)⩽3$$$$$$

Answered by mnjuly1970 last updated on 08/Dec/22

$$f\left(x\right)=1+cos\left(x\right)-\lfloor \frac{1}{3}+cos\left(x\right)\rfloor$$$$=\frac{1}{3}+cos\left(x\right)-\lfloor \frac{1}{3_{}}+cos\left(x\right)\rfloor +\frac{2}{3}$$$$\mathrm{\forall }u\in \mathbb{R},0⩽u-\lfloor u\rfloor <1\Rightarrow f\left(x\right)\in [\frac{2}{3},\frac{5}{3})$$

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