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Question Number 119634 by Ar Brandon last updated on 25/Oct/20

$$\mathrm{Q}1$$$$\mathrm{If}f:\mathbb{R}\to \mathbb{R}\mathrm{is}\mathrm{defined}\mathrm{by}$$$$f\left(\mathrm{x}\right)=\left[\mathrm{x}\right]+[\mathrm{x}+\frac{1}{2}]+[\mathrm{x}+\frac{2}{3}]-3\mathrm{x}+5$$$$\mathrm{where}\left[\mathrm{x}\right]\mathrm{is}\mathrm{the}\mathrm{integral}\mathrm{part}\mathrm{of}\mathrm{x},\mathrm{then}\mathrm{a}\mathrm{period}\mathrm{of}f\mathrm{is}$$$$\left(\mathrm{A}\right)1\left(\mathrm{B}\right)2/3\left(\mathrm{C}\right)1/2\left(\mathrm{D}\right)1/3$$$$$$$$\mathrm{Q}2$$$$\mathrm{Let}a<\mathrm{c}<\mathrm{b}\mathrm{such}\mathrm{that}\mathrm{c}-a=\mathrm{b}-\mathrm{c}.\mathrm{If}f:\mathbb{R}\to \mathbb{R}\mathrm{is}\mathrm{a}$$$$\mathrm{function}\mathrm{satisfying}\mathrm{the}\mathrm{relation}$$$$f(\mathrm{x}+a)+f(\mathrm{x}+\mathrm{b})=f(\mathrm{x}+\mathrm{c})\mathrm{for}\mathrm{all}\mathrm{x}\in \mathbb{R}$$$$\mathrm{then}\mathrm{a}\mathrm{period}\mathrm{of}f\mathrm{is}$$$$\left(\mathrm{A}\right)(\mathrm{b}-a)\left(\mathrm{B}\right)2(\mathrm{b}-a)$$$$\left(\mathrm{C}\right)3(\mathrm{b}-a)\left(\mathrm{D}\right)4(\mathrm{b}-a)$$

Answered by Olaf last updated on 26/Oct/20

$$$$$$\mathrm{Q}2.$$$$c-a=b-c$$$$c=\frac{a+b}{2}$$$$$$$$f(x+c)=f(x+a)+f(x+b)$$$$f\left(x\right)=f(x+a-c)+f(x+b-c)$$$$f\left(x\right)=f(x+a-\frac{a+b}{2})+f(x+b-\frac{a+b}{2})$$$$f\left(x\right)=f(x-\frac{1}{2}(b-a))+f(x+\frac{1}{2}(b-a))$$$$$$$$f(x-\frac{1}{2}(b-a))=f(x-(b-a))+f\left(x\right)$$$$f\left(x\right)-f(x+\frac{1}{2}(b-a))=f(x-(b-a))+f\left(x\right)$$$$f(x+\frac{1}{2}(b-a))=-f(x-(b-a))$$$$$$$$f(x+\frac{3}{2}(b-a))=-f\left(x\right)$$$$$$$$f(x+3(b-a))=-f(x+\frac{3}{2}(b-a))$$$$f(x+3(b-a))=-[-f\left(x\right)]=f\left(x\right)$$$$$$$$\mathrm{Period}\mathrm{is}3(b-a)$$

Answered by Olaf last updated on 26/Oct/20

$$\mathrm{Q}1.$$$$f\left(x\right)=\left[x\right]+[x+\frac{1}{2}]+[x+\frac{2}{3}]-3x+5$$$$f(x+1)=[x+1]+[(x+\frac{1}{2})+1]$$$$+[(x+\frac{2}{3})+1]-3(x+1)+5$$$$f(x+1)=\left[x\right]+1+[x+\frac{1}{2}]+1$$$$+[x+\frac{2}{3}]+1-3(x+1)+5$$$$f(x+1)=\left[x\right]+[x+\frac{1}{2}]+[x+\frac{2}{3}]-3x+5$$$$f(x+1)=f\left(x\right)$$$$f\mathrm{is}\mathrm{clearly}\mathrm{at}\mathrm{least}1-\mathrm{periodic}.$$$$$$$$f\left(0\right)=\left[0\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{3}\right]-3\left(0\right)+5=5$$$$f\left(\frac{1}{3}\right)=\left[\frac{1}{3}\right]+\left[\frac{5}{6}\right]+\left[1\right]-3\left(\frac{1}{3}\right)+5=5$$$$f\left(\frac{1}{3}\right)=f\left(0\right)$$$$\mathrm{By}\mathrm{extension},\mathrm{in}\mathrm{the}\mathrm{general}\mathrm{case}:$$$$f(x+\frac{1}{3})=f\left(x\right)$$$$f\mathrm{is}\frac{1}{3}-\mathrm{periodic}.$$

Commented by Ar Brandon last updated on 26/Oct/20

Thank you so very much Sir ��

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