Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 21247 by Tinkutara last updated on 17/Sep/17

$$\mathrm{If}\left[\right]\mathrm{represents}\mathrm{the}\mathrm{greatest}\mathrm{integer}$$$$\mathrm{function}\mathrm{and}f\left(x\right)=x-\left[x\right]\mathrm{then}$$$$\mathrm{number}\mathrm{of}\mathrm{real}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$f\left(x\right)+f\left(\frac{1}{x}\right)=1\mathrm{are}\mathrm{infinite}.$$$$\mathbf{True}/\mathbf{False}$$

Answered by dioph last updated on 21/Sep/17

$$x-\left[x\right]+\frac{1}{x}-\left[\frac{1}{x}\right]=1$$$$x=1\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{root}.$$$$\mathrm{if}x>1,\left[\frac{1}{x}\right]=0\mathrm{and}\mathrm{hence}:$$$$x-\left[x\right]+\frac{1}{x}=1$$$$\mathrm{For}\mathrm{some}\mathrm{fixed}\left[x\right]=k\mathrm{we}\mathrm{have}:$$$$x^2-(k+1)x+1=0$$$$x=\frac{k+1\pm \sqrt{k^2+2k-3}}{2}$$$$k=1\Rightarrow x=1\mathrm{which}\mathrm{we}\mathrm{have}\mathrm{already}$$$$\mathrm{considered}.$$$$k>1\Rightarrow 2k>3$$$$\Rightarrow k^2<k^2+2k-3<{(k+1)}^2$$$$\Rightarrow k+\frac{1}{2}<\frac{k+1+\sqrt{k^2+2k-3}}{2}<k+1$$$$\mathrm{Hence}\mathrm{there}\mathrm{is}\mathrm{one}\mathrm{real}\mathrm{root}x\mathrm{for}$$$$\mathrm{every}k=\left[x\right]>1$$$$\mathrm{Because}\mathrm{we}\mathrm{have}\mathrm{no}\mathrm{further}\mathrm{assumptions}$$$$\mathrm{about}k,\mathrm{the}\mathrm{function}\mathrm{does}\mathrm{indeed}$$$$\mathrm{have}\mathrm{infinite}\mathrm{real}\mathrm{roots}\left(\mathbf{True}\right)$$

Commented by dioph last updated on 21/Sep/17

$$k\in {\mathbb{Z}}^{+},\mathrm{then}:$$$$k>1\Rightarrow k⩾2\Rightarrow 2k⩾4>3$$

Commented by Tinkutara last updated on 22/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com