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Question Number 15082 by Tinkutara last updated on 07/Jun/17

$$\mathrm{The}\mathrm{domain}\mathrm{of}f\left(x\right)=\sqrt{x-2\left\{x\right\}}.(\mathrm{where}$$$$\{\cdot \}\mathrm{denotes}\mathrm{fractional}\mathrm{part}\mathrm{of}x)\mathrm{is}?$$

Answered by mrW1 last updated on 07/Jun/17

$$\mathrm{x}=\mathrm{n}+\mathrm{f}$$$$\left\{\mathrm{x}\right\}=\mathrm{f}$$$$\mathrm{x}-2\left\{\mathrm{x}\right\}=\mathrm{n}+\mathrm{f}-2\mathrm{f}=\mathrm{n}-\mathrm{f}⩾0$$$$\mathrm{n}⩾\mathrm{f}$$$$$$$$\mathrm{if}\mathrm{x}>0:$$$$\mathrm{n}⩾\mathrm{f}$$$$\mathrm{n}⩾1$$$$\Rightarrow \mathrm{x}⩾1$$$$$$$$\mathrm{if}\mathrm{x}⩽0:$$$$\mathrm{n}⩾\mathrm{f}$$$$\mathrm{n}=0$$$$\Rightarrow \mathrm{x}>-1$$$$$$$$\Rightarrow \mathrm{x}\in (-1,0]\mathrm{and}[1,+\mathrm{\infty })$$$$$$$$\mathrm{check}:$$$$\mathrm{x}=-0.5$$$$\mathrm{x}-2\left\{\mathrm{x}\right\}=-0.5-2(-0.5)=0.5>0$$$$$$$$\mathrm{x}=1.5$$$$\mathrm{x}-2\left\{\mathrm{x}\right\}=1.5-2\times 0.5=0.5>0$$

Commented by Tinkutara last updated on 07/Jun/17

$$\mathrm{But}\mathrm{answer}\mathrm{is}[2,\mathrm{\infty })$$

Commented by mrW1 last updated on 07/Jun/17

$$\mathrm{this}\mathrm{answer}\mathrm{is}\mathrm{wrong}.\mathrm{you}$$$$\mathrm{can}\mathrm{check}\mathrm{with}\mathrm{x}=1.5\mathrm{which}\mathrm{is}\mathrm{not}$$$$\mathrm{in}\mathrm{the}\mathrm{range},$$$$\mathrm{but}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{(1.5)-2(0.5)}=\sqrt{0.5}!\mathrm{ok}$$

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