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

The domain of f(x) = (√(x − 2{x})). (where {∙} denotes fractional part of x) is?

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

Answered by mrW1 last updated on 07/Jun/17

x=n+f {x}=f x−2{x}=n+f−2f=n−f≥0 n≥f if x>0: n≥f n≥1 ⇒x≥1 if x≤0: n≥f n=0 ⇒x>−1 ⇒x∈(−1,0] and [1,+∞) check: x=−0.5 x−2{x}=−0.5−2(−0.5)=0.5>0 x=1.5 x−2{x}=1.5−2×0.5=0.5>0

$$\mathrm{x}=\mathrm{n}+\mathrm{f} \\ $$$$\left\{\mathrm{x}\right\}=\mathrm{f} \\ $$$$\mathrm{x}−\mathrm{2}\left\{\mathrm{x}\right\}=\mathrm{n}+\mathrm{f}−\mathrm{2f}=\mathrm{n}−\mathrm{f}\geqslant\mathrm{0} \\ $$$$\mathrm{n}\geqslant\mathrm{f} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{x}>\mathrm{0}: \\ $$$$\mathrm{n}\geqslant\mathrm{f} \\ $$$$\mathrm{n}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\geqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{x}\leqslant\mathrm{0}: \\ $$$$\mathrm{n}\geqslant\mathrm{f} \\ $$$$\mathrm{n}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}>−\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\mathrm{x}\in\left(−\mathrm{1},\mathrm{0}\right]\:\mathrm{and}\:\left[\mathrm{1},+\infty\right) \\ $$$$ \\ $$$$\mathrm{check}: \\ $$$$\mathrm{x}=−\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{x}−\mathrm{2}\left\{\mathrm{x}\right\}=−\mathrm{0}.\mathrm{5}−\mathrm{2}\left(−\mathrm{0}.\mathrm{5}\right)=\mathrm{0}.\mathrm{5}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{x}=\mathrm{1}.\mathrm{5} \\ $$$$\mathrm{x}−\mathrm{2}\left\{\mathrm{x}\right\}=\mathrm{1}.\mathrm{5}−\mathrm{2}×\mathrm{0}.\mathrm{5}=\mathrm{0}.\mathrm{5}>\mathrm{0} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

But answer is [2, ∞)

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

Commented by mrW1 last updated on 07/Jun/17

this answer is wrong. you can check with x=1.5 which is not in the range, but f(x)=(√((1.5)−2(0.5)))=(√(0.5)) ! ok

$$\mathrm{this}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{you} \\ $$$$\mathrm{can}\:\mathrm{check}\:\mathrm{with}\:\mathrm{x}=\mathrm{1}.\mathrm{5}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{range}, \\ $$$$\mathrm{but}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{1}.\mathrm{5}\right)−\mathrm{2}\left(\mathrm{0}.\mathrm{5}\right)}=\sqrt{\mathrm{0}.\mathrm{5}}\:!\:\mathrm{ok} \\ $$

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