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Question Number 15084 by Tinkutara last updated on 07/Jun/17

The domain of f(x) = (1/(√(−x^2  + {x})));  (where {∙} denotes fractional part of x)  is?

$$\mathrm{The}\:\mathrm{domain}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\sqrt{−{x}^{\mathrm{2}} \:+\:\left\{{x}\right\}}}; \\ $$$$\left(\mathrm{where}\:\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{x}\right) \\ $$$$\mathrm{is}? \\ $$

Answered by mrW1 last updated on 07/Jun/17

x=n+f  −x^2 +{x}=−(n+f)^2 +f>0    (1) x<0  (n+f)^2 <f≤0 ⇒impossible    (2) x>0  (n+f)^2 <f<1  n+f<1  n<1, i.e. n=0  ⇒x∈(0,1)  y=(1/(√(−f^2 +f)))=(1/(√((1/4)−(f^2 −f+(1/4)))))=(1/(√((1/4)−(f−(1/2))^2 )))  ≥(1/(√(1/4)))=2  ⇒y∈[2,+∞)    check:  x=0.5  −x^2 +{x}=−0.5^2 +0.5=0.25>0  x=1.5  −x^2 +{x}=−1.5^2 +0.5<0

$$\mathrm{x}=\mathrm{n}+\mathrm{f} \\ $$$$−\mathrm{x}^{\mathrm{2}} +\left\{\mathrm{x}\right\}=−\left(\mathrm{n}+\mathrm{f}\right)^{\mathrm{2}} +\mathrm{f}>\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{x}<\mathrm{0} \\ $$$$\left(\mathrm{n}+\mathrm{f}\right)^{\mathrm{2}} <\mathrm{f}\leqslant\mathrm{0}\:\Rightarrow\mathrm{impossible} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}>\mathrm{0} \\ $$$$\left(\mathrm{n}+\mathrm{f}\right)^{\mathrm{2}} <\mathrm{f}<\mathrm{1} \\ $$$$\mathrm{n}+\mathrm{f}<\mathrm{1} \\ $$$$\mathrm{n}<\mathrm{1},\:\mathrm{i}.\mathrm{e}.\:\mathrm{n}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\sqrt{−\mathrm{f}^{\mathrm{2}} +\mathrm{f}}}=\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left(\mathrm{f}^{\mathrm{2}} −\mathrm{f}+\frac{\mathrm{1}}{\mathrm{4}}\right)}}=\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left(\mathrm{f}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$\geqslant\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}}}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{y}\in\left[\mathrm{2},+\infty\right) \\ $$$$ \\ $$$$\mathrm{check}: \\ $$$$\mathrm{x}=\mathrm{0}.\mathrm{5} \\ $$$$−\mathrm{x}^{\mathrm{2}} +\left\{\mathrm{x}\right\}=−\mathrm{0}.\mathrm{5}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}=\mathrm{0}.\mathrm{25}>\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{1}.\mathrm{5} \\ $$$$−\mathrm{x}^{\mathrm{2}} +\left\{\mathrm{x}\right\}=−\mathrm{1}.\mathrm{5}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}<\mathrm{0} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

But answer is (((1 − (√5))/2), 1) − {0}

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\left(\frac{\mathrm{1}\:−\:\sqrt{\mathrm{5}}}{\mathrm{2}},\:\mathrm{1}\right)\:−\:\left\{\mathrm{0}\right\} \\ $$

Commented by mrW1 last updated on 07/Jun/17

this answer is wrong.  you can check this answer with  x=−1.5 which is in the range.  but f(x)=(1/(√(−(−1.5)^2 +(−0.5))))=(1/(√(−2.75)))!

$$\mathrm{this}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{check}\:\mathrm{this}\:\mathrm{answer}\:\mathrm{with} \\ $$$$\mathrm{x}=−\mathrm{1}.\mathrm{5}\:\mathrm{which}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{range}. \\ $$$$\mathrm{but}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\sqrt{−\left(−\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} +\left(−\mathrm{0}.\mathrm{5}\right)}}=\frac{\mathrm{1}}{\sqrt{−\mathrm{2}.\mathrm{75}}}! \\ $$

Commented by Tinkutara last updated on 07/Jun/17

x = −1.5 is not in the range (((1 − (√5))/2), 1)  ≈ (−0.618, 1) − {0}  −1.5 ∉ (((1 − (√5))/2), 1) − {0}  Answer is correct Sir.

$${x}\:=\:−\mathrm{1}.\mathrm{5}\:\mathrm{is}\:\mathrm{not}\:\mathrm{in}\:\mathrm{the}\:\mathrm{range}\:\left(\frac{\mathrm{1}\:−\:\sqrt{\mathrm{5}}}{\mathrm{2}},\:\mathrm{1}\right) \\ $$$$\approx\:\left(−\mathrm{0}.\mathrm{618},\:\mathrm{1}\right)\:−\:\left\{\mathrm{0}\right\} \\ $$$$−\mathrm{1}.\mathrm{5}\:\notin\:\left(\frac{\mathrm{1}\:−\:\sqrt{\mathrm{5}}}{\mathrm{2}},\:\mathrm{1}\right)\:−\:\left\{\mathrm{0}\right\} \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{Sir}. \\ $$

Commented by mrW1 last updated on 07/Jun/17

I could also take x=−0.5 as example.  f(x)=(1/(√(−(0.5)^2 +(−0.5))))=(1/(√(−0.75))) !  but I think in your book it is an other  definition for {x} than I have learnt.  that is the reason for difference.

$$\mathrm{I}\:\mathrm{could}\:\mathrm{also}\:\mathrm{take}\:\mathrm{x}=−\mathrm{0}.\mathrm{5}\:\mathrm{as}\:\mathrm{example}. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\sqrt{−\left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} +\left(−\mathrm{0}.\mathrm{5}\right)}}=\frac{\mathrm{1}}{\sqrt{−\mathrm{0}.\mathrm{75}}}\:! \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{in}\:\mathrm{your}\:\mathrm{book}\:\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{other} \\ $$$$\mathrm{definition}\:\mathrm{for}\:\left\{\mathrm{x}\right\}\:\mathrm{than}\:\mathrm{I}\:\mathrm{have}\:\mathrm{learnt}. \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{for}\:\mathrm{difference}. \\ $$

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