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

The domain of f (x) = \frac{1}{\sqrt{- x^{2} + {x}}};

(where {\cdot} denotes fractional part of x)

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 \Rightarrow impossible

(2) x > 0

{(n + f)}^{2} < f < 1

n + f < 1

n < 1, i . e . n = 0

\Rightarrow x \in (0, 1)

y = \frac{1}{\sqrt{- f^{2} + f}} = \frac{1}{\sqrt{\frac{1}{4} - (f^{2} - f + \frac{1}{4})}} = \frac{1}{\sqrt{\frac{1}{4} - {(f - \frac{1}{2})}^{2}}}

⩾ \frac{1}{\sqrt{\frac{1}{4}}} = 2

\Rightarrow y \in [2, + \infty)

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

Commented by Tinkutara last updated on 07/Jun/17

But answer is (\frac{1 - \sqrt{5}}{2}, 1) - {0}

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) = \frac{1}{\sqrt{- {(- 1.5)}^{2} + (- 0.5)}} = \frac{1}{\sqrt{- 2.75}}!

Commented by Tinkutara last updated on 07/Jun/17

x = - 1.5 is not in the range (\frac{1 - \sqrt{5}}{2}, 1)

\approx (- 0.618, 1) - {0}

- 1.5 \notin (\frac{1 - \sqrt{5}}{2}, 1) - {0}

Answer is correct Sir .

Commented by mrW1 last updated on 07/Jun/17

I could also take x = - 0.5 as example .

f (x) = \frac{1}{\sqrt{- {(0.5)}^{2} + (- 0.5)}} = \frac{1}{\sqrt{- 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 .

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