find the range y=((x+[x])/(1−[x]+x))


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Question Number 85540 by M±th+et£s last updated on 22/Mar/20

$f i n d t h e r a n g e$ $y = \frac{x + [x]}{1 - [x] + x}$
	Answered by mr W last updated on 23/Mar/20

	$f o r x ⩾ 0 :$ $l e t x = n + d w i t h 0 ⩽ d < 1$ $y = \frac{x + [x]}{1 - [x] + x} = \frac{2 n + d}{1 + d} = 1 + \frac{2 n - 1}{1 + d}$ $f o r n = 0 :$ $y = 1 - \frac{1}{1 + d} ⩾ 1 - \frac{1}{1} = 0$ $y = 1 - \frac{1}{1 + d} < 1 - \frac{1}{2} = \frac{1}{2}$ $\Rightarrow 0 ⩽ y < \frac{1}{2}$ $f o r n ⩾ 1 :$ $y = 1 + \frac{2 n - 1}{1 + d} ⩽ 1 + \frac{2 n - 1}{1} = 2 n$ $y = 1 + \frac{2 n - 1}{1 + d} > 1 + \frac{2 n - 1}{1 + 1} = n + \frac{1}{2}$ $\Rightarrow n + \frac{1}{2} < y ⩽ 2 n$ $\Rightarrow 1.5 < y ⩽ 2$ $\Rightarrow 2.5 < y ⩽ 4$ $\Rightarrow 3.5 < y ⩽ 6$ $\Rightarrow 4.5 < y ⩽ 8$ $. . .$ $\Rightarrow y \in [0, \frac{1}{2}) \lor (1.5, 2] \lor (2.5, + \infty) . . . (i)$ $f o r x < 0 :$ $l e t x = - n - d w i t h 0 ⩽ d < 1$ $y = \frac{x + [x]}{1 - [x] + x} = 1 - \frac{2 n + 1}{1 - d}$ $y = 1 - \frac{2 n + 1}{1 - d} ⩽ 1 - \frac{2 n + 1}{1} = - 2 n$ $y = 1 - \frac{2 n + 1}{1 - d} > 1 - \frac{2 n + 1}{0} = - \infty$ $\Rightarrow - \infty < y ⩽ 2 n$ $\Rightarrow y \in (- \infty, 0] . . . (i i)$ $f r o m (i) a n d (i i) w e g e t r a n g e$ $y \in (- \infty, \frac{1}{2}) \lor (\frac{3}{2}, 2] \lor (\frac{5}{2}, + \infty)$
	Commented by M±th+et£s last updated on 23/Mar/20

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