find-the-range-y-x-x-1-x-x-

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

\dots

\Rightarrow y \in [0, \frac{1}{2}) \lor (1.5, 2] \lor (2.5, + \infty) \dots (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] \dots (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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