Solve for real x: (1/([x])) + (1/([2x])) = (x) + (1/3), where [x] is the greatest integer less than or equal to x and (x) = x − [x], [e.g. [3.4] = 3 and (3.4) = 0.4].


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Question Number 22463 by Tinkutara last updated on 18/Oct/17

$Solve for real x :$ $\frac{1}{[x]} + \frac{1}{[2 x]} = (x) + \frac{1}{3},$ $where [x] is the greatest integer less$ $than or equal to x and (x) = x - [x],$ $[e . g . [3.4] = 3 and (3.4) = 0.4] .$
	Answered by ajfour last updated on 18/Oct/17

	$l e t x = n + f$ $\Rightarrow \frac{1}{n} + \frac{1}{2 n + [2 f]} = f + \frac{1}{3}$ $C a s e I : f < 1 / 2$ $\Rightarrow \frac{3}{2 n} = f + \frac{1}{3}$ $\Rightarrow 0 ⩽ \frac{3}{2 n} - \frac{1}{3} < \frac{1}{2}$ $o r \Rightarrow \frac{1}{3} ⩽ \frac{3}{2 n} < \frac{5}{6}$ $\Rightarrow \frac{9}{5} < n ⩽ \frac{9}{2}$ $\Rightarrow n = 2, 3, 4$ $c o r r e s p o n d i n g f = \frac{3}{2 n} - \frac{1}{3}$ $f = \frac{5}{12}, \frac{1}{6}, \frac{1}{24}$ $S o x_{1} = 2 + \frac{5}{12} = \frac{29}{12}$ $x_{2} = 3 + \frac{1}{6} = \frac{19}{6}$ $x_{3} = 4 + \frac{1}{24} = \frac{97}{24}$ $C a s e I I : i f \frac{1}{2} ⩽ f < 1$ $\frac{1}{n} + \frac{1}{2 n + 1} = f + \frac{1}{3}$ $A s f ⩾ 1 / 2$ $\frac{1}{2} ⩽ \frac{3 n + 1}{n (2 n + 1)} - \frac{1}{3} < 1$ $\Rightarrow \frac{5}{6} ⩽ \frac{3 n + 1}{n (2 n + 1)} < \frac{4}{3}$ $c o n d i t i o n (a) :$ $\Rightarrow 10 n^{2} + 5 n ⩽ 18 n + 6$ $o r 10 n^{2} - 13 n - 6 ⩽ 0$ $10 {(n - \frac{13}{20})}^{2} ⩽ 6 + \frac{169}{40}$ $\Rightarrow \frac{13}{20} - \sqrt{\frac{409}{400}} ⩽ n ⩽ \frac{13}{20} + \sqrt{\frac{409}{400}}$ $\Rightarrow n = 0, 1$ $c o n d i t i o n (b) :$ $8 n^{2} + 4 n > 9 n + 3$ $\Rightarrow 8 n^{2} - 5 n - 3 > 0$ $\Rightarrow 8 {(n - \frac{5}{16})}^{2} > 3 + \frac{25}{32}$ $n < \frac{5}{16} - \frac{11}{16} a n d n > \frac{5}{16} + \frac{11}{16}$ $\Rightarrow n < - \frac{6}{16} a n d n > 1$ $I n t e r s e c t i o n o f t h e t w o c o n d i t i o n s$ $g i v e s n o s o l u t i o n f o r n f o r t h i s$ $c a s e .$ $S o f i n a l l y x = \frac{29}{12}, \frac{19}{6}, \frac{97}{24} .$
	Commented by Tinkutara last updated on 18/Oct/17

	$Thank you very much Sir!$
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