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-

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!