x-x-x-x-175-8-

Question Number 140329 by benjo_mathlover last updated on 06/May/21

x^{⌊ x ⌋} + x^{⌈ x ⌉} = \frac{175}{8}

Answered by john_santu last updated on 06/May/21

0 < x < 1 \to x^{0} + x^{1} = \frac{175}{8}

\Rightarrow x = \frac{167}{8} > 20

1 < x < 2 \Rightarrow x^{1} + x^{2} = \frac{175}{8}

\Rightarrow 8 x^{2} + 8 x - 175 = 0

\Rightarrow x = \frac{- 8 \pm \sqrt{64 + 32.175}}{16}

2 < x < 3 \Rightarrow x^{2} + x^{3} = \frac{175}{8}

\Rightarrow 8 x^{3} + 8 x^{2} - 175 = 0

\Rightarrow {(2 x)}^{3} + 2 {(2 x)}^{2} - 175 = 0

l e t 2 x = y \Rightarrow y^{3} + 2 y^{2} - 175 = 0

(y - 5) (y^{2} + 5 y + 35) = 0

\Rightarrow y = 5 = 2 x; x = \frac{5}{2} . (s o l u t i o n)

Answered by mr W last updated on 06/May/21

l e t x = n + f w i t h 0 ⩽ f < 1

{(n + f)}^{n} + {(n + f)}^{n + 1} = \frac{175}{8}

\frac{175}{8} = {(n + f)}^{n} + {(n + f)}^{n + 1} ⩾ n^{n} + n^{n + 1} = (n + 1) n^{n}

\Rightarrow n ⩽ 2

\frac{175}{8} = {(n + f)}^{n} + {(n + f)}^{n + 1} ⩽ {(n + 1)}^{n} + {(n + 1)}^{n + 1} = (n + 2) {(n + 1)}^{n}

\Rightarrow n ⩾ 2

\Rightarrow n = 2

x^{2} + x^{3} = \frac{175}{8}

(x - \frac{5}{2}) (x^{2} + \frac{7 x}{2} + \frac{35}{4}) = 0

\Rightarrow x = \frac{5}{2}

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