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Question Number 21247 by Tinkutara last updated on 17/Sep/17

If [] represents the greatest integer

function and f (x) = x - [x] then

number of real roots of the equation

f (x) + f (\frac{1}{x}) = 1 are infinite .

True / False

Answered by dioph last updated on 21/Sep/17

x - [x] + \frac{1}{x} - [\frac{1}{x}] = 1

x = 1 is not a root .

if x > 1, [\frac{1}{x}] = 0 and hence :

x - [x] + \frac{1}{x} = 1

For some fixed [x] = k we have :

x^{2} - (k + 1) x + 1 = 0

x = \frac{k + 1 \pm \sqrt{k^{2} + 2 k - 3}}{2}

k = 1 \Rightarrow x = 1 which we have already

considered .

k > 1 \Rightarrow 2 k > 3

\Rightarrow k^{2} < k^{2} + 2 k - 3 < {(k + 1)}^{2}

\Rightarrow k + \frac{1}{2} < \frac{k + 1 + \sqrt{k^{2} + 2 k - 3}}{2} < k + 1

Hence there is one real root x for

every k = [x] > 1

Because we have no further assumptions

about k, the function does indeed

have infinite real roots (True)

Commented by dioph last updated on 21/Sep/17

k \in Z^{+}, then :

k > 1 \Rightarrow k ⩾ 2 \Rightarrow 2 k ⩾ 4 > 3

Commented by Tinkutara last updated on 22/Sep/17

Thank you very much Sir!