Q27 / 60 · KVPY 2015 (SA) · Mathematics

For a real number $x$, let $[x]$ denote the largest integer less than or equal to $x$, and let $\{x\}=x-[x] .$ The number of solutions $x$ to be equation $[x]\{x\}=5$ with $0 \leq x \leq 2015$ is

A
0
B
3
C
2008
Correct answer
D
2009

Solution

$$ \begin{array}{l} \{x\}=x-[x] \\ {[x]\{x\}=5 \quad f \neq 0} \\ \quad \text { I } f=5 \\ \quad 0 < f < 1 \end{array} $$ Possible solutions $$ \left(6+\frac{5}{6}\right),\left(7+\frac{5}{7}\right), \ldots \ldots .,\left(2014+\frac{5}{2014}\right)=2009 $$

Hence, the correct answer is option: 'D'.

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