The value of the integral $\int_0^{10}\frac{\left|\frac{2[x]}{3x- [x]}\right|}{\left(\frac{2[x]}{3x- [x]}\right)}dx$, where [*] denotes the greatest integer function, is

none of these

Let $f(x) =\frac{2[x]}{3x- [x]}$

It is clear that f(x) is not defined if x = 0 and if 3x = [x]

So, in (–10, 0), f is not defined at $x =-\frac{1}{3}$

Case I : $x ∈(-10,-\frac{1}{3})$

$[x] < 0$ and $3x – [x] < 0$ so $\frac{[x]}{3x- [x]}>0$

$∴\left|\frac{2[x]}{3x- [x]}\right|=\frac{2[x]}{3x- [x]}$

$⇒\int_{-10}^{-1/3}1\,dx=-\frac{1}{3}+10=\frac{29}{3}$

Case II : $x ∈(-\frac{1}{3},0)$

$[x] < 0$ and $3x – [x] > 0$ so $\frac{[x]}{3x- [x]}<0$

$∴\left|\frac{2[x]}{3x- [x]}\right|=-\left(\frac{2[x]}{3x- [x]}\right)$

$⇒\int_{-\frac{1}{3}}^0(-1)dx=-(0+\frac{1}{3})=-\frac{1}{3}$

Hence, $\int_{-10}^0\frac{|f(x)|}{f(x)}dx=\int_{-10}^{-1/3}\frac{|f(x)|}{f(x)}dx+\int_{-1/3}^0\frac{|f(x)|}{f(x)}dx$

$=\frac{29}{3}-\frac{1}{3}=\frac{28}{3}$