Let $f\left(\mathrm{x}\right)$ be a function such that $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]$, where $\left[\mathrm{x}\right]$ is the greatest integer less than or equal to x. Then the number of solutions of the equation $\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=1 \mathrm{is}(\mathrm{are})$

Given, $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right],\mathrm{x}\in \mathrm{R}-\left\{0\right\}$
Now $\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=1$    $\therefore \mathrm{x}-\left[\mathrm{x}\right]+\frac{1}{\mathrm{x}}-\left[\frac{1}{\mathrm{x}}\right]=1$
$\Rightarrow \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)-\left(\left[\mathrm{x}\right]+\left[\frac{1}{\mathrm{x}}\right]\right)=1$ $\Rightarrow \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)=\left[\mathrm{x}\right]+\left[\frac{1}{\mathrm{x}}\right]+1$  .....(i)
Clearly, R.H.S is an integer                           $\therefore$L. H. S. is also an integer
Let $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{k}$ an integer.          $\Rightarrow {\mathrm{x}}^2-\mathrm{kx}+1=0$
$\therefore \mathrm{x}=\frac{\mathrm{k}\pm \sqrt{{\mathrm{k}}^2-4}}{2}$
For real values of $\mathrm{x},{\mathrm{k}}^2-4\ge 0\Rightarrow \mathrm{k}\ge 2\text{ or }\mathrm{k}\le -2$
We also observe that k=2 and -2 does not satisfy equation (i)
$\therefore$ The equation (i) will have solutions if k$>$ 2 or k$<-$2, where k$\in$z .
Hence equation (i) has infinite number of solutions.