If $$x$$ is an integer such that $$x +\frac{1}{x} = \frac{17}{4}$$, then the value of $$x - \frac{1}{x}$$:

We know that $$\left(a\ +\ b\right)^2\ -\ \left(a\ -\ b\right)^2\ =\ 4ab$$

So,

$$\left(x\ +\ \frac{1}{x}\right)^2\ -\ \left(x\ -\ \frac{1}{x}\right)^2\ =\ 4\times\ x\ \times\ \frac{1}{x}\ =\ 4$$

$$\left(x\ -\ \frac{1}{x}\right)^2\ =\ \left(\frac{17}{4}\right)^2\ -\ 4$$

$$\ \left(x\ -\ \frac{1}{x}\right)^2\ =\ \frac{\left(289\ -\ 64\right)}{16}\ =\ \frac{225}{16}$$

$$\ x\ -\ \frac{1}{x}\ =\ \frac{15}{4}$$

Hence, the correct answer is option B.