The inequality $$\log_{2} \frac{3x - 1}{2 - x} < 1$$ holds true for

Given, $$\log_{2} \dfrac{3x - 1}{2 - x} < 1$$

So, $$\dfrac{3x-1}{2-x}>0$$ and $$\dfrac{3x-1}{2-x}<2$$

Solving the first inequality:

$$\dfrac{3x-1}{2-x}>0$$

or, $$\left(3x-1\right)\left(x-2\right)<0$$

or, $$\dfrac{1}{3}$$ $$<$$ $$x$$ $$<$$ $$2$$

Also, $$\dfrac{3x-1}{2-x}<2$$

or, $$\dfrac{3x-1}{2-x}-2<0$$

or, $$\dfrac{3x-1-4+2x}{2-x}<0$$

or, $$\dfrac{5x-5}{2-x}<0$$

or, $$\left(x-1\right)\left(x-2\right)>0$$

so, $$x<1$$ and $$x>2$$

So, overall we can say, $$\dfrac{1}{3}$$ $$<$$ $$x$$ $$<$$ $$1$$

So, option A is the correct answer