Let $$x = -\frac{1}{1!}\cdot\frac{3}{4} + \frac{1}{2!}\cdot(\frac{3}{4})^2 -\frac{1}{3!}\cdot(\frac{3}{4})^3 +$$.... and $$y = x - \frac{x^2}{2} + \frac{x^3}{3} - $$..... then the value of y is

$$e^k = \sum_{n = 0}^\infty \frac{k^n}{n!} = 1 + k + \frac{k^2}{2!} + \frac{k^3}{3!}+ ....$$

We are given,

 $$x = -\frac{1}{1!}\cdot\frac{3}{4} + \frac{1}{2!}\cdot(\frac{3}{4})^2 -\frac{1}{3!}\cdot(\frac{3}{4})^3 +....$$

This can be written as,

$$e^{-\frac{3}{4}}\ =\ 1\ -\ \frac{1}{1!}\left(\frac{3}{4}\right)\ +\ \frac{1}{2!}\left(\frac{3}{4}\right)^2-\frac{1}{3!}\left(\frac{3}{4}\right)^3....$$

$$x\ =\ -\ \frac{1}{1!}\left(\frac{3}{4}\right)\ +\ \frac{1}{2!}\left(\frac{3}{4}\right)^2-\frac{1}{3!}\left(\frac{3}{4}\right)^3....\ \ =\ e^{-\frac{3}{4}}\ -\ 1$$

We also know that,

$$\log\left(1\ +\ k\right)\ =\ k\ -\ \frac{k^2}{2}\ +\ \frac{k^3}{3}+\ ....$$

We can also write,

$$y=x-\frac{x^2}{2}+\frac{x^3}{3}-...\ =\ \log\left(1\ +\ x\right)$$

Substituting the value of x, we get,

$$y\ =\ \log\left(1\ +\ e^{-\frac{3}{4}}\ -\ 1\right)\ =\ \log\left(e^{-\frac{3}{4}}\right)\ =\ -\frac{3}{4}$$

Hence, A is the correct answer.