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

$$\log (1- x)=$$

$$-x -\frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 - ..........$$

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

$$\therefore y = \log(1+x)$$

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

$$= e^{\frac{-3}{4}} - 1$$

$$\therefore y = \log(e^{\frac{-3}{4}})$$

$$= \frac{-3}{4}$$

Hence A is the correct answer.