If $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$,then $C$ is

Let $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$. Then,consider the following statements:
$(i)$ All the $A_i$'s are not distinct
(ii) There exists a pair,$A_p$ and $A_q$ such that $A_p^2 = A_q^2$ $(p \neq q)$
(iii) $\sum_{i=1}^4 A_i = \frac{1}{6}$
(iv) $\sum_{i=1}^4 A_i = 1$
Which one of the following is true?

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$

If $\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$,then $2A - 3B + 4C + 5D = $

If $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$,then the value of $a + b$ is:

If $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b}$ then $a \cdot A+b \cdot B=$