begin{aligned} & text{If } frac{x^3}{(2x-1)(x | vedclass

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(C) We have,$\frac{x^3}{(2x-1)(x-1)^2} = \frac{x^3}{2x^3-5x^2+4x-1}$.Performing long division,we get:$\frac{x^3}{2x^3-5x^2+4x-1} = \frac{1}{2} + \frac

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$\frac{17}{2}$

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(C) We have,$\frac{x^3}{(2x-1)(x-1)^2} = \frac{x^3}{2x^3-5x^2+4x-1}$.Performing long division,we get:$\frac{x^3}{2x^3-5x^2+4x-1} = \frac{1}{2} + \frac{1}{2} \left( \frac{5x^2-4x+1}{(2x-1)(x-1)^2} \right) \dots (i)$.Now,let $\frac{5x^2-4x+1}{(2x-1)(x-1)^2} = \frac{B}{2x-1} + \frac{C}{x-1} + \frac{D}{(x-1)^2}$.$5x^2-4x+1 = B(x-1)^2 + C(x-1)(2x-1) + D(2x-1)$.Putting $x=1$,we get $5-4+1 = D(2-1) \Rightarrow D=2$.Putting $x=1/2$,we get $5(1/4)-4(1/2)+1 = B(1/2-1)^2$ $\Rightarrow 5/4-2+1 = B(1/4)$ $\Rightarrow 1/4 = B/4$ $\Rightarrow B=1$.Comparing coefficients of $x^2$: $5 = B + 2C$ $\Rightarrow 5 = 1 + 2C$ $\Rightarrow 2C = 4$ $\Rightarrow C=2$.Substituting these into $(i)$:$\frac{x^3}{(2x-1)(x-1)^2} = \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2x-1} + \frac{2}{x-1} + \frac{2}{(x-1)^2} \right) = \frac{1}{2} + \frac{1/2}{2x-1} + \frac{1}{x-1} + \frac{1}{(x-1)^2}$.Comparing with the given expression,$A=1/2, B=1/2, C=1, D=1$.Then,$2A - 3B + 4C + 5D = 2(1/2) - 3(1/2) + 4(1) + 5(1) = 1 - 1.5 + 4 + 5 = 8.5 = \frac{17}{2}$.

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

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