Q59 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

If the equivalent partial fraction of $\frac{x^3}{(2 x-1)(x+2)(x-3)}$ is given by. $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$, then the value of $C$ is

  • A
    $\frac{1}{2}$
  • B
    $-\frac{1}{50}$
  • Correct answer
    C
    $-\frac{8}{25}$
  • D
    $\frac{27}{25}$

Solution

Here, $\begin{array}{l} \frac{x^3}{(2 x-1)(x+2)(x-3)}=A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3} \\ \Rightarrow x^3=A(2 x-1)(x+2)(x-3)+B(x+2)(x-3) . \\ \quad+C(2 x-1)(x-3)+D(2 x-1)(x+2) \end{array}$ $\begin{aligned} & \text {On putting } x=-2 \text {, } \\ & -8=C(-5)(-5) \\ & C=\frac{-8}{25} \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

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