Q59 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

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

  • A
    $-8 / 25$
  • Correct answer
    B
    $4 / 25$
  • C
    $-1 / 50$
  • D
    $1 / 2$

Solution

$\frac{x^3}{(2 x-1)(x+2)(x-3)}$ $=A+\frac{B}{(2 x-1)}+\frac{C}{(x+2)}+\frac{D}{(x-3)}$ $\begin{aligned} & x^3=A(2 x-1)(x+2)(x-3)+\mathrm{B}(x+2)(x-3)+\mathrm{C} \\ & (2 x-1)(x-3)+\mathrm{D}(2 x-1)(x+2) \ldots(\mathrm{i})\end{aligned}$ Put $x=1 / 2$ $\frac{1}{8}=B\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-3\right) \Rightarrow B=\frac{-1}{50}$ Put $x=-2$ $8=C(-4-1)(-2-3) \Rightarrow C=8 / 25$ Put $x=3$ $9=D(6-1)(3+2) \Rightarrow D=9 / 25$ Comparing coefficient of $x^3$ both sides in (i) $1=2 \mathrm{~A} \Rightarrow \mathrm{A}=1 / 2$ $A+B+C=\frac{1}{2}-\frac{1}{50}+\frac{8}{25}$ $=\frac{25-1-16}{50}=\frac{8}{50}=\frac{4}{25}$

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

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