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

• Correct answer
  A

  $\frac{1}{2}$
• B

  $-\frac{1}{50}$
• C

  $-\frac{8}{25}$
• D

  $\frac{27}{25}$

Solution

Given that $\begin{gathered} \frac{x^3}{(2 x-1)(x+2)(x-3)}= \\ A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3} \end{gathered}$ Let $\begin{aligned} f(x) & =\frac{x^3}{(2 x-1)(x+2)(x-3)} \\ & =\frac{x^3}{(2 x-1)\left(x^2-x-6\right)} \\ & =\frac{x^3}{2 x^3-3 x^2-11 x+6} \end{aligned}$ Here we see that the power of $x$ will be same in Nr and Dr. $\therefore$ First we divide the numerator by denominator
$\begin{array}{cc}\Rightarrow & \frac{x^3}{(2 x-1)(x+2)(x-3)} \\ = & \frac{1}{2}+\frac{\frac{3}{2} x^2+\frac{11}{2} x-3}{(2 x-1)(x+2)(x-3)} \\ \Rightarrow & A=\frac{1}{2}\end{array}$

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