Question

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

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

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Partial fraction:

$$\begin{gathered} \frac{x^3}{(2x-1)(x+2)(x-3)}=A+\frac{B}{(2x-1)}+\frac{C}{(x+2)}+\frac{D}{(x-3)} \\ \Rightarrow \frac{x^3}{(2x-1)(x+2)(x-3)}=\frac{A(2x-1)(x+2)(x-3)+B(x+2)(x-3)+C(2x-1)(x-3)+D(2x-1)(x+2)}{(2x-1)(x+2)(x-3)} \\ \Rightarrow x^3=A(2x-1)(x+2)(x-3)+B(x+2)(x-3)+C(2x-1)(x-3)+D(2x-1)(x+2) \end{gathered}$$

On comparing the coefficients of $x^3$, we get

$\begin{array}{rcl}1& =& 2A\\ & \Rightarrow & A=\frac{1}{2}\end{array}$

Hence, option A is correct.