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

Answer: (A)

$\frac{1}{2}$

$\frac{x^3}{(2x-1)(x+2)(x-3)}=$ $A+\frac{B}{(2x-1)}+\frac{C}{(x+2)}+\frac{D}{(x-3)}$
Cross multiplying,
$x^3=A(2x-1)(x+2)(x-3)+B(x+2)(x-3)+C(2x-1)(x-3)+D(2x-1)(x+2)$
Substitute $x=1/2$ both sides
$\Rightarrow 1/8=B\left(5/2\right)(-5/2)\Rightarrow B=-1/50$
Substitute $x=-2$ both sides,
$\Rightarrow -8=C(-5)(-5)\Rightarrow C=-8/25$
Substitute $=3$ both sides,
$\Rightarrow 27=D\left(5\right)\left(5\right)\Rightarrow D=27/25$
Now put $x=0$ both sides,
$\Rightarrow 0=6A-6B+3C-2D$
$\Rightarrow 6A=6B-3C+2D=-3/25+24/25+54/25=3\Rightarrow A=1/2$
Hence option 'A' is correct choice.