Question

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

• A. \(-\dfrac{8}{25}\)
• B. \(\dfrac{4}{25}\)
• C. \(-\dfrac{1}{50}\)
• D. \(\dfrac{1}{2}\)

Hint:

For improper rational functions, first divide or compare leading coefficients to find the constant term, then use the cover-up method for partial fraction constants.

Answer 1

The Correct Option is B

Step 1: Find $A$ (the integer part).
Since numerator and denominator both have degree 3, divide: leading term of numerator is $x^3$, denominator is $(2x)(x)(x) = 2x^3$. So $A = \frac{1}{2}$.

Step 2: Find $B$ using the cover-up method.
Set $2x-1=0 \Rightarrow x=\frac{1}{2}$. Then: \[ B = \frac{x^3}{(x+2)(x-3)}\bigg|_{x=1/2} = \frac{1/8}{(5/2)(-5/2)} = \frac{1/8}{-25/4} = \frac{1}{8} \times \frac{-4}{25} = -\frac{1}{50} \]

Step 3: Find $C$ using the cover-up method.
Set $x+2=0 \Rightarrow x=-2$. Then: \[ C = \frac{x^3}{(2x-1)(x-3)}\bigg|_{x=-2} = \frac{-8}{(-5)(-5)} = \frac{-8}{25} \]

Step 4: Compute $A+B+C$.
\[ A+B+C = \frac{1}{2} - \frac{1}{50} - \frac{8}{25} \] Convert to 50ths: $\frac{25}{50} - \frac{1}{50} - \frac{16}{50} = \frac{8}{50} = \frac{4}{25}$.

Step 5: Confirm the sign of $C$.
$(-2)^3 = -8$, $(2(-2)-1) = -5$, $(-2-3)=-5$. So denominator is $(-5)(-5)=25$. $C = -8/25$. Confirmed.

Step 6: State the answer.
\[ \boxed{\frac{4}{25}} \]