Question

If $\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)^{2}}+\frac{B}{9(x-1)}+\frac{C}{(x+2)}$ then $A+B=$

• A. 1/3
• B. 1/9
• C. -1/3
• D. 2/3

Hint:

Focus on solving $A$ first by substituting the value that makes the other terms zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This is another partial fraction problem. One of the constants is already given as \( 2/9 \).
We need to determine the values of \( A \) and \( B \) to find their sum.
Step 2: Key Formula or Approach:
Multiply both sides by \( (x-1)^2(x+2) \) to clear the denominator:
\[ x = A(x+2) + \frac{2}{9}(x-1)(x+2) + B(x-1)^2 \]
Step 3: Detailed Explanation:

Solve for A: Let \( x = 1 \).
\[ 1 = A(1+2) + \frac{2}{9}(0) + B(0) \]
\[ 1 = 3A \implies A = \frac{1}{3} \]

Solve for B: Let \( x = -2 \).
\[ -2 = A(0) + \frac{2}{9}(0) + B(-2-1)^2 \]
\[ -2 = B(-3)^2 \]
\[ -2 = 9B \implies B = -\frac{2}{9} \]

Calculate A + B:
\[ A + B = \frac{1}{3} + \left(-\frac{2}{9}\right) \]
To add these, convert to a common denominator of 9:
\[ A + B = \frac{3}{9} - \frac{2}{9} \]
\[ A + B = \frac{1}{9} \]

Step 4: Final Answer:
The value of \( A + B \) is \( 1/9 \).