Question

If \[ \frac{x^2}{(x^2 + 2)(x^4 - 1)} = \frac{A}{x^2 - 1} + \frac{B}{x^2 + 1} + \frac{C}{x^2 + 2}, \text{ then } A + B - C =\ ? \]

• A. \( 0 \)
• B. \( \frac{4}{3} \)
• C. \( \frac{3}{4} \)
• D. \( 2 \)

Hint:

When doing partial fraction decomposition involving quadratics, group and compare coefficients of like powers of \( x \) after clearing the denominator.

Answer 1

The Correct Option is B

Step 1: Factor the denominator: \[ (x^2 + 2)(x^4 - 1) = (x^2 + 2)(x^2 - 1)(x^2 + 1) \] Step 2: The partial fraction decomposition is already given: \[ \frac{x^2}{(x^2 + 2)(x^2 - 1)(x^2 + 1)} = \frac{A}{x^2 - 1} + \frac{B}{x^2 + 1} + \frac{C}{x^2 + 2} \] Step 3: Multiply both sides by the denominator: \[ x^2 = A(x^2 + 1)(x^2 + 2) + B(x^2 - 1)(x^2 + 2) + C(x^2 - 1)(x^2 + 1) \] Now expand each term: - \( A(x^2 + 1)(x^2 + 2) = A(x^4 + 3x^2 + 2) \) - \( B(x^2 - 1)(x^2 + 2) = B(x^4 + x^2 - 2) \) - \( C(x^2 - 1)(x^2 + 1) = C(x^4 - 1) \) Step 4: Combine and equate the coefficients: \[ x^2 = (A + B + C)x^4 + (3A + B)x^2 + (2A - 2B - C) \] Now compare coefficients with LHS \( x^2 \), i.e., - Coefficient of \( x^4 \): \( A + B + C = 0 \) - Coefficient of \( x^2 \): \( 3A + B = 1 \) - Constant term: \( 2A - 2B - C = 0 \) Step 5: Solve the system: From \( A + B + C = 0 \Rightarrow C = -A - B \) Substitute into the others: - \( 3A + B = 1 \) - \( 2A - 2B - (-A - B) = 0 \Rightarrow 2A - 2B + A + B = 0 \Rightarrow 3A - B = 0 \Rightarrow B = 3A \) Now, plug into second equation: \[ 3A + 3A = 1 \Rightarrow 6A = 1 \Rightarrow A = \frac{1}{6},
B = \frac{1}{2},
C = -\left(\frac{1}{6} + \frac{1}{2}\right) = -\frac{2}{3} \] So, \[ A + B - C = \frac{1}{6} + \frac{1}{2} + \frac{2}{3} = \frac{2 + 3 + 4}{6} = \frac{9}{6} = \frac{3}{2} \] Wait! Correction: Actually, \(\frac{1}{6} + \frac{1}{2} = \frac{2 + 6}{12} = \frac{8}{12} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \) Final answer: \[ A + B - C = \frac{4}{3} \]