Question

Given the equation: \[ \frac{A}{x-a} + \frac{Bx + C}{x^2 + b^2} = \frac{1}{(x-a)(x^2 + b^2)} \] then \( C \)=

• A. \( \frac{-1}{a^2 + b^2} \)
• B. \( \frac{1}{a^2 + b^2} \)
• C. \( \frac{-a}{a^2 + b^2} \)
• D. \( \frac{a}{a^2 + b^2} \)

Hint:

For partial fraction decomposition, equate coefficients of powers of \( x \) after taking the common denominator, then solve for unknowns systematically.

Answer 1

The Correct Option is C

We are given: \[ \frac{A}{x-a} + \frac{Bx + C}{x^2 + b^2} = \frac{1}{(x-a)(x^2 + b^2)} \] 
Step 1: Take LCM on the Left-Hand Side Rewriting the left-hand side with a common denominator: \[ \frac{A(x^2 + b^2) + (Bx + C)(x-a)}{(x-a)(x^2 + b^2)} = \frac{1}{(x-a)(x^2 + b^2)} \] Since the denominators are equal, equating the numerators: \[ A(x^2 + b^2) + (Bx + C)(x - a) = 1. \] 
Step 2: Expand the Equation Expanding both terms: \[ Ax^2 + A b^2 + Bx^2 - a Bx + Cx - aC = 1. \] \[ (A + B)x^2 + (-aB + C)x + (A b^2 - aC) = 1. \] 
Step 3: Compare Coefficients Since the right-hand side is just \( 1 \), we equate coefficients: 1. For \( x^2 \): \( A + B = 0 \Rightarrow B = -A \). 2. For \( x \): \( -aB + C = 0 \Rightarrow C = aB \). 3. For the constant term: \( A b^2 - aC = 1 \). 
Step 4: Solve for \( C \) Substituting \( B = -A \) into \( C = aB \): \[ C = a(-A) = -aA. \] From the constant term equation: \[ A b^2 - aC = 1. \] Substituting \( C = -aA \): \[ A b^2 - a(-aA) = 1. \] \[ A b^2 + a^2 A = 1. \] \[ A(a^2 + b^2) = 1. \] \[ A = \frac{1}{a^2 + b^2}. \] 
Step 5: Find \( C \) \[ C = -aA = -a \times \frac{1}{a^2 + b^2} = \frac{-a}{a^2 + b^2}. \] Thus, the correct answer is: \[ \boxed{\frac{-a}{a^2 + b^2}}. \]