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: \[ \boxed{\frac{-a}{a^2 + b^2}}. \]

Answer 2

Given the equation:

\[ \frac{A}{x - a} + \frac{B x + C}{x^2 + b^2} = \frac{1}{(x - a)(x^2 + b^2)}, \]

Step 1: Express the right side as a linear combination.
Rewrite the right-hand side as a sum of partial fractions with unknown coefficients \(A\), \(B\), and \(C\). Multiplying both sides by the denominator \((x - a)(x^2 + b^2)\), we have:

\[ A(x^2 + b^2) + (B x + C)(x - a) = 1. \]

Step 2: Rearrange the expression.
Expand and collect terms by powers of \(x\):

\[ A x^2 + A b^2 + B x^2 - a B x + C x - a C = 1. \]

Group like terms:

\[ (A + B) x^2 + (-a B + C) x + (A b^2 - a C) = 1. \]

Step 3: Write the left side as a quadratic polynomial and equate to constant.
The right side is constant \(1\), meaning the coefficients of \(x^2\) and \(x\) terms on the left must be zero:

\[ A + B = 0, \] \[ -a B + C = 0, \] \[ A b^2 - a C = 1. \]

Step 4: Express \(B\) and \(C\) in terms of \(A\).
From the first equation:

\[ B = -A. \]

From the second equation:

\[ C = a B = a (-A) = -a A. \]

Step 5: Substitute \(B\) and \(C\) back into the third equation and solve for \(A\).

\[ A b^2 - a (-a A) = 1 \implies A b^2 + a^2 A = 1, \] \[ A (a^2 + b^2) = 1, \] \[ A = \frac{1}{a^2 + b^2}. \]

Step 6: Find the value of \(C\) using the value of \(A\).

\[ C = -a A = -a \times \frac{1}{a^2 + b^2} = \frac{-a}{a^2 + b^2}. \]

Final result:

\[ \boxed{\frac{-a}{a^2 + b^2}}. \]