Question

If \[ \frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} \] then \( A + B - C + D = ? \)

• A. \( a \)
• B. \( 2a \)
• C. \( 3a \)
• D. \( 4a \)

Hint:

When solving equations involving partial fractions, first express them with a common denominator, expand the terms, and equate coefficients of like terms to find the unknown values.

Answer 1

The Correct Option is B

We start with the given equation: \[ \frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} \] 

Step 1: 
Find a common denominator for the right-hand side: \[ \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} = \frac{(Ax + B)(x^2 - ax + 1) + (Cx + D)(x^2 + ax + 1)}{(x^2 + ax + 1)(x^2 - ax + 1)} \] Expanding the numerators: \[ (Ax + B)(x^2 - ax + 1) = Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B \] \[ (Cx + D)(x^2 + ax + 1) = Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D \] Adding these, \[ (Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B) + (Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D) \] Grouping like terms: \[ (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \] Equating with the left-hand side: \[ 1 = (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \] Since there is no \( x^3 \), \( x^2 \), and \( x \) terms on the left-hand side, we set: \[ A + C = 0, \quad -Aa + Ca + B + D = 0, \quad A + C = 0, \quad -Ba + Da + B + D = 1 \] Solving these equations, we get: \[ A = -C, \quad B + D = Aa - Ca, \quad B + D = 1 \] From \( B + D = 1 \) and substituting the values, we find: \[ A + B - C + D = 2a. \] Thus, the correct answer is \( \mathbf{2a} \).