Question

If \( \frac{1}{x^4 + 1} = \frac{Ax + B}{x^2 + \sqrt{2}x + 1} + \frac{Cx + D}{x^2 - \sqrt{2}x + 1} \), then \( BD - AC = \):

• A. \( \frac{3}{8} \)
• B. \( \frac{1}{8} \)
• C. \( 1 \)
• D. \( 0 \)

Hint:

For partial fractions, multiply both sides by the common denominator and equate the coefficients of corresponding powers of \( x \).

Answer 1

The Correct Option is A

We are given: \[ \frac{1}{x^4 + 1} = \frac{Ax + B}{x^2 + \sqrt{2}x + 1} + \frac{Cx + D}{x^2 - \sqrt{2}x + 1} \] 

Step 1: Common Denominator 
The denominator on the right side is: \[ (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1) = x^4 + 1 \] Thus, \[ \frac{1}{x^4 + 1} = \frac{(Ax + B)(x^2 - \sqrt{2}x + 1) + (Cx + D)(x^2 + \sqrt{2}x + 1)}{x^4 + 1} \] Equating the numerators, \[ 1 = (Ax + B)(x^2 - \sqrt{2}x + 1) + (Cx + D)(x^2 + \sqrt{2}x + 1) \] 

Step 2: Expanding Both Terms 
Expanding the first term: \[ (Ax + B)(x^2 - \sqrt{2}x + 1) = Ax^3 - A\sqrt{2}x^2 + Ax + Bx^2 - B\sqrt{2}x + B \] Expanding the second term: \[ (Cx + D)(x^2 + \sqrt{2}x + 1) = Cx^3 + C\sqrt{2}x^2 + Cx + Dx^2 + D\sqrt{2}x + D \] Now combine like terms: \[ 1 = (A + C)x^3 + (-A\sqrt{2} + B + C\sqrt{2} + D)x^2 + (A + C)x + (B + D) \] 

Step 3: Equating Coefficients 
By comparing coefficients: - \( A + C = 0 \quad \Rightarrow \quad C = -A \) - \( -A\sqrt{2} + B + C\sqrt{2} + D = 0 \) - \( A + C = 0 \quad \Rightarrow \quad C = -A \) - \( B + D = 1 \) 

Step 4: Solving for \( A, B, C, D \) 
Since \( C = -A \), substitute this into the second equation: \[ -B\sqrt{2} + B - A\sqrt{2} + D = 0 \] Now from \( B + D = 1 \), let \( B = \frac{1}{2} \) and \( D = \frac{1}{2} \). 

 

Step 5: Calculate \( BD - AC \) 
\[ BD - AC = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \] \[ = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{3}{8} \] 

 

Final Answer: (1) \( \frac{3}{8} \)