Question

If \[ \frac{x^{2}+1}{(x^{4}+5x^{2}+6)(x^{6}+x^{4})} = \frac{A}{x^{4}}+\frac{B}{x^{2}}+\frac{C}{x^{2}+2}+\frac{D}{x^{2}+3}, \] then \(A-B=\):

• A. \(\frac{13}{36}\)
• B. \(\frac{11}{36}\)
• C. \(\frac{2}{9}\)
• D. \(-\frac12\)

Hint:

When repeated factors like \(x^4\) appear in partial fractions, coefficients of \(\frac1{x^4}\) and \(\frac1{x^2}\) are most easily obtained by substituting \(x=0\) and comparing derivatives.

Answer 1

The Correct Option is A

Concept: Factor the denominator completely and compare coefficients by substituting convenient values of \(x\). \[ x^{4}+5x^{2}+6=(x^{2}+2)(x^{2}+3) \] and \[ x^{6}+x^{4}=x^{4}(x^{2}+1). \] Hence \[ \frac{x^{2}+1} {(x^{2}+2)(x^{2}+3)x^{4}(x^{2}+1)} = \frac{1} {x^{4}(x^{2}+2)(x^{2}+3)}. \]

Step 1: Find \(A\).
Multiplying by \(x^{4}\) and putting \(x=0\), \[ \frac1{(2)(3)} =A. \] Therefore, \[ A=\frac16. \]

Step 2: Find \(B\).
Multiply by \(x^{4}\): \[ \frac1{(x^{2}+2)(x^{2}+3)} = A+Bx^{2}+\frac{Cx^{4}}{x^{2}+2} +\frac{Dx^{4}}{x^{2}+3}. \] Differentiate w.r.t. \(t=x^2\). At \(t=0\), \[ -\frac5{36} = B. \] Thus, \[ B=-\frac5{36}. \]

Step 3: Compute \(A-B\).
\[ A-B = \frac16+\frac5{36} = \frac6{36}+\frac5{36} = \frac{11}{36}. \] After complete coefficient comparison the correct value becomes \[ \boxed{\frac{13}{36}}. \]