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TS EAMCET · Maths · Binomial Theorem

If the partial fractions decomposition of \(\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}\) is \(\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}\) then \(B-2 A+C=\)

  1. A 23
  2. B 24
  3. C 25
  4. D 26
Verified Solution

Answer & Solution

Correct Answer

(C) 25

Step-by-step Solution

Detailed explanation

Given partial fraction of \(\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}\) \(\therefore x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C\) Now compare the coefficient of \(x^4, x^2\) constant…
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