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 \(B-2 A+C=\)
- A 23
- B 24
- C 25
- D 26
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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