AP EAMCET · Maths · Binomial Theorem
The absolute value of the difference of the coefficients of \(x^4\) and \(x^6\) in the expansion of \(\frac{2 x^2}{\left(x^2+1\right)\left(x^2+2\right)}\) is
- A \(\frac{13}{4}\)
- B \(\frac{1}{4}\)
- C \(\frac{9}{4}\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(\frac{13}{4}\)
Step-by-step Solution
Detailed explanation
Since \(\frac{2 x^2}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{-2}{x^2+1}+\frac{4}{x^2+2}\) \(=-2\left(1+x^2\right)^{-1}+\frac{4}{2}\left(1+\frac{x^2}{2}\right)^{-1}\) \(=-2\left(1+x^2\right)^{-1}+2\left(1+\frac{x^2}{2}\right)^{-1}\)…
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