AP EAMCET · Maths · Binomial Theorem
If \(\frac{x^4}{(x-1)^2(x+1)}=\mathrm{A} x+\mathrm{B}+\frac{\mathrm{P}}{(x-1)}+\frac{\mathrm{Q}}{(x-1)^2}+\frac{\mathrm{R}}{x+1}\), then \(2 \mathrm{AP}-\mathrm{BQ}+\mathrm{R}=\)
- A 3
- B \(\frac{13}{4}\)
- C \(-\frac{11}{4}\)
- D \(-\frac{7}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{13}{4}\)
Step-by-step Solution
Detailed explanation
\(\frac{x^4}{(x-1)^2(x+1)} = \frac{x^4}{x^3-x^2-x+1} = x+1 + \frac{2x^2-1}{(x-1)^2(x+1)}\) \(\mathrm{A}=1, \mathrm{B}=1\) \(\frac{2x^2-1}{(x-1)^2(x+1)} = \frac{\mathrm{P}}{x-1} + \frac{\mathrm{Q}}{(x-1)^2} + \frac{\mathrm{R}}{x+1}\)…
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