AP EAMCET · Maths · Basic of Mathematics
\(\frac{2 x^2+1}{x^3-1}=\frac{A}{x-1}+\frac{B x+C}{x^2+x+1} \Rightarrow 7 A+2 B+C=\)
- A \(8\)
- B \(9\)
- C \(10\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
Here, \(\frac{2 x^2+1}{x^3-1}=\frac{A}{x-1}+\frac{B x+C}{x^2+x+1}\) Resolving into partial fractions, \(\frac{2 x^2+1}{x^3-1}=\frac{A\left(x^2+x+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+x+1\right)}\)…
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