TS EAMCET · Maths · Binomial Theorem
If \(\frac{x-4}{x^2-5 x+6}\) can be expanded in the ascending powers of \(x\), then the coefficient of \(x^3\) is
- A \(\frac{-73}{648}\)
- B \(\frac{73}{648}\)
- C \(\frac{71}{648}\)
- D \(\frac{-71}{648}\)
Answer & Solution
Correct Answer
(A) \(\frac{-73}{648}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{x-4}{x^2-5 x+6}=\frac{x-4}{(x-2)(x-3)} \\ & =\frac{2}{(x-2)}-\frac{1}{(x-3)}=2(x-2)^{-1}-(x-3)^{-1} \\ & =2(-2)^{-1}\left(1-\frac{x}{2}\right)^{-1}-(-3)^{-1}\left(1-\frac{x}{3}\right)^{-1} \\ &…
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