AP EAMCET · Maths · Basic of Mathematics
If \(\frac{3 x^3-7 x+1}{(x-2)^5}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4}+\frac{E}{(x-2)^5}\), then \(\mathrm{A}(\mathrm{B}+\mathrm{C}+\mathrm{D}+\mathrm{E})=\)
- A \(0\)
- B \(64\)
- C \(348\)
- D \(256\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\( P(x) = 3x^3-7x+1 \) \( P'(x) = 9x^2-7 \) \( P''(x) = 18x \) \( P'''(x) = 18 \) \( P^{(4)}(x) = 0 \) \( A = \frac{P^{(4)}(2)}{4!} = \frac{0}{24} = 0 \) \( A(B+C+D+E) = 0 \cdot (B+C+D+E) = 0 \)
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