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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 \(\mathrm{A}(\mathrm{B}+\mathrm{C}+\mathrm{D}+\mathrm{E})=\)

  1. A \(0\)
  2. B \(64\)
  3. C \(348\)
  4. D \(256\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0\)

Step-by-step Solution

Detailed explanation

\( P(x) = 3x^3-7x+1 \) \( P'(x) = 9x^2-7 \) \( P''(x) = 18x \) \( P'''(x) = 18 \) \( P^{(4)}(x) = 0 \) \( A = \frac{P^{(4)}(2)}{4!} = \frac{0}{24} = 0 \) \( A(B+C+D+E) = 0 \cdot (B+C+D+E) = 0 \)