AP EAMCET · Maths · Indefinite Integration
\(\begin{aligned} & \text { If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text {, then } P(0)+A(a-b)(a-c)=\end{aligned}\)
- A \(a^4+b^4+c^4+a\)
- B \(a+b+c\)
- C \(a^4-a-b-c\)
- D \(a+b+c+a^4\)
Answer & Solution
Correct Answer
(D) \(a+b+c+a^4\)
Step-by-step Solution
Detailed explanation
If is given that \[ \begin{array}{r} \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} \\ \Rightarrow x^4=(x-a)(x-b)(x-c) P(x)+A(x-b)(x-c) \\ +B(x-c)(x-a)+C((x-a)(x-b) \end{array} \] At \(x=0, a b c P(0)=b c A+c a B+a b C\) At…
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