AP EAMCET · Maths · Basic of Mathematics
If \(\frac{x^2}{\left(x^2+2\right)\left(x^4-1\right)}=\frac{A}{x^2-1}+\frac{B}{x^2+1}+\frac{C}{x^2+2}\), then \(A+B-C=\)
- A \(0\)
- B \(\frac{4}{3}\)
- C \(\frac{3}{4}\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(x^2=A(x^2+1)(x^2+2)+B(x^2-1)(x^2+2)+C(x^2-1)(x^2+1)\) Let \(x^2=1\): \(1=A(1+1)(1+2)\) \(\Rightarrow\) \(1=6A\) \(\Rightarrow\) \(A=\frac{1}{6}\) Let \(x^2=-1\): \(-1=B(-1-1)(-1+2)\) \(\Rightarrow\) \(-1=-2B\) \(\Rightarrow\) \(B=\frac{1}{2}\) Let \(x^2=-2\):…
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