TS EAMCET · Maths · Quadratic Equation
If \(\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}\), then \(\cos ^{-1}(A+B+C+D)=\)
- A \(\frac{\pi}{2}\)
- B 0
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
Given, \(\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}\) \(\Rightarrow \frac{1}{x^4+x^2+1}\) \(=\frac{(A x+B)\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right)}{x^4+x^2+1}\) \(\Rightarrow \mathrm{l}=(A+C) x^3+(B-A+C+D) x^2\) \(+(A-B+C+D) x+(B+D)\) On comparing…
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