Q49 / 160 · AP EAMCET 2013 · Mathematics

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 $C+D$ is equal to

  • A
    $-1$
  • B
    $1$
  • C
    $2$
  • Correct answer
    D
    $0$

Solution

Given, $\begin{gathered} \frac{1}{x^4+x^2+1}=-\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1} \\ \Rightarrow 1=(A x+B)\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right) \\ \Rightarrow 1=\left(A x^3+A x^{2+} A x+B x^2-B x+B\right) \\ \quad+\left(C x^3+C x^2+C x+D x^2+D x+D\right) \\ \Rightarrow 1=(A+C) x^3+(-A+B+C+D) x^2 \\ \quad+(A-B+C+D) x+(B+D) \end{gathered}$ On comparing, the coefficient of like powers on both sides, we get $\begin{aligned} A+C & =0 \\ -A+B+C+D & =0 \\ A-B+C+D & =0 \\ B+D & =1 \end{aligned}$ and On adding Eqs. (ii) and (iii), we get $\begin{aligned} 2(C+D) & =0 \\ \Rightarrow \quad C+D & =0 \end{aligned}$

Hence, the correct answer is option: 'D'.

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