Q59 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{\left(x^2+1\right)}+\frac{B}{\left(x^2+1\right)^2}+$ $\frac{C}{\left(x^2+1\right)^3}$, then $A+C=$

  • A
    12
  • B
    10
  • C
    9
  • Correct answer
    D
    6

Solution

$$ \begin{aligned} & \text { If } \frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2} \\ & +\frac{C}{\left(x^2+1\right)^3} \\ & \Rightarrow x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C \\ & \end{aligned} $$ On comparing the coefficient of different terms $$ \begin{aligned} & A=1 ; 2 A+B=24 \text { and } A+B+C=28 \\ & \Rightarrow \quad A=1, B=22 \text {, so } A+C=6 . \end{aligned} $$

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

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