Q60 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

If $\frac{x^2-x+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{\mathrm{A} x+\mathrm{B}}{x^2+1}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+x+1}$ then $\mathrm{A}+2 \mathrm{~B}+\mathrm{C}+2 \mathrm{D}=$

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

Solution

We are given that $\begin{aligned} & \frac{x^2-x+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{A x+B}{x^2+1}+\frac{(x+D)}{x^2+x+1} \\ & \text { Now } x^2-x+1=(A x+B)\left(x^2+x+1\right)+(c x+D)\left(x^2+1\right) \\ & \Rightarrow x^2-x+1=(A+C) x^3+(A+B+D) x^2+(A+B+C) x+B+D \\ & A+C=0 \\ & A+B+D=1 \\ & A+B+C=-1 \\ & B+D=1 \\ & \Rightarrow A=0, \quad C=0, \quad B=-1, \quad D=2 \\ & \therefore A+2 B+C+2 D=0-2+0+4=2 \end{aligned}$

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

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