Q59 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

The coefficient of $x^6$ in the power series expansion of $\frac{x^4-12 x^2+7}{\left(x^2+1\right)^3}$ is

  • A
    149
  • B
    -253
  • Correct answer
    C
    -145
  • D
    253

Solution

We have, $$ \begin{aligned} & \frac{x^4-12 x^2+7}{\left(x^2+1\right)^3}=\left(x^4-12 x^2+7\right)\left(x^2+1\right)^{-3} \\ & =\left(x^4-12 x^2+7\right)\left(1-3 x^2+\frac{12 x^4}{2 !}-\frac{60}{3 !} x^6 \ldots\right) \\ & =\left(x^4-12 x^2+7\right)\left(1-3 x^2+6 x^4-10 x^6 \ldots\right) \end{aligned} $$ Coefficient of $x^6$ is $(-3-72-70)=-145$

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

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