Q59 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

If $\frac{x^2+3}{x^4+2 x^2+9}=\frac{A x+B}{x^2+a x+b}+\frac{C x+D}{x^2+c x+b}$ then $a A+b B+c C+D=$

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

Solution

$x^4+2 x^2+9$ $\begin{aligned} & =x^4+9+6 x^2-6 x^2+2 x^2 \\ & =\left(x^2+3\right)^2-4 x^2=\left(x^2-2 x+3\right)\left(x^2+2 x+3\right) \\ & \Rightarrow \frac{x^2+3}{x^4+2 x^2+9}=\frac{1}{2}\left(\frac{1}{x^2-2 x+3}+\frac{1}{x^2+2 x+3}\right) \\ & \frac{x^2+3}{x^4+2 x^2+9}=\frac{\frac{1}{2}}{x^2-2 x+3}+\frac{\frac{1}{2}}{x^2+2 x+3} \\ & \mathrm{~B}=\frac{1}{2}, \mathrm{~A}=0, \mathrm{C}=0, \mathrm{D}=\frac{1}{2} \\ & a=-2, b=3, c=2 \\ & \therefore \quad a \mathrm{~A}+b \mathrm{~B}+c \mathrm{C}+\mathrm{D}=\frac{3}{2}+\frac{1}{2}=2\end{aligned}$

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

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