Q59 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\begin{aligned} & \frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2} \\ & \Rightarrow 3 A+2 B+3 C=\end{aligned}$
- A$-D$
- B$D$
- Correct answerC$2 D$
- D$-2 D$
Solution
$\frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2}$
$\Rightarrow \frac{x^2+1}{x^4+4}$
$=\frac{(A x+B)\left(x^2+2 x+2\right)+(C x+D)\left(x^2-2 x+2\right)}{\left(x^2+2\right)^2-(2 x)^2}$
$\Rightarrow x^2+1=(A+C) x^3+(2 A+B-2 C+D) x^2$
$+(2 A+2 B+2 C-2 D) x+(2 B+2 D)$
On comparing both sides,
$A+C=0 \Rightarrow A=-C, 2 B+2 D=1$
and $2 A+B-2 C+D=1$
$\begin{aligned} & \Rightarrow \quad 2(A-C)+B+D=1 \\ & \Rightarrow \quad 4 A+\left(\frac{1}{2}\right)=1 \\ & \Rightarrow \quad 4 A=1-\frac{1}{2}=\frac{1}{2} \\ & \Rightarrow \quad A=\frac{1}{8} \text { and } C=-\frac{1}{8}\end{aligned}$
Now, $2 A+2 B+2 C-2 D=0$
$\begin{aligned} & \Rightarrow \quad 2(A+C)+2(B-D)=0 \\ & \Rightarrow \quad B=D \\ & \text { and } 2 B+2 D=1\end{aligned}$
$\Rightarrow \quad 4 B=1 \Rightarrow B=D=\frac{1}{4}$
Now, $3 A+2 B+3 C=2 B \quad[\because 3 A+3 C=0]$
$=2 D$
Hence, the correct answer is option: 'C'.