Q59 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$ then $B D-A C=$
- Correct answerA$\frac{3}{8}$
- B$\frac{1}{8}$
- C$1$
- D$0$
Solution
Since, given
$\begin{aligned}
& \frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1} \\
& \Rightarrow(A x+B)\left(x^2-\sqrt{2} x+1\right)+(C x+D)\left(x^2+\sqrt{2} x+1\right)=1
\end{aligned}$
If $x=0 \Rightarrow B+D=1$
Equating co-efficient of $x^3 \Rightarrow A+C=0$
Equating co-efficient of $x^2 \Rightarrow B-\sqrt{2} A+D+\sqrt{2} C=0$
$\Rightarrow 1+\sqrt{2}(C-A)=0 \Rightarrow A-C=\frac{1}{\sqrt{2}}$
Aftersolving $A+C=0$ and $A-C=\frac{1}{\sqrt{2}} \Rightarrow A=\frac{1}{2 \sqrt{2}}, C=\frac{-1}{2 \sqrt{2}}$
Equating co-efficient of $x \Rightarrow A-\sqrt{2} B+C+\sqrt{2} D=0$
$\Rightarrow B-D=0 \Rightarrow B=\frac{1}{2}, D=\frac{1}{2}$
Now, $B D-A C \Rightarrow \frac{1}{4}+\frac{1}{8}=\frac{3}{8}$.
Hence, the correct answer is option: 'A'.