Q59 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
$\begin{aligned} & \text { If } \frac{3 x^2+1}{\left(x^2+1\right)\left(x^2+2\right)^2}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2} \\ & +\frac{E x+F}{\left(x^2+2\right)^2}, \text { then } A+C+E=\end{aligned}$
- Correct answerA0
- B$\frac{7}{3}$
- C1
- D$\frac{4}{3}$
Solution
If
$$
\begin{array}{r}
\frac{3 x^2+1}{\left(x^2+1\right)\left(x^2+2\right)^2}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2}+\frac{E x+F}{\left(x^2+2\right)^2} \\
\Rightarrow \frac{3 x^2+1}{\left(x^2+1\right)\left(x^2+2\right)^2} \\
\end{array}
$$
$\begin{aligned} & A x+B\left(x^2+2\right)^2+(C x+D)\left(x^2+1\right) \\ = & \frac{\left(x^2+2\right)+(E x+F)\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2+2\right)^2}\end{aligned}$
$$
\begin{gathered}
\Rightarrow \quad 3 x^2+1=(A x+B)\left(x^2+2\right)^2+(C x+D) \\
\left(x^2+1\right)\left(x^2+2\right)+(E x+F)\left(x^2+1\right)
\end{gathered}
$$
On comparing the coefficient of different term, $A=0, C=0$ and $E=0$
So,
$$
A+C+E=0
$$
Hence, the correct answer is option: 'A'.