Q58 / 160 · TS EAMCET 2017 · Mathematics
If $\frac{x^2+5}{\left(x^2+1\right)(x-2)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1}$, then $A+B+C=$
- A-1
- B$\frac{2}{5}$
- Correct answerC-3
- D0
Solution
We have,
$$
\begin{aligned}
& \frac{x^2+5}{\left(x^2+1\right)(x-2)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1} \\
\Rightarrow & x^2+5=A\left(x^2+1\right)+(B x+C)(x-2) \\
\Rightarrow & x^2+5=A x^2+A+B x^2-2 B x+C x-2 C
\end{aligned}
$$
Equating the coefficient of $x^2, x$ and constant terms, we get
$$
1=A+B, 0=-2 B+C, 5=A-2 C
$$
Solving these equations, we get
$$
\begin{aligned}
& A=\frac{9}{5}, B=-\frac{4}{5}, C=-\frac{8}{5} \\
& \begin{aligned}
\therefore A+B+C & =\frac{9-4-8}{5} \\
& =\frac{-3}{5}
\end{aligned}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.