Question

If $\frac{x^{2} + 5}{\left(x^{2} + 1 \right)\left(x-2\right)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2} + 1 } , $ then $A + B + C = $

• A. -1
• B. $\frac{2}{5}$
• C. $\frac{-3}{5}$
• D. 0

Answer 1

Answer: (C)

We have,
$ \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) $
$\Rightarrow x^{2}+5=A\, x^{2}+A+B \,x^{2}-2 \,B x+C x-2 C$
Equating the coefficient of $x^{2}, x$ and constant
terms, we get
$41=A+B, 0=-2 B+C, 5=A-2 \,C$
Solving these equations, we get
$A=\frac{9}{5}, B=-\frac{4}{5}, C=-\frac{8}{5}$
$\therefore A+B+C= \frac{9-4-8}{5}$
$= \frac{-3}{5}$