Question

If $\frac{(x^2+1)}{(x^2+2)(x^2+3)}=\frac{-1}{x^2+2}+\frac{\alpha }{x^2+3}$, then $\alpha =$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Given, $\frac{(x^2+1)}{(x^2+2)(x^2+3)}=\frac{1}{x^2+2}+\frac{\alpha }{x^2+3}$ ....(1)
Resolving into partial fractions
$\frac{(x^2+1)}{(x^2+2)(x^2+3)}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+3}$ ....(2)
$\frac{(x^2+1)}{(x^2+2)(x^2+3)}=\frac{(Ax+B)(x^2+3)+(Cx+D)(x^2+2)}{(x^2+2)(x^2+3)}$
$\Rightarrow x^2+1=(A+C)x^3+(B+D)x^2+(3A+2C)x+(3B+2D)$
On comparing , we get
$A+C=0,B+D=1,3A+2C=0,3B+2D=1$
Solving these equations, we get
$A=C=0$ and $B=-1,D=2$
Put these values in (2)
$\frac{(x^2+1)}{(x^2+2)(x^2+3)}=\frac{-1}{(x^2+2)}+\frac{2}{(x^2+3)}$
On comparing with (1),
$\alpha =2$