Discussion Forum

1)

If   $\frac{x^{2}+5}{(x^{2}+1)(x-2)}=\frac{A}{x-2}+\frac{ Bx+C}{x^{2}+1}$

 then A+B+C=

Answer:

Option C

Explanation:

We have,

$\frac{x^{2}+5}{(x^{2}+1)(x-2)}=\frac{A}{x-2}+\frac{ Bx+C}{x^{2}+1}$

 $\Rightarrow$    $x^{2}+5=A(x^{2}+1)+(Bx+C)(x-2)$

$\Rightarrow$     $x^{2}+5=Ax^{2}+A+Bx^{2}-2Bx+Cx-2C$

 Equating  the coefficient of x² ,x and constant terms , we get

 1=A+B,0=-2B+C,5=A-2C

solving these equations , we get

 $A=\frac{9}{5},B=-\frac{4}{5},C=-\frac{8}{5}$

$\therefore$ A+B+C= $\frac{9-4-8}{5}$

                   =$\frac{-3}{5}$