Question

$If\frac{x^2}{(x^2+1)(x^2+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}$ then $(A,C)=$

• A. $(1,-1)$
• B. $(1,1)$
• C. $(0,0)$
• D. $(1,2)$

Answer 1

Answer: (C)

$(0,0)$

$\frac{x^2}{(x^2+1)(x^2+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}$
$\Rightarrow x^2=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)$
Put $x=1\Rightarrow 3A+3B+2C+2D=1$
Put $x=-1\Rightarrow -3A+3B-2C+2D=1$
$\Rightarrow 6A+4C=0$
$\Rightarrow A=-\frac{2}{3}C$ .....(i)
Put $x=2\Rightarrow 12A+6B+10C+5D=4$
Put $x=-2\Rightarrow -12A+6B-10C+D=$
$\Rightarrow 24A+20C=0$ .....(ii)
On solving , we get $A=0,C=0$
Hence, $(A,C)=(0,0)$