Question

$\frac{x^2-x-3}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\Rightarrow 2A+B+C=$

• A. $0$
• B. $4$
• C. $2$
• D. $6$

Answer 1

Answer: (C)

$2$

$LHS=\frac{x^2-x-3}{x^3-1};RHS=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

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

Comparing the co-efficients of $x^2,x,1$

on LHS and RHS
Co- efficients of $x^2$

$1=A+B$

Co- efficients of $x$

$A-B+C=-1$

$A-C=-3;A=-1;B=2;C=2$

So, $2A+B+C=2(-1)+2+2$
$=2$