Q51 / 160 · AP EAMCET 2006 · Mathematics

If $\frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3}$, then $A+C-B$ is equal to :

  • A
    0
  • Correct answer
    B
    2
  • C
    3
  • D
    5

Solution

$\because \quad \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3}$ $\Rightarrow \quad 3 x+2=A\left(2 x^2+3\right)+(B x+C)(x+1)$ On putting $x+1=0$ $3(-1)+2=A\left[2(-1)^2+3\right]$ $\Rightarrow \quad-3+2=A(5)$ $\Rightarrow \quad A=-\frac{1}{5}$ Now on comparing the coefficients of like power of $x$ Coefficient of $x^2$ $0=2 A+B$ $\Rightarrow \quad B=\frac{2}{5}$ and coefficient of $x$ $3=B+C$ $\Rightarrow \quad C=3-\frac{2}{5}=\frac{13}{5}$ $\therefore \quad A+C-B=-\frac{1}{5}+\frac{13}{5}-\frac{2}{5}$ $=\frac{13-2-1}{5}=\frac{10}{5}=2$

Hence, the correct answer is option: 'B'.

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