Q59 / 160 · TS EAMCET 2018 (04 May Shift 2) · 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 $A-B+C=$

  • A
    1
  • Correct answer
    B
    2
  • C
    3
  • D
    5

Solution

$$ \begin{aligned} & \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3} \\ & (3 x+2)=A\left(2 x^2+3\right)+(x+1)(B x+C) \end{aligned} $$ when $$ \begin{aligned} x & =-1 \\ 5 A & =-1 \quad \Rightarrow A=-\frac{1}{5} \end{aligned} $$ When $$ \begin{aligned} & x=0 \quad \Rightarrow 3 A+C=2 \\ & C=2+\frac{3}{5} \Rightarrow \frac{13}{5} \end{aligned} $$ When $$ x=1 $$ $$ \begin{aligned} 5 A+2 B+2 C & =5 \\ -1+2 B+\frac{26}{5} & =5 \Rightarrow 2 B=6-\frac{26}{5} \\ 2 B & =\frac{4}{5} \Rightarrow B=\frac{2}{5} \end{aligned} $$ $$ \therefore \quad A-B+C=-\frac{1}{5}-\frac{2}{5}+\frac{13}{5}=\frac{10}{5}=2 $$

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

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