Q59 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

If $\frac{x-2}{x^2(2 x-3)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{\mathrm{C}}{2 x-3}$ then $2(\mathrm{~A}-\mathrm{C})=$

  • A
    $3 \mathrm{~B}$
  • B
    $2 \mathrm{~B}$
  • C
    0
  • Correct answer
    D
    $\mathrm{B}$

Solution

$\begin{aligned} & \frac{\mathrm{x}-2}{\mathrm{x}^2(2 \mathrm{x}-3)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}^2}+\frac{\mathrm{C}}{2 \mathrm{x}-3} \\ & \Rightarrow(\mathrm{x}-2)=\mathrm{Ax}(2 \mathrm{x}-3)+\mathrm{B}(2 \mathrm{x}-3)+C \cdot \mathrm{x}^2 \end{aligned}$ Put $\mathrm{x}=0$ in $\mathrm{eq}^{\mathrm{n}}(1)$ $\Rightarrow \mathrm{B}=\frac{2}{3}$ Put $\mathrm{x}=\frac{3}{2}$ in $\mathrm{eq}^{\mathrm{n}}(1)$ $\mathrm{c}=\frac{-2}{9}$ Now compairing coefficients of $x$ in both sides of $\mathrm{eq}^{\mathrm{n}}$. $\begin{aligned} & -3 \mathrm{~A}+2 \mathrm{~B}=1 \Rightarrow \mathrm{A}=\frac{1}{9} \\ & \therefore 2(\mathrm{~A}-\mathrm{C})=2\left[\frac{1}{9}+\frac{2}{9}\right]=2 \times \frac{3}{9}=\frac{2}{3}=\mathrm{B} \end{aligned}$

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

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