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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 $2(\mathrm{~A}-\mathrm{C})=$
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Verified Answer
The correct answer is:
$\mathrm{B}$
$\begin{aligned}
& \frac{\mathrm{x}-2}{\mathrm{x}^2(2 \mathrm{x}-3)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}^2}+\frac{\mathrm{C}}{2 \mathrm{x}-3} \\
& \Rightarrow(\mathrm{x}-2)=\mathrm{Ax}(2 \mathrm{x}-3)+\mathrm{B}(2 \mathrm{x}-3)+C \cdot \mathrm{x}^2
\end{aligned}$
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 $\mathrm{eq}^{\mathrm{n}}$.
$\begin{aligned}
& -3 \mathrm{~A}+2 \mathrm{~B}=1 \Rightarrow \mathrm{A}=\frac{1}{9} \\
& \therefore 2(\mathrm{~A}-\mathrm{C})=2\left[\frac{1}{9}+\frac{2}{9}\right]=2 \times \frac{3}{9}=\frac{2}{3}=\mathrm{B}
\end{aligned}$
& \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 $\mathrm{eq}^{\mathrm{n}}$.
$\begin{aligned}
& -3 \mathrm{~A}+2 \mathrm{~B}=1 \Rightarrow \mathrm{A}=\frac{1}{9} \\
& \therefore 2(\mathrm{~A}-\mathrm{C})=2\left[\frac{1}{9}+\frac{2}{9}\right]=2 \times \frac{3}{9}=\frac{2}{3}=\mathrm{B}
\end{aligned}$
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