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If $\frac{x^2}{(2 x-1)(x+2)(x-3)}$ $=A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$, then $A$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Given that
$\begin{gathered}
\frac{x^3}{(2 x-1)(x+2)(x-3)}= \\
A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}
\end{gathered}$
Let
$\begin{aligned}
f(x) & =\frac{x^3}{(2 x-1)(x+2)(x-3)} \\
& =\frac{x^3}{(2 x-1)\left(x^2-x-6\right)} \\
& =\frac{x^3}{2 x^3-3 x^2-11 x+6}
\end{aligned}$
Here we see that the power of $x$ will be same in Nr and Dr.
$\therefore$ First we divide the numerator by denominator
$\begin{array}{cc}\Rightarrow & \frac{x^3}{(2 x-1)(x+2)(x-3)} \\ = & \frac{1}{2}+\frac{\frac{3}{2} x^2+\frac{11}{2} x-3}{(2 x-1)(x+2)(x-3)} \\ \Rightarrow & A=\frac{1}{2}\end{array}$
$\begin{gathered}
\frac{x^3}{(2 x-1)(x+2)(x-3)}= \\
A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}
\end{gathered}$
Let
$\begin{aligned}
f(x) & =\frac{x^3}{(2 x-1)(x+2)(x-3)} \\
& =\frac{x^3}{(2 x-1)\left(x^2-x-6\right)} \\
& =\frac{x^3}{2 x^3-3 x^2-11 x+6}
\end{aligned}$
Here we see that the power of $x$ will be same in Nr and Dr.
$\therefore$ First we divide the numerator by denominator
$\begin{array}{cc}\Rightarrow & \frac{x^3}{(2 x-1)(x+2)(x-3)} \\ = & \frac{1}{2}+\frac{\frac{3}{2} x^2+\frac{11}{2} x-3}{(2 x-1)(x+2)(x-3)} \\ \Rightarrow & A=\frac{1}{2}\end{array}$
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