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To find the coefficient of $x^4$ in the expansion of $\frac{3 x}{(x-2)(x-1)}$, the interval in which the expansion isvalid, is
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Verified Answer
The correct answer is:
$-1 < x < 1$
Given, $\frac{3 x}{(x-2)(x-1)}$ can be written as $\frac{6}{x-2}-\frac{3}{x-1}$
$\begin{aligned}
& \therefore \frac{3 x}{(x-2)(x-1)}=\frac{6}{x-2}-\frac{3}{x-1} \\
& =6(x-2)^{-1}-3(x-1)^{-1} \\
& =-3\left(1-\frac{x}{2}\right)+3(x-1)^{-1}
\end{aligned}$
It is valid iff $\left|\frac{x}{2}\right| < 1$ and $|x| < 1$
$\begin{aligned}
& \Rightarrow \quad|x| < 2 \text { and }|x| < 1 \\
& \Rightarrow \quad x \in(-2,2) \text { and } x \in(-1,1) \quad \therefore \quad x \in(-1,1)
\end{aligned}$
Hence, $-1 < x < 1$
$\begin{aligned}
& \therefore \frac{3 x}{(x-2)(x-1)}=\frac{6}{x-2}-\frac{3}{x-1} \\
& =6(x-2)^{-1}-3(x-1)^{-1} \\
& =-3\left(1-\frac{x}{2}\right)+3(x-1)^{-1}
\end{aligned}$
It is valid iff $\left|\frac{x}{2}\right| < 1$ and $|x| < 1$
$\begin{aligned}
& \Rightarrow \quad|x| < 2 \text { and }|x| < 1 \\
& \Rightarrow \quad x \in(-2,2) \text { and } x \in(-1,1) \quad \therefore \quad x \in(-1,1)
\end{aligned}$
Hence, $-1 < x < 1$
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