Q49 / 160 · AP EAMCET 2009 · Mathematics
For $|x| < 1$, the constant term in the expansion of $\frac{1}{(x-1)^2(x-2)}$ is
- A2
- B1
- C0
- Correct answerD$-\frac{1}{2}$
Solution
$\begin{aligned}
\frac{1}{(x-1)^2(x-2)} & =\frac{1}{-2(1-x)^2\left(1-\frac{x}{2}\right)} \\
& =-\frac{1}{2}\left[(1-x)^{-2}\left(1-\frac{x}{2}\right)^{-1}\right] \\
& =-\frac{1}{2}\left[(1+2 x+\ldots)\left(1+\frac{x}{2}+\ldots\right)\right]
\end{aligned}$
$\therefore$ Coefficient of constant term is $-\frac{1}{2}$.
Hence, the correct answer is option: 'D'.