Q. For $|x| < 1,$ the constant term in the expansion of $\frac{1}{(x - 1)^{2}(x - 2)}\ is$

A) 2
B) 1
C) 0
D) $- \frac{1}{2}$
Explanation

(d)
$\frac{1}{(x - 1)^{2}(x - 2)} = \frac{1}{- 2(1 - x)^{2}\left( 1 - \frac{x}{2} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2}\left\lbrack (1 - x)^{- 2}\left( 1 - \frac{x}{2} \right)^{- 1} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2}\left\lbrack (1 + 2x + \ldots)\left( 1 + \frac{x}{2} + \ldots \right) \right\rbrack$
$\therefore Coefficient\ of\ constant\ term\ is - \frac{1}{2}$.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.