The interval in which x must lie so that the numerically greatest term in the expansion of $(1 - x)^{21}$ has the numerically greatest coefficient, is

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$[\frac{4}{5}, \frac{5}{4}]$

$(\frac{5}{6}, \frac{6}{5})$

$[\frac{5}{6}, \frac{6}{5}]$

$(\frac{4}{5}, \frac{5}{4})$

answer is B.

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Detailed Solution

If $n$ is odd, then numerically the greatest

coefficient in the expansion of $(1 - x)^{n}$ is $^{n} C_{n - 1}$ or $, \frac{^{n} C_{n + 1} 2}{2} .$

Therefore, in case of ${(1 - x)}^{21}$ the numerically greatest

coefficient is $^{21} C_{10}$ or, $^{21} C_{11}$

Numerically greatest term

$=^{21} C_{11} x^{11} {or,}^{21} C_{10} x^{10}$

$^{21} C_{11} x^{11} >^{21} C_{12} x^{12}$ and $^{21} C_{10} x^{10} >^{21} C_{9} x^{9}$

$\Rightarrow \frac{21!}{10! 11!} > \frac{21!}{9! 12!} x$ and $\frac{21!}{11! 10!} x > \frac{21!}{9! 12!}$

$= \frac{6}{5} > x$ and $x < \frac{5}{6} \Rightarrow x \in (5 / 6, 6 / 5)$

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