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

If $n$ is odd, then numerically the greatest

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

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

coefficient is ${ }^{21}{C}_{10}$or, ${ }^{21}{C}_{11}$

Numerically greatest term 

${=}^{21}{C}_{11}{x}^{11}{\text{ or, }}^{21}{C}_{10}{x}^{10}$

${ }^{21}{C}_{11}{x}^{11}{>}^{21}{C}_{12}{x}^{12}$ 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 $x<\frac{5}{6}\Rightarrow x\in (5/6,6/5)$