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Q.
The interval in which $x$ must lies so that the numerically greatest term in the expansion of $(1- x )^{21}$ has the greatest coefficient is, $(x>0)$.
Binomial Theorem
Solution:
If n is odd, then numerically greatest coefficient in
the expansion of $(1-x)^{n}$ is $\frac{{ }^{n} C_{n-1}}{2}$ or $\frac{{ }^{n} C_{n+1}}{2}$.
Therefore in $(1- x )^{21}$, the numerically greatest coefficient is ${ }^{21} C _{10}$ or ${ }^{21} C _{11}$. So, the numerically greatest term
$={ }^{21} C _{11} x ^{11}$ or ${ }^{21} C _{10} x ^{10}$
So, $\left|{ }^{21} C _{11} x ^{11}\right|>\left|{ }^{21} C _{12} x ^{12}\right|$ and
$\left.\right|^{21} C_{10} x^{10}|>|^{21} C_{9} \cdot x^{9} \mid$
$\Rightarrow \frac{21 !}{10 ! 11 !}>\frac{21 !}{9 ! 12 !} \times$ and $\frac{21 !}{11 ! 10 !} x>\frac{21 !}{9 ! 12 !}\,\,\,\,\,(\because x>0)$
$\Rightarrow x<\frac{6}{5}$ and $x>\frac{5}{6} $
$\Rightarrow x \in\left(\frac{5}{6}, \frac{6}{5}\right)$