Question

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, $(x>0)$.

• A. $\left[\frac{5}{6},\frac{6}{5}\right]$
• B. $\left(\frac{5}{6},\frac{6}{5}\right)$
• C. $\left(\frac{4}{5},\frac{5}{4}\right)$
• D. $\left[\frac{4}{5},\frac{5}{4}\right]$

Answer 1

Answer: (B)

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
${|}^{21}{C}_{10}{x}^{10}|>{|}^{21}{C}_9\cdot x^9\mid$
$\Rightarrow \frac{21!}{10!11!}>\frac{21!}{9!12!}\times$ and $\frac{21!}{11!10!}x>\frac{21!}{9!12!}(\because x>0)$
$\Rightarrow x<\frac{6}{5}$ and $x>\frac{5}{6}$
$\Rightarrow x\in \left(\frac{5}{6},\frac{6}{5}\right)$