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) \[\left[ \frac{4}{5},\frac{5}{4} \right]\]
Correct Answer: B
Solution :
| [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 |
| \[|{{\,}^{21}}{{C}_{10}}{{x}^{10}}|\,\,>\,\,{{|}^{21}}{{C}_{9}}.{{x}^{9}}|\] |
| \[\Rightarrow \frac{21!}{10!11!}>\frac{21!}{9!12!}\times \] and \[\frac{21!}{11!10!}x>\frac{21!}{9!12!}\] |
| \[(\because \,\,\,\,x>0)\] |
| \[\Rightarrow x<\frac{6}{5}\] and \[x>\frac{5}{6}\Rightarrow x\in \left( \frac{5}{6},\frac{6}{5} \right)\] |
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