Binomial Theorem And Its Simple Applications Question 35
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).
Options:
A) $ [ \frac{5}{6},\frac{6}{5} ] $
B) $ ( \frac{5}{6},\frac{6}{5} ) $
C) $ ( \frac{4}{5},\frac{5}{4} ) $
D) $ [ \frac{4}{5},\frac{5}{4} ] $
Show Answer
Answer:
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, $ | ^{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!}\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 ( \frac{5}{6},\frac{6}{5} ) $
BETA
