Binomial Theorem And Its Simple Applications Question 36
Question: If the middle term in the expansion of $ {{( \frac{1}{x}+x,\sin ,x )}^{10}} $ equals to $ 7\frac{7}{8} $ then x is equal to; $ (n\in I) $
Options:
A) $ 2n\pi \pm \frac{\pi }{6} $
B) $ n\pi +\frac{\pi }{6} $
C) $ n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
D) $ n\pi +{{(-1)}^{n}}\frac{5\pi }{6} $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Middle term in the expansion is $ {{( \frac{10}{2}+1 )}^{th}} $ i.e., 6th tern. Thus $ T_6=7\frac{7}{8}\Rightarrow {{,}^{10}}C_5\frac{1}{x^{5}}.x^{5}{{\sin }^{5}}x=\frac{63}{8} $
$ \Rightarrow 252.{{\sin }^{5}}x=\frac{63}{8}\Rightarrow {{\sin }^{5}}x=\frac{1}{32} $
$ \Rightarrow \sin x=\frac{1}{2} $
$ \therefore x=n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
BETA
