Binomial Theorem And Its Simple Applications Question 38
Question: If the fourth term in the expansion of $ {{( \sqrt{{x^{( \frac{1}{\log ,x+1} )}}}+{x^{1/12}} )}^{6}} $ is equal to 200 and $ x > 1 $ , then x is equal to $ (log=log_{10}) $
Options:
A) $ {10^{\sqrt{2}}} $
B) 10
C) $ 10^{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given, $ T_4=200 $
$ \Rightarrow {{,}^{6}}C_3{{( \sqrt{{x^{( \frac{1}{\log x+1} )}}} )}^{3}}{{({x^{1/12}})}^{3}}=200 $
$ \Rightarrow 20.{x^{\frac{3}{2(log,x+1)}+\frac{1}{4}=}}200 $
$ \Rightarrow {x^{{ \frac{3}{2(log,x+1)}+\frac{1}{4} }}}=10 $
$ \Rightarrow ,\frac{3}{2(\log ,x+1)}+\frac{1}{4}=,{\log_{x}}10=\frac{1}{{\log_{10}}x} $
$ \Rightarrow \frac{3}{2(y+1)}+\frac{1}{4}=\frac{1}{y} $ where $ y={\log_{10}}x $
$ \Rightarrow y=-4 $ or $ y=1 $
$ \Rightarrow {\log_{10}}x=-4 $ or
$ \Rightarrow {\log_{10}}x=1 $
$ \Rightarrow x={10^{-4}} $ or $ 10\Rightarrow x=10(\because ,x>1) $
BETA
