Binomial Theorem And Its Simple Applications Question 8
Question: If the 7th term in the binomial expansion of $ {{( \frac{3}{\sqrt[3]{84}}+\sqrt{3},lnx )}^{9}},x>0 $ , is equal to 729, then x can be
Options:
A) $ e^{2} $
B) e
C) $ \frac{e}{2} $
D) 2e
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Let $ r+1=7\Rightarrow r=6 $ Given expansion is $ {{( \frac{3}{\sqrt[3]{84}}+\sqrt{3},ln,x )}^{9}},x>0 $ We have $ {T_{r+1}}={{,}^{n}}C_{r}{{(x)}^{n-r}}a^{r} $ for $ {{(x+a)}^{n}} $ .
$ \therefore $ According to the question $ 729={{,}^{9}}C_6{{( \frac{3}{\sqrt[3]{84}} )}^{3}}.{{(\sqrt{3}lnx)}^{6}} $
$ \Rightarrow 3^{6}=84\times \frac{3^{3}}{84}\times 3^{3}\times (6,ln,x) $
$ \Rightarrow ,{{(ln,x)}^{6}}=1\Rightarrow {{(ln,x)}^{6}}={{(ln,e)}^{6}}\Rightarrow x=e $
BETA
