Question

If the $7th$ term in the binomial expansion of $\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3}In\,x\right)^9, x > 0$, is equal to $729$, then $x$ can be:

• A. $e^2$
• B. $e$
• C. $\frac{e}{2}$
• D. $2e$

Answer 1

Answer: (B)

$Let \,r+1=7 \Rightarrow r=6$
Given expansion is
$\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3}ln\,x\right)^{^9}, x > 0$
We have
$T_{r+1}=^{n}C_{r}\left(x\right)^{n-r}\,a^{r} for \left(x+a\right)^{n}.$
$\therefore $ According to the question
$729=^{9}C_{6}\left(\frac{3}{\sqrt[3]{84}}\right)^{^3} .\left(\sqrt{3}ln\,x\right)^{6}$
$\Rightarrow 3^{6}=84\times\frac{3^{3}}{84}\times3^{3}\times\left(6\,ln\,x\right)$
$\Rightarrow \left(ln\,x\right)^{6}=1 \Rightarrow \left(ln\,x\right)^{6}=\left(ln\,e\right)^{6}$
$\Rightarrow x=e$