- Home
- Standard 11
- Mathematics
7.Binomial Theorem
normal
If $7^{th}$ term from beginning in the binomial expansion ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln \,x} \right)^9},\,x > 0$ is equal to $729$ , then possible value of $x$ is
A
$e^2$
B
$e$
C
$\frac {e}{2}$
D
$2e$
Solution
$\mathrm{T}_{7}=^{9} \mathrm{C}_{6}\left(\frac{3}{(84)^{1 / 3}}\right)^{3}(\sqrt{3} \ln \mathrm{x})^{6}=729$
$\Rightarrow(\ln \mathrm{x})^{6}=1$
$\Rightarrow \ln \mathrm{x}=\pm 1$
$\Rightarrow \mathrm{x}=\mathrm{e}, 1 / \mathrm{e}$
Standard 11
Mathematics
