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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
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
    $e^2$
  • ✓
    $e$
  • C
    $\frac {e}{2}$
  • D
    $2e$
✓

Answer

Correct option: B.
$e$
b
$\mathrm{T}_{7}=^{9} \mathrm{C}_{6}\left(\frac{3}{(84)^{1 / 3}}\right)^{3}(\sqrt{3} \ln \mathrm{x})^{6}=729$

$\Rightarrow(\ln \mathrm{x})^{6}=1$

$\Rightarrow \ln \mathrm{x}=\pm 1$

$\Rightarrow \mathrm{x}=\mathrm{e}, 1 / \mathrm{e}$

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