Question:
If the 7th term in the binomial expansion of \(\left(\dfrac{3}{\sqrt[3]{84}} + \sqrt{3}\ln x\right)^9\), \(x \gt 0\), is equal to 729 , then \(x\) can be:
If the 7th term in the binomial expansion of \(\left(\dfrac{3}{\sqrt[3]{84}} + \sqrt{3}\ln x\right)^9\), \(x \gt 0\), is equal to 729 , then \(x\) can be:
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Use \(T_7 = {}^9C_6 \cdot a^3 b^6\), where \(a = 3/\sqrt[3]{84}\) and \(b = \sqrt{3}\ln x\).
Updated On: Oct 31, 2026
- \(e^2\)
- \(e\)
- \(\dfrac{e}{2}\)
- \(2e\)
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The Correct Option is B
Solution and Explanation
\(T_7 = {}^9C_6\left(\dfrac{3}{\sqrt[3]{84}}\right)^{3}\left(\sqrt{3}\ln x\right)^6\).
\({}^9C_6 = 84\), \(\left(\dfrac{3}{\sqrt[3]{84}}\right)^3 = \dfrac{27}{84}\), \((\sqrt{3})^6 = 27\).
\(T_7 = 84\cdot\dfrac{27}{84}\cdot 27(\ln x)^6 = 729(\ln x)^6\).
Given \(T_7 = 729\): \((\ln x)^6 = 1\), so \(\ln x = \pm 1\) and \(x = e\) or \(x = \dfrac{1}{e}\).
Among the options, \(x = e\).
Answer: (2) \(e\)
\({}^9C_6 = 84\), \(\left(\dfrac{3}{\sqrt[3]{84}}\right)^3 = \dfrac{27}{84}\), \((\sqrt{3})^6 = 27\).
\(T_7 = 84\cdot\dfrac{27}{84}\cdot 27(\ln x)^6 = 729(\ln x)^6\).
Given \(T_7 = 729\): \((\ln x)^6 = 1\), so \(\ln x = \pm 1\) and \(x = e\) or \(x = \dfrac{1}{e}\).
Among the options, \(x = e\).
Answer: (2) \(e\)
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