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If the ratio of the 7 th term from the beginning to the 7 th term from the end in the expansion of  \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x is \,\,\frac{1}{6}, then x is

Option: 1

9


Option: 2

6


Option: 3

12


Option: 4

None of these.


Answers (1)

best_answer

T_7\, in \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x={ }^x C_6\left(2^{1 / 3}\right)^{x-6}\left(\frac{1}{3^{1 / 3}}\right)^6

7th term from the end in \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x

=T_7 \text { in }\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^x={ }^x C_6\left(\frac{1}{3^{1 / 3}}\right)^{x-6}\left(2^{1 / 3}\right)^6

\therefore \frac{{ }^x C_6\left(2^{1 / 2}\right)^{x-6}\left(\frac{1}{3^{1 / 3}}\right)^6}{{ }^x C_6\left(\frac{1}{3^{1 / 3}}\right)^{x-6}\left(2^{1 / 3}\right)^6}=\frac{1}{6} \\

\Rightarrow \frac{\left(2^{1 / 3}\right)^{x-12}}{\left(\frac{1}{3^{1 / 3}}\right)^{x-12}}=\frac{1}{6} \Rightarrow 2^{\frac{x-12}{3}} \cdot 3^{\frac{x-12}{3}}=\frac{1}{6} \\

 \Rightarrow 6^{\frac{x-12}{3}}=6^{-1} \Rightarrow \frac{x-12}{3}=-1 \Rightarrow x=9 .

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manish

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