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If the 4th term in the expansion of \left(2+\frac{3 x}{8}\right)^{10}has the maximum numerical value, then ‘x’ lies in the interval 

 

Option: 1

\left(\frac{-64}{21},-2\right) \cup\left(2, \frac{64}{4}\right)


Option: 2

\left(\frac{-60}{23},-2\right) \cup\left(2, \frac{64}{23}\right)


Option: 3

\left(\frac{-64}{21}, 2\right)


Option: 4

\left(-2, \frac{-64}{21}\right)


Answers (1)

best_answer

\begin{aligned} & \left|\frac{t_3}{t_4}\right|<1 \text { and }\left|\frac{t_5}{t_4}\right|<1 \text { i.e., }\left|\frac{2}{x}\right|<1 ;\left|\frac{21}{64} x\right|<1 \\ & \therefore x \in\left(\frac{-64}{21},-2\right) \cup\left(2, \frac{64}{21}\right) \end{aligned}

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