Question

Given that the $4$th term in the expansion of ${(2+\frac{3}{8}x)}^{10}$ has the maximum numerical value, find the range of value of x for which this will be true.

Answer 1

$(2+\frac{3}{8}x{)}^{10}$ ; Given $4^{th}$ term has maximum value

$(2+\frac{3}{8}x{)}^{10}$$=2^{10}$$(1+\frac{3}{16}x{)}^{10}$

$|\frac{T_4}{T_3}|\ge 1$ Since$|T_4|>|T_3|$

Also $|\frac{T_5}{T_4}|\le 1$

For $(1+x{)}^n;$ we have $\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}x$

For $(1+\frac{3}{16}x{)}^{10};$ $\frac{T_{r+1}}{T_r}=\frac{10-r+1}{r}\left(\frac{3x}{16}\right)$

$\frac{T_4}{T_3}=$ $\frac{10-3+1}{r}\left(\frac{3x}{16}\right)=\frac{x}{2}$

$|\frac{T_4}{T_3}|\ge 1$$⟹|\frac{x}{2}|\ge 1⟹|x|\ge 2$

$\frac{T_5}{T_4}=$ $\frac{10-4+1}{r}\left(\frac{3x}{16}\right)=\frac{21x}{64}$

$|\frac{T_5}{T_4}|\le 1$$⟹|\frac{21x}{64}|\le 1⟹|x|\le \frac{64}{21}$

Plotting on the no. line

We get

$xϵ[\frac{-64}{21},-2]\cup [2,\frac{64}{21}]$