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Given that $4^{th}$ term in the expansion of ${\left( {2 + \frac{3}{8}x} \right)^{10}}$ has the maximum numerical value, the range of value of $x$ for which this will be true is given by
$ - \frac{{64}}{{21}} < x < - 2$
$ - \frac{{64}}{{21}} < x < 2$
$\frac{{64}}{{21}} < x < 4$
None of these
Solution
(a) ${T_3},{T_4},{T_5}$in the given expansion are respectively $^{10}{C_2}{2^8}{\left( {\frac{{3x}}{8}} \right)^2},{\,^{10}}{C_3}{2^7}{\left( {\frac{{3x}}{8}} \right)^3},{\,^{10}}{C_4}{2^6}{\left( {\frac{{3x}}{8}} \right)^4}$
or $1620{x^2},810{x^3},\frac{{8505}}{{32}}{x^4}$
We are given that ${T_4}$ is numerically the greatest term so that $|{T_4}| > |{T_3}|$ and $|{T_4}| > |{T_5}|$
$\therefore \,\,\,\,|x| > 2$ and $\frac{{64}}{{21}} > |x|$
$2 < \,|x| < \frac{{64}}{{21}}$…..(i)
The above inequality (i) is equivalent to two inequalities
$2 < x < \frac{{64}}{{21}}$ and $ – \frac{{64}}{{21}} < x < – 2$
