Given that the 4th term in the expansion of 2+3x810has the maximum numerical value, then xlies in the interval.

Since $T_{4}$ is numerically the greatest term,

$|T_{3}| < |T_{4}|$ and $|T_{5}| < |T_{4}|$

$\Rightarrow |\frac{T_{3}}{T_{4}}| < 1$ and $|\frac{T_{5}}{T_{4}}| < 1$ .

But $\frac{T_{3}}{T_{4}} = \frac{^{10} C_{2} (2^{8}) (3 x / 8)^{2}}{^{10} C_{3} (2^{7}) (3 x / 8)^{3}} = \frac{2}{x}$

and $\frac{T_{5}}{T_{4}} = \frac{^{10} C_{4} (2^{6}) (3 x / 8)^{4}}{^{10} C_{3} (2^{7}) (3 x / 8)^{3}} = \frac{21 x}{64}$

Now, $|\frac{T_{3}}{T_{4}}| < 1 \Rightarrow |\frac{2}{x}| < 1 \Rightarrow x < - 2$ or $x > 2$ .

Next , $|\frac{T_{5}}{T_{4}}| < 1 \Rightarrow |\frac{21 x}{64}| < 1 \Rightarrow - \frac{64}{21} < x < \frac{64}{21}$

Hence, $x \in (- \frac{64}{21}, - 2) \cup (2, \frac{64}{21})$ .

Given that the 4th term in the expansion of ${(2 + \frac{3 x}{8})}^{10}$ has the maximum numerical value, then x
lies in the interval.