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(A) For the expansion $(a + bx)^n$,the term $T_{r+1}$ is numerically the greatest if $|T_{r+1}| \geq |T_r|$ and $|T_{r+1}| \geq |T_{r+2}|$.Here,$a = 2

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$-\frac{64}{21} < x < -2$

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(A) For the expansion $(a + bx)^n$,the term $T_{r+1}$ is numerically the greatest if $|T_{r+1}| \geq |T_r|$ and $|T_{r+1}| \geq |T_{r+2}|$.Here,$a = 2$,$bx = \frac{3x}{8}$,and $n = 10$.The condition $|T_4| \geq |T_3|$ implies $\frac{|T_4|}{|T_3|} \geq 1$.Using the formula $\frac{|T_{r+1}|}{|T_r|} = \frac{n-r+1}{r} \cdot |\frac{bx}{a}|$,we have $\frac{10-3+1}{3} \cdot |\frac{3x/8}{2}| \geq 1$.$\frac{8}{3} \cdot \frac{3|x|}{16} \geq 1 \implies \frac{|x|}{2} \geq 1 \implies |x| \geq 2$.Similarly,$|T_4| \geq |T_5|$ implies $\frac{|T_5|}{|T_4|} \leq 1$.$\frac{10-4+1}{4} \cdot |\frac{3x/8}{2}| \leq 1 \implies \frac{7}{4} \cdot \frac{3|x|}{16} \leq 1 \implies \frac{21|x|}{64} \leq 1 \implies |x| \leq \frac{64}{21}$.Combining these,$2 \leq |x| \leq \frac{64}{21}$.This corresponds to $x \in [-\frac{64}{21}, -2] \cup [2, \frac{64}{21}]$.Given the options provided,the range $ - \frac{64}{21} < x < -2$ is the correct interval.

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

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