If the 4th term in the expansion of ${\left(2+\frac{{\displaystyle 3}}{{\displaystyle 8}}\mathrm{x}\right)}^{10}$ has the maximum numerical value, then the range of values of x is

Since T₄   is the numerically greatest term in the expansion of 

$\begin{array}{rl} & 2^{10}{\left(1+\frac{3\mathrm{x}}{16}\right)}^{10}\\ & \therefore \left|\frac{{\mathrm{T}}_4}{{\mathrm{T}}_3}\right|\ge 1\text{and}\left|\frac{{\mathrm{T}}_4}{{\mathrm{T}}_5}\right|\ge 1\\ & \text{or,}\left|\frac{{\mathrm{T}}_4}{{\mathrm{T}}_3}\right|\ge 1\text{and}\left|\frac{{\mathrm{T}}_5}{{\mathrm{T}}_4}\right|\le 1\\ & \text{Now,}\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}}=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}\cdot \mathrm{x}\end{array}$

Taking r = 3 and r = 4 and replacing x by 3x/16 x and n by 10, in the above two, we get

$\left|\frac{{\displaystyle 11-3}}{{\displaystyle 3}}\cdot \frac{{\displaystyle 3\mathrm{x}}}{{\displaystyle 16}}\right|\ge 1\text{and} \left|\frac{{\displaystyle 11-4}}{{\displaystyle 4}}\cdot \frac{{\displaystyle 3\mathrm{x}}}{{\displaystyle 16}}\right|\le 1$

or, $\left|\mathrm{x}\right|\ge 2 \text{and }\left|\mathrm{x}\right|\le \frac{64}{21}$

If x is positive, then | x | = x

$\begin{array}{rl} & \therefore \mathrm{x}\ge 2 \text{and }\mathrm{x}\le \frac{64}{21}\\ & \therefore 2\le \mathrm{x}\le \frac{64}{21}----\left(1\right)\end{array}$

If x is negative, then | x | = – x

$\therefore -\mathrm{x}\ge 2\text{and}-\mathrm{x}\le \frac{64}{21}$

or,  $\mathrm{x}\le -2 \text{and }\mathrm{x}\ge -\frac{64}{21}$

$\therefore -\frac{64}{21}\le x\le -2$