Question

If $x>0$ and the $4^{th}$ term in the expansion of ${(2+\frac{3}{8}x)}^{10}$ has maximum value then find the range of x.

Answer 1

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

$\left|\frac{T_4}{T_3}\right|\ge 1$ $\because T_4$ has the maximum numerical value.

$\left|\frac{T_4}{T_5}\right|\ge 1\Rightarrow \left|\frac{T_5}{T_4}\right|\le 1$

${(1+x)}^n$ has $\left|\frac{T_{r+1}}{T_r}\right|=\frac{n-r+1}{r}\times x$

Consider the expansion

${(1+\frac{3}{16}x)}^{10}$ which is of the form ${(1+x)}^n$ where $x\to \frac{3x}{16}$ and $n=10$

$\Rightarrow \frac{T_{r+1}}{T_r}=\frac{10-r+1}{r}\times \frac{3x}{16}=\frac{8x}{16}=\frac{x}{2}$

We have $\left|\frac{T_4}{T_3}\right|\ge 1$

$\Rightarrow \left|\frac{x}{2}\right|\ge 1$

$\Rightarrow \left|x\right|\ge 2$ ......$\left(1\right)$

$\Rightarrow \left|\frac{T_5}{T_4}\right|\le 1$

$\Rightarrow |\frac{10-4+1}{4}\times \frac{3x}{16}|\le 1$

$\Rightarrow |\frac{7}{4}\times \frac{3x}{16}|\le 1$

$\Rightarrow 21\left|x\right|\le 64$

$\Rightarrow \left|x\right|\le \frac{64}{21}$ .......$\left(2\right)$

From $\left(1\right)\Rightarrow x^2\ge 4$

From $\left(2\right)\Rightarrow x^2\le {\left(\frac{64}{21}\right)}^2$

If $x^2\ge 4\Rightarrow x\ge 2$ or $x\le -2$

If $x^2\le {\left(\frac{64}{21}\right)}^2\Rightarrow x\le \frac{64}{21}$ or $\frac{-64}{21}\le x\le \frac{64}{21}$

$\therefore$ the range of $x$ is $2\le x\le \frac{64}{21}$ or $\frac{-64}{21}\le x\le -2$