Question

If $4^{th}$ term in the expansion of ${(2+\frac{3}{8}x)}^{10}$ is numerically greatest term . Then the range of $x$ is

• A. $[\frac{-64}{21},-2]\cup [2,\frac{64}{21}]$
• B. $[\frac{-53}{21},-2]\cup [2,\frac{53}{21}]$
• C. $[\frac{-64}{21},-1]\cup [1,\frac{64}{21}]$
• D. $[\frac{-53}{21},-1]\cup [1,\frac{53}{21}]$

Answer 1

The correct option is A $[\frac{-64}{21},-2]\cup [2,\frac{64}{21}]$

$\because T_4$ is the numerically greatest term in the expansion of ${(2+\frac{3}{8}x)}^{10}$
$\therefore \left|\begin{array}{c}\frac{T_4}{T_3}\end{array}\right|\ge 1$ and $\left|\begin{array}{c}\frac{T_5}{T_4}\end{array}\right|\le 1$
$\left|\frac{{}^{10}{C}_3{2}^7{\left(\frac{3}{8}x\right)}^3}{{}^{10}{C}_2{2}^8{\left(\frac{3}{8}x\right)}^2}\right|\ge 1$
and $\left|\frac{{}^{10}{C}_4{2}^6{\left(\frac{3}{8}x\right)}^4}{{}^{10}{C}_3{2}^7{\left(\frac{3}{8}x\right)}^3}\right|\le 1$
$\left|\begin{array}{c}\frac{11-3}{3}\cdot \frac{3x}{16}\end{array}\right|\ge 1$ and$\left|\begin{array}{c}\frac{11-4}{4}\cdot \frac{3x}{16}\end{array}\right|\le 1$
$\Rightarrow |x|\ge 2$ and $|x|\le \frac{64}{21}$
$\Rightarrow 2\le |x|\le \left(\frac{64}{21}\right)$
$\Rightarrow 2\le x\le \frac{64}{21}\text{or}-\frac{64}{21}\le x\le -2$