Question

For which of the following values of $x$, $5$th term will be the numerically greatest term in the expansion of ${(1+\frac{x}{3})}^{10}$.

• A. $-2$
• B. $1.8$
• C. $2$
• D. $-1.9$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $-2$
B $1.8$
C $2$
D $-1.9$

If $T_5$ is the numerical greatest term in the expansion of ${(1+\frac{x}{3})}^{10}.$ Then,
$\left|\begin{array}{c}\frac{T_5}{T_4}\end{array}\right|\ge 1\text{and}\left|\begin{array}{c}\frac{T_6}{T_5}\end{array}\right|\le 1$
For given binomial, $\frac{T_{r+1}}{T_r}=\frac{10-r+1}{r}\frac{x}{3}$
$\therefore \left|\begin{array}{c}\frac{7}{4}\times \frac{x}{3}\end{array}\right|\ge 1$ and $\left|\begin{array}{c}\frac{6}{5}\times \frac{x}{3}\end{array}\right|\le 1$
$\Rightarrow |x|\ge \frac{12}{7}$ and $|x|\le \frac{5}{2}$
$\Rightarrow \frac{12}{7}\le |x|\le \frac{5}{2}$
$\Rightarrow x\in \left[\begin{array}{c}-\frac{5}{2},-\frac{12}{7}\end{array}\right]\cup \left[\begin{array}{c}\frac{12}{7},\frac{5}{2}\end{array}\right]$