Question

Given that the 4th term in the expansion of ${(2+\frac{3x}{8})}^{10}$ has the maximum numerical value, then x lies in the interval

• A. $(2,\frac{64}{21})$
• B. $(-\frac{60}{23},-2)$
• C. $(-\frac{64}{21},-2)$
• D. $(2,-\frac{60}{23})$

Answer 1

Answer: (A), (C)

$(-\frac{64}{21},-2)$

Since $t_4$ is numerically the greatest term,
$|t_3|<|t_4|and|t_5|<|t_4|\Rightarrow |\frac{t_3}{t_4}|<1and|\frac{t_5}{t_4}|<1But\frac{t_3}{t_4}=\frac{{}^{10C_2\left(2^8\right){\left(\frac{3x}{8}\right)}^2}}{{}^{10}{C}_3\left(2^7\right){\left(\frac{3x}{8}\right)}^3}=\frac{2}{x}and\frac{t_5}{t_4}=\frac{{}^{10C_4\left(2^6\right){\left(\frac{3x}{8}\right)}^4}}{{}^{10}{C}_3\left(2^7\right){\left(\frac{3x}{8}\right)}^3}=\frac{21x}{64}Now,|\frac{t_3}{t_4}|<1\Rightarrow |\frac{2}{x}|<1$