Question

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

• 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)

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

Since $t_4$ is numerically the greatest term,
$\left|t_3\right|<\left|t_4\right|;\left|t_5\right|<\left|t_4\right|$
$\Rightarrow \left|\frac{t_3}{t_4}\right|<1;\left|\frac{t_5}{t_4}\right|<1$
but
$\frac{t_3}{t_4}=\frac{{{}^{10}C}_2\left(2^8\right){\left(\frac{3x}{8}\right)}^2}{{{}^{10}C}_3\left(2^7\right){\left(\frac{3x}{8}\right)}^3}=\frac{2}{x}$
and
$\frac{t_5}{t_4}=\frac{{{}^{10}C}_4\left(2^6\right){\left(\frac{3x}{8}\right)}^4}{{{}^{10}C}_3\left(2^7\right){\left(\frac{3x}{8}\right)}^3}=\frac{21x}{64}$
Now, $\left|\frac{t_3}{t_4}\right|<1\Rightarrow \left|x\right|>2$
and
$\left|\frac{t_5}{t_4}\right|<1\Rightarrow \left|x\right|<\frac{64}{21}$